Lecture Notes Functional Analysis (2014/15)

Home , Ball (mathematics), Banach space, Bounded operator, Complete metric space, Dual space, Functional (mathematics)

Roland Schnaubelt

These lecture notes are based on my course from winter semester 2014/15. Com- pared to the notes from three years ago, several details and very few subjects have been changed. I mostly kept the contents of the results discussed in the lectures, but the numbering has been shifted in some parts. Typically, the proofs and calculations in the notes are a bit shorter than those given in class. Moreover, the drawings and many additional, mostly oral remarks from the lectures are omitted here. On the other hand, in the notes I have added a few results (e.g., the Riesz–Thorin theorem) and a couple of proofs for peripheral statements not presented during the course. I am happy to thank Bernhard Konrad, JörgBäuerle and Johannes Eilinghoff for their careful proof reading of my lecture notes from 2009 and 2011.

Karlsruhe, December 30, 2019 Roland Schnaubelt Contents

1 Banach spaces3 1.1 Basic properties of Banach and metric spaces ...... 3 1.2 More examples of Banach spaces ...... 13 1.3 Compactness and separability ...... 20

2 Continuous linear operators 27 2.1 Basic properties and examples ...... 27 2.2 Standard constructions ...... 34 2.3 The interpolation theorem of Riesz and Thorin ...... 39

3 Two main theorems on linear operators 43 3.1 The principle of uniform boundedness ...... 43 3.2 The open mapping theorem and invertibility ...... 48

4 Duality 51 4.1 Hilbert spaces ...... 51 4.2 The duals of sequence and Lebesgue spaces ...... 54 4.3 The theorem of Hahn-Banach ...... 58 4.4 Reﬂexivity and weak convergence ...... 65 4.5 Adjoint operators ...... 70

5 The Fourier transform and generalized derivatives 75 5.1 The Fourier transform ...... 75 5.2 Sobolev spaces ...... 82 5.3 Tempered distributions ...... 88

Bibliography 92

2 Chapter 1

Banach spaces

1.1 Basic properties of Banach and metric spaces

Throughout, X 6= {0} and Y 6= {0} are vector spaces over the ﬁeld F ∈ {R, C}.

Deﬁnition 1.1. A seminorm on X is a mapping p : X → R+ satisfying a) p(αx) = |α| p(x) (homogeneity), b) p(x + y) ≤ p(x) + p(y) (triangle inquality) for all x, y ∈ X and α ∈ F. If in addition c) p(x) = 0 =⇒ x = 0 (deﬁniteness) holds for all x ∈ X, then p is called a norm. One usually writes p(x) = kxk and p = k · k. The pair (X, k · k) (or just X) is called a normed vector space. In view of Example 1.4(a), we interpret kxk as the length of x and kx − yk as the distance between x and y.

Deﬁnition 1.2. Let k·k be a seminorm on a vector space X. A sequence (xn)n∈N = (xn) in X converges to a limit x ∈ X if

∀ ε > 0 ∃ Nε ∈ N ∀ n ≥ Nε : kxn − xk ≤ ε.

We then write xn → x as n → ∞ or x = limn→∞ xn. Moreover, (xn) is a Cauchy sequence in X if

∀ ε > 0 ∃ Nε ∈ N ∀ n, m ≥ Nε : kxn − xmk ≤ ε. A normed vector space (X, k · k) is called a Banach space if each Cauchy sequence in (X, k · k) converges in X. Then one also calls k · k or (X, k · k) complete.

Remark 1.3. Let k · k be a seminorm on a vector space X and (xn) be a sequence in X. The following results are shown as in Analysis 2, see also Remark 1.7. a) |kxk − kyk| ≤ kx − yk for all x, y ∈ X. b) The vector 0 has the norm 0. c) If (xn) converges, then it is a Cauchy sequence. d) If (x ) converges or is Cauchy, then it is bounded, i.e., sup kx k < ∞. n n∈N n e) Limits are linear: If xn → x and yn → y in X as n → ∞ and α, β ∈ F, then αxn + βyn converges to αx + βy in X. f) Limits are unique in the norm case: Let k · k be a norm. If xn → x and xn → y in X as n → ∞ for some x, y ∈ X, then x = y. ♦

3 4 Chapter 1. Banach spaces

Let X be a vector space and M 6= ∅ be a set. For functions f, g : M → X and numbers α, β ∈ F one deﬁnes the functions f + g : M → X;(f + g)(t) = f(t) + g(t) and αf : M → X;(αf)(t) = αf(t). It is easily seen that the set {f : M → X} becomes a vector space if endowed with these operations. Function spaces are always equipped with this sum and scalar multiplication. Let X = F. Here one further puts fg : M → F;(fg)(t) = f(t)g(t). Let α, β ∈ R. We then write f ≥ α (f > α, respectively) if f(t) ≥ α (f(t) > α, respectively) for all t ∈ M. Similarly one deﬁnes α ≤ f ≤ β, f ≤ g, and so on.

Example 1.4. a) X = Fd is a Banach space with respect to the norms

d 1 X p p |x|p = |xk| , 1 ≤ p < ∞, and |x|∞ = max{|xk| | k = 1, . . . , d} k=1

d where x = (x1, . . . , xd) ∈ F , see Analysis 1+2 or the proof of Proposition 1.30. b) X = C([0, 1]) = {f : [0, 1] → F | f is continuous} with the supremum norm

kfk∞ := sup |f(t)| (= max |f(t)|). t∈[0,1] t∈[0,1] is a Banach space. We always endow X with this norm, unless something else is speciﬁed. In fact, it is clear that X is a vector space and that kfk∞ = 0 implies f = 0. We further compute

kαfk∞ = sup |α| |f(t)| = |α| sup |f(t)| = |α| kfk∞, t∈[0,1] t∈[0,1]

kf + gk∞ = sup |f(t) + g(t)| ≤ sup (|f(t)| + |g(t)|) ≤ kfk∞ + kgk∞, t∈[0,1] t∈[0,1] for all f, g ∈ X and α ∈ F, so that X is a normed vector space. Let (fn) be a Cauchy sequence in X; i.e., for all ε > 0 there is an index Nε ∈ N with

|fn(t) − fm(t)| ≤ kfn − fmk∞ ≤ ε for all n, m ≥ Nε and t ∈ [0, 1]. We stress that Nε does not depend on t. By this estimate, (fn(t))n∈N is a Cauchy sequence in F. Since F is complete, there exists f(t) := limn→∞ fn(t) in F for each t ∈ [0, 1]. Let t ∈ [0, 1] and ε > 0 be given. Take Nε from above and n ≥ Nε. We then deduce

|f(t) − fn(t)| = lim |fm(t) − fn(t)| ≤ lim sup kfm − fnk∞ ≤ ε. m→∞ m→∞

Since this inequality holds for all t ∈ [0, 1], it follows that kf − fnk∞ ≤ ε for all n ≥ Nε. Hence, fn converges uniformly (in t) to f, and so f is continuous by a convergence theorem from Analysis 1+2. Summing up, we have shown that fn converges in X to f ∈ X, as required. R 1 c) We further introduce kfk1 = 0 |f(t)| dt for f ∈ X = C([0, 1]). The number kf −gk1 corresponds to area between the graphs of f and g (if F = R), i.e., kf −gk1 is the distance in the mean; whereas kf − gk∞ corresponds to the maximal vertical distance between the graphs. We claim that k · k1 is an incomplete norm on X. In fact, let α ∈ F and f, g ∈ X. Clearly, kfk1 ≥ 0. If f 6= 0, then there are 0 ≤ a < b ≤ 1 and δ > 0 such that |f(t)| ≥ δ for all t ∈ [a, b], since f is continuous. It then follows Z b kfk1 ≥ |f(t)| dt ≥ (b − a)δ > 0. a 1.1. Basic properties of Banach and metric spaces 5

We also obtain Z 1 kαfk1 = |α| |f(t)| dt = |α| kfk1, 0 Z 1 Z 1 kf + gk1 = |f(t) + g(t)| dt ≤ (|f(t)| + |g(t)|) dt = kfk1 + kgk1. 0 0

As a result, k · k1 is a norm on X. Next, consider the functions given by

 1 1, 2 ≤ t ≤ 1,  n 1 1 1 fn(t) = nt − 2 + 1, 2 − n ≤ t ≤ 2 ,  1 1 0, 0 ≤ t ≤ 2 − n , for n ∈ N with n ≥ 3. For m ≥ n ≥ 3 we have 1 Z 2 1 kfn − fmk1 = |fn(t) − fm(t)| dt ≤ −→ 0 1 1 n 2 − n as n → ∞, so that (fn) is a Cauchy sequence with respect to k · k1. We suppose 1 that (fn) would converge to some f ∈ X with respect to k · k1. Fix any a ∈ (0, 2 ) 1 1 and take n ∈ N with 2 − n ≥ a. We then obtain that Z a Z a 0 ≤ |f(t)| dt = |f(t) − fn(t)| dt ≤ kfn − fk1 −→ 0 0 0 R a as n → ∞, i.e., 0 |f(t)| dt = 0. The continuity of f implies as above that f = 0 on 1 [0, a] for every a < 2 , and hence f(1/2) = 0. On the other hand, Z 1 Z 1 0 ≤ |f(t) − 1| dt = |f(t) − fn(t)| dt ≤ kf − fnk1 −→ 0 1 1 2 2 1 as n → ∞. As a consequence, f is equal to 1 on [ 2 , 1], which is a contradiction. d) Let X = C(R) and a < b in R. We see as in b) that p(f) = supt∈[a,b] |f(t)| deﬁnes a seminorm on X. Moreover, if p(f) = 0, then f = 0 on [a, b], but of course f does have to be the 0 function. ♦ Remark. The vector space X = C([0, 1]) is inﬁnite dimensional since the functions n pn ∈ X, n ∈ N, given by pn(t) = t are linearly independent. If one interprets a function 0 ≤ u ∈ X as a mass density, then kuk1 is the total mass and kuk∞ is the maximal density. ♦ Before discussing further examples, we want to investigate various ‘topological’ concepts in a more general framework without vector space structure.

Deﬁnition 1.5. A distance d on a set M 6= ∅ is a map d : M ×M → R+ satisfying a) d(x, y) = 0 ⇐⇒ x = y (deﬁniteness), b) d(x, y) = d(y, x) (symmetry), c) d(x, y) ≤ d(x, z) + d(z, y) (triangle inequality) for all x, y, z ∈ M. The pair (M, d) (or just M) is called a metric space. A sequence (xn) ⊆ M converges to a limit x ∈ M if

∀ ε > 0 ∃ Nε ∈ N ∀ n ≥ Nε : d(x, xn) ≤ ε, (1.1) and it is a Cauchy sequence if

∀ ε > 0 ∃ Nε ∈ N ∀ n, m ≥ Nε : d(xm, xn) ≤ ε. (M, d) or d is called complete if each Cauchy sequence converges in (M, d). 6 Chapter 1. Banach spaces

Example 1.6. a) Let X be a normed vector space and ∅= 6 M ⊆ X. Set d(x, y) = kx − yk for x, y ∈ M. Then Definition 1.5a) follows from Definition 1.1c), 1.5b) from 1.1a) with α = −1, and 1.5c) from 1.1b). As a result, (M, d) is a metric space (but not a vector space, in general). b) Let N ⊆ M and d be a metric on M. Then dN (x, y) = d(x, y) for x, y ∈ N defines the subspace metric dN on N. One often writes d instead of dN . c) Let M 6= ∅ be any set. One defines the discrete metric on M by setting d(x, x) = 0 and d(x, y) = 1 for all x, y ∈ M with x 6= y. It is easy to check that d is indeed a metric on M and that xn → x as n → ∞ in the discrete metric if and only if there is an m ∈ N such that xn = x for all n ≥ m. ♦

Remark 1.7. Let M be a metric space and (xn) be a sequence in M. Then the following assertions hold. a) If xn → x in M as n → ∞, then (xn) is Cauchy. (In fact, we have d(xn, xm) ≤ d(xn, x) + d(x, xm) ≤ 2ε for each ε > 0 and all n, m ≥ Nε with Nε from (1.1).) b) If xn → x and xn → y in M for some x, y ∈ X, then x = y. (In fact, we have d(x, y) ≤ d(x, xn) + d(xn, y) ≤ 2ε for each ε > 0 and all n ≥ max{Nε(x),Nε(y)}, where Nε(x) and Nε(y) are given by (1.1) for x and y, respectively.) c) If (xn) converges or is Cauchy, then it is bounded, i.e., there is a z ∈ M such that supn d(xn, z) < ∞. (In fact, there is an N ∈ N such that d(xn, xN ) ≤ 1 for all n ≥ N, and hence d(xn, xN ) ≤ 1 + max{d(xj, xN ) | j = 1,...,N} for all n ∈ N. ♦ The next result describes how to construct a distance from a given sequence of seminorms. This procedure is often used in analysis.

Proposition 1.8. Let X be a vector space and pj, j ∈ N, be seminorms on X such that for each x ∈ X \{0} there is a k ∈ N with pk(x) > 0. Then ∞ X pj(x − y) d(x, y) := 2−j , x, y ∈ X, 1 + p (x − y) j=1 j deﬁnes a metric on X such that d(xn, x) → 0 as n → ∞ if and only if pj(xn−x) → 0 as n → ∞ for each j ∈ N.

Proof. Note that the function ϕ(t) = t/(1+t) increases strictly on R+, that ϕ(0) = 0 and that ϕ(t) ∈ (0, 1) for t > 0. In particular, the series in the statement converges in R+. Let x, y, z, xn ∈ X for n ∈ N. We have d(x, y) = 0 if and only if pj(x−y) = 0 for all j ∈ N which is equivalent to x = y by the assumptions. Moreover, the identity d(x, y) = d(y, x) holds since pj(x − y) = pj(y − x) for each j ∈ N. Using the monotonicity of ϕ, we further estimate

∞ X pj(x − y) pj(y − z) d(x, z) ≤ 2−j + 1 + p (x − y) + p (y − z) 1 + p (x − y) + p (y − z) j=1 j j j j ≤ d(x, y) + d(y, z).

Thus, d is a metric on X. Assume that d(xn, x) → 0 as n → ∞. Fix any j ∈ N and let ε ∈ (0, 1/2). Set −j η = 2 ε. Then there is an Nε,j ∈ N such that, for all n ≥ Nε,j,

−j pj(x − xn) −j 2 ≤ d(x, xn) ≤ η = 2 ε, 1 + pj(x − xn) 1 pj(x − xn) ≤ ε(1 + pj(x − xn)) ≤ ε + 2 pj(x − xn), pj(x − xn) ≤ 2ε, so that pj(x − xn) → 0 as n → ∞. 1.1. Basic properties of Banach and metric spaces 7

Conversely, assume that pj(x − xn) → 0 as n → ∞ for each j ∈ N. Let ε > 0. Take an index Jε ∈ N with ∞ X 2−j ≤ ε.

j=Jε+1

We then ﬁnd an Nε ∈ N such that pj(x−xn) ≤ ε for all j ∈ {1,...,Jε} and n ≥ Nε. It follows that

Jε ∞ Jε X −j X −j X −j d(x, xn) ≤ 2 pj(x − xn) + 2 ≤ ε 2 + ε ≤ 2ε

j=1 j=Jε+1 j=1 for all n ≥ Nε.

Example 1.9. a) Let X = C(R) and pj(f) = max|t|≤j |f(t)| for j ∈ N. Since each pj is a seminorm and pj(f) = 0 for all j ∈ N yields f = 0, Proposition 1.8 shows that X possesses the metric

∞ X max|t|≤j |f(t) − g(t)| d(f, g) := 2−j , 1 + max |f(t) − g(t)| j=1 |t|≤j and that d(fn, f) → 0 as n → ∞ if and only if max|t|≤j |fn(t)−f(t)| → 0 as n → ∞ for each j ∈ N. In particular, (X, d) is endowed with the ‘uniform convergence on compact sets’ as introduced in Funktionentheorie. Moreover, (X, d) is complete. In fact, let (fn) be Cauchy in (X, d). As in the proof of Proposition 1.8 we see that

∀ ε ∈ (0, 1/2), j ∈ N ∃ Nε,j ∈ N ∀ n, m ≥ Nε,j : max |fn(t) − fm(t)| ≤ ε. |t|≤j

Since C([−j, j]) is Banach space for the supremum norm, for each j ∈ N there is (j) (j) a function g ∈ C([−j, j]) such that the restrictions fn|[−j, j] converge to g in C([−j, j]). This convergence implies that g(j)(t) = g(k)(t) whenever t ∈ [−j, j] ⊂ [−k, k]. We can thus define a function f ∈ C(R) by setting f(t) := g(j)(t) for any j ∈ N with |t| ≤ j. By construction, we have pj(fn − f) → 0 for each j ∈ N, and hence d(fn, f) → 0 as n → ∞. b) Let Y = Cb(R) = {f ∈ C(R) | f is bounded} and dY be the subspace metric of d in part a). Then (Y, dY ) is not complete. Indeed, take the functions fn ∈ Y given by fn(t) = |t| if |t| ≤ n and fn(t) = n otherwise. Then fn converges in (X, d) to the function f given by f(t) = |t|. Therefore (fn) is a Cauchy sequence in (Y, dY ). But it cannot have a limit g in Y , since then g would be equal to f∈ / Y . ♦ In a metric space (M, d) we write B(x, r) = {y ∈ X | d(x, y) < r} for the open ball with center x ∈ M and radius r > 0, and B(x, r) = {y ∈ X | d(x, y) ≤ r} for the closed ball. The next definitions belong to the most basic ones in analysis; below we characterize them in terms of sequences. Definition 1.10. Let M be a metric space. A subset O ⊆ M is called open if for each x ∈ O there is a radius rx > 0 such that B(x, rx) ⊆ O. Moreover, ∅ is open. Let x0 ∈ M. A subset N ⊆ M is called a neighborhood of x0 and x0 is an interior point of N if there is a radius r0 > 0 such that B(x0, r0) ⊆ N. Finally, a subset A ⊆ M is called closed if M \ A is open. Example 1.11. Let (M, d) be a metric space, x ∈ M, r > 0. a) The ball B(x, r) is open. In fact, for any given y ∈ B(x, r) we set ρ = r − d(x, y) > 0. Let z ∈ B(y, ρ). We then have

d(z, x) ≤ d(z, y) + d(y, x) < ρ + d(x, y) = r. 8 Chapter 1. Banach spaces

This estimate yields z ∈ B(x, r), and hence B(y, ρ) ⊆ B(x, r), as required. b) The ball B¯(x, r) is closed. Indeed, for any given y ∈ M \ B¯(x, r) we set R = d(x, y) > r. Let z ∈ B(y, R − r). We then have

R = d(x, y) ≤ d(x, z) + d(z, y) < d(x, z) + R − r, and thus d(x, z) > r. This fact implies that B(y, R − r) ⊂ M \ B¯(x, r), so that M \ B¯(x, r) is open, and hence B¯(x, r) is closed. c) The sets ∅ and M are both closed and open. As in part b) with r = 0, we see that the set {x} is closed for every x ∈ M. d) In the discrete metric of Example 1.6 we have {x} = B(x, 1) for each x ∈ M, and hence {x} is open. ♦ Proposition 1.12. In a metric space M the following equivalences hold. (a) A ⊆ M is closed ⇔ If xn ∈ A and xn → x in M as n → ∞, then x ∈ A. (b) O ⊆ M is open ⇔ For no x ∈ O there exists (xn) in M \ O converging to x. Proof. By deﬁnition, O ⊆ M is open if and only if for each x ∈ O there is an r > 0 such that for all y ∈ M with d(x, y) < r it follows that y ∈ O. This statement is equivalent to the fact that no x ∈ O can be the limit of a sequence in M \ O; i.e., assertion b) holds. Part a) follows from b) by taking complements in M.

Corollary 1.13. Let M be a complete metric space and A ⊆ M. If A is closed, then A is complete for the subspace metric dA. In particular, if X is a Banach space and Y is closed linear subspace, then Y is a Banach space for the restriction k · kY of the norm k · kX to Y .

Proof. Let (xn) be a Cauchy sequence in A with respect to d. Because of the completeness of M, there exists an x ∈ M with xn → x in (M, d) as n → ∞. Since A is closed, x belongs to A; and thus xn → x in (A, dA).

Example 1.14. Let X = C([0, 1]) be endowed with k · k∞. a) Let Y = {f ∈ X | f(0) = 0}. Clearly, Y is a linear subspace. Let fn ∈ Y converge to f in X. It follows that 0 = fn(0) → f(0) as n → ∞, hence f(0) = 0 and f ∈ Y . Therefore Y is closed and Y is a Banach space with respect to k · k∞. b) The set O = {f ∈ X | f > 0} is open in X. In fact, let f ∈ O. Then mint∈[0,1] f(t) =: δ > 0. Let g ∈ X with kf − gk∞ < δ. We now deduce that g(t) = f(t) + g(t) − f(t) ≤ δ − kf − gk∞ > 0 for all t ∈ [0, 1], so that g ∈ Y . This means that B(f, δ) ⊂ O, and hence O is open. c) The set C = {f ∈ X | f(0) > 0 and f(1) ≥ 0} is neither open nor closed in X. 1 1 In fact, the functions fn = n belong to C and converge in X to 0 6= C, so that 1 C is not closed. Moreover, the functions gn given by gn(t) = 1 − (1 + n )t do not belong to C, but they have the limit g(t) = 1 − t in X which is contained in C, so that C is not open. d) Let ` > 0. The set L = {f ∈ X | |f(t) − f(s)| ≤ ` |t − s| (∀ t, s ∈ [0, 1])} of functions with Lipschitz constant less or equal ` is closed in X. In fact, even if functions fn ∈ L converge pointwise to some f as n → ∞, we conclude

|f(t) − f(s)| = lim |fn(t) − fn(s)| ≤ ` |t − s| n→∞ for all t, s ∈ [0, 1], so that f ∈ L. 1 0 On the other hand, the set D = {f ∈ C ([−1, 1]) | kf k∞ ≤ 1} is not closed in 2 1 1/2 C([−1, 1]). In fact, the functions fn ∈ D given by fn(t) = (t + n ) for n ∈ N converge uniformly on [−1, 1] to the continuous function t 7→ f(t) = |t| which is not diﬀerentiable at 0. ♦ 1.1. Basic properties of Banach and metric spaces 9

Proposition 1.15. In a metric space M the following assertions hold. a) The union of an arbitrary collection of open sets in M is open. The intersection of finitely many open sets in M is open. b) The intersection of an arbitrary collection of closed sets in M is closed. The union of finitely many closed sets in M is closed. S Proof. Let be C be a collection of open sets O ⊂ M and let x ∈ V := O∈C O. Then there is a set O0 ∈ C containing x. Since O0 is open, we have an r > 0 such 0 that B(x, r) ⊂ O ⊂ V . Therefore V is open. Let O1, ··· ,On ⊂ M be open and x ∈ D := O1 ∩ · · · ∩ On. Again, there are rj > 0 such that B(x, rj) ⊂ Oj for each j ∈ {1, . . . , n}. Setting ρ := min{r1, . . . , rn} > 0, we arrive at B(x, ρ) ⊂ D, so that D is open. Assertion b) follows from a) by taking complements. The finiteness assumptions in the above result are needed, as seen by easy ex- 1 amples: The sets (0, 1 + n ) are open in R for each n ∈ N, but their intersection T (0, 1 + 1 ) = (0, 1] is not open in . The sets [0, 1 − 1 ] are closed in for each n∈N n R n R n ∈ , but their union S [0, 1 − 1 ] = [0, 1) is not closed in . N n∈N n R Definition 1.16. Let M be metric space and N ⊆ M. We define a) the interior N ◦ =int N of N in M by N ◦ =S{O ⊆ M | O open in M,O ⊆ N}, b) the closure N =cls N of N in M by N =T{A ⊆ M | A closed in M,A ⊇ N}, c) the boundary ∂N of N in M by ∂N = N \ N ◦ = N ∩ (M \ N ◦). The set N is called dense in M if N = M. The above concepts can be characterized in various ways. Proposition 1.17. Let M be a metric space M and N ⊆ M. We then have: a) (i) N ◦ in the largest open subset of N. (ii) N is open if and only if N = N ◦. ◦ (iii) N = {x ∈ M | ∃ r > 0 with B(x, r) ⊆ N} =: N1 = {x ∈ M | @ (xn) ⊆ M \ N with xn → x as n → ∞} =: N2. b) (i) N in the smallest closed subset of M containing N. (ii) N is closed if and only if N = N. (iii) N = ∂N∪˙ N ◦.

(iv) N = {x ∈ M | ∃ (xn) ⊆ N with xn → x as n → ∞} =: N3. c) (i) ∂N is closed.

(ii) ∂N = {x ∈ M | ∃ xn ∈ N, yn ∈/ N with xn → x, yn → x as n → ∞}.

d) N is dense in M if and only if for each x ∈ M there are xn ∈ N with xn → x as n → ∞. Proof. a) The inclusion N ◦ ⊂ N follows from the definition of N ◦, and N ◦ is open by Proposition 1.15. If N ◦ ⊂ O ⊂ N for an open set O, then O ⊂ N ◦ due to the definition of N ◦, so that O = N ◦. We have thus shown (i) which implies (ii). In ◦ ◦ ◦ (iii), we deduce N1 ⊂ N from the definition of N , and N ⊂ N2 is a consequence ◦ of Proposition 1.12 and the openness of N . If x∈ / N1, then there is a sequence (xn) in M \ N converging to x, i.e., x∈ / N2. Hence, assertion (iii) holds. b) Assertions (i) and (ii) can be shown as in part a), and Definition 1.16 yields ◦ (iii). Assertion (iii) in a) implies that N3 = M \ (M \ N) , so that N3 is closed. Clearly, N ⊆ N3. From Proposition 1.12 we infer N3 ⊆ N. Now, (i) yields (iv). c) and d) follow from parts a) and b) and the definition of ∂N. 10 Chapter 1. Banach spaces

Corollary 1.18. Let X be a normed vector space. a) If Y ⊂ X is a linear subspace, then Y is also a linear subspace. b) If Y ⊂ X is convex, then Y is also convex.

Proof. a) Let x, y ∈ Y and α, β ∈ F. By Proposition 1.17 there are xn, yn ∈ Y with xn → x and yn → y as n → ∞. Clearly, zn := αxn + βyn ∈ Y and

kzn − (αx + βy)k ≤ |α| kxn − xk + |β| kyn − yk −→ 0 as n → ∞. Proposition 1.17 yields assertion a). Part b) is shown similarly.

Let M be a metric space and f : M → X. The support of f is the set

supp f := clsM {t ∈ M | f(t) 6= 0}.

Example 1.19. a) Let F = R. The set P of polynomials is dense in X = C([0, 1]) by Weierstraß’ approximation theorem from Analysis 3. Since P ⊂ Ck([0, 1]), also the subspace Ck([0, 1]) is dense in C([0, 1]) for every k ∈ N. b) Let X be a normed vector space, x ∈ X, and r > 0. We then have B(x, r) = B(x, r) and ∂B(x, r) = ∂B(x, r) = {y ∈ X | kx − yk = r} =: S(x, r). In fact, Proposition 1.17 and Example 1.11 show that B(x, r)∪˙ ∂B(x, r) = B(x, r) ⊂ B(x, r). 1 Take y ∈ S(x, r) = B(x, r) \ B(x, r). Then yn = y − n (y − x) ∈ B(x, r) and 1 zn = y + n (y − x) ∈/ B(x, r) for all n ∈ N, and we have yn → y and zn → y as n → ∞. Consequently, y ∈ ∂B(x, r) and y ∈ ∂B(x, r) due to Proposition 1.17. These facts imply the assertions. c) Let M contain at least two elements and let d be the discrete metric on M from Example 1.6. We then have B(x, 1) = {x} and B(x, 1) = M, whence B(x, 1) = B(x, 1) 6= B(x, 1). d) Let X = C([0, 1]) and N = {f ∈ X | f ≥ 0}. Then N is closed in X, N ◦ = {f ∈ X | f > 0} =: O, and ∂N = {f ∈ N | ∃ t ∈ [0, 1] with f(t) = 0} =: R. Proof. Let fn ∈ N converge to some f in X. Then f(t) = limn→∞ fn(t) ≥ 0 for each t ∈ [0, 1], and so f ∈ N. Thus, N is closed. We have N = O∪˙ R. The set O 1 1 is open by Example 1.14. If f ∈ R, then fn := f − n ∈/ N and fn → f in X as n → ∞. Hence, R ∩ N ◦ = ∅. Proposition 1.17 now yields O = N ◦ and R = ∂N. 2 e) Let X = C([0, 1]) and put C(0, 1) := C((0, 1)). The closure of the set

Cc(0, 1) = Cc((0, 1)) := {f ∈ C(0, 1) | ∃ 0 < af ≤ bf < 1 : supp f ⊆ [af , bf ]} in X is given by

C0(0, 1) = C0((0, 1)) := {f ∈ C(0, 1) | ∃ lim f(t) = 0 = lim f(t)}. t→0 t→1

We consider Cc(0, 1) and C0(0, 1) as subspaces of X by extending functions by 0 to [0, 1]. In particular, C0(0, 1) is a Banach space for the supremum norm. Proof. As in Example 1.14 we see that C0(0, 1) is closed in X. Clearly, Cc(0, 1) ⊂ C0(0, 1), so that Cc(0, 1) ⊂ C0(0, 1) by Proposition 1.17. Let f ∈ C0(0, 1). Choose 1 1 functions ϕn ∈ Cc(0, 1) such that 0 ≤ ϕn ≤ 1 and ϕn(t) = 1 for n ≤ t ≤ 1 − n and n ≥ 2. Then fn := ϕnf ∈ Cc(0, 1) and

kf − fnk∞ = sup |1 − ϕn(t)| |f(t)| ≤ sup |f(t)|, 1 1 1 1 0≤t≤ n ,1− n ≤t≤1 0≤t≤ n ,1− n ≤t≤1 and the right hand side tends to 0 as n → ∞ since f ∈ C0(0, 1). As a result, C0(0, 1) ⊂ Cc(0, 1) and the assertion is proved. 2 f) The sets L and D from Example 1.14 have no interior points since for each ε > 0 there is a continuous, but not Lipschitz continuous f with kfk∞ < ε. ♦ 1.1. Basic properties of Banach and metric spaces 11

Remark 1.20. Let (M, d) be a metric space and N ⊆ M be endowed with the subspace metric dN . Then the open balls in (N, dN ) are given by

BN (x, r) = {y ∈ N | r > dN (x, y) = d(x, y)} = B(x, r) ∩ N.

A set C ⊆ N is called relatively open (closed) if it is open (closed) in (N, dN ). Similarly, one introduces the relative closure etc.. For instance, the open unit ball 2 2 2 2 in N = R+ for | · |2 is given by BN (0, 1) = {(x, y) ∈ R | x, y ≥ 0, x + y < 1}. The deﬁnitions further imply the following facts. a) S ⊆ N is relatively open (closed) in N if and only if there is an open (closed) subset S0 of M with S = S0 ∩ N. b) N is open (closed) in M if and only if openness (closedness) in N and M coincide. 0 S (In the proof of a), for an open S one may set S = x∈S BM (x, rx) where (BM (x, rx) ∩ N) ⊆ S.) ♦ 0 0 0 Deﬁnition 1.21. Let (M, d) and (M , d ) be metric spaces, f : M → M , x0 ∈ M, 0 0 and y0 ∈ M . We write y0 = limx→x0 f(x) if f(xn) → y0 in M as n → ∞ for every sequence (xn) ⊆ M with xn → x0 in M as n → ∞. The map f is called continuous at x0 if f(x0) = limx→x0 f(x). If f is continuous at every x0 ∈ M, then it is called continuous (on M). We write C(M,M 0) := {f : M → M 0 | f is continuous}, and put C(M) := C(M, F), where F is endowed with the usual metric. If f ∈ C(M,M 0) is bijective with f −1 ∈ C(M 0,M), then f is an homeomorphism. In other words, continuity means that f preserves convergence in the sense that: If xn → x0, then f(xn) → f(x0) as n → ∞.

Remark 1.22. Let (M1, d1) and (M2, d2) be metric spaces. On the product space M = M1×M2, we can deﬁne a metric by setting d((x, y), (u, v)) = d1(x, u)+d2(y, v). A sequence (xn, yn) in M converges with respect to d to (x, y) ∈ M if and only if xn → x in M1 and yn → y in M2 as n → ∞. These facts can be shown as for M1 = M2 = R in Analysis 2. ♦ Example 1.23. a) Every distance d : M × M → R is continuous, where M × M is endowed with the metric from Remark 1.22. In fact, let (xn, yn) → (x, y) in M ×M as n → ∞. Using

d(xn, yn) − d(x, y) ≤ d(xn, x) + d(x, yn) − d(x, y)

≤ d(xn, x) + d(x, y) + d(y, yn) − d(x, y) = d(xn, x) + d(y, yn),

d(x, y) − d(xn, yn) ≤ d(x, xn) + d(xn, yn) + d(yn, y) − d(xn, yn)

= d(xn, x) + d(y, yn), we deduce that |d(xn, yn) − d(x, y)| tends to 0 as n → ∞. b) Let X be a normed vector space. Then the maps F × X → X :(α, x) 7→ αx and X × X → X :(x, y) 7→ x + y are continuous (cf. the proof of Corollary 1.18). c) Let X = C([0, 1]) and ϕ : F → F be Lipschitz on B(0, r) ⊆ F with constant Lr, for every r ≥ 0. Set (F (g))(t) = ϕ(g(t)) for all t ∈ [0, 1] and g ∈ X. Clearly, F (g) ∈ X. Let g → g in X as n → ∞. Then R := sup kg k < ∞ and n n∈N n ∞ kgk∞ ≤ R. We thus deduce kϕ(gn)−ϕ(g)k∞ = sup |ϕ(gn(t))−ϕ(g(t))| ≤ sup LR |gn(t)−g(t)| = LR kgn−gk∞, t∈[0,1] t∈[0,1] so that F : X → X is continuous. (Similarly, one sees that F is Lipschitz on every ball of X.) ♦ One can characterize continuity in terms of open or closed sets. 12 Chapter 1. Banach spaces

0 0 00 00 Proposition 1.24. Let (M, d), (M , d ) and (M , d ) be metric spaces, x0 ∈ M, f : M → M 0 and g : M 0 → M 00. It then holds:

a) If f is continuous at x0 and g is continuous at f(x0), then also h = g ◦ f : 00 M → M is continuous at x0. b) The following assertions are equivalent.

(i) f is continuous at x0.

(ii) ∀ ε > 0 ∃ δ > 0 : ∀ x ∈ M with d(x, x0) < δ we have d(f(x), f(x0)) < ε. 0 −1 (iii) If V is a neighborhoood of f(x0) in M , then f (V ) is a neighborhood of x0 in M. c) The following assertions are equivalent.

(i) f is continuous on M. (ii) If O ⊆ M 0 is open, then f −1(O) is open in M. (iii) If A ⊆ M 0 is closed, then f −1(A) is closed in M.

Proof. a) follows from Definition 1.21 as in Analysis 1+2. b) (i) ⇒ (iii): If (iii) were false, then there would exist a neighborhood V of f(x0) 0 −1 in M and a sequence (xn) which does not belong to f (V ) and converges to x0 in M. Hence, f(xn) ∈/ V for all n ∈ N so that f(xn) cannot converge to f(x0). This fact contradicts (i). (iii) ⇒ (ii): Set V = B(f(x0), ε) for any given ε > 0. By (iii) there exists a −1 δ > 0 such that B(x0, δ) ⊆ f (V ). Thus, for every x ∈ B(x0, δ) we have that f(x) ∈ V = B(f(x0), ε), and so (ii) holds. (ii) ⇒ (i) can be deduced from Definition 1.21 as in Analysis 1+2. 0 −1 c) (i) ⇒ (iii): Let A ⊂ M be closed. Take xn ∈ f (A) with xn → x as n → ∞. Then A 3 f(xn) → f(x) by (i). Since A is closed, f(x) belongs to A; i.e., x ∈ f −1(A). Proposition 1.12 now implies that f −1(A) is closed. (iii) ⇒ (ii) follows by taking complements. (ii) ⇒ (i): Let x ∈ M. Take a neighborhood V of f(x) in M 0. Hence, there is an r > 0 such that B(x, r) ⊆ V . Due to (ii), f −1(B(f(x), r)) is open in M. As a result f −1(V ) is a neighborhood of x in M, so that f is continuous at x by part b). Definition 1.25. In the framework of Proposition 1.24, a function f : M → M 0 is called uniformly continuous if assertion (b.ii) holds for all x0 ∈ M with a δ = δ(ε) > 0 not depending on x0.

Example 1.26. Let X = C([0, 1]) and ϕ : X → F; ϕ(g) = g(0). If gn → g in X, then gn(0) → g(0) as n → ∞. Thus ϕ is continuous. So, if B ⊆ F is open (closed), −1 then ϕ (B) = {g ∈ X | g(0) ∈ B} is open (closed) in X. ♦ Quite often one has several norms on a vector space. The next deﬁnition is concerned with concepts that allow to compare norms. These concepts can then again be characterized by means of sequences and open or closed sets.

Deﬁnition 1.27. Let k · k and · be norms on a vector space X. If there is a constant C > 0 such that kxk ≤ C9 9x for all x ∈ X, then · is called ﬁner than k · k (or k · k coarser than · ).9 In9 this case one calls the9 norms9 comparable. If there are constants c, C > 09such9 that

c x ≤ kxk ≤ C x 9 9 9 9 for all x ∈ X, then the norms are called equivalent. 1.2. More examples of Banach spaces 13

Proposition 1.28. Let k · k and · be norms on a vector space X. Then the following assertions are equivalent.9 9 a) k · k is coarser than · . 9 9 b) If a sequence (xn) ⊆ X converges for · , then it converges for k · k. 9 9 c) If a set A ⊆ X is closed for k · k, then it is closed for · . 9 9 d) If a set O ⊆ X is open for k · k, then it is open for · . 9 9 In this case the limits in b) are equal. Moreover, the norms are equivalent if and only if one of the implications in b), c) or d) becomes an equivalence. Finally, let the norms be equivalent. Then (X, k · k) is complete if and only if (X, · ) is complete. 9 9 Proof. The implication ‘a) ⇒ b)’ and the equality of the limits are a consequence of the estimate kxn − xk ≤ C xn − x . To show ‘b) ⇒ c)’, let A be closed for k · k and let xn ∈ A converge to some9 x ∈ 9X for · . By b), xn also converge for k · k so that x ∈ A; i.e., A is closed for · . The9 implication9 ‘c) ⇒ d)’ follows by taking 9 9 complements. Let statement d) hold. The ball Bk·k(0, 1) is thus open for · , so 9 9 that we can ﬁnd an r > 0 with B · (0, r) ⊆ Bk·k(0, 1). This means that y < r implies that kyk < 1. Let x ∈ X. Since9 9 the vector y := r(2 x )−1 x has the9 9 triple norm r/2 < r, we conclude that 1 > kyk = r(2 x )−1 kxk9which9 yields a). The assertions concerning equivalence follow from9 9 the above results. One similarly shows the claim about completeness. Example 1.29. a) On a ﬁnite dimensional vector space X all norms are equivalent. (See Analysis 2.) On X = Fn we have the more precise result

1/q0 |x|∞ ≤ |x|p ≤ |x|1 ≤ n |x|q

n 1 1 for all x ∈ F and p, q ∈ [1, ∞], where q + q0 = 1. (See Analysis 2 or the next section.) b) Let X = C([0, 1]) and 0 < w ∈ X. Observe that δ := mint∈[0,1] w(t) > 0. Then the norms k · k∞ and kfkw := max0≤t≤1 w(t) |f(t)| on X are equivalent since δ kfk∞ ≤ kfkw ≤ kwk∞ kfk∞ for all f ∈ X. c) Let X = C([0, 1]). Since kfk1 ≤ kfk∞ for f ∈ X the supremum-norm is finer that the 1-norm on X. On the other hand, Example 1.4 and Proposition 1.28 show that these norms are not equivalent on X since only the first one is complete. This 1 fact can directly be seen using the functions fn given by fn(t) = 1−nt for 0 ≤ t ≤ n 1 1 and fn(t) = 0 for n ≤ t ≤ 1, since kfnk∞ = 1 and kfnk1 = 2n for all n ∈ N. The 1-norm and the sup-norm on Cc(R+) = {f ∈ C(R+) | supp f is bounded} are not comparable. Indeed, using functions as fn one sees that k · k1 cannot be finer than k·k∞. Conversely, take functions gn ∈ Cc(R+) with 0 ≤ gn ≤ and gn = 1 on [0, n] for n ∈ N. Then kgnk∞ = 1 and kgnkn ≥ n for all n, so that k · k∞ cannot be finer than k · k1. ♦

1.2 More examples of Banach spaces A) Sequence spaces

Let s = {x = (xj) = (xj)j∈N | xj ∈ F for all j ∈ N} be the space of all sequences in F. A distance on s is given by ∞ X |xj − yj| d(x, y) = 2−j , x = (x ) ∈ s, y = (y ) ∈ s. 1 + |x − y | j j j=1 j j 14 Chapter 1. Banach spaces

Moreover, for all vn, x ∈ s, we have d(vn, x) → 0 as n → ∞ if and only (vn)j → xj for each j ∈ N, as n → ∞. These assertions follow from Proposition 1.8 with pj(x) = |xj|. As in Example 1.9 one shows the completeness of (s, d). For x ∈ s, one deﬁnes the supremum norm by

kxk∞ := sup |xj| ∈ [0, ∞]} j∈N and introduces by

∞ ` = {x ∈ s | kxk∞ < ∞}, ∞ c = {x ∈ s | ∃ lim xj} ⊂ ` , j→∞

c0 = {x ∈ s | lim xj = 0} ⊂ c, j→∞

c00 = {x ∈ s | ∃ mx ∈ N such that xj = 0 for all j > mx} ⊂ c0 the spaces of bounded, converging, null and ﬁnite sequences, repectively. For 1 ≤ p < ∞ and x ∈ s, we further deﬁne

∞ p X p p kxkp := |xj| ∈ [0, ∞] and the space ` = {x ∈ s | kxkp < ∞}. j=1 Observe that

p |xk| ≤ kxkp for all k ∈ N, 1 ≤ p ≤ ∞, x = (xj) ∈ ` . (1.2) We also set  ∞, p = 1, 0  p p = p−1 , 1 < p < ∞, (1.3) 1, p = ∞.

We note that 1 1 + = 1; p00 = p; p0 = 2 ⇐⇒ p = 2; p ∈ (1, 2) ⇐⇒ p0 ∈ (2, ∞). (1.4) p p0

0 Proposition 1.30. Let p ∈ [1, ∞], x, y ∈ `p and z ∈ `p . We then have:

P∞ 1 a) kxzk1 = k=1 |xkzk| ≤ kxkp kzkp0 and xz ∈ ` (H¨older’sinequality). p b) kx + ykp ≤ kxkp + kykp and x + y ∈ ` (Minkowski’s inequality).

p ∞ c) ` is a Banach space; c and c0 are closed subspaces of ` . Proof. Assertions a) and b) are straightforward to verify for p = 1 and p = ∞, so that we only consider p, p0 ∈ (1, ∞) in a) and b). 0 a) Let x ∈ `p and z ∈ `p . We recall Young’s inequality

0 ap bp ab ≤ + for all a, b > 0 and p ∈ (1, ∞) (1.5) p p0 from Analysis 1+2. If kxkp = 0 or kzkp0 = 0, then xj = 0 or zj = 0 for all j ∈ N by (1.2), and a) holds. Otherwise, we ﬁx a j ∈ N and set a = |xj|/kxkp and b = |zj|/kzkp0 . Estimate (1.5) then yields

p p0 |xjzj| |xj| |zj| ≤ p + p0 kxk kzk 0 p kxk 0 p p p p kzkp0 1.2. More examples of Banach spaces 15

Since both terms on the right are summable, we have xz ∈ `1 and

∞ ∞ ∞ 1 X 1 X p 1 X p0 |xjzj| ≤ p |xj| + p0 |zj| = 1, kxk kzk 0 p kxkp 0 p p j=1 j=1 p kzkp0 j=1 using also (1.4). Thus, a) holds. p p p p p b) Let x, y ∈ ` . Observe that |xk + yk| ≤ 2 (|xk| + |yk| ) for all k ∈ N, so that p p−1 p0 x + y ∈ ` . By part a) and (1.3), the sequence w = (|xk + yk| )k belongs to ` p0 (p−1)p/(p−1) p since wk = |xk + yk| = |xk + yk| is summable. From a) we thus infer

Minkowski’s inequality now follows. ∞ c) The assertions on ` , c and c0 were shown in the exercises. So let p ∈ [1, ∞), p p x, y ∈ ` and α ∈ F. Clearly, αx ∈ ` and kαxkp = |α| kxkp. If kxkp = 0, then p xj = 0 for all j by (1.2). Hence, ` is a normed vector space in view of b). Let p vn = (vn)k∈N be a Cauchy sequence in ` for each n ∈ N. This means that for every ε > 0 there is an Nε ∈ N such that

|(vn)k − (vm)k| ≤ kvn − vmkp ≤ ε for all n, m ≥ Nε and each k ∈ N. So there exists xk = limn→∞(vn)k in F. Set x = (xk)k∈N. Using the second of the above inequalities, we obtain

K X p p p |(vn)k − (vm)k| ≤ kvn − vmkp ≤ ε k=1 for all n, m ≥ Nε and K ∈ N. Letting m → ∞, it follows

K X p p |(vn)k − xk| ≤ ε . k=1

Taking the supremum over K ∈ N, we deduce

∞ X p p |(vn)k − xk| ≤ ε k=1

p p for all n ≥ Nε. As a result, vn − x belongs to ` and converges to 0 in ` as n → ∞; p p i.e., x ∈ ` and vn → x in ` .

Proposition 1.31. For 1 < p < q < ∞ and x ∈ s, we have

1 p q ∞ c00 $ ` $ ` $ ` $ c0 $ ` and kxk∞ ≤ kxkq ≤ kxkp ≤ kxk1, p ` p k·k∞ c00 = ` for 1 ≤ p < ∞ and c00 = c0. 16 Chapter 1. Banach spaces

1 q ∞ − 1 Proof. It is clear that c00 $ ` and ` ⊆ c0 $ ` for all q < ∞. Set zk = k p . p q Then (zk) ∈/ ` , but (zk) ∈ ` ∩ c0 for all 1 ≤ p < q < ∞. It further holds q kxk∞ = |xN | ≤ kxkq for all x ∈ ` ⊆ c0 and some N = N(x) ∈ N, using also (1.2). p 1 Let x ∈ ` \{0} and 1 ≤ p < q < ∞, and set y = x. Then kykp = 1 and kxkp q p |yk| ≤ 1 by (1.2), so that |yk| ≤ |yk| for all k ∈ N. Summing over k, we obtain q p p q kykq ≤ kykp = 1. Thus, 1 ≥ kykq = kxkq/kxkp, whence kxkq ≤ kxkp and ` ⊆ ` . p For the ﬁnal claim, take x ∈ ` if p ∈ [1, ∞) and x ∈ c0 if p = ∞. The ﬁnite sequences vn = (x1, . . . , xn, 0,... ) converge to x in k · kp as n → ∞ showing the asserted density.

B) Lp spaces We have already discussed several Banach spaces with (variants of) the supremum 1 norm, partly in the exercises: C([0, 1]), C0(0, 1), Cb(R), C0(R), and C ([0, 1]). There are various natural combinations and generalizations, e.g., to compact K ⊆ Rd or open U ⊆ Rd or to higher derivatives. We add two related classes of supnorm spaces.

Example 1.32. Let D ⊆ C be open. Then H∞(D) = {f : D → C | f bounded and holomorphic} is a closed subspace of Cb(D) for the supremum norm, and thus a Banach space, since the uniform limit of holomorphic functions is holomorphic due to a convergence theorem by Weierstraß from Funktionentheorie. Moreover, A(D) = C(D, C) ∩ H∞(D) is a closed subspace of H∞(D), and for bounded D the maximum principle yields kfk∞ = supz∈∂D |f(z)| for f ∈ A(D). ♦

We look for a Banach space of functions with respect to the norm kfkp = (R |f|p dµ)1/p, where p ∈ [1, ∞). In the following we brieﬂy repeat the necessary material from Analysis 3 without proofs. Actually some of the statements below are more general than in Analysis 3 since they are formulated for every measure space, but they can be shown in the same way, see also [Rud87] or [Wer06]. A σ–algebra A on a set S 6= ∅ is a collection of subsets A of S satisfying the following properties.

a) ∅ ∈ A.

b) If A ∈ A, then S \ A ∈ A. S c) If Ak ∈ A for all k ∈ , then Ak ∈ A. N k∈N T Note that then S belongs to A and k Ak ∈ A if Ak ∈ A for all k ∈ N. We give a few basic examples for σ–algebras. a) The power set P(S) = {A | A ⊆ S} is a σ–algebra over S. b) Let A ⊂ S. Then {∅, A, S \ A, S} is a σ–algebra over S. c) Let M be a metric space and O = {O ⊂ M | O is open}. The smallest σ– algebra on M that contains O is given by

B(M) := {A ⊂ M | A ∈ A for each σ–algebra A ⊃ O}. and is called Borel σ–algebra on M. It contains all open and closed sets, all countable unions and intersections of closed or open sets, and so on. We write Bd instead d d d of B(R ) and endow C with B2d. For M = R one can replace O by the collection d of all compact sets or of all intervals. We stress that Bd is a strict subset of P(R ). 0 0 For B ∈ Bd one has B(B) = {A ∈ Bd | A ⊂ B} = {A ∩ B | A ∈ Bd}. d) On [0, ∞] one considers the metric discussed in the exercises for which also (a, ∞] for a ∈ [0, ∞) is an open set. One obtains B([0, ∞]) ⊃ B(R+). 1.2. More examples of Banach spaces 17

Let A be a σ–algebra on S.A (positive) measure µ on A is a map µ : A → [0, ∞] such that µ(∅) = 0 and

∞ [ X µ Ak = µ(Ak) for all pairwise disjoint Ak ∈ A, k ∈ N. k∈N k=1 The tripel (S, A, µ) is called a measure space. It is called finite if µ(S) < ∞, and S σ–finite if there are Sk ∈ A such that µ(Sk) < ∞ for all k ∈ N and k Sk = S. Let A, B, Ak ∈ A for k ∈ N with A ⊂ B and µ be a measure on A. We then obtain µ(A) ≤ µ(B), and for µ(A) < ∞ we have µ(B\A) = µ(B)−µ(A). Moreover, S P∞ µ( k Ak) ≤ k=1 µ(Ak). Continuing the above examples, we present a few measure spaces. a) Given s ∈ S, we define δs(A) = 1 if s ∈ A and δs(A) = 1 if s∈ / A. Then the point measure δs is a finite measure on P(S). b) For A ⊂ N, we define the counting measure ζ(A) as the number of elements of A. It is a σ–finite measure on P(N). c) On Bd there is exactly one measure λ = λd such that λ(J) = `1 · ... · `d for any interval J with side lengths `j. It is σ–finite and called Lebesgue measure. Let B ∈ Bd. The restriction λB = λ of λ to B(B) is a σ–finite measure, also called Lebesgue measure. Unless otherwise specified, we endow Borel sets in Rd with B(B) and λ. Let A and A0 be σ–algebras on S and S0, respectively. A map f : S → S0 is called measurable if f −1(A) ∈ A for every A ∈ A0. We collect the basic properties of measurable maps.

1 a) Let A ⊂ S. The characteristic function A : S → F is measurable if and only if A ∈ A. Linear combinations of measurable characteristic functions are called simple functions.

b) Compositions of measurable functions are measurable.

c) Continuous functions are measurable for the respective Borel σ–algebras.

d) A function with values in Fd is measurable if and only if the real and imaginary parts of all its components are measurable.

e) By F˜ we denote temporarily either F or [0, ∞] endowed with the Borel σ– ˜ algebras. Let f, g, fn : S → F be measurable for n ∈ N (for any σ–algebra on S) and let α ∈ F˜. Then the functions αf, f + g, fg, |f|, and (if it ˜ exists) limn→∞ fn : S → F; t 7→ limn→∞ fn(t) are measurable (where we set ˜ 0 · ∞ = 0). If F 6= C, then also the functions supn fn, infn fn, lim supn fn and lim infn fn are measurable (provided they exist if F = R). In particular, we obtain the measurability of the positive part f+ = max{0, f} and the negative part f− = max{0, −f} of a real–valued measurable function f.

Let (S, A, µ) be a measure space and f : S → [0, ∞] be measurable. Then f can be approximated monotonically by simple functions fn : S → [0, ∞). Using this R fact, one can then define an integral S f dµ which belongs to [0, ∞]. It is monotone, additive and positive homogeneous in f. The function f is called integrable if the integral is finite. A measurable function f : S → F is called integrable if the positive measurable maps (Re f)± and (Im f)± are integrable, and then one defines the integral of f by Z Z Z Z f dµ = f(s) dµ(s) := (Re f)+ dµ − (Re f)− dµ S S S 18 Chapter 1. Banach spaces

Z Z + i (Im f)+ dµ − i (Im f)− dµ. S S One writes dx instead of dλ or dλ(x) for the Lebesgue measure. The integral has the following properties. a) A measurable function f is integrable if and only if the (positive and measurable) function |f| is integrable. In this case, one has | R f dµ| ≤ R |f| dµ.

b) The space of integrable functions f : S → F is a vector space. The integral is linear and (if F = R) monotone in f. c) If f : S → [0, ∞] is measurable, the integral is equal to 0 if and only if the set {s ∈ S | f(s) 6= 0} ∈ A has measure 0.

p R p For p ∈ [1, ∞) and measurable f : S → F, we deﬁne kfkp := S |f| dµ and

p p p L (µ) = L (S) = L (S, A, µ) = {f : S → F | measurable, kfkp < ∞}.

One can show Minkowski’s inequality as in Proposition 1.30 so that k · kp is a seminorm. We list again a few basic examples. R a) For every function f : S → F and s ∈ S, we have S f dδs = f(s). p p R P∞ b) We have L (N, P(N), ζ)=` for p∈[1, ∞) and f dζ = k=1 f(k). p N c) Let B ∈ Bd. Here we usually write L (B) instead of L(B, B(B), λ). A measurable function f : B → F belongs to Lp(B) if and only if its 0–extension fe belongs to Lp(Rd), where p ∈ [1, ∞), and we have Z Z Z f dx = f dλB = fe dx. d B B R The crucial results for the Lebesgue integral are the following convergence theorems, where (S, A, µ) is a measure space and p ∈ [1, ∞).

a) Fatou’s lemma: For measurable fn : S → [0, ∞], n ∈ N, we have Z Z lim inf fn dµ ≤ lim inf fn dµ. S n→∞ n→∞ S

b) Monotone convergence: Let fn : S → [0, ∞] and fn ≤ fn+1 for all n ∈ N. Then the following limits exist in [0, ∞] and satisfy Z Z lim fn dµ = lim fn dµ. S n→∞ n→∞ S

p c) Dominated convergence, Lebesgue: Let fn, g ∈ L (µ) such that |fn(s)| ≤ g(s) p for all n ∈ N and fn(s) → f(s) as n → ∞, for every s ∈ S. Then f ∈ L (µ) and kfn − fkp → 0 as n → ∞. If p = 1, we also have Z Z f dµ = lim fn dµ. S n→∞ S

p d) Riesz–Fischer: Let (fn) be a Cauchy sequence in L (µ). Then there are p a subsequence (fnk )k, a set N ∈ A with µ(N) = 0 and f, g ∈ L (µ) such that

|fnk (s)| ≤ g(s) for all k ∈ N and fnk (s) → f(s) as k → ∞, for every s ∈ S \ N, and such that kfn − fkp as n → ∞. p It would be nice if L (µ) was a Banach space, but unfortunately kfkp = 0 does not imply f = 0, but only that the set {s ∈ S | f(s) 6= 0} has measure 0. This fact leads us to the following concept. 1.2. More examples of Banach spaces 19

Let (S, A, µ) be a measure space. A set N ∈ A is called a null set if µ(N) = 0. A property which holds for all s ∈ S \ N and a null set N is said to hold almost everywhere (a.e.). For instance, the Riesz–Fischer theorem says that the subsequence fnk converges pointwise a.e. to f. In the above convergence theorems, one replace pointwise properties by their counterparts holding a.e.. We note that a countable union of null sets is a null set and that M ∈ A is a null set if it is contained in a null set. Hyperplanes or each countable subset are null sets in Rd for the Lebesgue measure. Uncountable unions of null sets can have even inﬁnite measure; e.g., R is the union of all singletons {x}. We further introduce the set of null functions

N = {f : S → F | f is measurable, f = 0 a.e.} which is a linear subspace of Lp(µ) for every 1 ≤ p < ∞. We can thus define the vector space Lp(µ) := Lp(µ)/N = {fˆ = f + N | f ∈ Lp(µ)}. We further set Z Z ˆ kf + N kp := kfkp, resp. f dµ := f dµ, S S for each fˆ ∈ Lp(µ), resp. fˆ ∈ L1(µ), and any f ∈ fˆ. It can be seen that these definitions do not depend on the choice of the representative. By means of the p Riesz–Fischer theorem, one can show that (L (µ), k · kp) is a Banach space. In Lp(µ), we write f ≥ 0 if g ≥ 0 a.e. for some representative g of f. It then follows that g ≥ 0 a.e. for all representatives of f. Analogous statements hold for the relations = or >. One usually identifies fˆ with f, and works with Lp(µ) instead of Lp(µ). The set of simple functions in Lp(µ) is dense in Lp(µ) if p ∈ [1, ∞), see e.g. Satz IV.7.6 in [Wer06].

p Example 1.33. Let B ∈ Bd and p ∈ [1, ∞). The set P = {f ∈ L (B) | f ≥ 0} is closed in Lp(B) and contains no interior point.

Proof. Let fn ∈ P converge to f in X. Choose representatives gn of fn and g of f. The Riesz–Fischer theorem gives a subsequence and a null set N (possibly depending on the representatives) such that gnj (x) → g(x) for all x∈ / N as j → ∞. On the other hand, there are null sets Nj such that gn (x) ≥ 0 for all x∈ / Nj and j S j ∈ N. Therefore g(x) ≥ 0 for all x not contained in the nullset j Nj ∪ N. This means that f ∈ P , and so P is closed. Let f ∈ P . There is a constant c ≥ 0 such that the set B0 = {x ∈ B | 0 ≤ n f(x) ≤ c} is not a null set (otherwise f would be equal to ∞ a.e.). Let (Cj )j be a d countable disjoint covering of R with cubes of volume 1/n. Since λ(B0) > 0, for each n ∈ N there is an index j such that A = B ∩Cn has measure in (0, 1/n]. Set n n 0 jn 1 1/p fn = f −(c+1) An . Then fn does not belong to P and kf −fnkp ≤ (c+1)λ(An) tends to 0 as n → ∞; i.e., f is not an interrior point of P .

For a measure space (S, A, µ), we further introduce the space

∞ ∞ ∞ L (µ) = L (S) = L (S, A, µ) := {f : S → F | f measurable and bounded a.e.}

For a measurable function f : S → F we deﬁne the essential supremum norm by

kfk∞ = ess sup |f(s)| := inf{c ≥ 0 | |f(s)| ≤ c for a.e. s ∈ S} ∈ [0, ∞]. s∈S 20 Chapter 1. Banach spaces

Of course, f belongs to L∞(µ) if and only if its essential supremum norm is finite. One can see that for the Lebesgue measure on B ∈ Bd this definition coincides with the usual supremum norm if f is continuous and if λ(B ∩B(x, r)) > 0 füralle x ∈ B und r > 0. We also set

∞ ∞ L (µ) := L (µ)/N and kf + N k∞ := kfk∞

∞ Proposition 1.34. Let (S, A, µ) be a measure space. Then (L (µ), k · k∞) is a Banach space.

∞ Proof. Let fk ∈ L (µ) and αk ∈ F for k = 1, 2. Then there are ck ≥ 0 and null sets Nk such that |fk(s)| ≤ ck for all s ∈ S \ Nk. We thus obtain |α1f1(s) + α2f2(s)| ≤ |α1| c1 +|α2| c2 for s∈ / N1 ∪N2 = N, where N is a null set. Therefore, α1f1 +α2f2 ∈ ∞ ∞ L (µ), and so L (µ) is a vector space. Moreover, kf1 +f2k∞ ≤ kf1k∞ +kf2k∞. It ∞ is clear that kα1f1k∞ = |α1| kf1k∞. Observe that N is a linear subspace of L (µ) ∞ ˆ and that kf1k∞ = kf2k∞ if f1 = f2 a.e.. As result, L (µ) is a vector space, kfk∞ is well deﬁned, and it is a norm on L∞(µ). ˆ ∞ ˆ Let (fn) be Cauchy in L (µ). Fix functions fn ∈ fn. For every j ∈ N there is a 1 k(j) ∈ N such that kfn − fmk∞ ≤ j for all n, m ≥ k(j). Hence, the set

2 Nn,m,j = {s ∈ S | |fn(s) − fm(s)| > j } S is a null set for these n, m, j. Deﬁne the null set N := Nn,m,j. For n,m≥k(j),j∈N s ∈ S \ N, we then obtain

2 |f (s) − f (s)| ≤ for all n, m ≥ k(j) and j ∈ . n m j N

There thus exists f(s) = limn→∞ fn(s) in F for all s ∈ S \ N. We set f(s) = 0 for all s ∈ N. Then f is measurable. Let ε > 0 and take j ≥ 1/ε. It follows that

ˆ ˆ 2 kfn − fk∞ ≤ sup |fn(s) − f(s)| = sup lim |fn(s) − fm(s)| ≤ ≤ 2ε s∈S\N s∈S\N m→∞ j

ˆ ∞ ˆ ˆ ∞ if n ≥ k(j). Hence, f ∈ L (µ) and fn → f in L (µ) as n → ∞.

Proposition 1.35 (H¨older). Let (S, A, µ) be a measure space, p ∈ [1, ∞], f ∈ 0 Lp(µ), and g ∈ Lp (µ). Then the following assertions hold. 1 a) We have fg ∈ L (µ) and kfgk1 ≤ kfkp kgkp0 . b) If µ(S) < ∞ and 1 ≤ p < q ≤ ∞, then Lq(µ) ⊆ Lp(µ) and

1 Z 1 1 1 1 p p p − 1 p 1 p p q kfkp = |f| dµ ≤ k kr0 k |f| kr = µ(S) kfkq, S where r = q/p > 1 and r0 = q/(q − p) by (1.3).

The inclusion in Proposition 1.35b) is strict, as can be seen by the function (0, 1) 3 t 7→ f(t) = t−1/q. Similarly, one checks that there is no inclusion between Lq(Rd) and Lp(Rd) if p 6= q.

1.3 Compactness and separability

Compactness is one of the most important concepts in analysis, as one could already see in Analysis 1+2. We ﬁrst deﬁne this notion and some variants in a metric space. 1.3. Compactness and separability 21

Definition 1.36. Let M be a metric space and K ⊆ M. a) K is compact if every open covering C of K (i.e., a set C of open sets O ⊆ M S with K ⊆ {O | O ∈ C}) contains a finite subcovering {O1,...,Om} ⊆ C with K ⊆ O1 ∪ · · · ∪ Om. b) K is sequentially compact if every sequence (xn) in K possesses a subsequence converging to some x ∈ K. c) K is relatively compact if K is compact. d) K is totally bounded if for each ε > 0 there are m ∈ N and x1, . . . , xm ∈ M such that K ⊆ B(x1, ε) ∪ · · · ∪ B(xm, ε). In the definition of total boundedness, one can equivalently require that the points xj belong to K. In other words, if K is totally bounded in M, then it is also totally bounded in the metric space (K, dK ). To show this, take ε > 0 and the points x1, ··· , xm from Definition 1.36 with ε replaced by ε/2. If B(xj, ε/2) ∩ K is empty, we drop this element xj. Otherwise we replace it by some yj in B(xj, ε/2) ∩ K. Then B(xj, ε/2) ⊂ B(yj, ε), and hence also the balls B(yj, ε) cover K. If K is totally bounded, then K is bounded since K ⊆ B(x1, r) with r = ε + maxj∈{1,...,m} d(x1, xj) and the points xj from part d) of the above definition. As in Analysis 1 one sees that a sequence (xn) in a metric space M has a subsequence converging to some x in M if and only if x is an accumulation point of the sequence; i.e., for each r > 0 and m ∈ N there is an n ≥ m with xn ∈ B(x, r) (or equivalently, in each B(x, r) there are infinitely many xn). Theorem 1.37. A subset K of a metric space M is compact if and only if it sequentially compact.

Proof. (1) Let K be compact and suppose it is not sequentially compact. There thus exists a sequence (xn) in K without an accumulation point in K. In other words, for each y ∈ K, there is an ry > 0 such that B(y, ry) contains only finitely S many xn. Since K ⊆ y∈K B(y, ry) and K is compact, there are y1, . . . , ym ∈ K such that K ⊆ B(y1, ry1 ) ∪ · · · ∪ B(ym, rym ) =: B. But this inclusion cannot hold because B only contains finitely many xn; i.e., K must be sequentially compact. (2a) Claim: Let K be a subset of a metric space M such that each sequence in K has an accumulation point in K. Then K is totally bounded. Proof of the claim. Suppose that K were not totally bounded. There would thus exist an r > 0 such that K cannot be covered by finitely many balls of radius r. As a result, there exists an x1 ∈ K such that K 6⊆ B(x1, r). Take any x2 ∈ K \B(x1, r). Hence, d(x2, x1) ≥ r. Since K cannot be contained in B(x1, r) ∪ B(x2, r), we find an x3 ∈ K \ (B(x1, r) ∪ B(x2, r)) implying that d(x3, x1) ≥ r and d(x2, x1) ≥ r. Inductively, we obtain a sequence (xn) in K with d(xn, xm) ≥ r > 0 for all n > m. By assumption, this sequence has an accumulation point x. This means that there are infinitely xnj in B(x, r/2) and thus d(xnj , xni ) < r for all i 6= j. This is impossible so that K is totally bounded. ♦ (2b) Let K be sequentially compact and let C be an open covering of K. We suppose that no finite subset of C covers K. Due to part (2a), for each n ∈ N there are finitely many balls of radius 1/n covering K. For every n ∈ N, we thus find a ball Bn or radius 1/n such that Bn ∩ K is not covered by finitely many sets from C. Since K is sequentially compact, there is an accumulation point xb ∈ K of (xn). There further exists an open set Ob ∈ C with xb ∈ Ob, and hence B(x,b ε) ⊂ Ob for some ε > 0. We then obtain an index N ∈ N such that d(xN , xb) < ε/2 and N ≥ 2/ε. Each x ∈ BN thus satisfies

1 ε d(x, xb) ≤ d(x, xN ) + d(xN , xb) < N + 2 ≤ ε; i.e., BN ⊂ B(x,b ε) ⊂ Ob. This contradiction implies that K is compact. 22 Chapter 1. Banach spaces

Corollary 1.38. Let M be a complete metric space and N ⊆ M. Then the following assertions are equivalent.

a) N is relatively compact.

b) Each sequence in N has an accumulation point in N.

c) N is totally bounded.

Proof. The implication “a)⇒b)” follows from Theorem 1.37 applied to N, whereas “b)⇒c)” was shown in the proof of this theorem. Let N be totally bounded, and take a sequence (xn) in N. There are xen ∈ N such that d(xn, xen) ≤ 1/n for every n ∈ N. By assumption, N is covered by finitely 1 many balls Bj in M of radius 1. We can then find an index j1 and a subsequence ν1 such that x ∈ B1 for all k ∈ . There further exist finitely many balls B2 in eν1(k) j1 N j M of radius 1/2 which cover N. Again, there is an index j2 and a subsequence ν2 of ν such that x ∈ B2 for all k ∈ . By induction, for every m ∈ we obtain 1 eν2(k) j2 N N a subsequence ν of ν and a ball Bm of radius 1/m such that x ∈ Bm m m−1 jm eνm(k) jm for all k ∈ N. We now define the diagonal sequence nm = νm(m) for m ∈ N. Note that x , x ∈ Bm for p ≥ m since n and n belong to ν . Take ε > 0. Choose enm enp jm p m m m ∈ N with m, nm ≥ 1/ε. It follows that 1 2 1 d(x , x ) ≤ d(x , x ) + d(x , x ) + d(x , x ) ≤ + + ≤ 4ε nm np nm enm enm enp enp np nm m np

for all p ≥ m. Because M is complete, the Cauchy sequence (xnm )m has the limit x ∈ N. Theorem 1.37 now implies that N is compact; i.e., a) holds.

Corollary 1.39. Let K be a compact subset of a metric space M. Then K is closed and bounded.

Proof. The boundedness follows from Corollary 1.38. Let xn ∈ K converge to some x ∈ M. Since K is sequentially compact, there is a subsequence (xnk )k and a y ∈ K with xnk → y as k → ∞. Hence, x = y ∈ K and K is closed.

Example 1.40. a) A subset K of a ﬁnite dimensional normed vector space X is compact if and only if K is closed and bounded. (See Analysis 2.) In particular, B(0, 1) is compact in ﬁnite dimensions. p b) Let X = ` , 1 ≤ p ≤ ∞, and en = (0,..., 0, 1, 0,... ) ∈ B(0, 1) be the n–th 1 unit vector. We then have ken −emkp = 2 p if n 6= m, so that (en) has no converging subsequence. As a result, the closed (and bounded) unit ball in `p is not compact. c) Let X = C([0, 1]) and, for n ∈ N,  2n+1t − 2, 2−n ≤ t ≤ 3 · 2−n,  2 n+1 3 −n −n+1 fn(t) = 4 − 2 t, 2 · 2 ≤ t ≤ 2 , 0, otherwise.

Then kfnk∞ = 1 and kfn − fmk∞ = 1 for n 6= m, so that again the closed unit ball is not compact. ♦ The next theorem shows that the above simple examples are typical.

Theorem 1.41. Let X be a normed vector space. The closed unit ball B = B(0, 1) in X is compact if and only if dim X < ∞. 1.3. Compactness and separability 23

Proof. If dim X < ∞, then B is compact by Example 1.40 a). Let dim X = ∞. Take any x1 ∈ X with kx1k = 1. Set U1 = lin{x1}. Since U1 is finite dimensional, U1 is closed in X due to Lemma 1.42 and X 6= U1 because of dim X = ∞. Lemma 1.43 1 thus yields an x2 ∈ X with kx2k = 1 and kx2−x1k ≥ 2 . Then U2 := lin{x1, x2}= 6 X 1 is also closed, and Lemma 1.43 gives an x3 ∈ X with kx3k = 1, kx3 − x2k ≥ 2 , 1 and kx3 − x1k ≥ 2 . Inductively, we can now construct a sequence (xn) ⊆ B mit 1 kxn − xmk ≥ 2 for all n 6= m, so that B is not sequentially compact. Lemma 1.42. Let Y be a finite dimensional subspace of a normed vector space X. Then Y is closed in X. m Proof. Choose a basis B = {b1, . . . , bm} of Y . Lety ¯ = (y1, . . . , ym) ∈ F be the (uniquely determined) vector of coefficients with respect to B for a given y ∈ Y . m Set y¯ := kyk. This gives a norm on F , which is equivalent to |·|2 by Analysis 2, and9 thus9 it is complete. Let vn ∈ Y converge to some x in X. Then (¯vn) is Cauchy m m in F , and so there is a vectorz ¯ = (z1, . . . , zm) ∈ F such thatv ¯n → z¯ in · as n → ∞. Set z = z1b1 + ··· + zmbm ∈ Y . We then deduce that vn → z in9X 9and thus x = z ∈ Y ; i.e., Y is closed. Lemma 1.43 (F. Riesz). Let X be a normed vector space, Y 6= X be a closed linear subspace, and δ ∈ (0, 1). Then there exists an x¯ ∈ X with kx¯k = 1 and kx¯ − yk ≥ 1 − δ for all y ∈ Y .

Proof. Take any x ∈ X \ Y . Since X \ Y is open, we have d := infy∈Y kx − yk > 0. Hence, d < d/(1 − δ) and there is a vectory ¯ ∈ Y with kx − y¯k ≤ d/(1 − δ). Set 1 x¯ = kx−y¯k (x − y¯). We then obtain kx¯k = 1 and, using the above inequalities, 1 d kx¯ − yk = kx − (¯y + kx − y¯ky)k ≥ ≥ 1 − δ kx − y¯k kx − y¯k for all y ∈ Y . We note some important consequences of compactness essentially known from Analysis 2. Theorem 1.44. Let X be a normed vector space, K be a compact metric space, and f ∈ C(K,X). Then f is uniformly continuous and bounded. If X = R, then there are t± ∈ K such that f(t+) = maxt∈K f(t) and f(t−) = mint∈K f(t).

Proof. 1) Let ε > 0. Since f ∈ C(K,X), for every t ∈ K there is a δt > 0 such S 1 that kf(t) − f(s)k < ε for all s ∈ B(t, δt) ⊆ K. Since K = t∈K B(t, 3 δt) and K 1 1 is compact, there are t1, . . . , tm ∈ K such that K ⊆ B(t1, 3 δ1) ∪ · · · ∪ B(tm, 3 δm), 1 1 where δk := δtk . Set δ = min{ 3 δ1,..., 3 δm} > 0. Take any s, t ∈ K with 1 d(s, t) < δ. Then there are indices k, l ∈ {1, . . . , m} such that s ∈ B(tk, 3 δk) and 1 t ∈ B(tl, 3 δl), where we may assume that δk ≥ δl. We thus obtain 1 1 d(t , t ) ≤ d(t , s) + d(s, t) + d(t, t ) < δ + δ + δ ≤ δ k l k l 3 k 3 l k so that tl ∈ B(tk, δk), and hence

kf(s) − f(t)k ≤ kf(s) − f(tk)k + kf(tk) − f(tl)k + kf(tl) − f(s)k < 3ε.

2) Suppose there were tn ∈ K with kf(tn)k → ∞ as n → ∞. Since K is compact, there is a subsequence (tnj ) converging to some t ∈ K. Then kf(tnj )k → kf(t)k as j → ∞, due to the continuity of t 7→ kf(t)k, which contradicts kf(tn)k → ∞. 3) Let X = R. By part 2), the supremum s := supt∈K f(t) belongs to R. We can ﬁnd rn ∈ K with f(rn) → s as n → ∞. The compactness of K gives again a subsequence (rnj ) converging to some t+ ∈ K. Hence, f(t+) = limj→∞ f(rnj ) = s since f is continuous. The minimum is treated in the same way. 24 Chapter 1. Banach spaces

Theorem 1.41 shows that in an inﬁnite dimensional Banach space a closed and bounded subset does not need to be compact. The next two results give stronger suﬃcient conditions for (relative) compactness in `p and C(K). (It can be seen that the conditions are in fact necessary.)

Proposition 1.45. Let p ∈ [1, ∞). A set K ⊆ `p is relatively compact if it is bounded and ∞ X p lim sup |xj| = 0. N→∞ (xj )∈K j=N+1

Proof. Due to Corollary 1.38, it suﬃces to prove that K is totally bounded. So let ε > 0. By the assumption, there is an N ∈ N such that

∞ X p p |xj| < ε for all (xj) ∈ K. j=N+1

N ˆ For x = (xj) ∈ K, putx ˆ = (x1, . . . , xN ) ∈ F . Since |xˆ|p ≤ kxkp, the set K = {xˆ | x ∈ K} is bounded in FN , and thus it is totally bounded by Example 1.40 and m Corollary 1.38. So we obtain vectorsv ˆ1,..., vˆm ∈ F such that for all x ∈ K there p is an index l ∈ {1, . . . , m} with |xˆ − vˆl|p < ε. Set vk = (ˆvk, 0,... ) ∈ ` for all k ∈ {1, . . . , m}. The total boundedness of K now follows from

∞ p p X p p kx − vlkp = |xˆ − vˆl|p + |xj| < 2ε . j=N+1

Theorem 1.46 (Arzela–Ascoli). Let K be a compact metric space and F ⊆ C(K) be bounded and equicontinuous, i.e.,

∀ ε > 0 ∃ δε > 0 ∀ f ∈ F ∀ s, t ∈ K with d(s, t) ≤ δε : |f(s) − f(t)| ≤ ε. (1.6)

Then F is compact. If F is also closed, it is compact.

Proof. 1) For every εN := 1/N and N ∈ N, we have δN := δεN from (1.6). Since K is compact, Corollary 1.38 gives points t1,N , . . . , tmN ,N ∈ K with K ⊆ B(t1,N , δN )∪

· · · ∪ B(tmN ,N , δN ). We write {tk,N : k = 1, . . . , mN ; N ∈ N} = {sl : l ∈ N}. 2) Take a sequence (f ) in F . Since F is bounded, we have sup |f (s )| ≤ n n∈N n 1 supf∈F kfk < ∞. So there exists a converging subsequence (fν1(k)(s1))k in F.

Similarly, we obtain a subsequence ν2 of ν1 such that (fν2(k)(s2))k converges. Note that fν2(k)(s1) still converges in F as k → ∞. We iterate this procedure and set fnj = fνj (j) for all j ∈ N. Then fnj (sl) converges as j → ∞ for each ﬁxed l ∈ N, since fnj (sl) = fνl(kj )(sl) for all j ≥ l and a sequence kj → ∞ (depending on l).

3) Let ε > 0. Fix N ∈ N such that εN = 1/N ≤ ε. Take δN = δεN > 0 from (1.6) and the points tk,N from 1). By part 2) there is an index Jε ∈ N such that

|fni (tk,N )−fnj (tk,N )| ≤ ε for all i, j ≥ Jε and k ∈ {1, . . . , mN }. Let t ∈ K. There is an index l ∈ {1, . . . , mN } with d(t, tl,N ) < δN due to part 1). We can thus estimate

|fni (t) − fnj (t)| ≤ |fni (t) − fni (tl,N )| + |fni (tl,N ) − fnj (tl,N )| + |fnj (tl,N ) − fnj (t)|

≤ εN + ε + εN ≤ 3ε

for all i, j ≥ Jε and all t ∈ K; i.e., (fnj )j is Cauchy in C(K). This space is complete

(which can be shown as in Example 1.4) so that fnj converges in C(K) to some f in F as j → ∞. The compactness of F now follows from Corollary 1.38. 1.3. Compactness and separability 25

Corollary 1.47. Let α ∈ (0, 1] and K be a compact metric space. Let (fn) be bounded and uniformly H¨older(Lipschitz if α = 1) continuous, i.e.,

α ∃c ≥ 0 : |fn(t)| ≤ c, |fn(t)−fn(s)| ≤ c d(t, s) for all t, s ∈ K, n ∈ N. (1.7)

Then there is a subsequence fnj and a fucntion f ∈ C(K) such that kfnj −fk∞ → 0 as j → ∞. Moreover, f satisﬁes (1.7).

Proof. The ﬁrst part follows from Theorem 1.46 with F = {fn | n ∈ N}, noting that 1/α (1.6) holds with δε = (ε/c) . The last claim can be shown as in Example 1.14. Condition (1.7) is true for K = [0, 1], α = 1 and bounded sequences in C1([0, 1]).

Example 1.48. a) The sequence (fn) in Example 1.40c) is not equicontinuous since 3 −n −n 3 −n −n −n−1 fn( 2 2 ) − fn(2 ) = 1, but 2 2 − 2 = 2 → 0 as n → ∞. b) Arzela–Ascoli is wrong for Cb(R) or C0(R): Consider the sequence (fn) in Cb(R) given by fn(t) = 0 for |t − n| ≥ 1/2 and fn(t) = 1 − 2 |t − n| for |t − n| < 1/2, which is bounded and equicontinuous, but kfn − fmk∞ = 1 for n 6= m. 2 R 1 c) Let k ∈ C([0, 1] ) and set T f(t) = 0 k(t, s)f(s) ds for t ∈ [0, 1] and f ∈ X = C([0, 1]). Then the set F = {T f | f ∈ BX (0, 1)} is relatively compact.

Proof. Let f ∈ BX (0, 1). By Analysis 1+2 the function g = T f belongs to X. Clearly, kT fk∞ ≤ kkk∞ kfk∞ ≤ kkk∞, so that F is bounded. Let ε > 0. Since k is uniformly continuous, there is a δ > 0 such that |k(t, s) − k(t0, s)| ≤ ε for all t, t0, s ∈ [0, 1] with |t0 − t| ≤ δ. Arzela–Ascoli now implies the assertion since

Z 1 |T f(t0) − T f(t)| ≤ |k(t, s) − k(t0, s)| |f(s)| ds ≤ δ 0 for all f ∈ F and t, t0 ∈ [0, 1] with |t0 − t| ≤ δ. We add another concept that will be needed later in the course. Deﬁnition 1.49. A metric space is called separable if it contains a countable dense subset. In an exercise it will be shown that separability is preserved under homeomor- phisms. Lemma 1.50. Let X be a normed vector space and Y ⊆ X be a countable subset such that lin Y is dense in X. Then X is separable. Proof. The set

n n X o lin Y = y = q y y ∈ Y, q ∈ (or + i ), n ∈ Q j j j j Q Q Q N j=1 is countable, since Y is countable. Let x ∈ X and ε > 0. By assumption, there exists a y ∈ lin Y with kx − yk ≤ ε. We can also ﬁnd a z ∈ linQ Y with ky − zk ≤ ε. Hence, kx − zk ≤ 2ε.

p Example 1.51. a) The spaces ` , 1 ≤ p < ∞, and c0 are separable since c00 = lin{ek | k ∈ N} is dense in all of them by Proposition 1.31. n b) The space C([0, 1]) is separable since lin{pn | n ∈ N} with pn(t) = t is dense in C([0, 1]) by Weierstraß’ approximation theorem from Analysis 3. c) The space Lp(Rd) is separable for p ∈ [1, ∞). In fact, due to Korollar IV.7.8 1 d p d in [Wer06] the linear hull of Y1 := { I | I is an interval in R } is dense in L (R ). Let Y be the subset of Y1 where the intervals have rational vertices. For all ϕ ∈ Y1 26 Chapter 1. Banach spaces

and ε > 0 there is a ψ ∈ Y such that kϕ − ψkp ≤ ε. Hence, the linear hull of the countable set Y is dense in Lp(Rd). d) The space `∞ is not separable. In fact, the set Ω of {0, 1}–valued sequences is uncountable and two diﬀerent sequences in Ω have the distance 1. Suppose that ∞ S {vk | k ∈ N} were dense in ` . Then, Ω ⊆ k B(vk, 1/4) and thus each ω ∈ Ω belongs to exactly one of the balls, say to B(vk(ω), 1/4). This means that the map Ω → N; ω 7→ k(ω) is injective, which contradicts the uncountability of Ω. ♦ Chapter 2

Continuous linear operators

2.1 Basic properties and examples

Let X and Y be normed vector spaces. We denote their norms by k · kX and k · kY , or just by k · k. The space of linear mappings T : X → Y is designated by L(X,Y ). We write T x instead of T (x) and ST ∈ L(X,Z) instead of S ◦ T for all x ∈ X, T ∈ L(X,Y ), S ∈ L(Y,Z) and vector spaces Z. It is known from Linear Algebra that L(X,Y ) is a vector space with respect to the operations

(T + S)x := T x + Sx, (αT )x = αT x (∀ x ∈ X, α ∈ F,T,S ∈ L(X,Y )).

Note that T 0 = 0. If dim X < ∞ and dim Y < ∞, each operator T ∈ L(X,Y ) can be represented by a matrix, and it is thus continuous. In inﬁnite dimensional spaces there are discontinuous linear maps.

Example 2.1. a) By T (xk) = (kxk) we deﬁne a linear map T : c00 → c00. It 1/2 −1/2 −1/2 holds T (vn) = n en for vn = n en and n ∈ N. Since kvnkp = n → 0 and 1/2 kT (vn)kp = n → ∞ as n → ∞, T is not continuous for any p–norm on c00. b) The map T f = f 0 is linear from C1([0, 1]) to C([0, 1]), but discontinuous if we −1/2 endow both spaces with the norm kfk∞. (Consider fn(t) = n sin(nt), where −1/2 0 0 1/2 kfnk∞ ≤ n and kfnk∞ ≥ |fn(0)| = n .) On the other, T is continuous if we 1 0 equip C ([0, 1]) with the norm kfkC1 = kfk∞ + kf k∞ and C([0, 1]) with kfk∞.

Lemma 2.2. Let X and Y be normed vector spaces and T : X → Y be linear. The following assertions are equivalent.

a) T is Lipschitz continuous.

b) T is continuous.

c) T is continuous at x = 0.

d) T is bounded: There is a c > 0 with kT xkY ≤ c kxkX for every x ∈ X.

Proof. a) The implications a) ⇒ b) ⇒ c) are clear. c) ⇒ d): By c) and T 0 = 0 there exists a δ > 0 such that kT zk ≤ 1 for all z ∈ X δ with kzk ≤ δ. Let x ∈ X \{0}, and set z = kxk x. Since T is linear, we deduce δ 1 1 ≥ kT zk ≥ kxk kT xk and hence d) with c = δ . d) ⇒ a): The linearity of T and d) yield kT x − T zk = kT (x − z)k ≤ c kx − zk for all x, z ∈ X.

27 28 Chapter 2. Continuous linear operators

Deﬁnition 2.3. For normed vector spaces X and Y , we set

L(X,Y ) = {T : X → Y | T is linear and continuous}1 and put L(X,X) = L(X). The dual space L(X, F) of X is denoted by X?. An element x? ∈ X? is called linear functional on X, and one often writes hx, x?i instead of x?(x). For T ∈ L(X,Y ) the operator norm is given by

kT kL(X,Y ) = kT k = inf{c ≥ 0 | ∀ x ∈ X we have kT xkY ≤ c kxkX } < ∞.

Remark 2.4. Let X,Y,Z be normed vector spaces, x ∈ X,T ∈ L(X,Y ), and S ∈ L(Y,Z). Then the following assertions are true.

a) L(X,Y ) does not change if we replace the norms on X or Y by equivalent ones, though kT k may change.

b) kIk = 1, where I : X → X; Ix = x.

(1) kT xk (2) (3) c) kT k = supx6=0 kxk = supkxk≤1 kT xk = supkxk=1 kT xk =: s.

d) kT xk ≤ kT k kxk.

e) ST ∈ L(X,Z) and kST k ≤ kSk kT k.

Proof. a) and b) are clear. c) Let x 6= 0. For every ε > 0, Deﬁnition 2.3 yields kT xk ≤ (kT k + ε) kxk. This inequality gives ”≥” in (1). We clearly have ”≥” in (2) and (3). Further,

kT xk = T 1 x ≤ s, kxk kxk so that kT k ≤ s. kT xk d) is obvious for x = 0. For x 6= 0, from part c) we deduce kxk ≤ kT k. e) Part d) implies that kST xk ≤ kSk kT xk ≤ kSk kT k kxk, whence e) follows.

Proposition 2.5. Let X and Y be normed vector spaces. Then L(X,Y ) is a normed vector space with respect to the operator norm. If Y is a Banach space, then L(X,Y ) is also a Banach space. In particular, X? is a Banach space.

Proof. It is clear that L(X,Y ) is a vector space. If kT k = 0, then T x = 0 for all x ∈ X by Remark 2.4, and hence T = 0. Let T,S ∈ L(X,Y ), x ∈ X, and α ∈ F. We then deduce from Remark 2.4 that

kT + Sk = sup k(T + S)xk ≤ sup (kT xk + kSxk) ≤ kT k + kSk, kxk=1 kxk=1 kαT k = sup kαT xk = |α| sup kT xk = |α| kT k. kxk=1 kxk=1

Thus, L(X,Y ) is a normed vector space. Assume that Y is a Banach space. Let (Tn) be a Cauchy sequence in L(X,Y ); i.e., for every ε > 0 there is an Nε ∈ N such that kTn − Tmk ≤ ε for all n, m ≥ Nε. Let x ∈ X. Since Tn − Tm ∈ L(X,Y ), Remark 2.4 yields

kTnx − Tmxk ≤ kTn − Tmk kxk ≤ ε kxk

1One also uses the notation B(X,Y ) for this space. 2.1. Basic properties and examples 29

for all n, m ≥ Nε, so that (Tnx) is a Cauchy sequence in Y . Hence, there exists exactly one y ∈ Y with y = limn→∞ Tnx =: T x. Let α, β ∈ F and x, z ∈ X. We deduce from the linearity of Tn that

T (αx + βz) = lim Tn(αx + βz) = lim (αTnx + βTnz) = αT x + βT z. n→∞ n→∞

Since (Tn) is Cauchy, there is a c > 0 with kTnk ≤ c for all n, whence kT xk = limn→∞ kTnxk ≤ c kxk. So far we have shown that T ∈ L(X,Y ). The above displayed estimate further yields

k(T − Tn)xk = lim k(Tn − Tm)xk ≤ ε kxk, m→∞ i.e., kT − Tnk ≤ ε for all n ≥ Nε. Example 2.6 (Multiplication operators). a) Let X = C([0, 1]), and let m ∈ C([0, 1]) be given. Set T f = mf for every f ∈ X. 1) We then have T f ∈ X and T (αf + βg) = αmf + βmg = αT f + βT g for all f, g ∈ X and α, β ∈ F. Hence, T : X → X is linear. 2) Moreover, kT fk∞ = supt |m(t)| |f(t)| ≤ kmk∞ kfk∞, and thus T ∈ L(X) with kT k ≤ kmk∞. 3) Since k1k∞ = 1, we further obtain kT k ≥ kT 1k∞ = km1k∞ = kmk∞. So we have shown that kT k = kmk∞. b) Let X = Lp(Rd) and 1 ≤ p ≤ ∞, and letm ˆ ∈ L∞(Rd) be given. For fˆ ∈ X we deﬁne T fˆ = mf + N for any representatives f ∈ fˆ and m ∈ mˆ .2 ˆ 0 00 1) Take any f1 ∈ f and m1 ∈ mˆ . Then there are null sets N and N such 0 00 that f(t) = f1(t) for all t∈ / N and m(t) = m1(t) for all t∈ / N . Therefore, 0 00 m(t)f(t) = m1(t)f1(t) for all t∈ / N ∪ N , which is a null set. As a consequence, T fˆ does not depend on the representatives and it is thus well–deﬁned. 2) Clearly, mf is measurable. Let c > kmk∞. Then there is a null set N such p d that |m(t)f(t)| ≤ c |f(t)| for all t∈ / N. It follows that mf ∈ L (R ) and kmfkp ≤ ˆ ˆ kmk∞ kfkp. As a result, T maps X into X and kT fkp = kmfkp ≤ kmˆ k∞ kfkp. 3) For f, g ∈ L(Rd) and α, β ∈ F we have

T (αfˆ+ βgˆ) = T (αf + βg + N ) = αmf + βmg + N = αT fˆ+ βT g.ˆ

Consequently, T ∈ L(X) and kT k ≤ kmˆ k∞. 4) We now claim that kT k = kmˆ k∞. This claim holds ifm ˆ = 0. So let kmˆ k∞ = kmk∞ > 0 for any m ∈ mˆ . Take ε ∈ (0, kmk∞) and n ∈ N. We put Aε,n = {t ∈ B(0, n) | |m(t)| ≥ kmk∞ − ε}. By the deﬁnition of the essential supremum, there is 1 −11 p d a k ∈ N such that λ(Aε,k) > 0. We can thus set fε = k Aε,k kp Aε,k ∈ L (R ). It holds kfεkp = 1 and thus

1 Z ! p 1 p kT k ≥ kmfεkp = 1 |m(t)| dt ≥ kmk∞ − ε k Aε,k kp Aε,k for all ε, where we assume that p < ∞. Taking the supremum over ε > 0, we deduce kT k = kmˆ k∞. The case p = ∞ is treated in the same way. Example 2.7 (Integral operators). Let X = C([0, 1]), and let k ∈ C([0, 1]2) be given. For every f ∈ X we set

Z 1 (T f)(t) = k(t, s)f(s) ds, t ∈ [0, 1]. 0 2As an exception, in this example we explicitely take into acccount that Lq = Lq/N . 30 Chapter 2. Continuous linear operators

1) From Analysis 1+2, we know that T f ∈ X and that T : X → X is linear. R 1 2) Set κ := supt∈[0,1] 0 |k(t, s)| ds ≤ kkk∞. We then obtain Z 1 kT fk∞ ≤ sup |k(t, s)| |f(s)| ds ≤ κ kfk∞ t∈[0,1] 0 for all f ∈ X, whence T ∈ L(X) and kT k ≤ κ. R 1 3) By continuity there exists a t0 ∈ [0, 1] such that κ = 0 |k(t0, s)| ds. Set ¯ k(t0, s) fn(s) = 1 |k(t0, s)| + n for all s ∈ [0, 1] and n ∈ N. Since fn ∈ X and kfnk∞ ≤ 1, we conclude that Z 1 2 |k(t0, s)| kT k ≥ kT fnk∞ ≥ |T fn(t0)| = 1 ds −→ κ 0 |k(t0, s)| + n as n → ∞, using e.g. Lebesgue’s theorem. Consequently, kT k = κ. Note that for k ≥ 0 one obtains this equality more easily since kT k ≥ kT 1k∞ = κ in this case.

Example 2.8 (Linear functionals). a) Let X = C([0, 1]). For a fixed t0 ∈ [0, 1] we define ϕ(f) = f(t0) for all f ∈ X. It is clear that ϕ : X → F is linear and that ∗ |ϕ(f)| ≤ kfk∞. Hence, ϕ ∈ X and kϕk ≤ 1. On the other hand, k1k∞ = 1 and thus kϕk ≥ |ϕ(1)| = 1, implying kϕk = 1. 0 b) Let X = Lp(µ) for a measure space and 1 ≤ p ≤ ∞, and let g ∈ Lp (µ) R be fixed. By Hölder’sinequality ϕ(f) = fg dµ ∈ F exists for all f ∈ X, and ? |ϕ(f)| ≤ kfkp kgkp0 . Since linearity is clear, we see that ϕ ∈ X with kϕk ≤ kgkp0 . (Equality is shown in Section 4.2.) c) On X = C([0, 1]) with k·k1, the linear form f 7→ ϕ(f) = f(0) is not continuous. 1 For instance, the functions fn given by fn(t) = 1 − nt for 0 ≤ t ≤ n and fn(t) = 0 1 1 for n ≤ t ≤ 1 satisfy kfnk1 = 2n → 0 as n → ∞, but ϕ(fn) = 1 for all n ∈ N. p Example 2.9 (Shift operators). Let X ∈ {c0, c, ` | 1 ≤ p ≤ ∞}. The right and left shift operator on X are given by

Rx = (0, x1, x2,... ) and Lx = (x2, x3,... ).

Clearly, R,L : X → X are linear and kRxkp = kxkp and kLxkp ≤ kxkp for all x ∈ X, as well as kLe2kp = 1. We thus obtain that R,L ∈ L(X) with norm 1. Observe that LR = I, RLx = (0, x2, x3,...), and • R is injective, but not surjective, and it has a left inverse, but no right inverse; • L is surjective, but not injective, and it has a right inverse, but no left inverse. Recall that for T ∈ L(X) with dim X < ∞ injectivity and surjectivity are equivalent, and that a right or left inverse is automatically an inverse! Deﬁnition 2.10. Let X and Y be normed vector spaces and T : X → Y be linear. a)We denote kernel and range of T by N(T ) = {x ∈ X | T x = 0} and R(T ) = T (X) = TX = {y = T x | x ∈ X}, respectively. b) An injective operator T ∈ L(X,Y ) is called an embedding, and one then writes X,→ Y . c) A bijective operator T ∈ L(X,Y ) having a continuous inverse T −1 is called isomorphism or invertible. One then writes X ' Y . d) A map T ∈ L(X,Y ) is called isometric if kT xk = kxk for all x ∈ X, and contractive if kT k ≤ 1. 2.1. Basic properties and examples 31

Remark 2.11. a) Let X and Y be normed vector spaces and J : X → Y be an −1 isomorphism. Let xn ∈ X and yn = Jxn. Hence, xn = J yn and so (xn) converges if and only if (yn) converges. In particular, X is a Banach space if and only if Y is a Banach space. Let C ⊆ X and D = JC = (J −1)−1(C) ⊆ Y . Hence, C = J −1(D) and so C is open (closed) if and only if D is open (closed). b) N(T ) is closed for any T ∈ L(X,Y ). An isometry is contractive and injective, and a contraction is continuous. In Example 2.9, the right shift R is an isometry and the left shift L has norm 1, but L is not an isometry. c) Let T ∈ L(X,Y ) be an isometry. Then its inverse T −1 : R(T ) → X is linear and isometric. In fact, for z = T x ∈ R(T ) we compute kT −1zk = kxk = kT xk = kzk. d) Let X be a Banach space and let T ∈ L(X,Y ) satisfy kT xk ≥ c kxk for some c > 0 and all x ∈ X (e.g., if T is isometric). Then R(T ) is closed in Y . In fact, let −1 yn = T xn → y in Y as n → ∞. The assumption yields kxn −xmk ≤ c kyn −ymk → 0 as n, m → ∞. Since X is a Banach space, there exists x = limn→∞ xn. The continuity of T then implies that y = T x ∈ R(T ); i.e., R(T ) is closed. e) Let Y be a linear subspace of (X, k·kX ) with its own norm k·kY . The identity I :(Y, k · kY ) → (X, k · kX ) is continuous (and thus an embedding) if and only if kykX = kIykX ≤ c kykY for all y ∈ Y and a constant c ≥ 0 if and only if k · kY is p q 1 ﬁner than k · kX . Examples: ` ,→ ` if 1 ≤ p ≤ q ≤ ∞; C ([0, 1]) ,→ C([0, 1]). ♦

Example 2.12. a) Let ∅= 6 B ∈ Bd with λ(B) < ∞ such that for all x ∈ B and r > 0 we have λ(B∩B(x, r)) > 0. For f ∈ Cb(B) set Jf = f +N . This map is linear p 1/p and bounded from Cb(B) to L (B), p ∈ [1, ∞], since kJfkp = kfkp ≤ λ(B) kfk∞ for all f ∈ Cb(B). If Jf = 0, then f = 0 a.e.. Take any x ∈ B. The assumption gives xn ∈ B with f(xn) = 0 and xn → x. So f(x) = 0 by continuity. As a result, p J : Cb(B) → L (B) is an embedding. b) We deﬁne a map J on C([−1, 1]) by setting

0, |t| ≥ 2,  (2 + t)f(−2 − t), −2 < t < −1, Jf(t) = f(t), |t| ≤ 1,  (2 − t)f(2 − t), 1 < t < 2.

Clearly, Jf ∈ C0(R) and J is linear and isometric. Hence, J : C([−1, 1]) ,→ C0(R) d is an isometric embedding. (One can similarly embed C(K) into C0(R ) for any compact K ⊆ Rd using Tietze’s extension theorem, see Theorem 20.1 in [Rud87].) c) Let X = {f ∈ C1([0, 1]) | f(0) = 0} be endowed with the norm given by kfk = 0 0 kf k∞. Then the map D : X → C([0, 1]), Df = f , is an isometric isomorphism with inverse given by

Z t D−1g(t) = g(s) ds, t ∈ [0, 1], g ∈ C([0, 1]). 0 In the view of the previous Remark, the Banach space structures of X and C([0, 1]) are the ‘same’. However, this isomorphism destroys other properties such as posi- tivity (e.g., f(t) = t(1−t) is positive on [0, 1], in contrast to Df(t) = f 0(t) = 1−2t). Lemma 2.13. Let X be a normed vector space, Y be a Banach space, D ⊆ X be a dense linear subspace (endowed with the norm of X), and T0 ∈ L(D,Y ). Then there exists exactly one extension T ∈ L(X,Y ) of T0 i.e., T0x = T x for all x ∈ D. Moreover, kT0k = kT k and T is isometric if T0 is isometric.

Proof. Let x ∈ X. Choose xn ∈ D such that xn → x in X as n → ∞. Since kT0xn −T0xmk ≤ kT0kkxn −xmk, the sequence (T0xn) is Cauchy and thus converges to some T x in Y . Let also xen ∈ D tend to x. Because of kT0xn − T0xenk ≤ 32 Chapter 2. Continuous linear operators kT0k kxn − xenk → 0 as n → ∞, the vector T x indeed does not depend on the approximating sequence. It is clear that T x = T0x for x ∈ D (take xn = x). Let x, z ∈ D and α, β ∈ F. Pick xn, zn ∈ D with xn → x and zn → z. We then obtain

T (αx + βz) = lim T0(αxn + βzn) = lim (αT0xn + βT0zn) = αT x + βT z, n→∞ n→∞ and hence T : X → Y is linear. Since

kT xk = lim kT0xnk ≤ kT0k lim kxnk = kT0k kxk , n→∞ n→∞

T belongs to L(X,Y ) and kT k ≤ kT0k. (If T0 is isometric, one sees here that T is also isometric.) On the other hand, Remark 2.4 yields

kT k = sup kT xk ≥ sup kT0xk = kT0k , x∈X x∈D kxk=1 kxk=1 so that kT0k = kT k. Let S ∈ L(X,Y ) satisfy Sx = T0x for all x ∈ D. Let z ∈ X. Choose xn ∈ D with xn → z as n → ∞. The uniqueness assertion follows from

Sz = lim Sxn = lim T0xn = T z. n→∞ n→∞

Convolution and Young’s inequality In this paragraph we derive the important Young inequality for convolutions. As a preparation we recall Fubini’s theorem from Analysis 3, see also Theorem 8.8 in m+n [Rud87]. Let A ∈ Bm, B ∈ Bn, and (x, y) ∈ A × B ⊆ R . It can be seen that ˜ ˜ A × B ∈ Bm×n. Let f : A × B → F with F ∈ {[0, ∞], F} be measurable. We set

y y f : A → F˜, f (x) = f(x, y) for each ﬁxed y ∈ B, ˜ fx : B → F, fx(y) = f(x, y) for each ﬁxed x ∈ A.

These functions are measurable (for each ﬁxed y ∈ B, resp. x ∈ A) since e.g. fx = f ◦ jx for each x ∈ A and the continuous function jx : B → A × B, jx(y) = (x, y). Theorem 2.14 (Fubini). a) Let f : A × B → [0, ∞] be measurable. Then the functions F : A → [0, ∞] and G : B → [0, ∞], given by Z Z y F (x) := fx(y) dy for x ∈ A, G(y) := f (x) dx for y ∈ B, B A are measurable, and it holds Z Z Z Z Z f(x, y) d(x, y) = f(x, y) dy dx = f(x, y) dx dy. (2.1) A×B A B B A

1 b) Let f ∈ L (A × B). Then there are nullsets NA ⊆ A and NB ⊆ B such that y fx is integrable for all x ∈ A \ NA and f is integrable for all y ∈ B \ NB. We deﬁne F and G as above for x ∈ A \ NA and for y ∈ B \ NB, respectively, and we put F (x) = 0 and G(y) = 0 for x ∈ NA and y ∈ NB, respectively. Then F and G are integrable and formula (2.1) holds.

We next introduce and discuss the convolution f ∗ g of functions in Lp(Rd) for suitable p. Note that for measurable f, g : Rd → F the map

2d ϕ : R → [0, ∞); ϕ(x, y) = |f(x − y)g(y)|, 2.1. Basic properties and examples 33 is measurable as a combination of measurable maps. (1) Let f, g ∈ L1(Rd). By means of Fubini a) and the transformation z = x − y, we derive Z Z Z ϕ(x, y)d(x, y) = |f(x − y)| dx |g(y)| dy 2d d d R ZR ZR = |f(z)| dz |g(y)| dy = kfk1kgk1. d d R R Hence, ϕ ∈ L1(R2d) and Fubini b) shows that the convolution Z (f ∗ g)(x) := f(x − y)g(y) dy (2.2) d R is deﬁned for a.e. x ∈ Rd, that it is integrable, and that Z Z kf ∗ gk1 ≤ |f(x − y)g(y)| dy dx = kϕk1 = kfk1 kgk1. d d R R (2) Let q ∈ [1, ∞), f ∈ L1(Rd), and g ∈ Lq(Rd). Fubini a) again yields the existence of

Z Z 1 1 ψ(x) := |f(x − y)g(y)| dy = |f(x − y)| q0 |f(x − y)| q |g(y)| dy d d R R in [0, ∞] for a.e. x ∈ Rd and that ψ is measurable on Rd. From Hölder’sinequality, we then deduce q Z q0 Z ψ(x)q ≤ |f(x − y)| dy |f(x − y)| |g(y)|q dy d d R R Z q−1 q = kfk1 |f(x − y)| |g(y)| dy, d R also using q0 = q/(q − 1) and the transformation z = x − y. Step (1) now implies Z Z q q q−1 q q−1 q q q kψkq = ψ dx ≤ kfk1 k |f| ∗ |g| k1 ≤ kfk1 kfk1 |g| dx = kfk1 kgkq . d d R R This time we cannot directly deduce the measurability of f ∗ g from Fubini b) since we integrated ψq instead of ψ. To deal with this problem, we use Proposition 1.35 and Fubini a), and compute Z Z Z q q 1 kψkq ≥ ψ(x) dx ≥ δn ψ(x) dx = δn B(0,n)(x)ϕ(x, y) d(x, y) 2d B(0,n) B(0,n) R for a constant δn > 0 and every n ∈ N. Therefore, the function given by ϕn(x, y) = 1 2d B(0,n)(x)ϕ(x, y) is integrable on R . Fubini b) thus shows that f ∗ g is defined by (2.2) for a.e. x ∈ B(0, n) and each n ∈ N (and hence for a.e. x ∈ Rd) and that the 1 d map B(0,n)f ∗ g is measurable on R . Letting n → ∞, we see that the pointwise d q limit f ∗ g is measurable on R . The above estimate on kψkq finally yields Z Z q Z q q q q kf ∗ gkq = f(x − y)g(y) dy dx ≤ ψ dx ≤ kfk1 kgkq. d d d R R R We have thus proved a part of the next proposition, see Theorem 4.33 in [Bre11] for the remaining cases. A different proof for the full result is given in Section 2.3. 1 1 1 p d Proposition 2.15. Let 1 ≤ p, q, r ≤ ∞ with 1 + r = p + q . Take f ∈ L (R ) and g ∈ Lq(Rd). Then the convolution (f ∗ g)(x) in (2.2) is defined for a.e. x ∈ Rd and gives a function in Lr(Rd). We further have Young’s inequality

kf ∗ gkr ≤ kfkp kgkq. 34 Chapter 2. Continuous linear operators

2.2 Standard constructions A) Product spaces Let X and Y be normed vector spaces. The cartesian product X ×Y = {(x, y) | x ∈ X, y ∈ Y } is a normed vector space for each of the norms ( max{kxkX , kykY }, p = ∞, k(x, y)kp = p p 1/p (kxkX + kykY ) , p ∈ [1, ∞).

These norms are equivalent. We have (xn, yn) → (x, y) in X × Y if and only if xn → x in X and yn → y in Y , as n → ∞. Moreover, X ×Y is complete if X and Y are complete. These facts can be proved as in Analysis 2 for the case R = X = Y . There are obvious modiﬁcations for ﬁnite products.

B) Direct sums

Deﬁnition 2.16. Let X1 and X2 be closed linear subspaces of a normed vector space X such that X1 + X2 = X and X1 ∩ X2 = {0}. Then we say that X is the direct sum of X1 and X2 and that X2 is the complement of X1, and we write X = X1 ⊕ X2. Let Y be a vector space. A map P ∈ L(Y ) is called projection if P 2 = P . Lemma 2.17. Let X be a normed vector space and P ∈ L(X) be a projection. Then the operator I − P ∈ L(X) is also a projection, and we have

R(P ) = N(I − P ) =: X1,N(P ) = R(I − P ) =: X2, and X = X1 ⊕ X2. Moreover, kP k ≥ 1 if P 6= 0. Proof. From P = P 2 we deduce (I − P )2 = I − 2P + P 2 = I − P . If y ∈ R(P ), then there is an x ∈ X with y = P x and thus (I − P )y = P x − P 2x = 0; i.e. y ∈ N(I −P ). Conversely, if (I −P )x = 0, then x = P x ∈ R(P ). So we have shown the asserted equalities for X1, which yield those for X2 since I − P is a projection and I − (I − P ) = P . By Proposition 1.24 the subspaces X1 and X2 are closed as kernels of continuous maps. We can write x = P x + (I − P )x ∈ X1 + X2 for each x ∈ X. If x ∈ X1 ∩ X2, then P x = 0 and thus 0 = x − P x = x. Hence, 2 2 X = X1 ⊕ X2. The last assertion follows from kP k = kP k ≤ kP k .

Remark 2.18. a) Let X = X1 ⊕X2. For each x ∈ X we then have unique x1 ∈ X1 and x2 ∈ X2 with x = x1 + x2. Set P x = x1. Then P : X → X is the unique linear projection with R(P ) = X1 and N(P ) = X2. 2 Proof. The deﬁnition easily yields P = P , R(P ) = X1 and N(P ) = X2. Let x, y ∈ X and α, β ∈ F. Then there are xk, yk ∈ Xk such that x = x1 + x2 and y = y1 + y2. We now obtain

P (αx + βy) = P ((αx1 + βy1) + (αx2 + βy2)) = αx1 + βy1 = αP x + βP y,

2 so that P is linear. Let also Q ∈ L(X) satisfy Q = Q, R(Q) = X1 and N(Q) = X2. Let x ∈ X and write x = x1 + x2 as above. Then x1 = Qy for some y ∈ X, and so 2 Qx = Qx1 + Qx2 = Q y = Qy = x1 = P x. 2 b) Let X be a Banach space. We will see in Proposition 3.25 that the projection ∼ P from a) is continuous and that X1 ⊕ X2 = X1 × X2. ♦ 1 t Example 2.19. a) Let X = 2, t ∈ , and P = . Then P is a projection R R 0 0 with R(P ) = R × {0}, N(P ) = {(−tr, r) | r ∈ R}, and kP k = 1 + |t| for | · |1. 2.2. Standard constructions 35

1 1 b) Let X = L (R) and P f = R+ f for f ∈ X. Clearly, kP fk1 ≤ kfk1 and P 2 = P , so that P ∈ L(X) is a projection with kP k = 1. We further have (I−P )f = 1(−∞,0)f. To express the direct sum X = R(P ) ⊕ N(P ) more conveniently, we 1 introduce the isometric isomorphism J : R(P ) → L (R+), Jf = f|R+ with inverse −1 −1 given by J g = g on R+ and J g = 0 on (−∞, 0). On R− one proceeds similarly. 1 1 1 We can thus identify X with L (R+) ⊕ L (R−), considering L (R±) as subspaces of L1(R) by extending functions by 0. ∞ c) c0 has no complement in ` , see Satz IV.6.5 in [Wer05].

C) Quotient spaces Let X be a normed vector space, Y a linear subspace and

X/Y = {xˆ = x + Y | x ∈ X} be the quotient space. The quotient map

Q : X → X/Y, Qx =x, ˆ is linear and surjective with N(Q) = Y . One sets codim Y := dim X/Y . We deﬁne the quotient norm q by

q(x) = kxˆk = kQxk = inf kx − yk =: d(x, Y ). y∈Y forx ˆ = x+Y ∈ X/Y . Ifx ¯ +Y = x+Y , thenx ¯ −x ∈ Y and thus d(x, Y ) = d(¯x, Y ); i.e., kxˆk does not depend on the representative ofx ˆ. For α 6= 0, we have 1 kαxˆk = inf kα(x − y)k = |α| inf kx − zk = |α| kxˆk. y∈Y α z∈Y

Let x1, x2 ∈ X. Take ε > 0. There are yk ∈ Y mit kxk −ykk ≤ kxˆkk+ε for k = 1, 2. We then obtain

kxˆ1 +x ˆ2k = inf kx1 + x2 − yk ≤ kx1 − y1 + x2 − y2k ≤ kxˆ1k + kxˆ2k + 2ε. y∈Y

Since ε > 0 is arbitrary, the quotient norm is a seminorm. Now, let Y be closed. If kxˆk = 0 for somex ˆ ∈ X/Y , then there exist yn ∈ Y with kx − ynk → 0. From the closedness of Y it follows that x ∈ Y , and hencex ˆ = 0. So far we have established that X/Y is a normed vector space for the quotient norm. Because of kQxk = kxˆk ≤ kxk, the quotient map Q is continuous with kQk ≤ 1. Moreover, for each δ ∈ (0, 1), Lemma 1.43 gives ax ¯ ∈ X with kx¯k = 1 and

kQk ≥ kQx¯k = inf kx¯ − yk ≥ 1 − δ. y∈Y

Letting δ → 1, we deduce that kQk = 1. Proposition 2.20. Let X be a Banach space and Y be a linear subspace. a) Then X/Y is vector space with seminorm q : x + Y 7→ d(x, Y ). The map Q : X → X/Y ; Qx = x + Y , is a linear surjective contraction with N(Q) = Y . b) Let Y be closed. Then q is a norm and kQk = 1. c) Let X be a Banach space and Y be closed. Then X/Y is a Banach space

Proof. It remains to show the completeness of X/Y . Let (ˆxn) be a Cauchy sequence in X/Y . We ﬁnd a subsequence such that

−k kxˆm − xˆnk k ≤ 2 for all m ≥ nk. (∗) 36 Chapter 2. Continuous linear operators

−k Consequently, there are ynk ∈ Y with kxnk+1 −xnk −ynk k ≤ 2·2 for every k ∈ N. PN Set zk = xnk+1 − xnk − ynk and vN = xn1 + k=1 zk. Since X is a Banach space P∞ and k=1 kzkk < ∞, there exists x = limN→∞ vN in X. We further have

N N X X vN = xnN+1 − ynk and ynk ∈ Y, k=1 k=1

so thatv ˆN =x ˆnN+1 . Since Q is continuous, it followsx ˆnN+1 − xˆ = Q(vN − x) → 0 −N in X/Y as N → ∞. Finally, for ε > 0 there is an N = Nε ∈ N such that 2 ≤ ε and kxˆ − xˆnN k ≤ ε. Using (∗), we deduce

kxˆ − xˆmk ≤ kxˆ − xˆnN k + kxˆnN − xˆmk ≤ 2ε for all m ≥ nN . Example 2.21. a) Let X = Y ⊕ Z for a Banach space X. Then the map J : Z → X/Y ; Jz =z ˆ = z + Y , is linear and continuous. If Jz = 0, then z ∈ Y and thus z = 0. Letx ˆ = x + Y ∈ X/Y . There are y ∈ Y and z ∈ Z with x = y + z, so that xˆ =z ˆ = Jz. Hence, J is bijective. The continuity of J −1 follows from the open mapping theorem 3.21 below. As a result, Z ' X/Y via J. In Example 2.19b) we 1 1 1 thus obtain L (R)/L (R+) ' L (R−). b) Note that the quotient construction is more general than the direct sum. For ∞ ∞ instance, ` /c0 exists, though c0 has no complement in ` .

D) Completion Proposition 2.22. Let X be a normed vector space. Then there is a Banach space X˜ and a linear isometry J : X → X˜ such that JX is dense in X˜. Any other Banach space with this property is isometrically isomorphic to X˜.

Proof. Let E by the vector space of all Cauchy sequences v = (xn)n∈N ⊆ X. Note that for (xn) ∈ E the sequence (kxnk) is Cauchy in R and thus it exists p(v) := limn→∞ kxnk in R. It is easy to check that p is a seminorm on E and that its kernel is given by the linear subspace c0(X) of all null sequences in X. We now deﬁne the vector space X˜ = E/c0(X) and put v˜ = p(v) for any representative v ∈ E of v˜ ∈ X˜. If w ∈ E is another representative9 9 ofv ˜, then v − w ∈ c0(X) and we thus obtain p(v) ≤ p(v − w) + p(w) = p(w) as well as p(w) ≤ p(v). Hence, v˜ is well deﬁned and it gives a norm on X˜. We further introduce the map 9 9

J : X → X˜; x 7−→ (x, x, . . . ) + c0(X), which is linear and isometric. Letv ˜ ∈ X˜ and ε > 0. Choose a representative v = (xk) ∈ E. There is an N = Nε ∈ N such that kxk − xN k ≤ ε for all k ≥ N. Because of v˜ − JxN = limk→∞ kxk − xN k ≤ ε, the range of J is dense in X˜. Now let (˜9vm) be a9 Cauchy sequence in X˜ with representives vm = (xm,j)j ∈ E. For each m ∈ N, we can ﬁnd an index jm such that jm+1 > jm and 1 kx − x k ≤ for all j ≥ j . m,j m,jm m m

We deﬁne w = (ym)m = (xm,jm )m. From the above inequality we deduce

kyn − ymk = Jyn − Jym ≤ Jyn − vñ + vñ − v˜m + v˜m − Jym 9 9 9 9 9 9 9 9 = lim kxn,jn − xn,jk + vñ − v˜m + lim kxm,j − xm,jm k j→∞ 9 9 j→∞ 1 1 ≤ + vñ − v˜m + , n 9 9 m 2.2. Standard constructions 37 so that w ∈ E. Let ε > 0. Since w is a Cauchy sequence in X, there is an Nε ∈ N such that kyn − ymk ≤ ε for all n, m ≥ Nε. We can thus estimate

w˜−v˜m ≤ w˜−Jym + Jym−v˜m = lim kyn−ymk+ lim kxm,jm −xm,jk ≤ 2ε 9 9 9 9 9 9 n→∞ j→∞ for all m ≥ Nε. It remains to show uniqueness. Let X˜ 0 be a Banach space and J 0 : X → X˜ 0 be isometric and linear with dense range J 0X in X˜ 0. Remark 2.11 shows that the operator 0 −1 0 T0 = J ◦ (J ) : J X → X,˜ is well deﬁned and that it is isometric with dense range JX. Using Lemma 2.13 and 0 the completeness of X˜, we can extend T0 to an isometric linear map T : X˜ → X˜ which still has a dense range. From Remark 2.11 and the completeness of X˜ 0, we conclude that the range of T is closed, and hence it is equal to X˜. Consequently, T is the required isometric isomorphism. Remark 2.23. Usually one identiﬁes X with the subspace JX of X˜ (as one does with Q and R). Every T ∈ L(X,Y ) can uniquely be extended to an operator ˜ ˜ ˜ T ∈ L(X,Y ) by means of Lemma 2.13. Also, T is isometric if T is isometric. ♦ p Example 2.24. Let p ∈ [1, ∞). The map J :(C([0, 1]), k · kp) → L (0, 1); f 7→ f + N , is isometric. We see in Proposition 3.12 below that J has dense range. The ∼ p above remark yields a linear isometric map J˜ :(C([0, 1]), k · kp) → L (0, 1) with dense range. By Remark 2.11, the range of J˜ is closed and thus J˜ is an isometric isomorphism. In this way one can view Lp(0, 1) as the completion of C([0, 1]) with respect to the p–norm. ♦

E) Sum of Banach spaces Let X and Y be Banach spaces which are linear subspaces of a vector space Z that possesses a metric d for addition and scalar mulitplication of Z are continuous (we call such a metric compatible). We assume that the inclusion maps from X to Z and from Y to Z are continuous. We then deﬁne the sum X + Y = {z = x + y | x ∈ X, y ∈ Y } which is a linear subspace of Z. We can consider X and Y as linear subspaces of X + Y . A typical example is Lp(µ) + Lq(µ) for p, q ∈ [1, ∞] and a measure space (S, A, µ), where we may take Z as the space of measurable functions modulo null functions, endowed with the metric describing local convergence in measure. This space was used in Section 2.3. We point out that the sum X + Y does not need to be direct, i.e., for a given z ∈ X + Y there may be many pairs (x, y) ∈ X × Y such that z = x + y. Moreover, the norm in X does not need to be ﬁner or coarser than that of Y . We endow X +Y with the sum norm

kzkX+Y = inf{kxkX + kykY | z = x + y, x ∈ X, y ∈ Y } which turns out to be coarser those of X and of Y . Thus, X + Y can serve as a space where we can compare the convergence in X with that in Y . We will further need the linear subspace D = {(u, −u) | u ∈ X ∩ Y } of X × Y . Proposition 2.25. Let X and Y be Banach spaces which are linear subspaces of a vector space Z endowed with a compatible metric. We assume that the inclusion maps from X to Z and from Y to Z are continuous. Then (X + Y, k · kX+Y ) is a Banach space which is isometrically isomorphic to the quotient space (X × Y )/D, where X × Y is endowed with the norm kxkX + kykY . Moreover, kxkX ≤ kxkX+Y for x ∈ X and kykY ≤ kykX+Y for y ∈ Y . 38 Chapter 2. Continuous linear operators

Proof. Let z ∈ X + Y and α ∈ F. Note that kzkX+Y exists in R+. If kzkX+Y = 0, then there are xn ∈ X and yn ∈ Y such that z = xn + yn for all n ∈ N and kxnkX + kynkY → 0 as n → ∞. By continuity, xn and yn both tend to 0 in Z, and so z = 0 since the metric is compatible. We further have

kαzkX+Y = inf{kαxkX + kαykY | z = x + y, x ∈ X, y ∈ Y } = |α| kαzkX+Y .

Let z1, z2 ∈ X + Y . For any ε > 0, we can choose xj ∈ X and yj ∈ Y such that zj = xj + yj and kxjkX + kyjkY ≤ kzjkX+Y + ε for j = 1, 2. Since z1 + z2 = (x1 + x2) + (y1 + y2), we conclude

kz1 + z2kX+Y ≤ kx1 + x2kX + ky1 + y2kY ≤ kz1kX+Y + kz2kX+Y + 2ε.

As a result, X + Y is a normed vector space. The last assertion is clear. We next show that J :(X × Y )/D → X + Y ; J((x, y) + D) = x + y is linear, isometric and surjective and that D is closed in X × Y . In view of Re- mark 2.11 and Proposition 2.20, these facts imply that J is an isometric isomorphism and that X + Y is a Banach space. First, let vn = (xn, −xn) ∈ D converge to v in X × Y . Then xn tends to some x in X and −xn to some y in Y . Since both sequences also converge in Z, we obtain y = −x ∈ X ∩ Y and z ∈ D; i.e., D is closed in X × Y . We next treat J. If (x, y)+D = (x0, y0)+D in (X ×Y )/D, then (x−x0, y−y0) ∈ D and thus x − x0 = y0 − y. This means that J((x, y) + D) = x + y = x0 + y0 = J((x0, y0) + D), and J is in fact a map. It is clear that J is linear and surjective. Let (x, y) ∈ X × Y . The operator J is isometric since

k(x, y) + Dk(X×Y )/D = inf{k(x + u, y − u)kX×Y | u ∈ X ∩ Y }

= inf{kx + ukX + ky − ukY | u ∈ X ∩ Y } 0 0 0 0 0 0 = inf{kx kX + ky kY | x ∈ X, y ∈ Y, x + y = x + y}

= kx + ykX+Y , where we take u = x0 − x = y − y0.

Proposition 2.26. Let Xj and Yj (with j = 0, 1) be Banach spaces which are linear subspaces of vector spaces V and W with compatible metrics, respectively. We assume that these inclusion maps are continuous. Let T0 ∈ L(X0,Y0) and T1 ∈ L(X1,Y1) be operators such that T0u = T1u =: T u for all u ∈ X0 ∩ X1. Then T has a unique extension Te ∈ L(X0 + X1,Y0 + Y1) such that Te xj = Tjxj for all xj ∈ Xj and j = 0, 1.

Proof. Let x = x0 + x1 for xj ∈ Xj. We then deﬁne

Te x = T0x0 + T1x1 ∈ Y0 + Y1.

0 0 0 0 0 If x = x0 + x1 for xj ∈ Xj, then we obtain u := x0 − x0 = x1 − x1 ∈ X0 ∩ X1. It follows that

0 0 T0x0 + T1x1 = T0x0 + T0u + T1x1 − T1u = Te x + T u − T u = Te x, and thus Te : X0 + X1 → Y0 + Y1 is a map. Clearly, Te xj = Tjxj for all xj ∈ Xj and 0 0 0 0 j = 0, 1. Take any xj ∈ Xj and α, β ∈ F. Set x = x0 + x1. We then compute

0 0 0 Te(αx + βx ) = T0(αx0 + βx0) + T1(αx1 + βx1) 0 0 0 = α(T0x0 + T1x1) + β(T0x0 + T1x1) = αTe x + βTe x , 2.3. The interpolation theorem of Riesz and Thorin 39 so that Te is linear. Moreover,

kTe xkY0+Y1 ≤ kT0x0kY0 + kT1x1kY1 ≤ max{kT0k, kT1k} (kx0k + ky1k).

Taking the inﬁmum over all decompositions x = x0 + x1 in X0 + X1, we derive that Te is bounded. Let S ∈ L(X0 + X1,Y0 + Y1) be another extension of T0 and T1. Then Sx = Sx0 + Sx1 = T0x1 + T1x1 = Te x, and Te is unique.

2.3 The interpolation theorem of Riesz and Thorin

Interpolation theory is an important branch of functional analysis which treats the following problem. Let Xj and Yj (with j = 0, 1) be Banach spaces being linear subspaces of vector spaces V and W , respectively. Assume that T0 : X0 → Y0 and T1 : X1 → Y1 are bounded linear operators such that T0u = T1u =: T u for all u ∈ X0 ∩ X1. Due to Paragraph 2.2E) below, we can extend T to a bounded linear opeartor Te : X0 + X1 → Y0 + Y1 where the sum space

X + Y = {z = x + y | x ∈ X, y ∈ Y } is endowed with the complete norm

kzkX+Y = inf{kxkX + kykY | z = x + y, x ∈ X, y ∈ Y }, for Banach spaces X and Y being linear subspaces of vector space Z. One now wants to find Banach spaces X between X0 ∩ X1 and X0 + X1 and Y between Y0 ∩ Y1 and Y0 + Y1 such that Te can be restricted to a bounded linear map from X to Y which also extends T0. We refer to the lecture notes [Lun09] for an introduction to this area and its applications. Here we restrict ourselves to one of the seminal results in this subject due to Riesz and Thorin, which deals with Lp–spaces. Let (Ω, A, µ) and (Λ, B, ν) be σ-finite measure spaces and p0, p1, q0, q1 ∈ [1, ∞]. Set U := Lp0 (µ) ∩ Lp1 (µ) and V := Lq0 (ν) ∩ Lq1 (ν). Take θ ∈ [0, 1] and define

1 1 − θ θ 1 1 − θ θ = + and = + . p p0 p1 q q0 q1

Observe that any p ∈ [p0, p1] if p0 ≤ p1 and any p ∈ [p1, p0] if p0 ≥ p1 can be written in this way. The exponent q between q0 and q1 is then ﬁxed via θ ∈ [0, 1]. The spaces Lp0 (µ) and Lp1 (µ) need to be included in each other. We thus use the sum space Lp0 (µ) + Lp1 (µ) to express that operators on Lp0 (µ) and on Lp1 (µ) are restrictions of common operator. As seen in an exercise or in Analysis 3, H¨older’sinequality shows that U ⊆ Lp(µ) and V ⊆ Lq(ν) with

kfk ≤ kfk1−θ kfkθ and kgk ≤ kgk1−θ kgkθ (2.3) p p0 p1 q q0 q1 for all f ∈ U and g ∈ V . We recall that the space of simple functions

E(A) = lin{1A | A ∈ A, µ(A) < ∞} with a support of ﬁnite measure is dense in Lr(µ) if r ∈ [1, ∞) and that the space of simple functions E∞(A) = lin{1A | A ∈ A} ∞ is dense in L (µ). The proof given there also shows that for f ∈ U with p0 6= p1 p0 p1 we can ﬁnd a sequence fn ∈ E(A) such that fn → f in L (µ) and in L (µ), and thus also in Lp(µ) by (2.3). 40 Chapter 2. Continuous linear operators

We want to show that Lp(µ) with p as above is embedded into Lp0 (µ) + Lp1 (µ). p Let p0 ≤ p1. (The other case is treated analogously.) For f ∈ L (µ), we set −1 p fe = kfkp f so that kfekp = 1. Because of fe ∈ L (µ) the set {fe ≥ 1} := {ω ∈ Ω | f(ω) ≥ 1} has ﬁnite measure, and thus the function f = 1 f belongs to e 0 {fe≥1} e Lp0 (µ), cf. Proposition 1.35. The function f = 1 f is contained in L∞(µ) and 1 {f

p p p p kfkL 0 (µ)+L 1 (µ) = kfkp kfekL 0 (µ)+L 1 (µ) ≤ kfkp (kf0kp0 + kf1kp1 ) 1 1 Z p0 Z p1 p0 p1 = kfkp |fe| dµ + kfkp |fe| dµ {fe≥1} {f

≤ 2 kfkp , so that Lp(µ) ,→ Lp0 (µ) + Lp1 (µ). Similarly, one verifies Lq(ν) ,→ Lq0 (ν) + Lq1 (ν). Theorem 2.27 (Riesz–Thorin). Let (Ω, A, µ) and (Λ, B, ν) be σ-finite measure spaces, F = C, and p0, p1, q0, q1 ∈ [1, ∞]. Take θ ∈ [0, 1] and define p, q ∈ [1, ∞] via 1 1 − θ θ 1 1 − θ θ = + and = + . p p0 p1 q q0 q1

pj qj Assume there are Tj ∈ L(L (µ),L (ν)) such that T0u = T1u =: T u for all u ∈ U = p0 p1 p q L (µ) ∩ L (µ). Then T has a unique extension Tθ ∈ L(L (µ),L (ν)). Moreover, p0 p1 q0 q1 Tθ is the restriction of Te ∈ L(L (µ) + L (µ),L (ν) + L (ν)) which is the unique extension of T0 and T1 to this space (see Proposition 2.26), and we have

1−θ θ p q kTθkL(L (µ),L (ν)) ≤ kT0kL(Lp0 (µ),Lq0 (ν)) kT1kL(Lp1 (µ),Lq1 (ν)) .

We note that the theorem also holds for F = R with an additional multiplicative constant in the estimate, see Satz II.4.2 in [Wer05]. Proof. The result trivially holds if θ ∈ {0, 1}. So we can take θ ∈ (0, 1). First p let p0 = p1; i.e., p0 = p. Let f ∈ L (µ). The assumptions then yield T f ∈ V = Lq0 (ν) ∩ Lq1 (ν) and, using also (2.3),

kT fk ≤ kT fk1−θ kT fkθ ≤ kT k1−θ kfk1−θ kT kθ kfkθ = kT k1−θ kT kθ kfk q 0 q0 1 q1 0 p 1 p 0 1 p so that the theorem holds in this case. Next, let p0 6= p1. In this case, we have p < ∞, and thus E(A) is dense in Lp(µ). We consider

m X 1 p0 p1 f = aj Aj ∈ E(A) ⊆ U = L (µ) ∩ L (µ), j=1 where we may assume that the sets Aj are pairwise disjoint and have ﬁnite measure. The assumptions again yield T f ∈ V . Due to (4.6) (see also Satz II.2.4 in [Wer05]), it holds Z

kT fkq = sup (T f) g dν (2.4) g∈E(B),kgkq0 ≤1 Λ 0 0 if q < ∞. If q = ∞ (i.e., q = 1 which holds if and only if (q0, q1) = (1, 1)), then (2.4) Pn 1 is valid with E(B) replaced by E∞(B). Further, take g = k=1 bk Bk ∈ E(B) with kgkq0 ≤ 1 where we may assume that the sets Bk are pairwise disjoint and have ﬁnite 2.3. The interpolation theorem of Riesz and Thorin 41

0 measure. (If q = ∞, we allow for µ(Bk) = ∞.) Let z ∈ S := {ζ ∈ C | Re ζ ∈ [0, 1]}. We then deﬁne the function F (z) by

p 1−z + z −1 (F (z))(x) = |f(x)| p0 p1 f(x) for x ∈ Ω with f(x) 6= 0, and by (F (z))(x) = 0 if f(x) = 0. For q0 < ∞ we set

0 1−z z q 0 + 0 −1 (G(z))(y) = |g(y)| q0 q1 g(y) for y ∈ Λ with g(y) 6= 0, and put (G(z))(y) = 0 if g(y) = 0. If q0 = ∞, we simply 1−z z 0 1−z z set G(z) = g. We write p(z) = p( + )−1 and q(z) = q ( 0 + 0 )−1. Observe p0 p1 q0 q1 that F (θ) = f and G(θ) = g, as well as F (z) ∈ E(A), G(z) ∈ E(B) if q0 < ∞ 0 and G(z) ∈ E∞(B) if q = ∞. The assumptions lead to TF (z) ∈ V . We further introduce the function Z m n Z X X p(z) q(z) 1 1 ϕ(z) = T (F (z)) G(z) dν = aj|aj| bk|bk| (T Aj ) Bk dν Λ j=1 k=1 Λ for z ∈ S. The right hand side easily yields that ϕ ∈ C(S) is holomorphic on S◦. We want to apply the Three–Lines-Theorem, see Satz II.4.3 in [Wer05], and thus check the estimates assumed in this result. Writing z = s + it ∈ S with s ∈ [0, 1] and t ∈ R, we compute 1−s 1−s s q0 + s p + q0 q0 |F (z)(x)| = |f(x)| p0 p1 , |G(z)(y)| = |g(y)| 0 1 (2.5) for all s ∈ [0, 1], t ∈ R, x ∈ Ω and y ∈ Λ. The right hand sides are bounded in z ∈ S for ﬁxed x and y. In the same way one sees that ϕ is bounded on S. The Three–Lines-Theorem then yields Z 1−θ θ

(T f)g dν = |ϕ(θ)| ≤ sup |ϕ(it)| sup |ϕ(1 + it)| (2.6) Λ t∈R t∈R H¨older’sinequality, the assumptions and estimate (2.5) with s = 0 further imply

|ϕ(it)| ≤ kT F (it)k kG(it)k 0 ≤ kT k kF (it)k kG(it)k 0 0 q0 q0 0 p0 q0 1 0 0 1 p q q0 0 Z pp0 0 Z q0 q0 = kT0k |f| p0 dµ |g| 0 dν Ω Λ 0 p q p 0 p0 q0 p0 = kT0k kfkp kgkq0 ≤ kT0k kfkp , where we also used that kgkq0 ≤ 1. Similarly one sees that

|ϕ(1 + it)| ≤ kT F (1 + it)k kG(1 + it)k 0 ≤ kT k kF (1 + it)k kG(1 + it)k 0 1 q1 q1 1 p1 q1 1 0 0 1 p q q1 0 Z pp1 1 Z q1 q0 = kT1k |f| p1 dµ |g| 1 dν Ω Λ 0 p q p 0 p1 q1 p1 = kT1k kfkp kgkq0 ≤ kT1k kfkp . Formulas (2.4) and (2.6) now lead to

1−θ p(1−θ)/p0 θ pθ/p1 1−θ θ kT fkq ≤ kT0k kfkp kT1k kfkp = kT0k kT1k kfkp (2.7) for all f ∈ E(A). Let f ∈ U. As observed above the theorem, we can approximate f p0 p1 p by fn ∈ E(A) in L (µ), in L (µ) and in L (µ). By the assumptions, also T fn tends 42 Chapter 2. Continuous linear operators to T f in Lq0 (ν) and in Lq1 (ν), and hence in Lq(ν) due to (2.3). Inequality (2.7) thus holds for all f ∈ U. Lemma 2.13 then allows to extend T uniquely from its domain p q 1−θ θ U to an operator Tθ ∈ L(L (µ),L (µ)) with norm less or equal kT0k kT1k . Another observation before the theorem yields that Lp(µ) ,→ Lp0 (µ) + Lp1 (µ) and similarly for the range spaces. Hence, T fn = Te fn tends both to Tθf and Te f in q0 q1 L (ν) + L (ν) so that Tθ is an restriction of Te. We next use the Riesz–Thorin theorem to give a diﬀerent proof of Young’s inequality for convolutions from Proposition 2.15. (1a) Recall the deﬁnition (2.2) of the convolution f ∗ g for f, g ∈ L1(Rd). There we have also shown that

kf ∗ gk1 ≤ kϕk1 = kfk1 kgk1.

(1b) In a second step, we take f ∈ L1(Rd) and g ∈ L∞(Rd). We compute Z Z Z |ϕ(x, y)| d(x, y) ≤ |f(x − y)| kgk∞ dy dx d d B(0,n)×R B(0,n) R Z Z = kgk∞ |f(z)| dz dx = λ(B(0, n) kfk1 kgk∞. d B(0,n) R Fubini’s theorem now yields that (f ∗g)(x) is deﬁned for a.e. x ∈ B(0, n) and gives a measurable function on B(0, n). Letting n → ∞, the same holds on Rd. Replacing in the above estimate the integral over x ∈ B(0, n) by a supremum in x ∈ Rd, we further obtain kf ∗ gk∞ ≤ kfk1 kgk∞. 1 d 1 d (1c) Fix any f ∈ L (R ). We deﬁne T1g = f ∗g for g ∈ L (R ) and T∞g = f ∗g for ∞ d r d g ∈ L (R ). We have shown that that Tr ∈ L(L (R )) with kTrk ≤ kfk1 for r = 1, ∞. We can now extend the convolution to an operator T g = f ∗g := f ∗(g1 +g∞) 1 d ∞ d 0 for g = g1 + g∞ ∈ L (R ) + L (R ). Let q ∈ (1, ∞). Set θ = 1/q ∈ (0, 1). We then 1 1−θ θ have q = 1 + ∞ . Hence, the Riesz–Thorin theorem shows that we can restrict q d 1 d T to a bounded operator Tq ∈ L(L (R )) with kTqk ≤ kfk1. For all f ∈ L (R ), g ∈ Lq(Rd) and q ∈ [1, ∞], we have thus shown that f ∗ g ∈ Lq(Rd) and

kf ∗ gkq ≤ kfk1 kgkq.

(2a) We next ﬁx g ∈ Lq(Rd) and q ∈ [1, ∞], and vary f. For f ∈ L1(Rd), 1 d q d step (1c) allows to set S1f := f ∗ g giving an operator S1 ∈ L(L (R ),L (R )) q0 d with kS1k ≤ kgkq. Let now f ∈ L (R ). Due to H¨older’sestimate, the map y 7→ f(x − y)g(y) is integrable on Rd and Z f(x − y)g(y) dy ≤ kf(x − ·)kq0 kgkq = kfkq0 kgkq d R d for each x ∈ R . One sees as above that f ∗ g =: Sq0 g is a measurable function and

kSq0 gk∞ = kf ∗ gk∞ ≤ kfkq0 kgkq,

q0 d ∞ d i.e., Sq0 ∈ L(L (R ),L (R )) with norm less or equal kgkq. We can thus deﬁne 0 the convolution Sf = f ∗ g for all g ∈ Lq(Rd), f ∈ L1(Rd) + Lq (Rd) and q ∈ [1, ∞]. 0 1 1 1 0 (2b) Finally, take p ∈ [1, q ] and r ∈ [1, ∞] with 1+ r = p + q . Choose θ = q/p ∈ [0, 1]. Observe that 1 1 − θ θ 1 1 − θ θ = + and = + . p 1 q0 r q ∞

p d r d Riesz–Thorin now allows to restrict S to an operator Sp ∈ L(L (R ),L (R )) with norm less or equal kgkq. We have thus proved Proposition 2.15. Chapter 3

Two main theorems on bounded linear operators

We discuss two fundamental results of functional analysis, the principle of uniform boundedness and the open mapping theorem, which both rely on a corollary to Baire’s theorem.

3.1 The principle of uniform boundedness and strong convergence

Theorem 3.1 (Baire). Let M be a complete metric space and On ⊆ M be open T and dense for each n ∈ . Then O := On is dense in M. N n∈N

Proof. For every x0 ∈ M and δ > 0 we must ﬁnd an x ∈ B0 ∩ O, where B0 := B(x0, δ). So let x0 ∈ M and δ > 0. Since O1 is open and dense, there is an 1 x1 ∈ O1 ∩ B0 and a δ1 ∈ (0, 2 δ] with B(x1, δ1) ⊆ O1 ∩ B0. Iteratively, one ﬁnds 1 xn ∈ On ∩ Bn−1, δn ∈ (0, 2 δn−1] and Bn := B(xn, δn) such that

Bn ⊆ On ∩ Bn−1 ⊆ On ∩ (On−1 ∩ Bn−2) ⊆ · · · ⊆ On ∩ ... ∩ O1 ∩ B0. (∗)

−m −m Since δm ≤ 2 δ, we obtain xn ∈ Bm ⊆ B(xm, 2 δ) for all n ≥ m. Hence,

(xn)n∈N is a Cauchy sequence and there exists x = limn→∞ xn which belongs to Bm for each m ∈ N. The inclusions (∗) now yield x ∈ O ∩ B0.

The set R2 \ R is open and dense in R2. A property that holds for all x in an open and dense subset is often called generic. S Corollary 3.2. Let M be a complete metric space and M = n∈ An for closed ◦ N subsets An ⊆ M. Then there exists an N ∈ N such that AN 6= ∅.

◦ Proof. Suppose that An = ∅ for all n ∈ N. Then On = M \ An were open and T dense. Theorem 3.1 thus implies that On is dense in M. But, this impossible T S n∈N since n On = M \ n An = ∅ by the assumption.

Example 3.3. One needs the completeness in the above corollary. Consider for instance, (c00, k · kp) and the sets An = {x = (x1, . . . , xn, 0,... ) | xj ∈ F}. These S sets are closed for all n ∈ N by Lemma 1.42, and n An = c00. But, each An has 1 empty interior since for x ∈ An the vectors ym = x + m en+1 ∈/ An tend to x as m → ∞. ♦

43 44 Chapter 3. Two main theorems on linear operators

Theorem 3.4 (Principle of uniform boundedness). Let X be a Banach space, Y be a normed vector space, and T ⊆ L(X,Y ). If the set of operators T is pointwise bounded (i.e., ∀ x ∈ X ∃ cx ≥ 0 ∀ T ∈ T : kT xk ≤ cx), then T is uniformly bounded (i.e., ∃ c > 0 ∀ T ∈ T : kT k ≤ c).1 S Proof. Set An = {x ∈ X | kT xk ≤ n for all T ∈ T }. By asssumption, An = n∈N X. Let xk ∈ An converge to x in X. We then have kT xk = limk→∞ kT xkk ≤ n for all T ∈ T so that An is closed for every n ∈ N. Corollary 3.2 now yields N ∈ N, x0 ∈ AN and ε > 0 with B(x0, ε) ⊆ AN . Let z ∈ B(0, ε). We have x0 − z, x0 + z ∈ B(x0, ε) ⊆ AN , and hence

1 1 1 1 N N kT zk = kT ( 2 (z +x0)+ 2 (z −x0))k ≤ 2 kT (z +x0)k+ 2 kT (x0 −z))k ≤ 2 + 2 = N.

Finally, let x ∈ X with kxk ≤ 1. Set z = εx ∈ B(0, ε). It follows N ≥ kT zk = N ε kT xk; i.e., kT k ≤ ε for all T ∈ T . Corollary 3.5 (Banach–Steinhaus). Let X be a Banach space, Y be a normed vector space, and Tn belong to L(X,Y ) for every n ∈ N. Assume that (Tnx)n converges in Y for each x ∈ X. Then supn kTnk < ∞.

Proof. This result follows from the principle of uniform boundedness since cx := supn kTnxk is ﬁnite for each x ∈ X, by the assumption.

Example 3.6. Let X = c00 and Y = c0 be endowed with the supremum norm. Set Tnx = (x1, 2x2, . . . , nxn, 0,... ) for n ∈ N and x ∈ X. Then Tn belongs to L(X,Y ) with kTnk = n since kTnxk∞ ≤ n kxk∞ and kTnk ≥ kTnenk∞ = n for all n ∈ N and x ∈ X. Hence, the sequence (Tn) is unbounded, but (Tnx)n converges for each x ∈ X because there is an index m = mx ∈ N with xk = 0 if k > m and thus Tnx = (x1, 2x2, . . . , mxm, 0,... ) for all n ≥ m. So the completeness of X is needed in Theorem 3.4. ♦

Deﬁnition 3.7. Let X and Y be normed vector spaces and Tn,T ∈ L(X,Y ) for n ∈ . We say that Tn converges strongly to T as n → ∞ if Tnx → T x in Y as N s n → ∞ for each x ∈ X. One then writes Tn −→ T .

Remark 3.8. Let X and Y be normed vector spaces and Tn,T,Sn,S ∈ L(X,Y ) for n ∈ N. Then the following assertions hold. a) The strong limit is uniquely determined. b) If Tn tends to T in operator norm as n → ∞, then Tn converges to T strongly since kTnx − T xk ≤ kTn − T k kxk for all x ∈ X. The converse is wrong in general. 2 Consider e.g. the operators given by Pnx = (x1, . . . , xn, 0,... ) on X = Y = ` which converge strongly to I but kPn − Ik = 1 for all n ∈ N. c) If Tn tends strongly to T , Sn tends strongly to S and α, β ∈ F, then αTn +βSn converges strongly to αT + βS as n → ∞. ♦ Lemma 3.9. Let X be a normed vector space, Y be a Banach space, S ⊆ X such that D := lin S is dense in X, and Tn ∈ L(X,Y ) for all n ∈ N. Assume that sup kT k =: M < ∞ and that (T x) converges for every x ∈ S as n → ∞. n∈N n n n Then there is a (unique) operator T ∈ L(X,Y ) such that Tn converges strongly to T and kT k ≤ lim infn→∞ kTnk =: M0. If D = X, these assertions also hold if Y is a normed vector space.

Proof. By linearity, the limit T0x := limn→∞ Tnx exists for every x ∈ D. One checks as in the proof of Proposition 2.5 that T0 is linear. Choose a subsequence with M0 = limj→∞ kTnj k. We then obtain kT0xk = limj→∞ kTnj xk ≤ M0 kxk for

1Observe that uniform boundedness is equivalent to: ∃ c > 0 ∀ T ∈ T , x ∈ B(0, 1) : kT xk ≤ c. 3.1. The principle of uniform boundedness 45

each x ∈ D, hence T0 ∈ L(D,Y ) and kT0k ≤ M0. So far we have not used that Y is Banach space, and the addendum is thus proved if D = X. In the general case, Lemma 2.13 gives an extension T ∈ L(X,Y ) of T0 with kT k = kT0k ≤ M0 ≤ M, since Y is a Banach space and D is dense. Let ε > 0 and x ∈ X. Choose z ∈ D with kx − zk ≤ ε. We estimate

lim sup kT x − Tnxk ≤ lim sup (kT (x − z)k + kT z − Tnzk + kTn(z − x)k) n→∞ n→∞

≤ 2Mε + lim kT0z − Tnzk = 2Mε. n→∞

Since ε > 0 is arbitrary, Tn tends strongly to T .

Example 3.10. a) Let X = Y = c0, D = c00 and Tnx = (x1, 2x2, . . . , nxn, 0,... ) for all x ∈ c0 and n ∈ N. As in Example 3.6 we see that Tn ∈ L(X) with kTnk = n → ∞. Due to Corollary 3.5, the operators Tn do not have a strong limit. However, for each x ∈ c00 the vectors Tnx tend to (x1, . . . , mxm, 0,...) as n → ∞ where m = mx ∈ N is given by Example 3.6. The uniform bound on kTnk is thus needed in Lemma 3.9. b) Let X = Y = c0 and Tnx = xnen for all x ∈ c0 and n ∈ N. Then kTnxk∞ = |xn| tend to T := 0 as n → ∞ for each x ∈ X; i.e., Tn → 0 strongly. However, kTnenk∞ ≥ 1 and kTnxk∞ ≤ kxk∞ for all n ∈ N and x ∈ X. Here the limit limn→∞ kTnk = 1 exists, but it is strictly larger than kT k = 0, despite the strong convergence. ♦ Example 3.11 (Pointwise divergent Fourier series). We endow X = {f ∈ C([−π, π]) | f(−π) = f(π)} with the supremum norm and f ∈ X. By Analysis 3 there are ak, bk ∈ F such that the Fourier sum

n a0 X S (f, t) = + (a cos(kt) + b sin(kt)), t ∈ [−π, π], n 2 k k k=1 converges to f in L2(−π, π) as n → ∞. Extend f to a 2π–periodic function on R. It is straightforward to check that

( sin(n+ 1 )t Z π 2 1 2 sin t , t ∈ [−π, π] \{0}, Sn(f, t) = f(s + t)Dn(s) ds with Dn(t) = 2 1 π −π n + 2 , t = 0,

? see Werner [Wer05, S.146]. Set ϕn(f) = Sn(f, 0). Then ϕn ∈ X for each n ∈ N. We suppose that ϕn(f) would converge for every f ∈ X. Corollary 3.5 would then imply supn kϕnk < ∞. However, the proof of Satz IV.2.10 in [Wer05] yields 1 Z π kϕnk = |Dn(t)| dt −→ ∞ as n → ∞. π −π Consequently, there must exist a function f ∈ X whose Fourier series diverges at the point t = 0. (A more or less concrete example of a pointwise divergent Fourier series is given in Section 18 of [Koe89].) ♦ To apply the above lemma, one often needs dense subspaces of ‘good’ functions. The next proposition is one of the fundamental results in this direction. It holds in much greater generality, see Theorem 3.14 in [Rud87].

d Proposition 3.12. Let U ⊆ R be open and p ∈ [1, ∞). Then the set Cc(U) = {f ∈ p C(U) | supp(f) ⊆ U is bounded} is dense in L (U). (More precisely, Cc(U) + N is dense in Lp(U).) 46 Chapter 3. Two main theorems on linear operators

Proof. Let f ∈ Lp(U), p ∈ [1, ∞), and ε > 0. There is a simple function g ∈ Lp(U) with kf −gkp ≤ ε due to Satz IV.7.6 in [Wer06]. We next employ a cutoﬀ argument d to obtain functions with a compact support in U. Put Un = B(0, n) if U = R . Otherwise, we set

1 Un = {x ∈ U ∩ B(0, n) | d(x, ∂U) < n } 1 for all n ∈ N. Observe that the functions Un g converge pointwise to g as n → ∞ 1 1 and | Un g| ≤ |g| for all n ∈ N. Lebesgue’s theorem then shows that Un g tends to g in Lp(U). We can thus ﬁx an index N ∈ N so that the simple function m 1 X 1 h := UN g = aj Aj j=1 satisﬁes kg − hkp ≤ ε, for some m ∈ N, aj ∈ F and Aj ∈ B(U) with λ(Aj) < ∞ and Aj ⊆ UN . Since λ is ‘regular’ (see Satz IV.3.15 in [Wer06]) there are compact Kj and open −1 −1 p Oj such that Kj ⊆ Aj ⊆ Oj and λ(Oj \ Kj) ≤ (ε|aj| m ) for each j. Replacing Oj by Oj ∩UN if necessary, we may assume that Oj ⊆ UN . By an exercise there is a function ϕj ∈ C(U) such that ϕj = 1 on Kj, supp ϕj ⊆ UN ⊆ UN+1 and 0 ≤ ϕj ≤ 1 Pm for all j = 1, . . . , m. Hence, ϕ := j=1 ajϕj belongs to Cc(U) and

m m X X 1 1 p kϕ − hkp ≤ |aj| kϕj − Aj kp ≤ |aj| λ(Oj \ Kj) ≤ ε. j=1 j=1

It follows that kf − ϕkp ≤ 3ε.

Example 3.13 (Left translations). Let X ∈ Lp(R) for some 1 ≤ p < ∞. For every t ∈ R we deﬁne (T (t)f)(s) = f(s + t) for s ∈ R and f ∈ X. It is clear that T (t): X → X is linear, Z p p p kT (t)fkp = |f(s + t)| ds = kfkp , R and T (t) has the inverse T (−t). Hence, T (t) ∈ L(X) is an isometric isomorphism. (This assertion can similarly be shown for p = ∞.) For any ﬁxed t > 0 we have ( ) 1, −t ≤ s ≤ 0 T (t)1 (s) = 1 (s + t) = = 1 (s) [0,t] [0,t] 0, otherwise [−t,0]

−1/p1 p for all s ∈ R. Setting f = t [0,t] we thus obtain kfkp = 1 and kT (t)f − fkp = −1 R t 1p t −t ds = 2. Therefore the map R 3 t 7→ T (t) ∈ L(X) is not continuous with respect to the operator norm. However, it is strongly continuous, i.e., the maps R 3 t 7→ T (t)f ∈ X are continuous for every f ∈ X. In fact, because kT (t)k = 1 for all t ∈ R and Cc(R) is dense in X by Proposition 3.12, it suﬃces to consider f ∈ Cc(R) due to Lemma 3.9. Let t, t0 ∈ R. Since f ∈ Cc(R) is uniformly continuous, we derive

kT (t)f − T (t0)fk∞ = sup |f(s + t) − f(s + t0)| −→ 0 as t → t0. s∈R

Note that there is a compact interval J ⊆ R such that supp(T (t)f − T (t0)f) ⊆ J for |t − t0| ≤ 1. We now obtain

1/p kT (t)f − T (t0)fkp ≤ λ(J) kT (t)f − T (t0)fk∞ −→ 0 as t → t0.

These results also hold for X = C0(R) with slightly modiﬁed proofs. ♦ 3.1. The principle of uniform boundedness 47

Molliﬁer

d ∞ Let U ⊂ R be open. We deﬁne the space of test functions on U by Cc (U) := ∞ d C (R ) ∩ Cc(U). Observe that the function

1 ( − 2 e 1−|x|2 , |x| < 1, ϕ(x) = 2 0, |x|2 ≥ 1,

∞ d −1 −d 1 belongs to Cc (R ). Set k = kϕk1 ϕ and kε(x) = ε k( ε x) for all x ∈ R and ∞ d ε > 0. Then kε ∈ Cc (R ) with supp kε = B(0, ε), kε ≥ 0 and Z Z −d 1 kkεk1 = ε k( ε x)dx = k(y) dy = 1 d d R R

1 1 for all ε > 0, where we have used the transformation y = ε x. We write f ∈ Lloc(U) if f|B ∈ L1(B) for any bounded Borel set B ⊆ U. The extension of f by 0 is ˜ denoted by f. We define the mollifier Gε by Z Z ˜ ˜ (Gεf)(x) = (kε ∗ f)(x) = kε(x − y)f(y) dy = kε(x − y)f(y) dy (3.1) d ¯ R U∩B(x,ε) for all x ∈ U. (The integral exists for all x ∈ U since f is integrable on U ∩ B¯(x, ε) and kε is bounded.) The next result indicates that the operators Gε are a powerful tool to approximate functions by smooth ones. As the proof indicates, they are often used in combination with cutoff arguments. We improve the proposition considerably in Section 5.2.

d 1 Proposition 3.14. Let U ⊆ R be open, f ∈ Lloc(U), ε > 0 and 1 ≤ p ≤ ∞. Deﬁne Gε by (3.1). The following assertions hold. ∞ a) Gεf ∈ C (U). If there is a compact set K ⊆ U with f(x) = 0 for a.e. ∞ x ∈ U \ K, then Gεf ∈ Cc (U) for all suﬃciently small ε > 0. p p b) Gε ∈ L(L (U)) with kGεk ≤ 1. Let 1 ≤ p < ∞ and f ∈ L (U). Then Gεf → f in Lp(U) as ε → 0. ∞ p c) Let 1 ≤ p < ∞. Then Cc (U) is dense in L (U). More precisely, if f ∈ p q ∞ L (U) ∩ L (U) for some 1 ≤ p, q < ∞ then there are fn ∈ Cc (U) converging to f in Lp(U) and in Lq(U).

1 Proof. a) Fix ε > 0 and f ∈ Lloc(U). Let x0 ∈ U. There is an r > 0 such that B(x0, r) ⊆ U. Let x ∈ B(x0, r), j ∈ {1, . . . , d}, and l = 0, 1. We then have

|∂l k (x − y)f(y)| ≤ k∂l k k 1 (y) |f(y)| =: h(y) xj ε xj ε ∞ B(x0,r+ε)∩U for all y ∈ U. The function h is integrable. A corollary to the theorem of dominated convergence now yields that kε ∗ f is continuous and partially diﬀerentiable at x0. ∞ Iterating this argument, one concludes that Gεf ∈ C (U). Let K ⊆ U be compact such that f(x) = 0 for a.e. x ∈ U \ K. Since K ∩ ∂U = ∅ 1 and K is compact, ε0 := 2 inf{dist(x, ∂U) | x ∈ K} > 0 and hence S = K + B(0, ε0) ⊆ U. So (3.1) yields supp(Gεf) ⊂ S for ε ∈ (0, ε0]. Note that S is compact. Indeed, let xn ∈ S. Then xn = yn +zn for some yn ∈ K and zn ∈ B(0, ε0).

Compactness implies that ynk → y in K as k → ∞ and znkj → z in B(0, ε0) as j → ∞. Therefore, xnkj tends to x = y + z ∈ S. p b) Let f ∈ L (U). Young’s inequality (see Proposition 2.15) implies that Gεf ∈ p ˜ p L (U) and kGεfkp ≤ kkεk1 kfkp = kfkp. As a result, Gε ∈ L(L (U)) with kGεk ≤ 1 for all ε > 0. Let 1 ≤ p < ∞. To prove the convergence, we only have to consider p g ∈ Cc(U) due to Lemma 3.9 since kGεk ≤ 1 and Cc(U) is dense in L (U) by 48 Chapter 3. Two main theorems on linear operators

Proposition 3.12. Let g ∈ Cc(U) and K = supp(g). Again, S := K + B(0, ε0) ⊆ U is compact. We then derive Z Z

sup |Gεg(x) − g(x)| = sup kε(x − y)˜g(y) dy − kε(x − y) dy g(x) d d x∈U x∈U R R Z ≤ sup kε(x − y) |g˜(y) − g(x)| dy x∈U B¯(x,ε)

≤ kkεk1 sup |g˜(y) − g˜(x)| −→ 0 as ε → 0, |x−y|≤ε using thatg ˜ is uniformly continuous and kkεk1 = 1. Because of supp Gεg ⊆ S for 1/p ε ∈ (0, ε0] and supp g ⊆ S, the distance kGεg − gkp ≤ λ(S) kGεg − gk∞ tends to 0 as ε → 0. p q d c) Let f ∈ L (U) ∩ L (U) for some 1 ≤ p, q < ∞. Put Un = B(0, n) if U = R . 1 Otherwise, we set Un = {x ∈ U ∩ B(0, n) | d(x, ∂U) < n } for all n ∈ N. Using assertion a), for each n ∈ N we can choose an εn ∈ (0, 1/n] such that fn := 1 ∞ 1 1 Gεn ( Un f) ∈ Cc (U). Since Un f → f pointwise and | Un f| ≤ |f|, Lebesgue’s 1 p q theorem yields Un f → f in L (U) and in L (U) as n → ∞. Employing kGεn k ≤ 1 and part b), we then derive

1 1 kGεn ( Un f) − fkr ≤ kGεn k k Un f − fkr + kGεn f − fkr −→ 0 as n → ∞, where r ∈ {p, q}.

3.2 The open mapping theorem and invertibility

The invertibility of T ∈ L(X,Y ) means that for each y ∈ Y there is a unique solution x = T −1y of the equation T x = y which depends continuously on y. We start with a few simple properties of invertible operators and then establish the automatic continuity of T −1 in Banach spaces.

Lemma 3.15. Let X,Y,Z be normed vector spaces, T ∈ L(X,Y ) and S ∈ L(Y,Z) be invertible. Then ST ∈ L(X,Z) is invertible with the inverse T −1S−1 ∈ L(Z,X).

−1 −1 −1 −1 Proof. The operators ST and T S are continuous. Moreover, T S ST = IX −1 −1 and STT S = IZ .

Lemma 3.16. Let Z be a Banach space and zj ∈ Z, j ∈ N0, satisfy s := P∞ Pn j=0 kzjk < ∞. Then the partial sums Sn = j=0 zj converge in Z as n → ∞. P∞ Their limit is denoted by j=0 zj and has norm less or equal s. Pn Proof. Let n > m in N0. Then kSn − Smk ≤ j=m+1 kzjk tends to 0 as n, m → ∞. Since Z is Banach space, (Sn) has a limit S, and kSk = limn→∞ kSnk ≤ s.

Proposition 3.17 (Neumann series). Let X be Banach space, Y be a normed vector space and T,S ∈ L(X,Y ). Assume that T is invertible and that kSk < kT −1k−1. Then S + T is invertible and

∞ X (S + T )−1 = (−T −1S)nT −1 =: R (convergence in L(Y,X)), n=0 kT −1k k(S + T )−1k ≤ . 1 − kT −1Sk

In particular, the set of invertible operators is open in L(X,Y ). 3.2. The open mapping theorem and invertibility 49

−1 P∞ −1 n Proof. We have q := kT SkL(X) < 1 by assumption. Then n=0 k(T S) k ≤ 1/(1 − q). By Proposition 2.5, L(X) is a Banach space, and hence Lemma 3.16 yields the convergence of

∞ ∞ X X R = (−T −1S)nT −1 = T −1(−ST −1)n n=0 n=0 in L(Y,X) and that kRk ≤ kT −1k/(1 − q). Moreover,

∞ ∞ ∞ X X X R(S + T ) = (−T −1S)n(T −1S + I) = − (−T −1S)j + (−T −1S)n = I, n=0 j=1 n=0 ∞ X (S + T )R = (ST −1 + I)(−ST −1)n = I. n=0

1 1 Example 3.18. Let X = c00 with k · k∞ and T x = (x1, 2 x2, 3 x3,...). Then −1 −1 T ∈ L(X), T is bijective with T y = (y1, 2y2, 3y3,...). But T : c00 → c00 is not continuous by Example 2.1. ♦ Deﬁnition 3.19. Let M and M 0 be metric spaces. A map f : M → M 0 is called open if f(O) is open in M 0 for each open set O ⊆ M. Remark 3.20. a) A bijective map f : M → M 0 is open if and only if f(O) = {y ∈ M 0 | f −1(y) ∈ O} = (f −1)−1(O) is open in M 0 for each open O ⊆ M if and only if f −1 is continuous. −1 b) In Example 3.18 the maps T,T : c00 → c00 are bijective, T is continuous but not open, and T −1 is open but not continuous, due to part a). Observe that c00 is not a Banach space. ♦ Theorem 3.21 (Open mapping theorem). Let X and Y be Banach spaces and T ∈ L(X,Y ) be surjective. Then T is open. If T is even bijective, then is T invertible. Proof. The second assertion is a consequence of the ﬁrst one because of Remark 3.20. So let T be surjective. Set Ur = BX (0, r) and Vr = BY (0, r) for every r > 0.

Claim A). There is an ε > 0 with Vε ⊆ TU2. Assume that Claim A) has been shown. Let O ⊆ X be open, x ∈ O, and y = T x. Then there is an r > 0 with BX (x, r) ⊆ O. From A) and the linearity of T we deduce εr r r BY (y, 2 ) = y + 2 Vε ⊆ T x + 2 TU2 = TBX (x, r) ⊆ TO. Hence, T is open. S∞ Proof of A). The surjectivity of T yields Y = n=1 TUn. Since Y is complete, Corollary 3.2 gives N ∈ N, y0 ∈ Y and r > 0 such that

BY (y0, r) ⊆ TUN = (2N) TU 1 = 2N TU 1 . 2 2

−1 Hence, BY (z, ε) ⊆ TU 1 where z = (2N) y0 and ε = r/(2N). Observe that TU 1 2 2 is convex and TU 1 = −TU 1 . By approximation, these facts also hold for TU 1 , cf. 2 2 2 Corollary 1.18. Therefore

1 1 Vε = BY (z, ε) − z ⊆ TU 1 − TU 1 = 2( TU 1 + TU 1 ) ⊆ 2 TU 1 = TU1. 2 2 2 2 2 2 2 For later use, we also note that the above inclusion yields

Vαε = αVε ⊆ α TU1 = TUα for all α > 0. (∗) 50 Chapter 3. Two main theorems on linear operators

Now, Claim A) and thus the theorem follow from the next assertion.

Claim B). It holds TU1 ⊆ TU2. Proof of B) Let y ∈ TU1. Then there is an x1 ∈ U1 with ky − T x1k < ε/2; i.e., y − T x1 ∈ Vε/2 and so y − T x1 ∈ TU1/2 due to (∗). Similarly, we obtain

x2 ∈ U1/2 with y − T x1 − T x2 = y − T (x1 + x2) ∈ V ε ⊆ TU 1 . 4 4

Inductively, we ﬁnd xn ∈ U21−n such that

y − T (x1 + ··· + xn) ∈ Vε2−n . (+) P∞ Since X is a Banach space and n=1 kxnk < 2, by Lemma 3.16 there exists the P∞ vector x := n=1 xn ∈ U2. Hence, T x ∈ TU2. Letting n → ∞ in (+), we thus obtain

∞ X T x = T xn = y ∈ TU2. n=1

Corollary 3.22. Let k · k and · be complete norms on the vector space X such that kxk ≤ c x for some c >90 and9 all x ∈ X. Then these norms are equivalent. 9 9 Proof. The map I :(X, · ) → (X, k·k) is continuous by assumption, and it is linear and bijective. Due to the9 completeness9 and Theorem 3.21, the map I−1 :(X, k·k) → (X, · ) is also continuous. The assertion now follows from I−1x = x. 9 9 1 Example 3.23. On X = C ([0, 1]) we have the complete norm f = kfk∞ + 0 kf k∞ and the non complete norm kfk = kfk∞ ≤ f which are9 not9 equivalent 9 9 0 (e.g., the functions fn(t) = sin(nt) satisfy kfnk∞ ≤ 1, but fn ≥ |fn(0)| = n). So 9 9 in Theorem 3.21 also Y must be complete. ♦ Corollary 3.24. Let X and Y be Banach spaces and T ∈ L(X,Y ) be injective. The following assertions are equivalent. a) The operator T −1 : R(T ) → X is continuous. b) There is a constant c > 0 such that kT xk ≥ c kxk for every x ∈ X. c) The range R(T ) is closed. Proof. a) ⇒ b): For all x ∈ X we have kxk = kT −1T xk ≤ kT −1k kT xk. The implication ‘b)⇒c)’ was shown in Remark 2.11; and ‘c)⇒ a)’ follows from Theorem 3.21 because R(T ) is complete by c). Proposition 3.25. Let X be a Banach space and X = Y ⊕ Z. Then the following asserions hold. a) X =∼ Y × Z. b) The projection P with R(P ) = Y and N(P ) = Z is continuous (cf. Re- mark 2.18). Proof. a) Since Y and Z are Banach spaces, their product Y × Z is a Banach space for the norm k(y, z)k = kyk + kzk, see Paragraph 2.2A). Further, the map T : Y ×Z → X, T (y, z) = y +z, is linear, continuous (since kT (y, z)k ≤ kyk+kzk = k(y, z)k and bijective (since X = Y ⊕ Z). Theorem 3.21 now yields a). b) Let x = y + z ∈ Y ⊕ Z. Then P x = y. The operator T from a) satisﬁes (y, z) = T −1x. Hence, part a) yields

−1 −1 kP xk = kyk ≤ kyk + kzk = kT xkY ×Z ≤ kT k kxk. Chapter 4

Duality

Let X be a normed vector space. We study the ‘duality pairing’

? ? ? ? ? X × X → F, (x, x ) 7→ x (x) =: hx, x i = hx, x iX . This map is linear in both components, and it is continuous since |hx, x?i| ≤ ? ? kxkX kx kX? (see an exercise). We will ﬁrst determine the dual space X if X is a Hilbert space and if X = Lp(µ) for p ∈ [1, ∞).

4.1 Hilbert spaces

Deﬁnition 4.1. A scalar product on a vector space X is a map (·|·): X2 → F with

a) (αx1 + βx2|y) = α (x1|y) + β (x2|y), b) (x|y) = (y|x),

c) (x|x) ≥ 0, (x|x) = 0 ⇐⇒ x = 0, for all x1, x2, x, y ∈ X and α, β ∈ F. The map is called sesquilinear form if a) and b) hold, and positive deﬁnite if c) holds. The pair (X, (·|·)) is called a pre Hilbert space. We set kxk = p(x|x).1

Remark 4.2. Let (X, (·|·)) be a pre Hilbert space. (i) Properties a) and b) yield (0|y) = 0, (x|0) = 0, (x|x) ∈ R, and ¯ ¯ (x|αy1 + βy2) = (αy1 + βy2|x) =α ¯ (y1|x) + β (y2|x) =α ¯ (x|y1) + β (x|y2) for all α, β ∈ F and x, y1, y2 ∈ X. (ii) We have the Cauchy–Schwarz inequality:

|(x|y)| ≤ kxk kyk for all x, y ∈ X. (CS)

Equality holds if and only if x and y are linearly dependent. See Lineare Algebra or Proposition I.1.4 in [Con90]. (iii) (X, k · k) is a normed vector space, where k · k is given by Deﬁnition 4.1. In fact, let x, y ∈ X and α ∈ F. Using (CS), we compute kxk = 0 ⇐⇒ (x|x) = 0 ⇐⇒ x = 0, kαxk = pαα¯ (x|x) = |α| kxk,

1Here one also uses the notions ‘inner product’ and ‘inner product space’.

51 52 Chapter 4. Duality

kx + yk2 = (x + y|x + y) = kxk2 + (x|y) + (y|x) + kyk2 = kxk2 + 2 Re (x|y) + kyk2 ≤ (kxk + kyk)2 . (4.1)

(iv) The scalar product is Lipschitz on each ball of X2 and thus continuous from X2 to F. In fact, (CS) yields

|(x1|y1) − (x2|y2)| ≤ |(x1 − x2|y1)| + |(x2|y1 − y2)| ≤ r kx1 − x2k + r ky1 − y2k √ ≤ 2r k(x1, y1) − (x2, y2)k

2 2 1 for all xk, yk ∈ X with k(xk, yk)k := (kxkk + kykk ) 2 ≤ r. (v) From (4.1) we deduce the parallelogramm identity

kx + yk2 + kx − yk2 = kxk2 + 2 Re (x|y) + kyk2 + kxk2 − 2 Re (x|y) + kyk2 = 2 kxk2 + 2 kyk2 for all x, y ∈ X. (4.2)

Deﬁnition 4.3. Let (·|·) be a scalar product on X. Then the norm k · k given in Deﬁnition 4.1 is called Hilbert norm. If k · k is complete, then (X, (·|·)) is called Hilbert space.

Example 4.4. a) X = Fd is a Hilbert space with

d d X 2 X 2 (x|y) = xky¯k and kxk2 = |xk| . k=1 k=1

Note that (X, | · |p) is not induced by a scalar product if d ≥ 2 and p 6= 2, since 2 2 2/p 2/p 2 2 |e1 + e2|p + |e1 − e2|p = 2 + 2 6= 2(|e1|p + |e2|p) = 4 contradicting (4.2). b) X = `2 is a Hilbert space with

∞ ∞ X 2 X 2 (x|y) = xky¯k and kxk2 = |xk| . k=1 k=1

The first series converges absolutely due to Hölder’sinequality with p = p0 = 2. c) Let (S, A, µ) be a measure space. Then X = L2(µ) is a Hilbert space with Z Z 2 2 (f|g) = fg dµ and kfk2 = |f| dµ. S S The product fg is integrable by Hölder’sinequality with p = p0 = 2. d) A linear subspace of a pre Hilbert space is a pre Hilbert space with the restricted scalar product. ♦ In pre Hilbert spaces one can also define angles, and not only distances as in normed vector spaces. We restrict ourselves to the angle π/2.

Deﬁnition 4.5. Two elements x and y of a pre Hilbert space X are called orthogonal if (x|y) = 0. Two subsets A, B ⊆ X are called orthogonal if a ⊥ b for all a ∈ A and b ∈ B. One then writes x ⊥ y and A ⊥ B, and also x ⊥ A instead of {x} ⊥ A. The orthogonal complement of A is given by

A⊥ = {x ∈ X | x ⊥ a for every a ∈ A}.

A projection P ∈ L(X) is called orthogonal if R(P ) ⊥ N(P ).

Example 4.6. a) Functions f, g ∈ L2(µ) with disjoint support are orthogonal since then (f|g) = R fg dµ = 0. 4.1. Hilbert spaces 53

int 2 b) The functions fn(t) = e for n 6= m in Z are orthogonal in L (0, 2π) since Z 2π 1 2π int −imt i(n−m)t (fn|fm) = e e dt = e = 0. 0 i(n − m) 0

c) Let f ∈ L2(R) be even and g ∈ L2(R) be odd. Then f ⊥ g, since fg is odd R and hence fg dt = 0. ♦ Remark 4.7. Let X be a pre Hilbert space, A, B ⊆ X, and x, y ∈ X. It holds: a) x ⊥ x holds if and only if x = 0; and thus X⊥ = {0}. x ⊥ y holds if and only if y ⊥ x. {0}⊥ = X. b) If x ⊥ y, then kx + yk2 = kxk2 + kyk2 due to (4.1). (Pythagoras) c) A ∩ A⊥ ⊆ {0} and A ⊆ (A⊥)⊥ =: A⊥⊥, because of a). d) A⊥ is a closed linear subspace. In fact, it is easy to check that it is a linear subspace. Take xn ⊥ A with xn → x. Then (x|a) = limn→∞ (xn|a) = 0 for all a ∈ A, so that x ⊥ A and A⊥ is closed. e) If A ⊆ B, then B⊥ ⊆ A⊥. As in d) one sees that A⊥ = (lin A)⊥. f) If (x|y) = (z|y) for some x, z ∈ X and all y from a dense subset D ⊆ X, then x = z. In fact, by approximation we obtain x − z ⊥ X, and thus x = z by a). ♦ Many of the special properties of Hilbert spaces rely on the next theorem. Theorem 4.8 (Projection theorem). Let X be a Hilbert space and Y ⊆ X be a closed linear subspace. It then holds X = Y ⊕ Y ⊥, and there is a unique orthogonal projection P ∈ L(X) with R(P ) = Y , N(P ) = Y ⊥, and kP k = 1 (if Y 6= {0}). We further have Y ⊥⊥ = Y , X/Y ' Y ⊥, and

kx − P xk = inf kx − yk for every x ∈ X. y∈Y

Proof. Due to Remark 4.7, Y ⊥ is a closed linear subspace of X and Y ∩ Y ⊥ = {0}. Let x ∈ X. To construct P , we look for yx ∈ Y with kx−yxk = infy∈Y kx−yk =: δ. 1 There are yn ∈ Y with kx − ynk → δ as n → ∞. From (4.2) and 2 (yn + ym) ∈ Y we deduce

1 2 1 1 2 0 ≤ k 2 (yn − ym)k = k 2 (yn − x) − 2 (ym − x)k 1 2 1 2 1 2 = 2 kyn − xk + 2 kym − xk − k 2 (yn + ym) − xk 1 2 1 2 2 ≤ 2 kyn − xk + 2 kym − xk − δ −→ 0 as n, m → ∞.

Since X is complete, it exists yx = limn→∞ yn in Y and we obtain kyx − xk = limn→∞ kyn − xk = δ. ⊥ Set zx := x − yx. We want to show that zx ∈ Y . Take w ∈ Y \{0} and put −2 α = kwk (w|zx). We then obtain αw + yx ∈ Y and thus

2 2 2 δ ≤ kx − (αw + yx)k = kzx − αwk |(z |w)|2 = kz k2 − 2 Re α(z |w) + |α|2kwk2 = δ2 − x , x x kwk2 using (4.1) and the deﬁnition of zx and α. Therefore, (zx|w) = 0 and thus zx ⊥ ⊥ Y . Taking into account x = yx + zx, we have shown that Y + Y = X, and hence X = Y ⊕ Y ⊥. Remark 2.18 gives a unique operator P = P 2 ∈ L(X) with ⊥ R(P ) = Y and N(P ) = Y , where P x = yx. Pythagoras further implies that 2 2 2 2 kP xk ≤ kyxk + kzxk = kxk since yx ⊥ zx, i.e., P ∈ L(X) and kP k ≤ 1. Lemma 2.17 then yields kP k = 1 if Y 6= {0}. ⊥⊥ ⊥ It holds Y ⊆ Y due to Remark 4.7. If x ⊥ Y , then 0 = (x|zx) = (yx+zx|zx) = 2 ⊥⊥ kzxk so that x = yx ∈ Y ; i.e., Y = Y . Example 2.21 further implies that Y ⊥ = X/Y . 54 Chapter 4. Duality

2 ⊥ 1 0 Example 4.9. a) Let X = F and Y = F×{0}. Then Y = {0}×F and P = ( 0 0 ). 2 2 ∼ 2 b) Let X = L (R) and Y = {f ∈ L (R) | f = 0 a.e. on R−} = L (R+). Then ⊥ 2 ∼ 2 1 Y = {f ∈ L (R) | f = 0 a.e. on R+} = L (R−) with P f = R+ f. Compare Example 2.19 for a) and b). c) Let X = L2(R) and Y = {f ∈ L2(R) | f(t) = f(−t) for a.e. t ∈ R} be the subspace of even functions. As in Example 1.33 one sees that the linear subspace 1 Y is closed in X. Set P f(t) = 2 (f(t) + f(−t)) for a.e. t ∈ R and f ∈ X. Then 2 2 P f ∈ Y , P = P and kP fk2 ≤ 2 kfk2 for f ∈ X. As a result, P ∈ L(X) with kP k = 1, R(P ) = Y , and N(P ) = {f ∈ L2(R) | f(t) = −f(−t) for a.e. t ∈ R} is the space of odd functions in X. The projection P is orthogonal by Example 4.6. d) Let A be a subset of a Hilbert space X. Set Y = lin A. Remark 4.7 and the ⊥⊥ ⊥⊥ projection theorem then yield that A = Y = Y . ♦ Let X be a Hilbert space. For every y ∈ X we deﬁne Φ(y): X → F by setting [Φ(y)](x) := (x|y) for all x ∈ X. (4.3)

The map Φ(y) is linear, and it holds |[Φ(y)](x)| ≤ kxk kyk by (CS). Hence, Φ(y) ∈ ? X with kΦ(y)kX? ≤ kykX .

? Theorem 4.10 (Riesz). Let X be a Hilbert space and deﬁne ΦX = Φ : X → X by (4.3). Then Φ is bijective, isometric and antilinear.2 With y := Φ−1(x?), we thus obtain for each x? ∈ X? exactly one y ∈ X such that hx, x?i = (x|y) for all x ∈ X, ? where kykX = kx kX? . Proof. Equation (4.3) and Remark 4.2 imply that Φ is antilinear. Let y ∈ X \{0}, 1 and set x = kyk y. Then kxk = 1, and we thus obtain 1 kΦ(y)k ? ≥ |[Φ(y)](x)| = (y|y) = kyk X kyk so that kΦ(y)kX? = kykX . This means that Φ is isometric and hence injective. Let ϕ ∈ X? \{0}. Then Z := N(ϕ) 6= X is a closed linear subspace of X. Theorem 4.8 ⊥ ⊥ −1 ⊥ yields that Z 6= {0}. We take any y0 ∈ Z \{0} and set y1 = ϕ(y0) y0 ∈ Z \{0}. Let x ∈ X. We calculate ϕ(x−ϕ(x)y1) = ϕ(x)−ϕ(x)ϕ(y1) = 0 using that ϕ(y1) = 1. ⊥⊥ As a result, x − ϕ(x)y1 ∈ Z = Z leading to

2 0 = (x − ϕ(x)y1|y1) = (x|y1) − ϕ(x) ky1k , −2 ϕ(x) = (x| ky1k y1) for all x ∈ X.

We have shown that ϕ ∈ R(Φ) and so Φ is bijective. The other claims are clear.

Usually, one identiﬁes a Hilbert space X with its dual X?; i.e., one omits the Riesz isomorphism ΦX in the notation. We stress that this can only be done for one Hilbert space at the same time.

4.2 The duals of sequence and Lebesgue spaces

p 0 Let Xp = ` for 1 ≤ p < ∞ and X∞ = c0 for p = ∞, as well as p ∈ [1, ∞] with 1 1 p0 p + p0 = 1. For each y ∈ ` , we deﬁne

∞ X Φp(y): Xp → F, (Φp(y))(x) = xkyk. k=1

2 This means that Φ(αy + βz) =α ¯Φ(y) + β¯Φ(z) for all α, β ∈ F and y, z ∈ X. 4.2. The duals of sequence and Lebesgue spaces 55

H¨older’sinequality shows that this series converges absolutely and |Φp(y)(x)| ≤ ? kxkp kykp0 . Since Φp(y) is linear, we obtain Φp(y) ∈ Xp with

kΦ (y)k ? ≤ kyk 0 . p Xp p As a result, the map

∞ p0 ? X p0 Φp : ` → Xp , hx, Φp(y)i = xkyk (∀ x ∈ Xp, y ∈ ` ) (4.4) k=1 is contractive, and it is clearly linear. Since (Φp(y))(ek) = yk for all k ∈ N, the operator Φp is injective.

p0 Proposition 4.11. Equation (4.4) deﬁnes an isometrical isomorphism Φp : ` → ? ? ? p0 Xp for all p ∈ [1, ∞]. We thus obtain for each x ∈ Xp exactly one y ∈ ` such that ? P ? hx, x i = x y for all x ∈ X , where kyk 0 = kx k ? . Via this isomorphism, k k k p p Xp

? 1 1 ? ∞ p ? p0 c0 ' ` , (` ) ' ` , and (` ) = ` for 1 < p < ∞.

Proof. In view of the above considerations it remains to show the surjectivity of Φp −1 −1 and that Φ is contractive since then kyk 0 = kΦ Φ yk 0 ≤ kΦ yk ? ≤ kyk 0 for p p p p p p Xp p 0 all y ∈ `p . We restrict ourselves to the case p = 1, where p0 = ∞. (The remainder can be proved similarly, see Theorem 4.14 and the exercises.) ? 1 ? ? Let x ∈ (` ) . We define yk := x (ek) for every k ∈ N and y = (yk). Since ? ? ∞ ? |yk| ≤ kx k kekk1 = kx k for all k ∈ N, we obtain y ∈ ` and kyk∞ ≤ kx k. ? ? Equation (4.4) also implies that [Φ1(y)](ek) = yk = x (ek). Since x and Φ1(y) ? are linear, we arrive at Φ1(y)(x) = x (x) for all x ∈ c00. Using continuity and the 1 ? density of c00 in ` , we conclude that Φ1(y) = x and so Φ1 is bijective. It also −1 follows that Φ1 is contractive. We will see in Example 4.25 that the dual of ìnfty is not isomorphic to `1. Next, let (S, A, µ) be a σ–finite measure space, 1 ≤ p < ∞, and X = Lp(µ). For each 0 g ∈ Lp (µ) we define the map Z p Φp(g): L (µ) → F, [Φp(g)](f) = fg dµ. (4.5) S Hölder’sinequality (cf. Example 2.8) yields that

p ? Φp(g) ∈ L (µ) and kΦp(g)kX? ≤ kgkp0 .

p0 p ? Consequently, Φp : L (µ) → L (µ) is linear and contractive. To show that Φp is in fact an isometric isomorphism, we need some preparations. A map ν : A → F is called an F-valued measure3 if

∞ ∞ X [ ∃ ν(An) = ν An in F n=1 n=1 P∞ for all pairwise disjoint An ∈ A, n ∈ N. Since then ν(∅) = ν(∅∪∅∪... ) = n=1 ν(∅) in F, we obtain that ν(∅) = 0. A measure µ in the sense of Paragraph 1.2B) is also called positive measure. (It is an F–valued measure if and only if it is ﬁnite.) An F-valued measure ν is called µ–continuous if ν(A) = 0 for all A ∈ A with µ(A) = 0. One then writes ν µ.

3 One usually says signed measure instead of ‘R–valued measure’. 56 Chapter 4. Duality

Example 4.12. Let µ be a positive measure on A and ρ ∈ L1(µ). We set ν(A) = R 1 R S Aρ dµ = A ρ dµ ∈ F for all A ∈ A. Then ν is an F-valued measure on A with ν µ. (In this case one writes dν = ρ dµ and calls ρ the density of ν with respect to µ.) If also ρ ≥ 0, then ν is positive measure. For a measurable f with f ≥ 0 or ρf ∈ L1(µ), we have f ∈ L1(ν) and R f dν = R fρ dµ (provided that ρ ≥ 0). S∞ Proof. Let Aj ∈ A, j ∈ N, be pairwise disjoint. Set A = j=1 Aj ∈ A and Sn 1 1 1 Bn = j=1 Aj ∈ A for n ∈ N. Then Bn ρ tends to Aρ pointwise and | Bn ρ| ≤ |ρ| ∈ L1(µ). Dominated convergence thus yields

n ∞ Z Z X Z X ν(A) = 1Aρ dµ = lim 1B ρ dµ = lim 1A ρ dµ = ν(Aj). n→∞ n n→∞ j S S j=1 S j=1 It is clear that ν µ and that ν is positive if ρ ≥ 0. The other assertions can easily be checked for simple functions and then follow by approximation. There is a converse to the above example which is one of the fundamental results of measure theory. We give a proof based on Riesz’ Theorem 4.10, where we restrict ourselves to the main special case. Theorem 4.13 (Radon–Nikodym). Let A be a σ–algebra on S, µ be a σ–ﬁnite positive measure on A and ν be an F–valued measure on A with ν µ. Then there is a unique w ∈ L1(µ) such that dν = w dµ. If ν is a positive measure, then w ≥ 0. Proof. We show the theorem if µ and ν are positive and bounded. For the general case we refer to Theorem 6.10 in [Rud87] or Appendix C in [Con90]. 1) Consider the ﬁnite positive measure τ = µ+ν on A. Observe that µ(A), ν(A) ≤ Pm 1 τ(A) for all A ∈ A. For simple functions f = j=1 ak Aj with m ∈ N, aj ∈ F and Aj ∈ A (which are pairwise disjoint, without loss of generality), we deduce

m m 2 X 2 X 2 2 kfkL2(µ) = |aj| µ(Aj) ≤ |aj| τ(Aj) = kfkL2(τ). j=1 j=1

1/2 1/2 H¨older’sinequality further yields kfkL1(µ) ≤ µ(S) kfkL2(µ) ≤ µ(S) kfkL2(τ), see Proposition 1.35. If ϕ = 0 a.e. for τ, then also for µ. Hence, kfkL1(µ) ≤ 1/2 2 µ(S) kfkL2(τ) for all simple functions f ∈ L (τ). By approximation, each f ∈ 2 1 1/2 L (τ) belongs to L (µ) and kfkL1(µ) ≤ µ(S) kfkL2(τ). As a result, by Z 2 ϕ : L (τ) → F, ϕ(f) = f dµ, S we can deﬁne a linear map which is continuous since |ϕ(f)| ≤ kfkL1(µ) ≤ 1/2 2 µ(S) kfkL2(τ) for all f ∈ L (τ). 2) Theorem 4.10 now yields a function g ∈ L2(τ) ,→ L1(τ) such that Z Z Z 0 ≤ µ(A) = 1A dµ = ϕ(1A) = 1Ag dτ = g dτ S S A 1 1 for all A ∈ A; i.e, dµ = g dτ. Set An = {g ≤ − n } := {s ∈ S | g(s) ≤ − n } ∈ A for n ∈ N. The above inequality implies that Z τ(A ) 0 ≤ g dτ ≤ − n , An n S and hence τ(An) = 0 for all n ∈ N. We deduce that {g < 0} = n An is τ–null set, thus a µ–null set. Similarly, one sees that g is real–valued a.e. Hence, g ≥ 0 a.e. In the same way, one obtains a function 0 ≤ h ∈ L1(τ) with dν = h dτ. 4.2. The duals of sequence and Lebesgue spaces 57

R 3) Set N = {g = 0}. Then µ(N) = N g dτ = 0. From the assumption ν µ we infer that ν(N) = 0. We now deﬁne the measurable function ( h(s) , s ∈ S \ N, 0 ≤ w(s) = g(s) 0, s ∈ N.

For every A ∈ A, we then compute Z Z Z Z ν(A) = ν(A ∩ N c) = h dτ = wg dτ = w dµ = w dµ, A∩N c A∩N c A∩N c A R using Example 4.12. This means that dν = w dµ. Moreover, kwk1 = S w dµ = ν(S) is ﬁnite. To show uniqueness, take w ∈ L1(µ) with ν(A) = R w dµ for all A ∈ A. e A e 1 As in 2), one sees that we ≥ 0. Set Bn = {we ≥ w + n } for n ∈ N. Since Z µ(Bn) 0 = ν(Bn) − ν(Bn) = (w − we) dµ ≤ − , Bn n we deduce we ≤ w µ–a.e. as in 2). By symmetry, one als has w ≤ we µ–a.e., and so 1 w = we in L (µ). Theorem 4.14. Let 1 ≤ p < ∞ and (S, A, µ) be a measure space which is σ– p0 p ? ﬁnite if p = 1. Then the map Φp : L (µ) → L (µ) from (4.5) is an isometric 0 isomorphism, and thus Lp(µ)? =∼ Lp (µ) via

0 Z ∀ ϕ ∈ Lp(µ)? ∃! g ∈ Lp (µ) ∀ f ∈ Lp(µ): hf, ϕi = fg dµ. S p Proof. Set X = L (µ). It remains to prove that Φp is surjective and kΦp(g)kX? ≥ p0 0 p kgkp0 for all g ∈ L (µ). We assume that µ(S) < ∞ and 1 < p < ∞. Then p = p−1 . (The general case is treated in Appendix B of [Con90], see also Theorem II.2.4 of [Wer05] for σ–finite measure spaces.) p ? p 1) Let ϕ ∈ L (µ) . Since µ(S) < ∞, we have 1A ∈ L (µ) for all A ∈ A. Define 1 ν(A) = ϕ( A). Take Aj ∈ A with Aj ∩ Ak = ∅ for all j, k ∈ N with j 6= k. Set S∞ Sn 1 1 A = j=1 Aj ∈ A and Bn = j=1 Aj ∈ A for all n ∈ N. Observe that Bn → A 1 1 p 1 pointwise as n → ∞ and 0 ≤ Bn ≤ A ∈ L (µ) for all n ∈ N. Hence, Bn tends p to 1A in L (µ) as n → ∞ by the theorem of dominated convergence. Moreover, 1 1 1 Bn = A1 + ··· + An for all n ∈ N. Using the continuity and linearity of ϕ, we then conclude that n ∞ X X ν(A) = ϕ( lim 1B ) = lim ϕ(1A ) = ν(Ak). n→∞ n n→∞ k k=1 k=1 1 p As a result, ν is an F–valued measure. If µ(A) = 0, then A = 0 in L (µ) and so ν(A) = ϕ(1A) = 0; i.e., ν µ. The Radon–Nikodym Theorem thus gives a function g ∈ L1(µ) with Z ϕ(1A) = ν(A) = 1A g dµ for all A ∈ A. S The linearity of ϕ yields Z ϕ(f) = fg dx for every simple function f : S → F. (∗) S Since we only know that g ∈ L1(µ), at first we can extend this equation to f ∈ L∞(A) only. By Analysis 3 (see also Satz IV.4.6 in [Wer06]) there are simple 58 Chapter 4. Duality functions fn : S → F converging to f uniformly as n → ∞. Since µ(S) < ∞, p Hölder’sinequality (Proposition 1.35) implies that fn → f in L (µ), and hence ϕ(fn) → ϕ(f) as n → ∞. Moreover, Z

(f − fn)g dµ ≤ kf − fnk∞ kgk1 −→ 0 S as n → ∞. As a result, (∗) holds for all f ∈ L∞(A). 0 2) To show g ∈ Lp (A), we set

(0, g(s) = 0, 0 h(s) = |g(s)|p g(s) , g(s) 6= 0. Then h is measurable and

0 0 |h|p = |g|p(p −1) = |g|p = gh.

1 ∞ Let An = {|g| ≤ n} ∈ A for n ∈ N. Then the functions An g belong to L (µ) ,→ p0 1 ∞ p L (µ) and An h to L (µ) ,→ L (µ), again owing to µ(S) < ∞. Employing (∗), we can now compute Z Z p0 1 1 1 ? 1 An |g| dµ = An hg dµ = ϕ( An h) ≤ kϕkX k An hkp S S 1 1 Z p Z p p p0 ? 1 ? 1 = kϕkX An |h| dµ = kϕkX An |g| dµ , S S 1 Z p0 p0 1 ? An |g| dµ ≤ kϕkX . S

p0 Letting n → ∞, we arrive at g ∈ L (µ) and kgkp0 ≤ kϕkX? . p p 3) Let f ∈ L (µ). There are simple functions fn converging to f in L (µ). Hence, ϕ(fn) → ϕ(f) and Z 0 (f − fn)g dµ ≤ kf − fnkp kgkp −→ 0 S

p as n → ∞. We deduce that (∗) holds for all f ∈ L (µ), i.e., Φp(g) = ϕ and Φp is surjective. Finally, kgkp0 ≤ kϕkX? = kΦp(g)kX? by 2).

0 Remark. Usually one identiﬁes Lp (µ) with Lp(µ)? for 1 ≤ p < ∞ and writes hf, gi = R fg dµ for the duality, and analogously for the sequence spaces.

4.3 The theorem of Hahn-Banach

Many of the non–trivial properties of duality rely on the Hahn-Banach theorem proved below. We start with a more algebraic version. Let X be a vector space. A map p : X → R is called sublinear if p(λx) = λp(x) and p(x + y) ≤ p(x) + p(y) for all x, y ∈ X and λ ∈ R+. A typical example is a seminorm. Theorem 4.15 (Order theoretic Hahn-Banach). Let X be a vector space with F = R, p : X → R be sublinear, Y ⊆ X be a linear subspace, and ϕ0 : Y → R be linear with ϕ0(y) ≤ p(y) for all y ∈ Y . Then there exists a linear ϕ : X → R with ϕ(y) = ϕ0(y) for all y ∈ Y and ϕ(x) ≤ p(x) for all x ∈ X. 4.3. The theorem of Hahn-Banach 59

Proof. 1) We deﬁne

M = {(Z, ψ) | Z is a linear subspace of X,Y ⊆ Z, ψ : Z → R linear with ψ|Y = ϕ0, ψ ≤ p|Z },

0 0 0 0 which contains (Y, ϕ0). On M we set (Z, ψ) ≤ (Z , ψ ) if Z ⊆ Z and ψ |Z = ψ. Straightforward calculations show that this defines a partial order on M. Let K be a totally ordered non–empty subset of M; i.e., if (Z, ψ), (Z0, ψ0) ∈ K, then (Z, ψ) ≤ (Z0, ψ0) or (Z0, ψ0) ≤ (Z, ψ). We want to check that U := S{Z | (Z, ψ) ∈ K} is a linear subspace of X, that by setting f(x) := ψ(x) for every x ∈ U and any (Z, ψ) ∈ K with x ∈ Z we define a linear map f : U → R, and that (U, f) belongs to M and is an upper bound for K. Take x ∈ U. There is a pair (Z, ψ) in K with x ∈ Z. If also x ∈ Zˆ for some (Z,ˆ ψˆ) ∈ K, then e.g. (Z,ˆ ψˆ) ≤ (Z, ψ) and so ψˆ(x) = ψ(x). Hence, f : U → R is well defined. Take further y ∈ U and α, β ∈ R. Choose a pair (Z0, ψ0) ∈ K with y ∈ U 0 and (Z, ψ) ≤ (Z0, ψ0), say. The asserted linearity follows from αx + βy ∈ Z0 and

f(αx + βy) = ψ0(αx + βy) = αψ0(x) + βψ0(y) = αf(x) + βf(y).

Similarly one checks that f|Y = ϕ0 and f ≤ p|U so that (U, f) ∈ M. By construction, (Z, ψ) ≤ (U, f) for all (Z, ψ) ∈ K. Zorn’s Lemma (see Theorem 1.2.7 of [DuS57]) now gives a maximal element (V, ϕ) in M. We next show that V = X; i.e., the linear form ϕ has the required properties. 2) Assume that V 6= X and ﬁx some x0 ∈ X \ V . We deﬁne the linear subspace V˜ := V + lin{x0}. Since V ∩ lin{x0} = {0}, for each x ∈ V˜ we have unique v ∈ V ˜ and t ∈ R with x = v + tx0. We will construct a linear mapϕ ˜ : V → R such that (V, ϕ) (V,˜ ϕ˜) ∈ M. This fact contradicts the maximality of (V, ϕ), so that V = X and the theorem will be established. Let v, w ∈ V . We then obtain

ϕ(v) + ϕ(w) = ϕ(v + w) ≤ p(v + w) ≤ p(v + x0) + p(w − x0), and thus ϕ(w) − p(w − x0) ≤ p(v + x0) − ϕ(v) for all v, w ∈ V . Consequently, there exists a number α ∈ sup (ϕ(w) − p(w − x0)), inf (p(v + x0) − ϕ(v)) . w∈V v∈V ˜ ˜ We now deﬁneϕ ˜(x) = ϕ(v) + αt for every x = v + tx0 ∈ V . Thenϕ ˜ : V → R is linear. For y ∈ Y ⊆ V , we haveϕ ˜(y) = ϕ(y) = ϕ0(y). For x = v ∈ V (i.e., t = 0), it holdsϕ ˜(x) = ϕ(v) ≤ p(v) = p(x). Using the deﬁnition of α and the sublinearity of p, we obtain for t > 0 that 1 1 ϕ˜(x) ≤ ϕ(v) + t(p( v + x ) − ϕ( v)) = p(v + tx ) = p(x). t 0 t 0 and for t < 0 that 1 1 ϕ˜(x) ≤ ϕ(v) + t(ϕ(− v) − p(− v − x )) = p(v + tx ) = p(x). t t 0 0 ˜ Hence,ϕ ˜ ≤ p|V˜ . Summing up, we have shown that (V, ϕ) (V, ϕ˜) ∈ M. Lemma 4.16. Let X be a normed vector space with F = C. a) Let x? ∈ X?. Set ξ?(x) = Re x?(x) for all x ∈ X. Then ξ? : X → R is ? ? ? R–linear with kξ k := supkxk≤1 |ξ (x)| = kx k. b) Let ξ? : X → R be continuous and R–linear. Set x?(x) = ξ?(x) − iξ?(ix) for all x ∈ X. Then x? ∈ X? with kx?k = kξ?k and Re x? = ξ?. 60 Chapter 4. Duality

Proof. The R–linearity of ξ? in a) and the C–linearity of x? in b) can be checked in a straightforward way. Clearly, ξ? = Re x? in b), and hence kξ?k ≤ kx?k in a) and b). Take x ∈ X with kxk = 1. Set α = 1 if x?(x) = 0 and α = x?(x)/ |x?(x)| otherwise. Then kα−1xk ≤ 1 and so the C–linearity of x? yields

? ? 1 ? 1 ? 0 ≤ |x (x)| = x ( α x) = ξ ( α x) ≤ kξ k.

Taking the supremum over x with kxk ≤ 1, we obtain kx?k ≤ kξ?k in a) and b).

Due to their importance, the Hahn–Banach theorem, the principle of uniform boundedness and the open mapping theorem are called the three basic principles of functional analysis.

Theorem 4.17 (Main version of Hahn-Banach). Let X be a normed vector space, Y ⊆ X be a linear subspace (endowed with the norm of X), and y? ∈ Y ?. Then there exists an x? ∈ X? such that hy, x?i = hy, y?i for all y ∈ Y and kx?k = ky?k.

Proof. 1) Let F = R. Set p(x) = ky?k kxk for all x ∈ X. The map p is sublinear and y?(y) ≤ p(y) for all y ∈ Y . Theorem 4.15 yields a linear x? : X → R with ? ? ? x |Y = y and x (x) ≤ p(x) for all x ∈ X. We further have

−x?(x) = x?(−x) ≤ p(−x) = p(x) so that |x?(x)| ≤ p(x) = ky?k kxk, for all x ∈ X. As a result, kx?k ≤ ky?k and x? ∈ X?. The equality kx?k = ky?k now follows from the estimate

kx?k ≥ sup |hy, x?i| = sup |hy, y?i| = ky?k. kyk=1,y∈Y kyk=1,y∈Y

2) Let F = C. We consider X as a normed vector space XR over R by restricting the scalar multiplication to real scalars; i.e., to the map R × X → X,(α, x) 7→ αx. Lemma 4.16a) shows that the real part η? = Re y? belongs to Y ? and kη?k = ky?k. ? ? ? R ? ? Due to Step 1), the functional η has an extension ξ ∈ (XR) with kξ k = kη k = ky?k. Lemma 4.16b) then provides an x? ∈ X? with kx?k = kξ?k = ky?k and x?(y) = ξ?(y) − iξ?(iy) = Re y?(y) − i Re y?(iy) = Re y?(y) − i Re(iy?(y)) = y?(y) for all y ∈ Y , where we used the C–linearity of y?.

∞ ? Example 4.18. Let Y = c ⊆ X = ` and y (y) = limn→∞ yn for y ∈ Y . Clearly, y? belongs to Y ? and has norm 1. The Hahn–Banach theorem yields an extension ? ∞ ? ? ? ? x ∈ (` ) of y with norm 1. Note that x (y) = y (y) = 0 for y ∈ c0. The functional x? cannot be represented by a sequence z ∈ `1 as in (4.4) since otherwise it would follow both z 6= 0 and

∞ ? X 0 = hen, x i`∞ = (en)jzj = zn for all n ∈ N. ♦ j=1

The next result uses a basic construction of functionals y? and allows to distin- guish between a linear subspace Y and a vector x0 ∈/ Y . This fact leads to several fundamental consquences of the Hahn–Banach theorem.

Proposition 4.19. Let X be a normed vector space, Y & X be a closed linear ? ? ? subspace, and x0 ∈ X \ Y . Then there exists an x ∈ X such that x (y) = 0 for ? ? all y ∈ Y , x (x0) = d(x0,Y ) := infy∈Y kx0 − yk > 0, and kx k = 1. 4.3. The theorem of Hahn-Banach 61

Proof. We deﬁne the linear subspace Z = Y + lin{x0} of X and the linear map ? ? z : Z → F by z (y + tx0) = t d(x0,Y ) for all y ∈ Y and t ∈ F, using that ? ? Y ∩ lin{x0} = {0}. Clearly, z |Y = 0 and z (x0) = d(x0,Y ). We further compute

? kz k = sup |t| inf kx0 − y˜k ≤ sup ktx0 + yk ≤ 1, y˜∈Y ky+tx0k≤1 ky+tx0k≤1

1 where we have choseny ˜ = − t y assuming that t 6= 0 without loss of generality. Take yn ∈ Y with kx0 − ynk → d(x0,Y )(> 0, since otherwise x0 ∈ Y¯ = Y ). The properties of z? now yield ? 1 ? d(x0,Y ) − 0 kz k ≥ (x0 − yn), z = −→ 1 as n → ∞, kx0 − ynk kx0 − ynk so that kz?k = 1. A Hahn–Banach extension of z? yields the required x? ∈ X?.

? Corollary 4.20. Let X be a normed vector space, x, x1, x2 ∈ X, and D? ⊆ X be dense. Then the following assertions hold. a) If x 6= 0, then there exists an x? ∈ X? with hx, x?i = kxk and kx?k = 1.

? ? ? ? b) If x1 6= x2, then there exists an x ∈ X with x (x1) 6= x (x2).

? ? c) kxk = max ? ? ? |hx, x i| = sup ? ? |hx, x i|. x ∈X ,kx kX? ≤1 x ∈D?,kx kX? ≤1 Proof. Assertion a) is a consequence of Proposition 4.19 with Y = {0}, and a) ? implies b) taking x = x1 − x2. For x in X we have supkx?k≤1 |hx, x i| ≤ kxk, and so assertion c) follows from a) and an approximation argument. Example 4.21. Let 1 ≤ p < ∞ and (S, A, µ) be a measure space, which is σ–ﬁnite if p = 1. Given f ∈ Lp(µ) \{0}, we set g = kfk1−pf¯|f|p−2 1 (i.e., g = 1 f¯ p {f6=0} kfk2 p0 for p = 2). One can then check that g ∈ L (µ) with kgkp0 = 1 and hf, gi = kfkp. 0 Let D be a dense subset of Lp (µ). Corollary 4.20 and Theorem 4.14 further yield Z

kfkp = sup fg dµ . (4.6) g∈D,kgkp0 ≤1 S

Corollary 4.22. Let X be a normed vector space and M ⊆ X. The set M is bounded in X if and only if the sets x?(M) are bounded in F for each x? ∈ X?. ? ? Proof. The implication “⇒” is clear. To show the converse, set Tx(x ) = hx, x i ? ? ? for each ﬁxed x ∈ M and all x ∈ X . The assumption yields Tx ∈ L(X , F) with ? ? ? ? ? ? |Tx(x )| ≤ c(x ) := supx∈M |x (x)| < ∞ for all x ∈ X and x ∈ M. Since X is a Banach space, the principle of uniform boundedness gives a constant C such that

? C ≥ kTxk = sup |hx, x i| = kxk kx?k≤1 for all x ∈ M, where we have used Corollary 4.20c). Corollary 4.23. Let X be a normed vector space and Y ⊆ X be a linear subspace. Then, Y is not dense in X if and only if there exists an x? ∈ X?\{0} with hy, x?i = 0 for all y ∈ Y . Proof. The implication “⇒” follows from Proposition 4.19. If Y is dense and x? ∈ X? vanishes on Y , then x? must be 0 by continuity. Corollary 4.24. Let X be a normed vector space and X? be separable. Then X is separable. 62 Chapter 4. Duality

? Proof. By an exercise, we have a dense subset {xn | n ∈ N} in ∂BX? (0, 1). There are ? 1 yn ∈ X with kynk = 1 and |hyn, xni| ≥ 2 for every n ∈ N. Set Y = lin{yn | n ∈ N}. Suppose that Y¯ 6= X. Corollary 4.23 yields an x? ∈ X? such that kx?k = 1 and ? ? ? 1 hy, x i = 0 for all y ∈ Y . There exists a j ∈ N with kx − xj k ≤ 4 . We then deduce

1 ? ? ? 1 ≤ hyj, x i = hyj, x − x i ≤ 2 j j 4 which is a contradiction.

1 ? ∞ 1 ? Example 4.25. The spaces c0 and ` = c0 are separable, whereas ` = (` ) is not separable, see Example 1.51. (Since separability is preserved under isomorphisms by an exercise, we can omit here the isomorphisms from Proposition 4.11.) The above result implies that also (`∞)? is not separable. In particular, `1 cannot be ∞ ? isomorphic to (` ) , cf. Proposition 4.11. ♦

Example 4.26 (Operators with ﬁnite rank). Let X and Y be normed vector spaces with dim X ≥ n, x1, . . . , xn ∈ X be linearly independent, and y1, . . . , yn ∈ Y . For each k ∈ {1, . . . , n}, we put Zk = lin{x1, . . . , xk−1, xk+1, . . . , xn} if n ≥ 2 and ? ? Z1 = {0} if n = 1. Proposition 4.19 provides us with an xk ∈ X such that ? ? ? xk|Zk = 0 and xk(xk) = 1; i.e., hxj, xki = δjk for j, k ∈ {1, . . . , n}. We now deﬁne

n X ? T x = hx, xki yk ∈ lin{y1, . . . , yn} =: Y0 k=1

Pn ? for all x ∈ X. Clearly, T ∈ L(X,Y ) with kT k ≤ k=1 kxkkkykk. Since also T xj = yj, we have R(T ) = Y0. If X = Y and xk = yk for all k, we further deduce

n n 2 X X ? ? T x = hx, xkihxk, xj ixj = T x. ♦ j=1 k=1

Proposition 4.27. For a normed vector space X the following assertions hold.

a) Let U ⊆ X be a ﬁnite dimensional linear subspace. Then U is closed and there exists a closed linear subspace Z with X = U ⊕ Z.

b) Let Y be a closed linear subspace with ﬁnite codimension dim X/Y . Then there exists a closed linear subspace V with dim V = dim X/Y and X = Y ⊕ V .

Proof. a) Let {x1, . . . , xn} be a basis of U. We deﬁne T as in Example 4.26 with xk = yk. Then T ∈ L(X) is a projection with range U and thus Lemma 2.17 implies assertion a). b) Let Q : X → X/Y , Qx = x + Y , be the quotient map (see Proposition 2.20) and let {b1, . . . , bn} a basis of X/Y . Since Q is surjective, there are xk ∈ X Pn with Qxk = bk. Set V = lin{x1, . . . , xn}. If k=1 αkxk = 0 for some αk ∈ F, Pn then k=1 αkbk = Q0 = 0, and hence αk = 0 for all k = 1, . . . , n. Therefore, {x1, . . . , xn} is linearly independent and dim V = dim X/Y . By part a), the space Pn V is closed. If x ∈ Y ∩ V , then x = k=1 βkxk for some βk ∈ F since x ∈ V . Pn On the other hand, Qx = 0 which yields 0 = k=1 βkbk and thus βk = 0 for all k = 1, . . . , n. As a result, x = 0. Take x ∈ X. There are αk ∈ F such that Qx = α1b1 + ... + αnbn. Set v = α1x1 + ... + αnxn. We then obtain Q(x − v) = 0, and thus x − v ∈ Y and x = x − v + v ∈ Y + V . Consequently, X = Y ⊕ V . 4.3. The theorem of Hahn-Banach 63

Geometric version of Hahn–Banach Let X be a normed vector space, A, B ⊆ X, A ∩ B = ∅ and A, B 6= ∅. A functional x? ∈ X? separates A and B if

∀ a ∈ A, b ∈ B : Reha, x?i < Rehb, x?i, and it separates A and B strictly if

? ? sA := sup Reha, x i < iB := inf Rehb, x i. a∈A b∈B Observe that N(x?) is a closed linear subspace with codimension 1 (see exercises). ? ? Hence, H0 = x0 + N(x ) is a closed affine hyperplane. Let F = R, x separate A ? and B, and x0 ∈ X satisfy γ := hx0, x i ∈ [sA, iB]. Then A and B are contained ? ? in different halfspaces H± := {x ∈ X | x (x) R γ} separated by H0, since x |A ≤ ? ? sA ≤ x |H0 = γ ≤ iB ≤ x |B. Let A ⊆ X. We define the Minkowski functional pA : X → [0, ∞] by setting

1 pA(x) = inf{λ > 0 | λ x ∈ A} for x ∈ X, where inf ∅ = ∞. Note that pB(0,1)(x) = kxk.

Remark 4.28. If A, B ⊆ X are convex and α ∈ F, then also αA and A + B are convex. To check the second assertion, take ai ∈ A, bi ∈ B, and t ∈ [0, 1] for i = 1, 2. We then have t(a1 + b1) + (1 − t)(a2 + b2) = (ta1 + (1 − t)a2) + (tb1 + (1 − t)b2) which belongs to A + B. The ﬁrst assertion is shown similarly. ♦ Lemma 4.29. Let X be a normed vector space and A ⊆ X convex with 0 ∈ A◦. (Thus there exists a δ > 0 with B(0, δ) ⊆ A.) Then the following assertions hold.

1 a) We have pA(x) ≤ δ kxk for all x ∈ X.

b) The map pA is sublinear. −1 c) If A is also open, we have A = pA ([0, 1)). δ Proof. a) The assumption implies that kxk x ∈ A for all x ∈ X \{0}, so that 1 pA(x) ≤ δ kxk. b) Let t > 0, x, y ∈ X, and ε > 0. Clearly, pA(0x) = 0pA(x). Moreover,

t 1 pA(tx) = inf{λ > 0 : λ x ∈ A} = inf{tµ > 0 : µ x ∈ A} = t pA(x).

1 1 Further, take 0 < λ ≤ pA(x) + ε and 0 < µ ≤ pA(y) + ε with λ x, µ y ∈ A. Since A is convex, the vector 1 λ 1 µ 1 (x + y) = x + y λ + µ λ + µ λ λ + µ µ belongs to A, so that pA(x + y) ≤ λ + µ ≤ pA(x) + pA(y) + 2ε. Letting ε → 0, we deduce assertion b). c) Now, let A be open. First take x with pA(x) < 1. Then there exists a λ ∈ (0, 1) 1 1 with λ x ∈ A. The convexity of A yields that x = λ λ x + (1 − λ)0 ∈ A. Conversely, 1 take x with pA(x) ≥ 1. Then λ x is contained in X \ A for all λ < 1. Since X \ A is closed, we obtain x ∈ X \ A letting λ → 1. Theorem 4.30 (Separation theorems). Let X be a normed vector space, A, B ⊆ X be convex and non–empty, and A ∩ B = ∅. Then the following assertions hold. a) If A and B are open, there exists an x? ∈ X? separating A and B. 64 Chapter 4. Duality

b) If A is closed and B is compact, there exists an x? ∈ X? separating A and B strictly. (Special case: B = {x}.)

Proof. We only treat F = R. The case F = C can be deduced from the case F = R by means of Lemma 4.16. a) Let A and B be open. Fix some x0 ∈ A − B and set C = A − B − x0 = S b∈B A − b − x0. Then C is open because of Proposition 1.15, C is convex by Remark 4.28, and we have 0 ∈ C as well as y0 := −x0 ∈/ C (since 0 ∈/ A − B). Thanks to Lemma 4.29, the Minkowski functional pC is sublinear and pC (y0) ≥ 1. ? ? We deﬁne y (ty0) = tpC (y0) for all t ∈ R. The map y : lin{y0} → R is linear ? and y (y) ≤ pC (y) for all y = ty0. (If t < 0, we have pC (y) ≥ 0 ≥ tpC (y0).) ? ? Theorem 4.15 now gives a linear x : X → R such that x (x) ≤ pC (x) for all x ∈ X ? ? and x (y0) = y (y0) = pC (y0) ≥ 1. Let x ∈ X. Lemma 4.29 a) implies that 1 |x?(x)| = max{x?(x), x?(−x)} ≤ max{p (x), p (−x)} ≤ kxk C C δ ? ? for some δ > 0, hence x ∈ X . Let x = a − b − x0 ∈ C for any a ∈ A and b ∈ B. Since y0 = −x0, Lemma 4.29 c) yields

? ? ? ? 1 > pC (x) ≥ hx, x i = ha, x i − hb, x i + hy0, x i.

? ? ? Using hy0, x i ≥ 1, we deduce that ha, x i < hb, x i, and thus assertion a) holds. b) Let A be closed and B be compact. We can thus ﬁnd an ε > 0 such that ka − bk ≥ 3ε for all a ∈ A and b ∈ B. (See the exercises.) Therefore, the sets S Aε = A + B(0, ε) = a∈A B(a, ε) and Bε = B + B(0, ε) are disjoint and, as above, also open and convex. Part a) gives an x? ∈ X? with ha + x, x?i < hb + y, x?i for all a ∈ A, b ∈ B and x, y ∈ B(0, ε). Choosing y = 0 and x = ±εz with z ∈ B(0, 1), we conclude that ε |hz, x?i| < hb − a, x?i for all a ∈ A, b ∈ B, and z ∈ B(0, 1). Taking the supremum over z ∈ B(0, 1), we obtain 0 < ε kx?k ≤ hb − a, x?i for all a ∈ A and b ∈ B which implies assertion b).

? Let X be a normed vector space, A ⊆ X and B? ⊆ X be non–empty. The annihilators of A and B? are given by A⊥ = {x? ∈ X? | ∀ a ∈ A we have ha, x?i = 0} ⊆ X?, ⊥ ? ? B? = {x ∈ X | ∀ b ∈ B? we have hx, b i = 0} ⊆ X. In view of Riesz’ Theorem 4.10, in a Hilbert space X =∼ X? these two sets are isomorphic to the othorgonal complement.

? Remark 4.31. Let X be a normed space, A ⊆ X and B? ⊆ X be non–empty. ⊥ ⊥ a) As in Remark 4.7 one veriﬁes that A and B? are closed linear subspaces of ? ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ X and X, respectively. Also, A ⊆ (A ), (lin A) = A and (lin B?) = B?. b) Corollary 4.23 shows that A⊥ = {0} if and only if lin A = X. From Corol- lary 4.20 we deduce that A⊥ = X? if and only if A = {0}. ⊥ c) By deﬁnition, we have B? = X if and only if B? = {0}. Due to Corollary 4.20, ? ⊥ lin B? = X implies that B? = {0}. 1 ? ∞ d) Let X = ` and B? = {en | n ∈ N} ⊆ X = ` . We then have lin B? = c0, but ⊥ 1 ? B? = {0} since ` = c0. ♦ Proposition 4.32. For a normed vector space X and a non–empty subset A ⊆ X, we have lin A = ⊥(A⊥).

⊥ ⊥ ⊥ ⊥ Proof. Remark 4.31 yields lin A ⊆ (A ). Suppose there were an x0 ∈ (A )\lin A. Theorem 4.30 b) then gives an x? ∈ X? with

? ? s := sup Rehx, x i < Rehx0, x i. x∈lin A 4.4. Reﬂexivity and weak convergence 65

If r := Rehx, x?i= 6 0 for some x ∈ lin A, then we deduce for t ∈ R that ( as t → ∞, if r > 0, Rehtx, x?i = tr → ∞ as t → −∞, if r < 0.

This contradicts the above estimate for s, and thus Rehx, x?i = 0 for all x ∈ lin A. Hence, s = 0. Using it instead of t, we similarly obtain Imhx, x?i = 0 for all x ∈ lin A ? ⊥ ? so that x ∈ A , which yields the contradiction 0 = s < Rehx0, x i = 0.

1 ? ∞ Let X = ` and B? = {en | n ∈ N} ⊆ X = ` . In Remark 4.31d), we have seen ⊥ ⊥ ⊥ ? ∞ that lin B? = c0 and that B? = {0}, hence ( B?) = X = ` 6= lin B?. See Theorem 4.7 in [Rud91] for more information.

Proposition 4.33. Let X be a normed vector space and Y ⊆ X be a closed linear subspace. Then the maps

? ⊥ ? ? ⊥ ? T : X /Y → Y ,T (x + Y ) = x |Y , S :(X/Y )? → Y ⊥, Sϕ = ϕ ◦ Q, are isometric isomorphisms, where Q : X → X/Y is the quotient map Qx = x + Y .

This result is shown in the exercises.

4.4 Reﬂexivity and weak convergence

Let X be a normed vector space. The bidual of X is X?? = (X?)?. For each x ∈ X ? we deﬁne the map JX (x): X → F by

? ? ? ? hx ,JX (x)iX? = hx, x iX for all x ∈ X . (4.7)

? ? ? Clearly, JX (x) is linear in x and we have |[JX (x)](x )| ≤ kxk kx k, so that JX (x) ∈ ?? X . Moreover, the map JX is linear in x, and by means of Corollary 4.20 and equation (4.7) we obtain

? kxkX = sup |hx, x i| = kJX (x)kX?? . kx?k≤1

Proposition 4.34. Let X be a normed vector space. Then equation (4.7) deﬁnes ?? a linear isometry JX : X → X .

Deﬁnition 4.35. X is called reﬂexive if JX from (4.7) is surjective.

Remark 4.36. a) If X is reflexive, then X =∼ (X?)? and thus is X a Banach space. However, there are non–reflexive Banach spaces which are isomorphic to its bidual (with an isomorphism different from JX ), see Example 1.d.2 in [LiT96]. ? ⊥ ⊥ b) If X reflexive and B? ⊆ X , then B? = JX ( B?) by (4.7). Here (and in some other points below) one sees that reflexive Banach spaces share some properties of Hilbert spaces which are not true in a general Banach space. c) In Corollary 4.61 we will show that reflexivity is preserved under isomorphisms. ?? d) Usually one identifies X with R(JX ) and reflexive spaces X with X . ♦ Example 4.37. a) Hilbert spaces X are reflexive. Proof. Recall from Theorem 4.10 that a Hilbert space Y is isomorphic to its dual ? space via the isomorphism ΦY : Y → Y given by hz, ΦY (y)i = (z|y) for all z ∈ Y . The dual space X? of X is also a Hilbert space equipped with the scalar product 66 Chapter 4. Duality

? ? −1 ? −1 ? ?? ?? ? (x |y )X? := (ΦX y |ΦX x )X (check!). Now take any x ∈ X . Set x = −1 ?? ? −1 ? ΦX? x ∈ X and x = ΦX x ∈ X. We then obtain

? ?? ? ? −1 ? −1 ? ? hz , x iX? = (z |x )X? = (ΦX x |ΦX z )X = hx, z iX

? ? ?? for every z ∈ X ; i.e., JX (x) = x as asserted. 2 b) Let 1 < p < ∞ and (S, A, µ) be a measure space. Then X = Lp(µ) is reﬂexive (e.g., X = `p). r0 ? Proof. Theorem 4.14 says that the map Φr : L (µ) → X deﬁned by hϕ, Φr(ψ)i = 0 R ϕψ dµ for all ϕ ∈ Lr(µ) and ψ ∈ Lr (µ) is an isometric isomorphism, where ?? p0 ? 00 r ∈ (1, ∞). Take φ ∈ X . Then φ ◦ Φp belongs L (µ) . Since p = p, Theorem p p0 4.14 yields an f ∈ L (µ) with Φp0 (f) = φ ◦ Φp. Let Φp(g) with g ∈ L (µ) be an arbitrary element of X?. We then compute Z hΦp(g), ϕiX? = φ(Φp(g)) = hg, Φp0 (f)iLp0 = gf dµ = hf, Φp(g)iLp . S

Hence, φ = JX (f) as asserted. 2 ?? ∼ ∞ ∞ c) Proposition 4.11 shows that c0 = ` . Moreover, ` c0 since c0 is separable ∞ and ` is not separable (see Example 1.51). Hence, c0 is not reflexive. ♦ In the above example and also below, we use that separability is preserved under isomorphisms (see exercises). Proposition 4.38. For a normed vector space X, the following assertions hold. a) If X is reflexive, then each closed linear subspace Y of X is reflexive (where Y is endowed with the norm of X). b) Let X be a Banach space. Then X is reflexive if and only if X? is reflexive. c) Let X be reflexive. Then is X separable if and only if X? separable. Proof. a) Let Y ⊆ X be a closed linear subspace. Take y?? ∈ Y ??. Each x? ∈ X? ? ? ? ? ?? satisfies x |Y ∈ Y with kx |Y kY ? ≤ kx kX? . We define the linear map x by ?? ? ? ?? ? ? ?? ? ? ?? x (x ) = hx |Y , y iY ? for all x ∈ X . As |x (x )| ≤ kx kX? ky kY ?? , the functional x?? belongs to X??. By the reflexivity of X, there is a vector y ∈ X with

? ?? ? ?? ? ? ? hx |Y , y iY ? = hx , x iX? = hy, x iX for all x ∈ X .

Suppose that y∈ / Y . Since Y is closed, Proposition 4.19 yields anx ˜? ∈ X? with ? ? x˜ |Y = 0 and hy, x˜ iX 6= 0. This fact contradicts the above equation in display and we thus obtain y ∈ Y . Take any y? ∈ Y ?. Let x? ∈ X? be a Hahn–Banach extension ? ? ? ? ?? of y . Then, hy, y iY = hy, x iX = hy , y iY ? by the above considerations, and ?? thus JY (y) = y . ??? ??? ? ??? b) Let X be reflexive. Take x ∈ X . We set x (x) := hJX (x), x iX?? for all ? ? ?? ?? ?? x ∈ X. Clearly, x belongs to X . Every x ∈ X can be written as x = JX (x) ? ?? ?? ??? ?? ?? ? for some x ∈ X, so that hx , x iX? = hx , x iX?? for all x ∈ X . Thus, X is reflexive. Conversely, assume that X is not reflexive. By Proposition 4.19 there exists an ??? ??? ??? ? x ∈ X \{0} with hJX (x), x iX?? = 0 for all x ∈ X. Suppose that X was ? ? ??? ? reflexive. Then there must exist an x ∈ X with x = JX? (x ). Hence,

? ? ? 0 = hJX (x),JX? (x )iX?? = hx ,JX (x)iX? = hx, x iX for all x ∈ X, which means that x? = 0, contradicting x??? 6= 0. c) The implication “⇐” was shown in Corollary 4.24. If X is separable, then X?? =∼ X is also separable. Hence, X? is separable by Corollary 4.24. 4.4. Reﬂexivity and weak convergence 67

Example 4.39. a) The space X = `1 is not reflexive, because it is separable and ? ∼ ∞ its dual X = ` is non-separable. Since c0 is not reflexive by Example 4.37 and it is a closed subspace of `∞, also the space `∞ fails to be reflexive. ∞ 1 b) The spaces C([0, 1]), L (0, 1) and L (0, 1) are not reflexive (see exercises). ♦ In order to obtain fundamental compactness results, one introduces new convergence concepts.

Deﬁnition 4.40. Let X be a normed vector space.

a) A sequence (xn) in X converges weakly to x ∈ X if

? ? ? ? ∀ x ∈ X : hxn, x i → hx, x i as n → ∞.

? ? ? ? ? b) A sequence (xn) in X converges weakly to x ∈ X if

? ? ∀ x ∈ X : hx, xni → hx, x i as n → ∞.

σ ? We then write xn * x or xn −→ x or σ–limn→∞ xn = x, respectively, xn * x or ? ? σ ? ? ? ? xn −→ x or σ –limn→∞ xn = x . One often replaces here the letter ‘σ’ by ‘w’.

Remark 4.41. (Positive results.) Let X be a normed vector space, xn, x, y ∈ X, ? ? ? xn, x ∈ X and n → ∞. a) The weak? convergence in X? = L(X, F) is just the strong convergence of a sequence of operators in L(X, F). b) Weak and weak? convergence is linear. σ ? ? c)If xn → x in the norm of X, then xn −→ x (since |hxn −x, x i| ≤ kxn −xk kx k). ? ? ? ? ? σ ? d ? If xn → x in the norm of X , then xn −→ x . In X = F weak or weak ? convergence are equivalent to componentwise convergence (take x = ek or x = ek), and thus to convergence in norm. d) Weak and weak? limits are unique. σ σ ? ? Proof. Let xn −→ x and xn −→ y. If x 6= y, then Corollary 4.20 yields an x ∈ X ? ? ? ? ? with hx, x i 6= hy, x i, but hxn, x i converges to both hx, x i and hy, x i, which is impossible. The second part follows from a). 2 ? ? ? e) For x ∈ X we have hx, x iX = hx ,JX (x)hX? so that the σ-convergence on X implies the σ?-convergence. If X is reflexive, then the two types of convergence on X? coincide. ? ? ? f) If (hx, xni)n is Cauchy in F for each x ∈ X and a sequence (xn) in X , then ? ? ? ? σ ? there exists an x ∈ X such that xn −→ x as n → ∞, due to Lemma 3.9. In this sense, X? is ‘weakly? sequentially complete’. g) Reflexive spaces are ‘weakly sequentially complete’ due to e) and f). ♦ Remark 4.42. (Negative results.) Let n → ∞. a) Weakly or weakly? convergent sequences may diverge in norm. Weak or weak? limits may have a strictly smaller norm. 2 2 σ Proof. In X = ` we have hen, xi = xn → 0 for all x ∈ ` . This means that en −→ 0 ? σ 1/2 2 and en −→ 0. But ken − emk2 = 2 for all n 6= m so that (en) diverges in ` . Moreover, kenk = 1 and the weak limit 0 has a strictly smaller norm. 2 b) A weakly? convergent sequence may have no weakly converging subsequence. ? 1 ? Proof. In X = ` = c0 we have hx, enic0 = xn → 0 for each x ∈ c0 so that ? σ ∞ 1 ? en −→ 0. Take any divergent subsequence nj → ∞ and choose y ∈ ` = (` ) such 1 that (ynj )j diverges. Then henj , yi` = ynj does not converge as j → ∞. 2 c) There are (nonreflexive) spaces which are not ‘weakly sequentially complete’. 68 Chapter 4. Duality

∞ 1 Proof. Let X = c0 and vn = e1 + ··· + en ∈ c0 ⊆ ` . For each y ∈ ` , we have n ∞ X X 1 1 1 hvn, yic0 = hy, vni` = yk −→ yk = hy, i` as n → ∞. k=1 k=1

? σ 1 ∞ This means that vn −→ in ` and that (hvn, yic0 )n is a Cauchy sequence in F 1 for every y ∈ ` . But, if (vn) had a weak limit x in c0, then x would also be a ? ∞ ? weak limit of (vn) in ` and thus x = 1 by the uniqueness of weak limits, which is impossible. 2 ? ? ? Proposition 4.43. Let X be a normed vector space, xn, x ∈ X and xn, x ∈ X ? ? for n ∈ N, D? ⊆ X with lin D? = X , and D ⊆ X with lin D = X. We then have: σ ? ? a) xn −→ x as n → ∞ if and only if supn kxnkX < ∞ and hxn, y i → hx, y i as ? n → ∞ for all y ∈ D?.

? ? σ ? ? b) Let X be complete. Then xn −→ x as n → ∞ if and only if supn kxnkX? < ∞ ? ? and hy, xni → hy, x i as n → ∞ for all y ∈ D. ? ? If a) holds, then kxk ≤ limn→∞kxnk; and if b) holds, then kx k ≤ limn→∞kxnk. Moreover, in b) the implication ‘⇐’ holds for all normed vector spaces X. Proof. The proposition follows from Corollary 3.5 and Lemma 3.9 using in a) that ?? ? JX (xn) ∈ X = L(X , F). p Example 4.44. a) Let X = c0 oder X = ` for 1 < p < ∞,(vn) ⊆ X be bounded, and x ∈ X. Then Proposition 4.43 with D? = {en | n ∈ N} yields σ vn −→ x ⇐⇒ [vn]k = hvn, eki → hx, eki = xk for all k ∈ N (n → ∞). (On the right-hand side one has ‘componentwise convergence’.) For sequences in `1 or `∞ one obtains an analogous result for the σ?-convergence 1 taking D = {en | n ∈ N} ⊆ c0 or ⊆ ` b) The implication ‘⇐’ fails in a) for p = 1. In fact, the sequence (en) converges componentwise to 0, but it is boundad and diverges weakly in `1 by the proof of Remark 4.42b). c) The example vn = nen shows that the assumption of boundedness cannot be omitted in Proposition 4.43. p d) Let (S, A, µ) be a measure space, X = L (µ), 1 < p < ∞, and (fn) be bounded in X. Then

σ Z Z fn −→ f ⇐⇒ fn dµ → f dµ for all A ∈ A with µ(A) < ∞ (n → ∞) A A p0 due to Proposition 4.43 since the space D? of simple functions is dense in L (µ). An analogous result holds for the weak? convergence in L∞(µ), if the measure space is σ–ﬁnite. ♦ Remark 4.45. If X? is separable, then the weak sequential convergence in bounded subsets of X is given by the metric

∞ ? X hx − y, xj i d(x, y) = 2−j 1 + hx − y, x?i j=1 j

? ? where D? = {xj | j ∈ N} is dense in X . This result follows Propositions 4.43 and ? 1.8 using the seminorms pj(x) = hx, xj i . Observe that for each x ∈ X \{0} there ? is an index k ∈ N such that hx, xki= 6 0 due to Corollary 4.20 and the density of ? D?. Similarly, if X is separable then the weak sequential convergence in bounded ? subsets of X is given by an analogous metric. ♦ 4.4. Reﬂexivity and weak convergence 69

The next theorem says that closed convex sets are ‘weakly sequentially closed’. It is a consequence of the separation theorem.

Theorem 4.46 (Mazur). Let X be a normed vector space, K be a closed and convex subset of X, and let xn ∈ K converge weakly to some x ∈ X as n → ∞. Then x ∈ K, and there is a sequence (yN ) of convex combinations of {xn | n ≥ N} converging to x in norm as N → ∞.

Proof. 1) Suppose that x were not contained in K. Theorem 4.30b) (with A = K ? ? ? ? and B = {x}) then gives an x ∈ X with supn Rehxn, x i < Rehx, x i. But this ? ? inequality cannot hold since hxn, x i converges to hx, x i as n → ∞. Hence, x ∈ K. 2) Let N ∈ N and deﬁne the (convex!) set

n m o X CN = y = tjxj m ∈ N, m ≥ N, tj ≥ 0, tN + ··· + tm = 1 . j=N

Then CN is closed (in norm) and convex (see Corollary 1.18), and xn ∈ CN for all n ≥ N. By part 1), x belongs to each CN . So we can choose yN ∈ CN such that kx − yN k ≤ 1/N for every N ∈ N. Remark 4.47. a) One needs convexity in Mazur’s theorem: For instance, the 2 2 vectors en in the (closed) unit sphere S of ` converge weakly in ` to 0 ∈/ S. b) Mazur’s theorem can fail for the weak? convergence: For instance, the vectors ? ? 1 en converge weakly to 0 in X = ` , and they belong to the closed and affine (thus 1 convex) subspace A = {y ∈ ` | hy, 1i = 1}, but 0 ∈/ A. ♦ We now prove an important extension of the Bolzano–Weierstraß theorem to infinite dimensional Banach spaces: In dual spaces the balls B(0, r) are ‘weakly? sequentially compact’ (instead of being compact as in finite dimensions).

Theorem 4.48 (Banach–Alaoglu, simpliﬁed version). Let X be a separable normed ? ? ? ? vector space. Let (xn) be a bounded sequence in X . Then there is an x ∈ X and a subsequence x? converging weakly? to x? as j → ∞, where kx?k ≤ lim kx? k. nj n→∞ n

? Proof. Let {xk | k ∈ N} be dense in X. Since (hx1, xni)n∈N is bounded in F, there exists a subsequence (hx , x? i) converging in as j → ∞. Since also (hx , x? i) 1 ν1(j) j F 2 ν1(j) j is bounded, there is a converging subsequence (hx , x? i) . Iteratively, we obtain 2 ν2(j) j subsequences ν ⊆ ν such that (hx , x? i) converges for each k ∈ . We k k−1 k νk(j) j N deﬁne y? = x? for each m ∈ . Then (hx , y? i) converges as m → ∞ for all m νm(m) N k m m k ∈ N because of (νm(m))m≥k ⊆ (νk(j))j∈N. Since {xk | k ∈ N} is dense and the ? ? subsequence (ym) of (xn) is also bounded, Lemma 3.9 implies the assertion.

Example 4.49. a) The above version of the theorem of Banach–Alaoglu can fail ∞ if X is not separable. As an example consider X = ` and the map ϕn(x) = xn ? for x ∈ X. Then ϕn ∈ X for all n ∈ N and kϕnk = 1. Take any subsequence k (ϕnk )k. For j ∈ N, we set xj = (−1) if j = nk and xj = 0 otherwise, and put ∞ k ? x = (xj) ∈ ` . Then hx, ϕnk i = (−1) diverges, and thus (ϕn) has no weakly convergent subsequence. We refer to Theorem V.3.1 in [Con90] for the full version of the theorem of Banach–Alaoglu which holds without the separability assumption. b) The theorem of Banach–Alaoglu can fail for the weak convergence. As an 1 example, consider en ∈ ` = X for all n ∈ N. Take any subsequence enk and choose ∞ 1 ? y ∈ ` = (` ) such that ynk diverges. Then henk , yi = ynk does not converge. ♦ Theorem 4.50. Let X be a reﬂexive Banach space. Then every bounded sequence (xn) in X has a weakly convergent subsequence. 70 Chapter 4. Duality

Proof. Let Y = lin{xn | n ∈ N}. By Proposition 4.38, the space Y is reﬂexive, and it is separable by Lemma 1.50. Proposition 4.38 now shows that Y ? is separable. Applying Theorem 4.48 to Y ?, we obtain a weakly? convergent subsequence ?? (JY (xnk ))k of (JY (xn))n in Y . Since Y is reﬂexive, there exists an x ∈ Y with

? ? ? ? ? ? hxnk , y iY = hy ,JY (xnk )iY −→ hy ,JY (x)iY = hx, y iY

? ? ? ? ? ? for all y ∈ Y , as k → ∞. Let x ∈ X . Then the restriction x |Y belongs to Y . As a result,

? ? ? ? hxnk , x iX = hxnk , x |Y iY −→ hx, x |Y iY = hx, x iX ,

σ which means that xnk −→ x as k → ∞. Example 4.51. Let X be reﬂexive and C ⊆ X be closed and convex. Then there exists a vector x ∈ C with minimal norm.

Proof. If 0 ∈ C, then we can take x = 0. Otherwise, d := inf{kxk | x ∈ C} > 0, and there are xn ∈ C such that kxnk → d as n → ∞. Since the sequence (xn) is bounded, it has a subsequence (xnj ) weakly converging to some x ∈ X, due to Theorem 4.50. Moreover, kxk ≤ lim infn→∞ kxnk = d. Mazur’s Theorem 4.46 further yields that x ∈ C, and thus kxk ≥ d.

4.5 Adjoint operators

Let X and Y be normed vector spaces and T ∈ L(X,Y ). For each y? ∈ Y ? we ? deﬁne a map ϕy? : X → F by setting ϕy? (x) = hT x, y iY . It is clear that ϕy? is ? ? linear in x ∈ X and that |ϕy? (x)| ≤ kT k ky k if kxk ≤ 1. So we obtain ϕy? ∈ X ? ? ? ? with kϕy? k ≤ kT k ky k. We now deﬁne T y := ϕy? ∈ X , which means that

? ? ? ? ? hT x, y iY = hx, T y iX (∀ x ∈ X, y ∈ Y ). (4.8)

Equation (4.8) determines T ?y? uniquely as an element of X?. Therefore (4.8) deﬁnes a map T ? : Y ? → X? which is called the adjoint of T . If X and Y are pre Hilbert spaces, analogously we introduce the Hilbert space adjoint T 0 : Y → X of T by

0 (T x|y)Y = (x|T y)X (∀ x ∈ X, y ∈ Y ). (4.9)

0 −1 ? Let X and Y be Hilbert spaces. Then T = ΦX T ΦY for the Riesz isomorphisms from Theorem 4.10.

Proposition 4.52. Let X,Y,Z be normed vector spaces, S, T ∈ L(X,Y ),R ∈ L(Y,Z), and α ∈ F. The following assertions hold. a) T ? ∈ L(Y ?,X?) with kT ?k = kT k.

b) (T + S)? = T ? + S?, (αT )? = αT ?, and (RT )? = T ?R?.

The analogous assertions (with (αT )0 = αT 0) and T = (T 0)0 =: T 00 hold for pre Hilbert spaces and the Hilbert space adjoints.

Proof. Let α, β ∈ F, x ∈ X, y?, u? ∈ Y ? and z? ∈ Z?. For a), we compute

hx, T ?(αy? + βu?)i = hT x, αy? + βu?i = αhT x, y?i + βhT x, u?i = αhx, T ?y?i + βhx, T ?u?i = hx, αT ?y? + βT ?u?i. 4.5. Adjoint operators 71

Since x ∈ X is arbitrary, this means that T ?(αy? + βu?) = αT ?y? + βT ?u? and thus T ? is linear. Moreover, Corollary 4.20 yields

kT k = sup kT xk = sup |hT x, y?i| = sup |hx, T ?y?i| kxk≤1 kxk≤1,ky?k≤1 kxk≤1,ky?k≤1 = sup kT ?y?k = kT ?k, ky?k≤1 and assertion a) holds. We further calculate

hRT x, z?i = hT x, R?z?i = hx, T ?R?z?i so that (RT )? = T ?R?. The remaing parts of b) and the Hilbert space variants of a) and b) are shown similarly. In the Hilbert setting, we ﬁnally compute

(T x|y) = (x|T 0y) = (T 0y|x) = (y|T 00x) = (T 00x|y) for all y ∈ X; i.e., T = T 00. (Note that T 00 exists due to a).) In view of the above result, each operator T ∈ L(X,Y ) possesses its bi-adjoint T ?? := (T ?)? ∈ L(X??,Y ??) with kT k = kT ??k. Deﬁnition 4.53. Let X and Y be Hilbert spaces and T ∈ L(X,Y ). The operator T 0 0 −1 0 is called unitary if T T = IX and TT = IY (i.e., it exists T = T ). Let X = Y . Then T is called self adjoint if (T x|y) = (x|T y) for all x, y ∈ X (i.e., T = T 0).

d ? Example 4.54. a) For X = F and a matrix T = [akl], the adjoint T is given by 0 [alk] and T by [alk]. p b) On X = c0 or X = ` with 1 ≤ p < ∞ consider the shift operators Lx = ? (xn+1)n and Rx = (0, x1, x2,... ). Take x ∈ X and y ∈ X . We calculate

∞ ∞ X X hLx, yi = xk+1yk = xnyn−1 = hx, Ryi k=1 n=2 so that L? = R. Similarly one derives R? = L and, for p = 2, L0 = R and R0 = L. c) Let X = L2(R) and (T (t)f)(s) = f(s + t) for f ∈ X and s, t ∈ R, see Example 3.13. For f, g ∈ X we compute Z Z (T (t)f|g) = f(s + t)g(s) ds = f(τ)g(τ − t) dτ = (f|T (−t)g), R R i.e., T (t)0 = T (−t) = T (t)−1. Hence, T (t) is unitary. Analogously one sees that T (t)? = T (−t) on Lp(R) for p ∈ [1, ∞) and t ∈ R. d) Let A ∈ Bd, 1 ≤ p < ∞, and k : A × A → F be measurable with

Z Z p/p0 !1/p p0 κp = |k(x, y)| dy dx < ∞, if p > 1, A A Z κ1 = ess sup |k(x, y)| dx < ∞, if p = 1. y∈A A

0 For p = 2 this means that k ∈ L2(A×A). Let p ∈ (1, ∞), f ∈ Lp(A) and g ∈ Lp (A). The function (x, y) 7→ ϕ(x, y) = k(x, y)f(y)g(x) is measurable on A×A as a product of measurable functions. Using Fubini a) and H¨older(ﬁrst in the y and then in the x integral), we deduce Z Z Z |ϕ| d(x, y) = |k(x, y)| |f(y)| dy |g(x)| dx A×A A A 72 Chapter 4. Duality

1 1 Z Z 0 0 Z p ≤ |k(x, y)|p dy p |f(y)|p dy |g(x)| dx A A A Z 1 p0 p0 ≤ κp kfkp |g(x)| dx = κp kfkp kgkp0 < ∞. A

This means that ϕ is integrable on A × A, and so Fubini b) yields that Z Z hg(x) := k(x, y) f(y) g(x) dy = g(x) k(x, y) f(y) dy A A exists for x∈ / Nfg and a null set Nfg, and that hg is measurable on A (after setting 1 hg(x) = 0 for all x ∈ Nfg). We now take gn = A∩B(0,n) for every n ∈ N and define S the null set Nf = n Nfgn . In this way we see that the function ( R k(x, y) f(y) dy, x ∈ A \ N , T f(x) := A f 0, x ∈ Nf , exists and that it is measurable. Note that this definition does not depend on the representative of f. Using again Hölderin the y–integral, we further estimate

p p Z Z p Z Z 0 0 Z p p p p p kT fkp = k(x, y) f(y) dy dx ≤ |k(x, y)| dy |f(y)| dy dx A A A A A p p = κp kfkp .

As a consequence, T f ∈ Lp(A). The linearity of T is clear, and so T ∈ L(Lp(A)) with norm less or equal κp. Since ϕ is integrable, Fubini b) ﬁnally yields Z Z Z Z hT f, gi = k(x, y)f(y) dy g(x) dx = f(y)k(x, y)g(x) dx dy A A A A Z Z = f(y) k(x, y)g(x) dx dy = hf, T ?gi. A A This means that Z T ?g(y) = k(x, y)g(x) dx for a.e. y ∈ A A

p0 0 R and for all g ∈ L (A). For p = 2 one derives analogously T g(y) = A k(x, y)g(x) dx for a.e. y ∈ A and all g ∈ L2(A). Hence, T is self adjoint if k(x, y) = k(y, x) for all 1 x, y ∈ A. For p = 1 one can further show that T ∈ L(L (A)) with kT k ≤ κ1. ♦ Proposition 4.55. For normed spaces X and Y , the following assertions hold.

?? a) For T ∈ L(X,Y ) we have T ◦ JX = JY ◦ T .

−1 ?? b) If Y is reﬂexive, then T = JY T JX . Proof. Let x ∈ X and y? ∈ Y ?. Employing (4.8) and (4.7), we compute

? ?? ? ? ? ? ? ? hy ,T JX (x)iY ? = hT y ,JX (x)iX? = hx, T y iX = hT x, y iY = hy ,JY (T x)iY ? .

These equalities yields assertion a) which implies assertion b).

We mostly identify T and T ?? if X and Y are reﬂexive. 4.5. Adjoint operators 73

Range and kernel of bounded linear operators Let T ∈ L(X,Y ). Then the kernel N(T ) is closed, but R(T ) need not to be closed. An example is the Volterra operator given by

Z t T f(t) = f(s)ds (t ∈ [0, 1]) 0 for f ∈ X = C([0, 1]). Here, T ∈ L(X) and R(T ) = {g ∈ C1([0, 1]) | g(0) = 0} is diﬀerent from R(T ) = {g ∈ X | g(0) = 0}.

Proposition 4.56. Let X and Y be normed vector spaces and T ∈ L(X,Y ). Then the following assertions hold.

a) R(T )⊥ = N(T ?).

b) R(T ) = ⊥N(T ?). Hence, R(T ) is dense if and only if T ? is injective.

c) N(T ) = ⊥R(T ?). Hence, T is injective if R(T ?) is dense.

d) R(T ?) ⊆ N(T )⊥.

e) Let X be reﬂexive. Then R(T ?) = N(T )⊥. Hence, R(T ?) is dense if and only if T is injective.

Proof. a) Let y? ∈ Y ?. The vector y? belongs to R(T )⊥ if and only if for all x ∈ X we have 0 = hT x, y?i = hx, T ?y?i, which is equivalent to y? ∈ N(T ?). b) Proposition 4.32 and part a) show that R(T ) = ⊥(R(T )⊥) = ⊥N(T ?). The second assertion now follows from Remark 4.31. c) Let x ∈ X. Due Corollary 4.20, the vector x belongs to N(T ) if and only if for all y? ∈ Y ? we have 0 = hT x, y?i = hx, T ?y?i, which is equivalent to x ∈ ⊥R(T ?). Remark 4.31 implies the second assertion. d) Proposition 4.32 and part a) imply that R(T ?) = ⊥(R(T ?)⊥) = ⊥N(T ??). We thus have to prove that ⊥N(T ??) ⊆ N(T )⊥. So let y? ∈ ⊥N(T ??) and take ?? any x ∈ N(T ). Then Proposition 4.55 yields T JX (x) = JY T x = 0 so that ?? ? ? ? ⊥ JX (x) ∈ N(T ) and thus hx, y i = hy ,JX (x)i = 0. This means that y ∈ N(T ) , as required. e) Let X be reﬂexive. It remains to show that N(T )⊥ ⊆ ⊥N(T ??) in view of the proof of part d). So let y? ∈ N(T )⊥ and take any x?? ∈ N(T ??). Because X ?? is reﬂexive, there exists an x ∈ X such that JX (x) = x . Proposition 4.55 yields ?? that JY T x = T JX x = 0. Since JY is injective by Proposition 4.34, the vector x belongs to N(T ), and hence hy?, x??i = hx, y?i = 0, as required.

Remark 4.57. In c) and d) of the above result, the converse implication and inclusion, respectively, do not hold in general. In fact, let X = c0 and T = I − L for the left shift L. If T x = 0 then xn = xn+1 for all n ∈ N and thus N(T ) = {0}. In Example 1.25 of the lecture notes [ST] it is shown that the range of T ? = I − R ? 1 ? ⊥ in X = ` is not dense, hence R(T ?) 6= X = N(T ) . ♦ The next result is an easy consequence of Proposition 4.56 b).

Corollary 4.58. Let X and Y be normed vector spaces, y ∈ Y , and let T ∈ L(X,Y ) have closed range. Then the equation T x = y has a solution x0 ∈ X if and only ? ? ? if hy, y i = 0 for all y ∈ N(T ), and every other solution is given by x = x0 + z for any z ∈ N(T ). Hence, T is surjective if and only if R(T ) is closed and T ? is injective. 74 Chapter 4. Duality

Corollary 4.59. Let X and Y be Banach spaces and T ∈ L(X,Y ). The operator T is invertible if and only if a) T ? injective and b) there is a constant c > 0 such that kT xk ≥ c kxk for all x ∈ X. Proof. Note that statement b) implies that T is injective. From Corollary 3.24 we deduce that b) holds if and only if T is injective and R(T ) is closed. Hence, Corollary 4.58 implies that the validity of a) and b) is equivalent to the bijectivity of T , and thus to its invertibility. Theorem 4.60. Let X and Y be Banach spaces and T ∈ L(X,Y ). The operator T is invertible if and only if T ? ∈ L(Y ?,X?) is invertible. In this case we have (T −1)? = (T ?)−1.

−1 ? Proof. Let T be invertible. We then have IX = T T , and thus IX? = (IX ) = ? −1 ? −1 ? ? ? −1 ? T (T ) . Similarly, IY ? = (T ) T . Thus T has the inverse (T ) ∈ L(X?,Y ?). Let T ? be invertible, and thus injective. Then T ?? is invertible. Let x ∈ X. Proposition 4.55 now implies that

?? −1 ?? ?? −1 ?? ?? −1 kxk = kJX (x)k = k(T ) T JX (x)k ≤ k(T ) k kT JX (x)k = k(T ) k kT xk.

From Corollary 4.59 we now deduce the invertibility of T . The above results will be used and extended in the lecture Spectral Theory. Corollary 4.61. Let X be reﬂexive and Φ: X → Y be an isomophism. Then Y is reﬂexive.

?? ?? −1 ?? −1 ?? Proof. Let y ∈ Y . Set y := ΦJX (Φ ) y ∈ Y , using the easy part of the above theorem. Take y? ∈ Y ?. By means of (4.7) and (4.8), we compute

? −1 ?? −1 ?? ? −1 ?? −1 ?? ? ? hy, y iY = hΦJX (Φ ) y , y iY = hJX (Φ ) y , Φ y iX ? ? ?? −1 ?? ? ?? ?? −1 ?? ? ?? = hΦ y , (Φ ) y iX? = hy , Φ (Φ ) y iY ? = hy , y iY ? .

?? Hence, JY (y) = y . Proposition 4.62. Let X and Y be Hilbert spaces and T ∈ L(X,Y ). It then holds: a) T is a isometry if and only if we have (T x|T z)Y = (x|z)X for all x, z ∈ X. b) T is unitary if and only if T is bijective and isometric if and only if T is bijective and preserves the scalar product.

Proof. (a) The implication ‘⇐’ is shown by setting x = z. To verify ‘⇒’, take α ∈ F and x, z ∈ X. Using (4.1) and that T is isometric, we calculate

(T (x + αz)|T (x + αz)) = kT xk2 + 2 Re (T x|αT z) + kαT zk2 = kxk2 + 2 Reα ¯ (T x|T z) + |α|2 kzk2, kT (x + αz)k2 = (x + αz|x + αz) = kxk2 + 2 Reα ¯ (x|z) + |α|2 kzk2.

Since (T (x + αz)|T (x + αz)) = kT (x+αz)k2, it follows Reα ¯ (T x|T z) = Reα ¯ (x|z) . Choosing α = 1 and α = i (if F = C), we deduce assertion a). b) The second equivalence is a consequence of part a). To show the ﬁrst equivalence, take x, z ∈ X. If T is unitary we obtain (T x|T z) = (x|T −1T z) = (x|z), so that T is isometric by a). If T is isometric, a) yields (T 0T x|z) = (T x|T z) = (x|z). Since z ∈ X is arbitrary, we conclude that T 0T x = x for all x ∈ X and hence T 0T = I. Now, the bijectivity of T implies that T 0 = T −1. This proposition is needed in the next section. Chapter 5

The Fourier transform and generalized derivatives

The Fourier transform is a fundamental tool in many branches of mathematics and its applications. In the ﬁrst section we study its basic properties in an L2 context. If one wants to treat partial diﬀerential equations in L2 (or Lp) spaces, one needs weak derivatives and the Sobolev spaces W k,p. The second section gives a brief introduction in these topics and establishes important links between the Fourier transform and the spaces W k,2(Rd). In the last section we take a glimpse into the theory of (tempered) distributions.

5.1 The Fourier transform

Deﬁnition 5.1. Let f ∈ L1(Rd) and ξ ∈ Rd. Then Z ˆ 1 −i ξ·x f(ξ) = (Ff)(ξ) := d e f(x) dx (5.1) d (2π) 2 R

Pd d is the Fourier transform of f, where ξ · x := k=1 ξkxk for ξ = (ξk)k ∈ C and d 2 d x = (xk)k ∈ C . Hence, |x|2 = x · x for x ∈ R . We set ϕ(x, ξ) = e−i ξ·xf(x). Observe that |ϕ(x, ξ)| = |f(x)| is integrable in x ∈ Rd for every ξ ∈ Rd and that Rd 3 ξ 7→ ϕ(ξ, x) is continuous for a.e. x ∈ Rd. Using a corollary to the theorem of dominated convergence, we thus conclude

d ˆ d ˆ − 2 1 d f is continuous on R and kfk∞ ≤ (2π) kfk1 for each f ∈ L (R ). (5.2) √ ˆ Example 5.2. a) Let d = 1 and f = 1[a,b]. We then have f(0) = (b − a)/ 2π and

1 Z b i(e−ibξ − e−iaξ) fˆ(ξ) = √ e−iξx dx = √ , ξ 6= 0. 2π a 2π ξ

b) Let d = 1 and f(x) = (1 + x2)−1. Using complex curve integrals, it can be p shown that fˆ(ξ) = π/2 e−i |ξ| for ξ ∈ R, see also §9.7 in [Rud87]. In the above two examples (non-)rapid decay and (non-)smoothness on f corre- spond to (non-)smoothness and (non-)rapid decay of fˆ, respectively, cf. Lemma 5.7. In the next example, both f and fˆ are smooth and decay rapidly. 1 2 d c) Let γ(x) = exp(− 2 |x|2) for x ∈ R be the standard Gaussian. We show that d γ is a ﬁx vector of the Fourier transform, i.e., γb = γ. Let ξ ∈ R . Observe that

75 76 Chapter 5. The Fourier transform and generalized derivatives

1 1 2 1 2 2 (x + iξ) · (x + iξ) = 2 |x|2 + iξ · x − 2 |ξ|2. We then obtain Z Z − d −(iξ·x+ 1 |x|2) − d − 1 |ξ|2 − 1 (x+iξ)·(x+iξ) γˆ(ξ) = (2π) 2 e 2 2 dx = (2π) 2 e 2 2 e 2 dx d d R R d d 1 2 1 Z 1 2 1 2 1 Z 1 2 − |ξ| Y − (xk+iξk) − |ξ| Y − z = e 2 2 √ e 2 dxk = e 2 2 √ e 2 dz k=1 2π R k=1 2π iξk+R d Z − 1 |ξ|2 Y 1 − 1 t2 = e 2 2 √ e 2 dt = γ(ξ), k=1 2π R √ R − 1 t2 using the formula e 2 dt = 2π from Analysis 3. In the penultimate equality R we shifted the path of integration in the complex plane. To justify this shift, we ﬁx ξ ∈ R \{0} and use the rectangular path Γn with vertices −n, n, n + iξ and R − 1 z2 ± −n + iξ. Cauchy’s theorem yields e 2 dz = 0. The two vertical lines S in Γn n ± Γn have length |ξ|, and on Sn it holds

− 1 z2 − 1 Re(±n+iτ)2 − 1 n2 1 |ξ|2 e 2 = e 2 = e 2 e 2 ,

R − 1 z2 where 0 ≤ |τ| ≤ |ξ|. Hence ± e 2 dz tends to 0 as n → ∞, and above used Sn equality is true. ♦ Let f ∈ Lp(Rd), t, x ∈ Rd, and 1 ≤ p ≤ ∞. To describe important mapping properties of F, we set i t·x et(x) = e , and introduce the translation operator Tt by

(Ttf)(x) = f(x + t).

p d p d As in Example 3.13 one sees that Tt : L (R ) → L (R ) is an isometric isomorphism with inverse T−t. For a > 0 we further deﬁne the dilation operator Da by

(Daf)(x) = f(ax).

Observe that D1/aDa = DaD1/a = I, and that the substitution y = ax yields Z Z p p −d p −d p kDafkp = |f(ax)| dx = a |f(y)| dy = a kfkp d d R R −d/p p d p d for p < ∞ (and analogously for p = ∞). As a result, a Da : L (R ) → L (R ) is an isometric isomorphism. Finally, also the reﬂection operator

p d p d R : L (R ) → L (R ); Rf(x) = f(−x), is an isometric isomorphism with R2 = I, i.e., R−1 = R.

Proposition 5.3. Let f, g ∈ L1(Rd), t ∈ Rd and a > 0. The following formulas hold. ˆ a) F(Ttf) = etf. ˆ b) F(etf) = T−tf. −d ˆ c) F(Daf) = a D 1 f. a d d) F(f ∗ g) = (2π) 2 fˆgˆ.

Proof. Let f, g ∈ L1(Rd), t ∈ Rd, a > 0, and ξ ∈ Rd. Using the substitutions y = x + t and z = ax, we check the assertions a), b) and c) by the calculations Z Z 1 −iξ·x 1 −iξ·(y−t) iξ·t ˆ F(Ttf)(ξ) = d e f(x + t) dx = d e f(y) dy = e f(ξ), d d (2π) 2 R (2π) 2 R 5.1. The Fourier transform 77

Z 1 −iξ·x it·x ˆ F(etf)(ξ) = d e e f(x) dx = f(ξ − t), d (2π) 2 R 1 Z 1 Z 1 −iξ·x −d −i a ξ·z −d ˆ 1 F(Daf)(ξ) = d e f(ax) dx = d a e f(z) dz = a f( a ξ). d d (2π) 2 R (2π) 2 R

To prove d), we ﬁrst recall from (the proof of) Proposition 2.15 that f ∗ g ∈ L1(Rd) and that the map R2d 3 (x, y) 7→ f(y−x)g(x) is integrable. Hence, also the function (x, y) 7→ e−i ξ·yf(y − x)g(x) belongs to L1(R2d). Fubini’s theorem thus yields Z Z − d −i ξ·y F(f ∗ g)(ξ) = (2π) 2 e f(y − x)g(x) dx dy d d ZR ZR − d −i ξ·(y−x) −i ξ·x = (2π) 2 e f(y − x)e g(x) dy dx d d ZR ZR − d −i ξ·z −i ξ·x d = (2π) 2 e f(z) dz e g(x) dx = (2π) 2 fˆ(ξ)ˆg(ξ), d d R R where we also employed the subtitution z = y − x in one of y–integrals.

a 2 d Example 5.4. We set f(x) = exp(− 2 |x − v| ) for all x ∈ R and some a > 0 and v ∈ Rd. The Fourier transform of this Gaussian function is given by fˆ(ξ) = −d/2 1 2 d √ a exp(−i v · ξ) exp(− 2a |ξ|2) for all ξ ∈ R . In fact, we have f = T−vD aγ. Proposition 5.3 and Example 5.2 thus yield

d d ˆ √ − 2 − 2 f = e−vF(D aγ) = e−va D √1 γˆ = a e−vD √1 γ, a a as asserted. ♦ As one of its main properties, the Fourier transform maps derivatives into multiplication by polynomials, and vice versa. To state this fact concisely, we use the n n multi-index notation: For α = (α1, . . . , αn) ∈ N0 and x = (x1, . . . , xn) ∈ R , we set ∂|α| α α1 αn α α1 αn |α| := α1 + ··· + αn, x := x1 · ... · xn , ∂ := ∂1 ··· ∂n = α1 αn . ∂x1 ··· ∂xn

We further denote the function Rd 3 x → xαf(x) by xαf. Observe that

α α1 αn |α| m |x | = |x1| · ... · |xn| ≤ |x|2 ≤ 1 + |x|2 for x ∈ Rn and |α| ≤ m. To relate the Fourier transform with derivatives we need a space of smooth func- ∞ d tions. Unfortunately, the space Cc (R ) is not invariant under the Fourier transform. Instead one uses the (somewhat less convenient) ‘Schwartz space’ on which F becomes a bijection, as seen below.

∞ d d Deﬁnition 5.5. For f ∈ C (R ), m ∈ N0 and α ∈ N0, we set

m α pm,α(f) = sup |x|2 |∂ f(x)|. d x∈R

We deﬁne the Schwartz space Sd by

∞ d d Sd = {f ∈ C (R ) | pm,α(f) < ∞ for all m ∈ N0, α ∈ N0}.

Notice that Sd is a vector space and that all derivatives of f ∈ Sd decay faster than −m |x|2 for any m ∈ N, as |x|2 → ∞. One thus calls f ∈ Sd rapidly decreasing. It is −|x|2/2 straightforward to check that the function γ(x) = e 2 belongs to Sd. Moreover, one can replace pm,α(f) by p2m,α(f) in the deﬁnition of Sd without changing Sd. 78 Chapter 5. The Fourier transform and generalized derivatives

d Remark 5.6. a) Let f ∈ Sd, m ∈ N0 and α ∈ N0. We estimate

m α d+1 −1 m m+d+1) α |x|2 |∂ f(x)| = (1 + |x|2 ) (|x|2 + |x|2 ) |∂ f(x)| d+1 −1 ≤ (1 + |x|2 ) (pm,α(f) + pd+m+1,α(f))

d d+1 −1 d for all x ∈ R . Since the function x 7→ (1 + |x|2 ) is integrable on R (see m α 1 d d m α Analysis 3), we deduce |x|2 ∂ f ∈ L (R ) ∩ C0(R ), and hence |x|2 ∂ f belongs to Lp(Rd) for all p ∈ [1, ∞] by (2.3). ∞ d p d b) Because Cc (R ) ⊂ Sd ⊂ L (R ), Proposition 3.14 yields that Sd is dense in Lp(Rd) for every p ∈ [1, ∞). d c) Observe that pm,α is a seminorm on Sd for all m ∈ N0 and α ∈ N0, where p00 is the supnorm. We order these seminorms as a sequence (pj)j∈N. Due to Proposition 1.8, the Schwartz space Sd has the metric

∞ X pj(f − g) d(f, g) = 2−j , 1 + p (f − g) j=1 j and d(fn, f) → 0 as n → ∞ if and only if pm,α(f −fn) → 0 as n → ∞ for all m ∈ N0 d and α ∈ N0. One can verify that Sd is complete for this metric, cf. Example 1.9. ♦ The next lemma deals with the announced relation between Fourier transform 2 2 and derivatives. We use the Laplace operator given by ∆ = ∂1 + ··· + ∂d and the space of smooth, polynomially bounded functions

∞ d d −nα α Ed = {f ∈ C (R ) | ∀ α ∈ N0 ∃ nα ∈ N0 : sup |x|2 |∂ f(x)| < ∞}. |x|2≥1

Note that Schwartz functions and polynomials belong to Ed.

d Lemma 5.7. Let f ∈ Sd, g ∈ Ed,and α ∈ N0. Then the following assertions hold. a) fb ∈ C∞(Rd), ∂αfb = (−i)|α| F(xαf), F(∂αf) = i|α| ξαfb. 2 2 2 2 2 2 b) F∆f = F∂1 f + ··· + F∂df = i (ξ1 + ··· ξd)Ff = −|ξ|2Ff. α c) The mappings f 7→ gf and f 7→ ∂ f are continuous from Sd to Sd. d) The Fourier transform is continuous from Sd to Sd.

d d Proof. Let ξ, x ∈ R , f, fn ∈ Sd, g ∈ Ed, α ∈ N0, and m ∈ N0. We show a) for α = ek, the assertion then follows by induction. There exists

∂ −i ξ·x −i ξ·x e f(x) = −ixke f(x) =: ϕk(ξ, x), ∂ξk

d and R 3 x 7→ |ϕk(ξ, x)| = |xkf(x)| is integrable by Remark 5.6. A corollary of the theorem of dominated convergence thus shows that

∂ d Z ˆ − 2 −i ξ·x ∃ f(ξ) = (2π) −ixke f(x) dx = −iF(xkf)(ξ). ∂ξ d k R k k 0 0 For the second part we write [−n, n] = Cn and x = (x , xk) with x = d−1 1 d (x1, . . . , xk−1, xk+1, . . . , xd) ∈ R . Using that ∂kf ∈ L (R ) and integrating by parts in xk, we compute n d Z Z 0 − −i ξ·(x ,xk) 0 0 F(∂kf)(ξ) = (2π) 2 lim e ∂kf(x , xk) dxk dx n→∞ d−1 Cn −n Z Z n − d −i ξ·x −i ξ·x 0 0 2 = (2π) lim iξke f(x) dx + e f(x , xk) dx n→∞ d d−1 xk=−n Cn Cn

= iξk fb(ξ). 5.1. The Fourier transform 79

Here the second integral Jn in the second line tends to 0 as n → ∞ since Z X 0 −d 0 d 0 0 d d−1 −d |Jn| ≤ |(x ,N)|2 |(x ,N)|2 |f(x ,N)| dx ≤ 2 n n pd,0(f). d−1 N=±n Cn

Assertion b) is a consequence of a). Thanks to Leibniz’ rule, the function m α m+nγ β −nγ γ |x|2 ∂ (fg) is a linear combination of terms |x|2 (∂ f) |x|2 ∂ g for all β, γ ∈ d N0 with β + γ = α. Since g ∈ Ed, we obtain pm,α(fg) ≤ c (pm+k,α(f) + pm,α(f)), where k = max|γ|≤|α| nγ and c only depends on m, α, d and g. Hence, fg belongs to Sd. The asserted continuity follows by replacing f by f − fn. Similarly, one checks the second part of c). Using a) and b), we further compute

2m α |α| 2 2 m α |α| 2m m α |ξ|2 ∂ fb = (−i) (ξ1 + ··· + ξd) F(x f) = (−i) (−i) F(∆ (x f)).

m α 1 d Due to c) and Remark 5.6, the function ∆ (x f) belongs to Sd ⊆ L (R ) so that its Fourier transform is bounded by (5.2). This means that fb ∈ Sd and

m α p2m,α(F(f − fn)) ≤ c k∆ (x (f − fn))k1.

Using again c) and Remark 5.6, the term on the right hand side can be bounded by a linear combination of certain seminorms pk,β(f − fn); i.e., Fd : Sd → Sd is continuous.

1 d ˆ d Corollary 5.8 (Riemann-Lebesgue lemma). If f ∈ L (R ), then f ∈ C0(R ). 1 d d Hence, F ∈ L(L (R ),C0(R )).

1 d Proof. Let f ∈ L (R ). Due to Remark 5.6, there are fn ∈ Sd converging to f in 1 d ˆ d ˆ L (R ). Lemma 5.7 yields fn ∈ Sd ⊂ C0(R ). By (5.2), the functions fn converge to ˆ ˆ d f in supnorm so that f ∈ C0(R ). The second assertion then follows from (5.2).

The next lemma is the crucial step towards the main results of this section. Observe that in its second part a double integral disappears due to cancellations of the highly oscillating integrands.

Lemma 5.9. The following assertions hold. R ˆ R a) d fg dx = d fgˆ dx for all f, g ∈ Sd. R R 2 d b) F = R, i.e., (FFf)(x) = f(−x) for all f ∈ Sd and x ∈ R .

−i y·x 2d Proof. Let f, g ∈ Sd. Since (x, y) 7→ e f(x)g(y) is integrable on R , Fubini’s theorem yields Z Z Z − d −i y·x fˆ(y)g(y) dy = (2π) 2 e f(x)g(y) dx dy d d d R ZR R Z Z − d −i y·x = f(x) (2π) 2 e g(y) dy dx = f(x)ˆg(x) dx. d d d R R R In the second assertion one is led to the integrand e−i ξ·ye−i y·xf(x) which is not integrable for (y, x) ∈ R2d. So Fubini’s theorem does not apply directly, and one d has to use a regularization. To that purpose, ﬁx ξ ∈ R and a > 0. Set ha = −i ξ·y a2 2 d e−ξDaγ ∈ Sd, i.e., ha(y) = e exp(− 2 |y|2) for y ∈ R . Due to the theorem of dominated convergence with the majorant |fˆ|, the integral Z Z ˆ ˆ −i ξ·y Ja := f(x)ha(x) dx = f(y)e γ(ay) dy d d R R 80 Chapter 5. The Fourier transform and generalized derivatives

d converges to (2π) 2 (Ffˆ)(ξ) as a → 0. On the other hand, part a), Proposition 5.3 and Example 5.2 imply that Z Z −d Ja = f(x) F(e−ξDaγ)(x) dx = f(x) a (TξD 1 γ)(x) dx d d a ZR ZR −d 1 = f(x) a γ( a (x + ξ)) dx = f(az − ξ)γ(z) dz, d d R R

1 where we also use the substitution z = a (x + ξ). We can now apply the theorem of dominated convergence with the majorant kfk∞γ to conclude that d/2 Ja → f(−ξ)kγk1 = (2π) f(−ξ) as a → 0, which shows assertion b).

4 Proposition 5.10. The Fourier transform F : Sd → Sd is bijective with F = I. d For all f, g ∈ Sd and x ∈ R , we have Z −1 1 i x·ξ F g(x) = d e g(ξ) dξ, (5.3) d (2π) 2 R (Ff|Fg)L2 = (f|g)L2 , (5.4)

f ∗ g ∈ Sd, (5.5) − d F(fg) = (2π) 2 fˆ∗ g.ˆ (5.6)

2 4 3 3 Proof. Lemma 5.9 shows that I = R = F = FF = F F on Sd so that F : 3 Sd → Sd has the inverse F = RF. This fact already gives (5.3). Let f, g ∈ Sd and x ∈ Rd. Equation (5.3) further yields Z Z − d −i x·ξ − d F(Fg)(ξ) = (2π) 2 e gˆ(ξ) dξ = (2π) 2 ei x·ξgˆ(ξ) dξ d d R R = (F −1Fg)(x) = g(x).

So we can deduce from Lemma 5.9a) that Z Z Z ˆ ˆ (f|gˆ)L2 = f gˆ dξ = f F(Fg) dx = fg dx = (f|g)L2 . d d d R R R For the ﬁnal two assertions, Proposition 5.3 and Lemma 5.7 imply that F(f ∗ g) = d/2 ˆ −1 (2π) fgˆ =: ϕ belongs to Sd. This fact yields f ∗ g = F ϕ ∈ Sd and further

F(fˆ∗ gˆ) = (2π)d/2F 2(f)F 2(g) = (2π)d/2R(fg) = (2π)d/2F 2(fg) using that R = F 2. If we apply F −1, we derive (5.6).

The equality (5.4) shows that kFfk2 = kfk2 for all f ∈ Sd. Since Sd is dense 2 d 2 d 2 d in L (R ) by Remark 5.6, we can extend F to an isometry F2 : L (R ) → L (R ) which is also called Fourier transform (use Lemma 2.13). Let f ∈ L2(Rd) ∩ L1(Rd). ∞ d In Proposition 3.14 we have constructed functions fn ∈ Cc (R ) ⊂ Sd which con- 2 d 1 d 2 d verge to f in L (R ) and in L (R ). Since Ffn → F2f in L (R ), there is a subsequence Ffnj converging to F2f a.e. as j → ∞ due to Riesz–Fischer. On the other hand, Ffnj converges uniformly to Ff by (5.2). Thus, F2f = Ff a.e.. We 2 d 2 d ˆ now write F : L (R ) → L (R ) instead of F2, and also F2f = f. Warning: Ff is not given by the formula (5.1) if f ∈ L2(Rd) \ L1(Rd). In the next theorem we collect the main properties of F on L2(Rd), except for its behavior under derivatives which will be dealt with in the following two sections. 5.1. The Fourier transform 81

Theorem 5.11. The Fourier transform on Sd extends to a unitary operator F : L2(Rd) → L2(Rd) which is given by (5.1) on L2(Rd) ∩ L1(Rd). Let f, g ∈ L2(Rd), h ∈ L1(Rd), ϕ = F −1ψ for some ψ ∈ L1(Rd), t ∈ Rd, and a > 0. Then the following assertions hold. a) F 2 = R, F 4 = I, F −1 = F 3 = RF. d −1 − 2 R i x·ξ d b) F h(x) = (2π) d e h(ξ) dξ for x ∈ (inversion formula). R R c) (Ff|Fg)L2 = (f|g)L2 (Plancherel identity). R ˆ R d) d fg dx = d fgˆ dx (jumping hat lemma). R R ˆ ˆ −d ˆ e) F(Ttf) = etf, F(etf) = T−tf, F(Daf) = a D 1 f. a d − d f) F(h ∗ f) = (2π) 2 hˆfˆ, F(ϕf) = (2π) 2 ϕˆ ∗ fˆ (convolution theorem). 0 g) F ∈ L(Lp(Rd),Lp (Rd)) for p ∈ [1, 2] (Hausdorﬀ–Young inequality). Proof. Recall that F is an isometry on L2(Rd). The equations F 2 = R, F 4 = I, and those in c)–e) hold on the dense subspace Sd as shown in Proposition 5.3, Lemma 5.9 and Proposition 5.10. Since the maps F, R, Tt, Da, f 7→ etf and the scalar product are continuous from L2(Rd), resp. from L2(Rd) × L2(Rd), to L2(Rd), these identities can be extended to L2(Rd) by approximation. The equation F 4 = I implies I = FF 3 = F 3F so that F : L2(Rd) → L2(Rd) has the inverse F −1 = F 3 = RF which also yields b). As a bijective isometry on a Hilbert space, F is unitary by Proposition 4.62. For f), note that f 7→ h ∗ f is continuous on L2(Rd) by Proposition 2.15 (Young’s inequality) and h ∈ L1(Rd). Hence the ﬁrst part of f) can be deduced from Proposi- tion 5.3 by approximation. For the second part, take fn, ψn ∈ Sd so that fn → f in 2 d 1 d −1 −1 L (R ) and ψn → ψ in L (R ) as n → ∞. Then ϕn := F ψn tends to ϕ = F ψ ∞ d ˆ ˆ ˆ in L (R ) by (5.2). Hence, ϕnfn converges to ϕf and ψn ∗ fn to ψ ∗ f =ϕ ˆ ∗ f in L2(Rd). Equation (5.6) now implies the second part of f). Assertion g) is a consequence of the Riesz–Thorin Theorem 2.27 with θ = 2/p0 ∈ [0, 1], since F ∈ L(L1(Rd),L∞(Rd)) and F ∈ L(L2(Rd)). 2 d 1 1 d 2 d Let f ∈ L (R ). Set fn = B(0,n)f ∈ L (R ) ∩ L (R ) for n ∈ N. Since 2 d 2 d |fn| ≤ |f| ∈ L (R ), we have fn → f in L (R ) as n → ∞ by the theorem of ˆ ˆ 2 d dominated convergence. Therefore, fn → f in L (R ) as n → ∞, where

d Z ˆ − 2 −iξ·x d fn(ξ) = (2π) e f(x) dx, ξ ∈ R , B(0,n) is the truncated Fourier transform. Let J ⊆ R be an interval and X be a Banach space. A function u : J → X is diﬀerentiable at t ∈ J if the limit u0(t) := lim 1 (u(t + h) − u(t)) h→0,t+h∈J h exists in X. Then u is continuous at t. We write u ∈ C1(J, X) if u0(t) exists for all t ∈ J and u0 : J → X is continuous. Example 5.12. We consider the diﬀusion equation

d ∂tu(t, x) = ∆u(t, x), t ≥ 0, x ∈ R , (5.7) d u(0, x) = u0(x), x ∈ R , for a given initial value u0 ∈ Sd. Let us assume for moment that we have a function d u : R+ × R 7→ C such that the function u(t): x 7→ u(t, x) belongs to Sd for all 1 2 d t ≥ 0, u ∈ C (R+,L (R )) and u satisﬁes (5.7). We setu ˆ(t, ξ) = (Fu(t))(ξ) for all t ≥ 0 and ξ ∈ Rd. Since F is continuous on L2(Rd), we have Fu0(t) = F lim 1 (u(t + h) − u(t)) = lim 1 (u(t + h) − u(t)) h→0 h h→0 h b b 82 Chapter 5. The Fourier transform and generalized derivatives

1 2 d for all t ≥ 0, so that ub ∈ C (R+,L (R )) and ∂tub = F∂tu. Applying F to (5.7), we then deduce from Lemma 5.7 that 2 ∂tuˆ(t) = F∂tu(t) = F∆u(t) = −|ξ|2 uˆ(t). For each ﬁxed ξ ∈ Rd we thus arrive at the ordinary diﬀerential equation 2 d ∂tuˆ(t, ξ) = −|ξ|2 uˆ(t, ξ), t ≥ 0, ξ ∈ R , d uˆ(0, ξ) = uc0(ξ), ξ ∈ R , which has the solution 2 −t |ξ|2 √ uˆ(t, ξ) = e uc0(ξ) = (D 2tγ)(ξ) uc0(ξ). Using the properties of F, we compute

d d −1 −1 √ − 2 −1 √ − 2 √ u(t) = F [(FF D 2tγ)uc0] = (2π) (F D 2tγ) ∗ u0 = (4πt) (D1/ 2tγ) ∗ u0 for all t > 0. This gives

Z |x−y|2 − d − 2 u(t, x) = (4πt) 2 e 4t u0(x) dx =: (γt ∗ u0)(x) d R for all t > 0 and x ∈ Rd. Deﬁne u by this equation. Since γt ∈ Sd, we can reverse the above arguments (or check directly) that this u solves (5.7) and has the asserted regularity. As a result, we have found a unique solution of (5.7) in the class indicated above. Moreover, ku(t)k = kuˆ(t)k2 ≤ kuc0k2 = ku0k2 since the Fourier transform is an √ isometry and |D 2tγ| ≤ 1. ♦

The above result is restricted to initial values u0 in Sd. At the end of the next d 2 d section we solve the diﬀusion equation on R for u0 ∈ L (R ) again using the Fourier transform. To that purpose we have to extend the diﬀerentiation formulas in Lemma 5.7a) to a much larger class of functions, the Sobolev spaces.

5.2 Sobolev spaces

The classical (partial) derivative does not fit well to Lp spaces since it is defined via a pointwise limit. For a treatment of partial differential equations in an L2 or Lp context one needs the more general concept of ‘weak derivatives’. Here we restrict ourselves to some basic results on the spatial domain U = Rd. For an introduction to this area we refer to the books [Bre11] and [Dob06], and also to [ST]. 1 d ∞ d To motivate the definition below, we consider u ∈ C (R ). Take any ϕ ∈ Cc (R ). There is an a > 0 such that supp ϕ ⊆ (−a, a)d. Integrating by parts, we obtain Z Z Z a 0 0 0 ∂1u ϕ dx = ∂1u(x1, x ) ϕ(x1, x ) dx1 dx d R Ca −a Z Z a a 0 0 h 0 0 i 0 = − u(x1, x ) ∂1ϕ(x1, x ) dx1 + u(x1, x )ϕ(x1, x ) dx x =−a Ca −a 1 Z = − u ∂1ϕ dx, (5.8) d R d−1 where Ca = [−a, a] . Other partial derivatives can be treated in the same way. The following definition now relies on the observation that the right hand side of (5.8) is defined for all u in the vector space 1 d d 1 d Lloc(R ) = {f : R → C | f is measurable, f|K ∈ L (K) for all compact K ⊆ R }.

It thus can serve as deﬁnition of ∂1u on the left hand side of (5.8). Observe that p d 1 d L (R ) ⊆ Lloc(R ) for all 1 ≤ p ≤ ∞. 5.2. Sobolev spaces 83

1 d d d Deﬁnition 5.13. Let u ∈ Lloc(R ), α ∈ N0, k ∈ N and 1 ≤ p ≤ ∞. If there is a 1 d function f ∈ Lloc(R ) such that Z Z fϕ dx = (−1)|α| u ∂αϕ dx, d d R R ∞ d α for all ϕ ∈ Cc (R ), then f =: ∂ u is called α-th weak derivative of u. We write 1 d α 1 d Dα for the space of u ∈ Lloc(R ) such that a weak derivative ∂ u ∈ Lloc(R ) exists. One further deﬁnes the Sobolev spaces by

k,p d p d α p d W (R ) = {u ∈ L (R ) | u ∈ Dα and ∂ u ∈ L (R ) for all |α| ≤ k} and endows them with  1 p  X α p  k∂ uk , 1 ≤ p < ∞,  p  0≤|α|≤k kukk,p =   α  max k∂ uk∞, p = ∞, 0≤|α|≤k

0 ej where ∂ f := f. We write ∂j instead of ∂ .

1 d k,p d As usually, the spaces Lloc(R ) and W (R ) are spaces of equivalence classes modulo the space of null functions N . In the above concepts one can replace Rd by any nonempty open subset U. The above definition first raises the question whether the weak derivative is uniquely determined. Already to settle this basic question one needs the mollifier Gε from Proposition 3.14 whose properties are used below intensively.

1 d Lemma 5.14. Let f ∈ Lloc(R ) satisfy Z ∞ fϕ dx = 0 for all ϕ ∈ Cc (U). d R

Then f = 0 a.e.. Hence, a function u ∈ Dα has exactly one α-th weak derivative. Proof. Assume that f 6= 0 on a Borel set B with λ(B) > 0. By Satz IV.3.15 of [Wer06], there is a compact set K ⊆ B with λ(K) > 0. The set K1 = K + B(0, 1) 1 ∞ d is also compact. Proposition 3.14 then shows that ϕ = G1 K1 belongs to Cc (R ). Moreover, the deﬁnition (3.1) of Gε yields Z Z 1 ϕ(x) = k1(x − y) K1 (y) dy = k1(x − y) dy = 1 B(x,1) B(x,1)

1 d 1 d for all x ∈ K. Since ϕf ∈ L (R ), the functions Gε(ϕf) converge to ϕf in L (R ) as ε → 0, due to Proposition 3.14. There thus exist a nullset N and a subsequence

εj → 0 such that (Gεj (ϕf))(x) → ϕ(x)f(x) = f(x) 6= 0 as j → ∞ for each x ∈ K\N. Fix any x ∈ K\N. For every j ∈ N, we further deduce Z

(Gεj (ϕf))(x) = kεj (x − y)ϕ(y)f(y) dy = 0 d R ∞ d from the assumption, since the function y 7→ kεj (x − y)ϕ(y) belongs to Cc (R ). This contradiction implies the assertions.

k d Remark 5.15. a) In view of formula (5.8), we have C (R ) + N ⊆ Dα for all d k d α ∈ N0 mit |α| ≤ k, and weak and classical derivatives coincide for u ∈ C (R ). This fact justiﬁes to use the same notation for both them. 84 Chapter 5. The Fourier transform and generalized derivatives

α 1 d d b) The set Dα is a vector space and ∂ : Dα → Lloc(R ) is linear for all α ∈ N0. k,p d c) Let 1 ≤ p ≤ ∞ and k ∈ N. It is straightforward to check that (W (R ), k·kk,p) k,p d is a normed vector space. Moreover, a sequence (un) converges in W (R ) if and α p d only if every sequence (∂ un) converges in L (R ) for |α| ≤ k. d) Let 1 ≤ p ≤ ∞ and k ∈ N. The map k,p d p d m α J : W (R ) → L (R ) ; u 7→ (∂ u)|α|≤m, is a linear isometry, where m = 1 + d + ... + dk and Lp(Rd)m is endowed with the m norm k(fj)j=1k := | (kfjkp)j |p. (We see in the proof of the next proposition that W k,p(Rd) is isometrically isomorphic to a closed subspace of Lp(Rd)m.) Since the m p-norm and the 1-norm on R are equivalent, there are constants Ck, ck > 0 with X α X α ck k∂ ukp ≤ kukk,p ≤ Ck k∂ ukp 0≤|α|≤k 0≤|α|≤k

k,p d for all u ∈ W (R ). ♦ 1 d The following lemma first gives a convergence result in Lloc(R ) for weak derivatives which is analogous to a well-known result for uniform limits and the classical derivative. Second, it makes clear that the mollifiers fit perfectly well to weak derivatives. Here and below we use the fact that the map Z p d p0 d L (R ) × L (R ) → F;(f, g) 7→ fg dx, d R is continuous for all p ∈ [1, ∞], due to Hölder’sinequality. 1 d α Lemma 5.16. a) Let un ∈ Dα and u, f ∈ Lloc(R ) such that un → u and ∂ un → f 1 d α in Lloc(R ) as n → ∞. Then u ∈ Dα and ∂ u = f. d α α b) Let u ∈ Dα for some α ∈ N0 and ε > 0. Then ∂ Gεu = Gε∂ u. ∞ d 1 Proof. a) Let ϕ ∈ Cc (R ). In a) we have convergence on L (supp ϕ). Using also Definition 5.13, we compute Z Z Z α α |α| α u ∂ ϕ dx = lim un ∂ ϕ dx = lim (−1) (∂ un) ϕ dx d n→∞ d n→∞ d R RZ R = (−1)|α| f ϕ dx, d R so that a) holds. d b) Fix ε > 0, u ∈ Dα and x ∈ R . The function y 7→ ϕ(y) := kε(x − y) belongs ∞ d to Cc (R ). From a corollary to Lebesgue’s theorem and Definition 5.13 it follows Z Z α α |α| α (∂ Gεu)(x) = ∂x kε(x − y)u(y) dy = (−1) (∂ ϕ)(y)u(y) dy d d ZR R α α = ϕ(y) ∂ u(y) dy = (Gε∂ u)(x). d R

The above convergence result is a crucial tool when extending properties from classical to weak derivatives, see Proposition 5.19 below. We ﬁrst use it to show the completeness of Sobolev spaces. Proposition 5.17. Let 1 ≤ p ≤ ∞ and k ∈ N. Then W k,p(Rd) is a Banach space. It is separable if 1 ≤ p < ∞ and reﬂexive if 1 < p < ∞. Moreover, W k,2(Rd) is a Hilbert space endowed with the scalar product Z X α α (f|g)k,2 = (∂ f) ∂ g dx. d |α|≤k R 5.2. Sobolev spaces 85

k,p d α Proof. Let (un)n be a Cauchy sequence in W (R ). Then (∂ un)n is a Cauchy p d d α p d sequence in L (R ) for every α ∈ N0 with |α| ≤ k, and thus ∂ un → fα in L (R ) p d p d for some fα ∈ L (R ) as n → ∞. Setting u := f0, we see that un → u in L (R ). α k,p d Lemma 5.16 now implies that u ∈ Dα and fα = ∂ u. Hence, W (R ) is a Banach space. Using Remark 2.11, we then deduce from Remark 5.15d) that W k,p(Rd) is isometrically isomorphic to a closed subspace of Lp(Rd)m. The remaining assertions now follow by isomorphy from known results.

Example 5.18. a) Let u ∈ C( ) be such that u := u belong to C1( ). We R ± |R± R± then have u ∈ D1 with

0 u+ on (0, ∞) ∂1u = 0 =: f. u− on (−∞, 0)

For u(x) = |x|, we thus obtain ∂1f = 1(0,∞) − 1(−∞,0).

∞ Proof. For every ϕ ∈ Cc (R), integration by parts yields Z Z 0 Z ∞ 0 0 0 uϕ dt = u−ϕ dt + u+ϕ dt R −∞ 0 Z 0 Z ∞ 0 0 0 0 ∞ = − u−ϕ dt + u−ϕ|−∞ − u+ϕdt + u+ϕ |0 −∞ 0 Z = − f ϕ dt, R since u+(0) = u−(0) by the continuity of u. 1 b) The function u = R+ does not belong to D1.

1 ∞ Proof. Assume there would exist f = ∂1u ∈ Lloc(R). For every ϕ ∈ Cc (R) we then obtain Z Z Z ∞ 1 0 0 fϕ dt = − R+ ϕ dt = − ϕ (t) dt = ϕ(0). R R 0 Taking ϕ with supp ϕ ⊆ (0, ∞), we deduce from a variant of Lemma 5.14 that f = 0 on (0, ∞). Similarly, it follows that f = 0 on (−∞, 0). Hence, f = 0 and so ∞ ϕ(0) = 0 for all ϕ ∈ Cc (R), which is wrong.

d β c) Let p ∈ [1, ∞) and β ∈ (1 − p , 1]. Set u(x) = |x|2 for 0 < |x|2 < 1 and fj(x) = β−2 βxj|x|2 for 0 < |x|2 < 1 and j ∈ {1, . . . , d}, as well as and u(0) = fj(0) = 0. 1,p Then u ∈ W (B(0, 1)) and ∂ju = fj. Observe that u is unbounded and has no continuous extension at x = 0 if β < 0 (which is possible if d ≥ 2).

Proof. The functions r 7→ rβprd−1 and r 7→ r(β−1)prd−1 are integrable on (0, 1) d p since β > 1 − p . Using polar coordinates, we infer that u, fj ∈ L (B(0, 1)) for −2 2 β/2 j ∈ {1, . . . , d}. We introduce the regularized functions un(x) = (n + |x|2) 1 d 1,p for n ∈ N and x ∈ B(0, 1). Then un ∈ Cb (R ) ⊂ W (B(0, 1)) and ∂jun(x) = −2 2 β −1 βxj (n + |x|2) 2 . Observe that un(x) and ∂jun(x) tend to u(x) and fj(x) for β/2 x 6= 0 as n → ∞, respectively. Moreover, |un| ≤ max{|u|, 2 } and |∂jun| ≤ |fj| p a.e. on B(0, 1) for all n and j. Hence, un → u and ∂jun → fj in L (B(0, 1)) as n → ∞ due to dominated convergence. The assertion thus follows from Lemma 5.16.

Under suitable regularity and integrability assumptions, the product and substitution rules also hold for weak derivatives. Here we only present the basic version of the product rule that is needed below. 86 Chapter 5. The Fourier transform and generalized derivatives

0 Proposition 5.19. Let p ∈ [1, ∞), u ∈ W 1,p(Rd) and v ∈ W 1,p (Rd). Then, 1,1 d uv ∈ W (R ) and ∂j(uv) = (∂ju)v + u∂jv for every j ∈ {1, . . . , d}. 1 d Proof. Hölder’sinequality implies that uv,(∂ju)v and u∂jv belong to L (R ). Set ∞ d p d ∞ d p0 d un = G1/nu ∈ C (R ) ∩ L (R ) and vn = G1/nv ∈ C (R ) ∩ L (R ) for n ∈ N. 0 First, let p > 1 so that p < ∞. Due to Proposition 3.14, the functions un converge p d p0 d to u in L (R ) and vn converge to v in L (R ) as n → ∞. Lemma 5.16 further yields p d p0 d ∂jun = G1/n∂ju and ∂jvn = G1/n∂jv. Since ∂ju ∈ L (R ) and ∂jv ∈ L (R ), p d p0 d Proposition 3.14 also implies that ∂jun → ∂ju in L (R ) and ∂jvn → ∂jv in L (R ) as n → ∞, for each j ∈ {1, . . . , d}. In view of (5.8) and using the product rule for 1 ∞ d C –functions, for each ϕ ∈ Cc (R ) we can now conclude that Z Z Z uv ∂jϕ dx = lim unvn ∂jϕ dx = − lim [(∂jun)vn + un(∂jvn)] ϕ dx d n→∞ d n→∞ d R Z R R = − [(∂ju)v + u(∂jv)] ϕ dx. (5.9) d R 1 d Hence, the weak derivative ∂j(uv) = (∂ju)v + u(∂jv) ∈ L (R ) exists. 0 Let p = 1. The above used convergence in Lp (Rd) may now fail since p0 = ∞. 1 d The functions un and ∂jun still converge to u and ∂ju in L (R ). Passing to a subsequence, we can thus assume that they converge pointwise a.e. and that 1 d |un|, |∂jun| ≤ g a.e. for some g ∈ L (R ) and all n and j. We further use that kvnk∞ ≤ kvk∞ and k∂jvnk∞ ≤ k∂jvk∞ by Proposition 3.14 and that the restric- 1 tions of v and ∂jv belong L (B(0, k)) for each k > 0. If we apply Proposition 3.14 for each L1(B(0, k)) and iteratively choose subsequences, we obtain (diagonal) subsequences (vnm )m and (∂jvnm )m which converge to v and ∂jv pointwise a.e. as m → ∞. By means of the dominated convergence theorem, we can now conclude the assertion as in (5.9). We come now to an important density result. In the proof we first use a cutoff argument to obtain a compact support and then perform a mollification. The cutoff must be chosen so that the extra terms caused by the W k,p-norm vanish in the limit. ∞ d k,p d Theorem 5.20. The spaces Cc (R ) and Sd are dense in W (R ) for all p ∈ [1, ∞) and k ∈ N. k,p d Proof. Remark 5.6a) yields Sd ⊆ W (R ) so that we only have to consider ∞ d Cc (R ). We restrict ourselves to k = 1, the general case is treated similarly. 1) Let u ∈ W 1,p(Rd). Take any φ ∈ C∞(R) with 0 ≤ φ ≤ 1, φ = 1 on [0, 1] and φ = 0 on [2, ∞). Set

1 ϕn(x) = φ n |x|2 (“cut-oﬀ function”) d ∞ d for n ∈ N and x ∈ R . We then have ϕn ∈ Cc (R ), 0 ≤ ϕn ≤ 1 and k∂jϕnk∞ ≤ 0 1 d kφ k∞ n for all n ∈ N, as well as ϕn(x) → 1 for all x ∈ R as n → ∞. Thus kϕnu − ukp → 0 as n → ∞ by Lebesgue’s convergence theorem with majorant |u|. Let j ∈ {1, . . . , d}. Proposition 5.19 further implies that

1 0 k∂j(ϕnu − u)kp = k(ϕn∂ju − ∂ju) + (∂jϕn)ukp ≤ kϕn∂ju − ∂jukp + n kφ k∞ kukp Again by Lebesgue, the right hand side tends to 0 as n → ∞. Given ε > 0, we can thus ﬁx an m ∈ N such that kϕmu − uk1,p ≤ ε. ∞ d 2) Proposition 3.14 implies that the functions G 1 (ϕmu) belong to C ( ) for n c R p d all n ∈ and that G 1 (ϕmu) → ϕmu in L ( ) as n → ∞. Moreover, Lemma 5.16 N n R yields ∂jG 1 (ϕmu) = G 1 ∂j(ϕmu), so that ∂jG 1 (ϕmu) converges to ∂j(ϕmu) in n n n p d L ( ) as n → ∞. As a result, kG 1 (ϕmu)−uk1,p ≤ kG 1 (ϕmu)−ϕmuk1,p +ε ≤ 2ε R n n for all suﬃciently large n. 5.2. Sobolev spaces 87

We now come to main result of this section which describes the space W k,2(Rd) via the Fourier transform in a very convenient way. Recall that kuk2 = kubk2 for u ∈ L2(Rd) by Theorem 5.11. d Theorem 5.21. Let k ∈ N and α ∈ N0 with |α| ≤ k. We then have k,2 d 2 d k 2 d k W (R ) = {u ∈ L (R ) | |ξ|2 ub ∈ L (R )} =: H k,2 d 2 k 2 1 k,2 d and the norm of W (R ) is equivalent to (kuk2 + k |ξ|2 ubk2) 2 . For u ∈ W (R ) it further holds α |α| α F(∂ u) = i ξ u.b (5.10)

Proof. Due to Lemma 5.7, Schwartz functions u ∈ Sd satisfy (5.10). We thus obtain Z 2 X α 2 X α 2 X α 2 2 kukk,2 = kF∂ uk2 = kξ uk2 = |ξ | |u|2 dξ b d b |α|≤k |α|≤k R |α|≤k ( 2 k 2 ≤ c1 (kuk2 + k |ξ|2 ubk2) 2 k 2 (5.11) ≥ c2 (kuk2 + k |ξ|2 ubk2) for u ∈ Sd and constants cj > 0. k,2 d k,2 d Let u ∈ W (R ). Theorem 5.20 gives un ∈ Sd which converge to u in W (R ) 2 d 2 d as n → ∞. Since F is continuous on L (R ), the functions ucn tend to ub in L (R ) and (possibly after passing to a subsequence) pointwise a.e., as n → ∞. Hence, the α α functions ξ ucn converge pointwise a.e. to ξ ub. On the other hand, equation (5.11) α 2 d α yields that (ξ ucn)n is Cauchy in L (R ) for each |α| ≤ k, and thus ξ ucn converge α 2 d k,2 d to ξ ub in L (R ) as n → ∞. We conclude that every u ∈ W (R ) fulﬁlls (5.11) and that W k,2(Rd) ⊆ Hk. k ∞ d Conversely, take u ∈ H . Let ϕ ∈ Cc (R ) and |α| ≤ k. From Theorem 5.11 and Lemma 5.7 we deduce Z α α α |α| α |α| α u ∂ ϕ dx = (u|∂ ϕ) = (Fu|F∂ ϕ) = (u|i ξ ϕb) = ((−i) ξ u|Fϕ) d b b R Z = (F 0((−i)|α| ξα u)|ϕ) = (−1)|α| ϕ F −1(i|α| ξα u) dx. b d b R α −1 |α| α 2 d k,2 d Therefore u ∈ Dα and ∂ u = F (i ξ ub) ∈ L (R ), and hence u ∈ W (R ); i.e., W k,2(Rd) = Hk. Applying F to the equation in the previous sentence, we also derive (5.10) for all u ∈ W k,2(Rd). k d α 2 There are u ∈ W2 (R ) such that x u∈ / L (R) for |α| = k, e.g., the function ( |t|−3/2, |t| ≥ 1, u(t) = 1, |t| < 1,

0 for k = 1, cf. Example 5.18. Moreover, one can show directly that here ub is even not integrable near 0. Therefore the other equation in Lemma 5.7a), namely α |α| α k d ∂ ub = (−i) F(x u), cannot be extended from u ∈ Sd to every u ∈ W2 (R ) in the present framework, but we obtain it in a more general setting in Theorem 5.27. Example 5.22. We consider the diﬀusion equation

d ∂tu(t, x) = ∆u(t, x), t ≥ 0, x ∈ R , (5.12) d u(0, x) = u0(x), x ∈ R ,

2 d for a given initial value u0 ∈ L (R ). We show that there is a unique solution of (5.12); i.e., a function u ∈ C([0, ∞),L2(Rd)) ∩ C1((0, ∞),L2(Rd)) such that u(t) 88 Chapter 5. The Fourier transform and generalized derivatives

2 d 2 d belongs to W2 (R ) for all t > 0 and u satisﬁes (5.12) as equations in L (R ). To that purpose, we deﬁne

−1 −t |ξ|2 u(t) = F (mtuc0) with mt(ξ) = e 2 d k k for t > 0 and ξ ∈ R as in Example 5.12. Since |ξ|2 ub(t) = |ξ|2 mtuc0 belongs to L2(Rd), Theorem 5.21 implies that u(t) ∈ W k,2(Rd) for all k ∈ N and t > 0. From (5.10) we then infer

2 2 −1 2 F∆u(t) = −|ξ|2 ub(t) = −|ξ|2 mtuc0, ∆u(t) = −F (|ξ|2 mtuc0). 1 Let v(t) = Fu(t) = mtuc0 for t > 0. Clearly, h (v(t+h)−v(t)) converges pointwise to 2 1 2 1 d −|ξ|2 mtuc0 as h → 0 and | h (v(t + h) − v(t))| ≤ |ξ|2 mt ∈ L (R ). Dominated con- 0 2 vergence then implies that v has the (continuous) derivative v (t) = −|ξ|2 mtuc0 in L2(Rd) for t > 0. The continuity of F −1 on L2(Rd) thus yields that u ∈ C1((0, ∞),L2(Rd)) and

0 −1 2 u (t) = −F (|ξ|2 mtuc0) = ∆u(t) 2 d for t > 0. Finally, mtuc0 tends to uc0 in L (R ) as t → 0 by Lebesgue’s theorem with 2 d majorant |uc0|. Hence, u ∈ C([0, ∞),L (R )) with u(0) = u0. As in Example 5.12 one sees that every solution is given by the above formula so that solutions are unique. Moreover, ku(t)k2 ≤ ku0k2 for all t ≥ 0. ♦

5.3 Tempered distributions

The theory of distributions allows to deﬁne derivatives of any order for rather general objects such as locally integrable functions or measures on open subsets of Rd, see e.g. [Rud91]. Here we only discuss the subclass of tempered distributions on Rd to which one can extend the Fourier transform in a very convenient way.

Deﬁnition 5.23. Tempered distributions are continuous linear functionals on Sd ∗ One writes Sd := {u : Sd → F | u is linear and continuous} for the space of tempered ∗ distributions and hϕ, uiSd = u(ϕ) for u ∈ Sd and ϕ ∈ Sd. m α Recall that ϕn → ϕ in Sd means that pm,α(ϕn − ϕ) = k|x| ∂ (ϕn − ϕ)k∞ → 0 d for all m ∈ N0 and α ∈ N0, as n → ∞. It is possible to deﬁne weak* convergence ∗ in Sd , but we will not deal with such questions. 1 d Example 5.24. a) (regular tempered distributions) Let f ∈ Lloc(R ) satisfy Z κ ak(f) := |f(x)| dx ≤ ck k≤|x|2

∞ X Z |uf (ϕ − ϕn)| ≤ |ϕ − ϕn| |f(x)| dx k=0 k≤|x|2

∞ X κ−m ≤ kfkL1(B(0,1))p0,0(ϕ − ϕn) + pm,0(ϕ − ϕn) ck k=1 0 ≤ c (p0,0(ϕ − ϕn) + pm,0(ϕ − ϕn))

0 for a constant c > 0. Hence, uf : Sd → F is continuous. The linearity of uf is clear, ∗ and so it belongs to Sd . One often writes f instead of uf . b) (measures) Let µ be a measure on Bd with µ(B(0, 1)) < ∞ and µ(B(0, k + 1) \ B(0, k)) ≤ ckκ for all k ∈ N and some κ, c ≥ 0. Then one sees as in a) that Z uµ(ϕ) = ϕ dµ, ϕ ∈ Sd, d R defines a tempered distribution uµ, which is often simply denoted by µ. d d α α c) (Dirac distributions) Let y ∈ R and α ∈ N0. We set δy (ϕ) = ∂ ϕ(y) for α α ϕ ∈ Sd. Let ϕn converge to ϕ in Sd. Then |δy (ϕ) − δy (ϕn)| ≤ p0,α(ϕ − ϕn) tends α ∗ to 0, so that δy ∈ Sd . ♦ ∗ d Definition 5.25. Let u ∈ Sd , g ∈ Ed, and α ∈ N0. For all ϕ ∈ Sd we define

a) (gu)(ϕ) = hϕ, guiSd := hgϕ, uiSd = u(gϕ), α α |α| α |α| α b) (∂ u)(ϕ) = hϕ, ∂ uiSd := (−1) h∂ ϕ, uiSd = u((−1) ∂ ϕ), c) u(ϕ) := (Fu)(ϕ) = hϕ, Fui := hFϕ, ui = u(Fϕ), b Sd Sd d) (Ru)(ϕ) = hϕ, RuiSd := hRϕ, uiSd = u(Rϕ), d e) (ϕ ∗ u)(x) := hT−xRϕ, uiSd = u(T−xRϕ) for every x ∈ R . In view of Lemma 5.7, the maps gu, ∂αu, Fu, and Ru are continuous and linear ∗ from Sd to F, and hence they belong to Sd . Moreover, T−xRϕ ∈ Sd. Observe that we multiply and convolve tempered distributions only with the (very regular) functions in Ed and Sd, respectively. The following examples and the theorem below indicate that the above deﬁnitions extend the known concepts in a natural way and that they allow to generalize several main properties of the Fourier ∗ transform to the space Sd . d d Example 5.26. Let ϕ ∈ Sd, g ∈ Ed, α ∈ N0, and x, y ∈ R . 1 d a) Let f ∈ Lloc(R ) be as in Example 5.24a). Then guf = ugf since Z (guf )(ϕ) = ϕgf dx = ugf (ϕ). d R k,p d α b) Let f ∈ W (R ) for some p ∈ [1, ∞] and |α| ≤ k ∈ N. Then ∂ uf = u∂αf since the deﬁnitions yield Z Z α |α| α |α| α α hϕ, ∂ uf iSd = (−1) h∂ ϕ, uf iSd = (−1) (∂ ϕ)f dx = ϕ ∂ f dx d d R R α = hϕ, u∂ f iSd .

2 d c) Let f ∈ L (R ). Then Fuf = uFf since Theorem 5.11 implies Z Z

hϕ, Fuf iSd = hFϕ, uf iSd = ϕf dx = ϕfb dx = hϕ, uFf iSd . d b d R R α |α| α d) We have ∂ δy = (−1) δy since

α |α| α |α| α |α| α hϕ, ∂ δyiSd = (−1) h∂ ϕ, δyiSd = (−1) ∂ ϕ(y) = (−1) δy (ϕ).

−d/2 e) We have Fδy = (2π) e−y since 1 Z hϕ, Fδ i = hFϕ, δ i = e−iy·xϕ(x) dx = hϕ, (2π)−d/2e i . y Sd y Sd d/2 −y Sd (2π) d R 90 Chapter 5. The Fourier transform and generalized derivatives

d/2 f) We have Fey = (2π) δy since Proposition 5.10 implies Z iy·ξ d/2 −1 d/2 hϕ, FeyiSd = hFϕ, eyiSd = ϕ(ξ)e dξ = (2π) (F ϕ)(y) = (2π) ϕ(y). d b b R

2 ∗ Assertion f) can also be deduced from e) since F is equal to R in Sd , too, as shown in the next theorem (with a similar proof as above). 1 d g) Let f ∈ L (R ). Then ϕ ∗ uf = ϕ ∗ f, since Z

ϕ ∗ uf (x) = hT−xRϕ, uf iSd = ϕ(−(z − x))f(z) dz = ϕ ∗ f(x). ♦ d R

∗ d Theorem 5.27. Let u ∈ Sd , ϕ, ψ ∈ Sd, and α ∈ N0. The following assertions hold. ∗ ∗ 4 −1 3 a) F : Sd → Sd is bijective with F = I and F = F = RF. b) F(∂αu) = i|α|ξαFu and ∂α(Fu) = (−i)|α|F(xαu). c) ϕ ∗ u ∈ Ed, and hence ϕ ∗ u induces a tempered distribution. d) ∂α(ϕ ∗ u) = (∂αϕ) ∗ u = ϕ ∗ ∂αu. d/2 −d/2 e) F(ϕ ∗ u) = (2π) ϕbub and F(ϕu) = (2π) ϕb ∗ u.b ∗ d Proof. Let u ∈ Sd , ϕ ∈ Sd, and α ∈ N0. In a), Proposition 5.10 yields

4 3 4 hϕ, F uiSd = hFϕ, F uiSd = ··· = hF ϕ, uiSd = hϕ, uiSd ,

4 ∗ ∗ ∗ −1 3 so that F = I on Sd and F : Sd → Sd is bijective with inverse F = F . Similarly, we show the remaining equality F 2 = R by computing

2 2 hϕ, F uiSd = hF ϕ, uiSd = hRϕ, uiSd .

The ﬁrst equality in b) follows from Lemma 5.7 and

α α |α| α |α| α hϕ, F∂ uiSd = hFϕ, ∂ uiSd = (−1) h∂ Fϕ, uiSd = i hF(x ϕ), uiSd |α| α = hϕ, i ∂ FuiSd .

The second part of b) established in the same way. For the proof of assertions c) and d) we refer to Theorem 7.19 in [Rud91]. To show ∞ d d e), ﬁrst take ψ ∈ Cc (R ). There is a closed interval I ⊂ R such that supp ψ ⊂ I. Using the deﬁnitions, part a) and Proposition 5.3 in the last step, we compute Z Z hψ,b F(ϕ ∗ u)iSd = hRψ, ϕ ∗ uiSd = ψ(−x)(ϕ ∗ u)(x) dx = ψ(−x)u(T−xRϕ) dx d R −I Z Z = u(ψ(z)TzRϕ) dz = u ψ(z)TzRϕ dz (5.13) I I = hR(ψ ∗ ϕ), ui = hF(ψ ∗ ϕ), Fui = (2π)d/2hψϕ, Fui . Sd Sd bb Sd

Here the second integral in (5.13) is understood as an Sd–valued Riemann integral on I; i.e., as the limit in Sd of Riemann sums such as

mn X d Sn(y) = ψ(zj,n)(Rϕ)(y + zj,n) vol(Qj,n), y ∈ R , j=1 where zj,n ∈ Qj,n, the rectangles Qj,n, j ∈ {1, . . . , mn}, subdivide I and maxj vol(Qj,n) tends to 0 as n → ∞. Clearly, Sn belongs to Sd. We omit the somewhat tedious, but elementary proof that Sn indeed converges in Sd. Hence, u can be taken out of the approximating Riemann sums by its linearity and out of 5.3. Tempered distributions 91 the limit by its continuity. This fact justiﬁes that we have interchanged u and the integral in (5.13). So far we have shown

hψ, F(ϕ ∗ u)i = (2π)d/2hψ, ϕui (5.14) b Sd b bb Sd ∞ d for all ψ ∈ Cc (R ). Arguing as in the ﬁrst part of the proof of Theorem 5.20, ∞ d one can show that Cc (R ) is dense in Sd, see Theorem 7.10 in [Rud91]. Since the Fourier transform is continuous on Sd by Lemma 5.7, the identity (5.14) is thus valid for all ψb with ψ ∈ Sd due to an approximation argument. We can now replace here ψb by ψ ∈ Sd using that F is bijective on Sd thanks to Proposition 5.10. So the ﬁrst part of e) is shown. For the second part, observe that

2 hψ, R(ϕ)R(u)iSd = hψRϕ, RuiSd = hR(ψ)R ϕ, uiSd = hRψ, ϕuiSd = hψ, R(ϕu)iSd , for all ψ ∈ Sd; i.e., R(ϕ)R(u) = R(ϕu). Employing also a), we then calculate

d/2 2 2 d/2 d/2 2 F(ϕb ∗ ub) = (2π) F (ϕ)F (u) = (2π) R(ϕu) = (2π) F (ϕu), which yields the second part of e) if we apply F −1. Bibliography

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[DuS57] N. Dunford and J.T. Schwartz, Linear Operators Part I: General Theory. John Wiley & Sons, 1957. [Koe89] T.W. K¨orner, Fourier Analysis. Cambridge University Press, 1989. [LiT96] J. Lindenstraus and L. Tzafriri, Classical Banach Spaces 1: Sequence spaces. Springer–Verlag, 1996.

[Lun09] A. Lunardi, Interpolation Theory. 2nd Edition. Edizioni della Normale, Pisa, 2009. [Rud87] W. Rudin, Real and Complex Analysis. McGraw–Hill, 1987.

[Rud91] W. Rudin, Functional Analysis. 2nd edition. McGraw–Hill, 1991. [ST15] R. Schnaubelt, Spectral Theory. Class Notes. Karlsruhe, 2015. [Wer05] D. Werner, Funktionalanalysis. 5te erweiterte Auﬂage. Springer–Verlag, 2005.

[Wer06] D. Werner, Einf¨uhrungin die h¨ohere Analysis. Springer–Verlag, 2006.