209: Honors Analysis in Rn Diﬀerentiation and Function Spaces

209: Honors Analysis in Rn Diﬀerentiation and function spaces

A vector (linear) space E over a ﬁeld K is a set endowed with two operations + : E × E → E and · : K × E → E that satisfy the familiar axioms [(1) : f + g = g + f], [(2) : ∃ 0, f + 0 = f], [(3) : (f + g) + h = f + (g + h)], [(4) : ∃ − f , f + (−f) = 0], [(5) : 1 · f = f], [(6) : ∀λ, µ ∈ K, (λµ) · f = λ · (µ · f)], [(6) : λ · (f + g) = λ · f + λ · g]. Deﬁnition 1 Let E be a real or complex vector space. A function k·k : E → R+ is a norm on E if   kfk = 0 ⇔ f = 0 kλfk = |λ|kfk, ∀ λ ∈ K (K = R or K = C  kf + gk ≤ kfk + kgk.

A vector space endowed with a norm is called a normed vector space. The norm defines a topology. An set U is open if for any f ∈ U there exists r > 0 so that B(f, r) ⊂ U where B(f, r) = {g | kf − gk < r}. The norm defines a distance d(f, g) = kf − gk. Definition 2 We say that E is a Banach space if it is a normed vector space that is complete, i.e. every Cauchy sequence converges.

Basic examples of Banach spaces are: Rn, C(K), Lp(dµ), for µ a complete measure on a measurable space (X, Σ), with special case `p(Z), 1 ≤ p ≤ ∞. For K ⊂ Rn a compact set

C(K) = {f : K → R | f continuous} with norm kfkC = supx∈K |f(x)|. Exercise 1 C(K) is a real Banach space. Tha analogous space of complex valued functions is a complex Banach space. Lp(dµ) is a real or complex Banach space.

1 Exercise 2 Let T : X → Y be a linear operator between two normed linear spaces. Linear means that T (f + g) = T f + T g and T (λf) = λT f. The following are equivalent: 1.T is continuous. 2.T is continuous at 0. 3. ∃C ≥ 0 such that kT (x)k ≤ Ckxk ho lds ∀ x ∈ X. Exercise 3 If T : X → Y is linear and continuous then kT xk kT xk sup = sup = sup kT xk x6==0 kxk {x;kxk≤1} kxk {x;kxk=1} holds. The common value is denoted kT k.

Exercise 4 If T : X → Y is linear and continuous and if S : Y → Z is linear and continuous, then ST : X → Z deﬁned by ST (x) = S(T (x)) is linear and continuous.

n m Exercise 5 Let E = R , F = R , ej = (δjk)j,k=1,...n and fα = (δαβ)α,β=1,...,m the canonical bases. Let T : E → F be a linear operator. Prove that it is continuous. Associate to T the matrix (tαi)α=1,...m,i=1,...n deﬁned by Pm β=1 tβifβ = T (ei). Prove that the matrix associated to the composition of two operators TS is the product of the corresponding matrices.

Proposition 1 Let L(E,F ) = {T : E → F | T linear and continuous}. Then T 7→ kT k is a norm on L(E,F ) and L(E,F ) is a Banach space.

Theorem 1 (Hahn-Banach, real). Let T : S → R be a linear map, deﬁned on a linear subspace of the real vector space X. Assume that there exists a real valued function p : X → R+ such that p(x + y) ≤ p(x) + p(y) and p(tx) = tp(x) for all x, y in X and all t ≥ 0. Assume that T (x) ≤ p(x) holds for all x ∈ S. Then, there exists a linear functional F : X → R that satisﬁes F (x) = T (x) for all x ∈ S and F (x) ≤ p(x) for all x ∈ X.

Theorem 2 (Hahn-Banach, complex). Let X be a complex vector space, S a linear subspace, p a nonnegative real valued function that satsiﬁes p(x + y) ≤ p(x)+p(y) and p(zx) = |z|p(x) for all x, y in X and all z ∈ C. Let T : S → C be a linear functional satisfying |T (x)| ≤ p(x) for all x ∈ S. Then there exists a linear functional F deﬁned on X such that F (x) = T (x) for allx ∈ S and |F (x)| ≤ p(x) for all x ∈ X

2 Proofs: Please see Kolmogorov and Fomin, Chapter 4, section 14.

Deﬁnition 3 A scalar product on a real (complex) vector space H is a bi- linear (sesquilinear) map H × H → C, denoted (f, g), that is, a map with the properties

(f, g) = (g, f), (λf1 + µf2, g) = λ(f1, g) + µ(f2, g), λ µ ∈ C (f, f) ≥ 0, (f, f) = 0 ⇔ f = 0.

The scalar product (hermitian product, dot product) deﬁnes a norm

kfk = p(f, f).

The Schwarz inequality holds

|(f, g)| ≤ kfkkgk.

Deﬁnition 4 We say that H is a Hilbert space if it is a real or complex vector space endowed with a scalar product such that H is complete, i.e. any Cauchy sequence converges.

A basic example of a Hilbert space is L2(dµ) = {f | R |f|2dµ < ∞} where µ is a complete measure on a measurable space (X, Σ). The scalar product is Z (f, g) = f(x)g(x)dµ X In any Hilbert space we have the parallelogram identity

kf + gk2 + kf − gk2 = 2(kfk2 + kgk2).

A set C ⊂ H is convex if, whenever f, g ∈ C then (1 − t)f + tg ∈ C holds for all t ∈ [0, 1].

Proposition 2 Let C ⊂ H be a closed convex subset of a Hilbert space. Then the minimum norm is attained: ∃c ∈ C such that

kck = inf kfk f∈C

3 Proof. Let d = inff∈C kfk. Let fn ∈ C be such that limn→∞ kfnk = d. By the parallelogram identity " 2# 2 1 2 2 fn + fn+p kfn − fn+pk = 4 kfnk + kfn+pk − 2 2 Suppose that > 0 is given, and we choose N such that, for n ≥ N we have 2 2 4 1 kfnk ≤ d + 4 . Then, because of C’s convexity, we have 2 (fn + fn+p) ∈ C, 1 2 2 hence, by the deﬁnition of d, k 2 (fn + fn+p)k ≥ d . It follows then that

kfn − fn+pk ≤ holds for all n ≥ N, p ≥ 0. Thus, the sequence fn is Cauchy. Because H is Hilbert, the sequence converges, fn → c, and because C is closed and fn ∈ C, the limit c ∈ C. Because the norm is continuous we obtain kck = d. . If S ⊂ H is any set, we deﬁne S⊥ = {f ∈ H | (f, s) = 0, ∀s ∈ S} Exercise 6 S⊥ is a closed, linear subspace of H. (That means that S is a closed set and that it is a vectorial subspace of H). Proposition 3 Let M ⊂ H be a closed linear sunbspace. Then for every h ∈ H there exists a unique m ∈ M such that h − m ∈ M ⊥. The identity khk2 = kmk2 + kh − mk2 holds. The map P : H → M that assigns P h = m is linear, continuous and onto M. P is idempotent, i.e. composition with itself does not change it, P 2 = P . P is selfadjoint, i.e. (P f, g) = (f, P g) holds. The map P is called the orthogonal projector onto M. The proof follows from the previous proposition by observing that h + M is closed and convex. Then, let q ∈ x + M be of minimum norm. We can write q = x − m with m ∈ M (because M is a linear space). Assume by contradiction that there exists n ∈ M such that (q, n) 6= 0. Without loss of generality we may assume that (q, n) > 0. (Indeed, this is achieved by setting n0 = zn with z = (q, n).) Then a simple calculation shows that kqk2 > kq − tnk2 can be achieved by taking 0 < t small enough. This contradicts the minimality.

4 Theorem 3 (Riesz) Let L : H → C be a continuous linear map from the Hilbert space H. Then there exists a unique f ∈ H such that Lh = (h, f) holds for all h ∈ H.

Proof. Consider kerL = {h | Lh = 0} and assume it is not the entire H. (If it is then f = 0 will do the job). Then let g ∈ (kerL)⊥. The existence of g 6= 0 is guaranteed by the orthogonal projector theorem, by taking h∈ / kerL and setting g = h − P h. Then, for any element f ∈ H we note that h = L(g)f −(Lf)g ∈ kerL, and thus (h, g) = 0 holds. This implies −2 that Lf = (f, ge) holds for all f with ge = [kgk L(g)]g. . A family {ea}a∈A of elements in a Hilbert space is said to be orthonormal if (ea, eb) = δab where δab = 1 if a = b and δab = 0 if a 6= b. We deﬁne for any f ∈ H, fb(a) = (f, ea). Let F ⊂ A be a ﬁnite set. Consider M to be the linear subspace generated by {ea}a∈F . Then the orthogonal projector onto M is P given by P f = a∈F fb(a)ea. Note that one can think of P f as a solution to a variational problem: Find the element m of M that best approximates f, i.e. that minimizes the distance kf − mk. An orthonormal set {ea}A is said to be complete if it is maximal (under inclusion).

Theorem 4 TFAE

1. {ea}A is maximal orthonormal 2. {ea}A is orthonormal and the linear space it generates is dense. 2 P 2 3. kfk = a∈A |fb(a)|

The sum in 3 is the integral with respect to counting measure H0. i.e. the supremum of all ﬁnite sums. Complete orthonormal sets are called orthonormal bases. The cardinality of the base is an invariant of the Hilbert space. Hilbert spaces with countable bases are called separable Hilbert spaces.

2 Diﬀerentiation

Deﬁnition 5 Let U ⊂ E be an open set in the Banach space E and let f : U → F where F is another Banach space. We say that f is Fr´echet

5 diﬀerentiable at x0 ∈ U if there exists a bounded linear operator A : E → F such that, for every > 0 there exists δ > 0 such that

kf(x) − f(x0) − A(x − x0)k ≤ kx − x0k holds for all x ∈ U with kx − x0k ≤ δ. The map A is called the derivative of n m f at at x0. When E = R and F = R then this is the usual deﬁnition of derivative. We will use the notation

A = (Df)(x0)

The chain rule:

Proposition 4 Let E,F,G be Banach spaces, U open in E, x0 ∈ U, V open in F , f(x0) ∈ V , and f : U ⊂ E → F , g : V → G. Assume that f is Fréchet differentiable at x0, g is continuous on V and and Fréchetdifferentiable at −1 f(x0). Then h = g ◦ f defined on U ∩ g (V ) is is differentiable at x0 and Dh(x0) = (Dg(f(x0)))(Df)(x0)

Definition 6 The function f : U → F is Gateaux differentiable at x0 ∈ U if, for any v ∈ E, with kvk = 1, the map defined on a neighborhood of 0 in R

t 7→ f(x0 + tv) is diﬀerentiable at t = 0, i.e.

f(x0 + tv) − f(x0) (Dvf)(x0) = lim t→0, t6=0 t exists. The derivative is called the directional derivative of f in the direction n v at x0. When E = R , v = ei = (δij)j=1,...n then the directional derivative is called the partial derivative. We will use the notation

f(x + tei) − f(x) (∂if)(x) = lim t→0, t6=0 t

Proposition 5 If f : U → F is Fréchetdifferentiable at x0 then it is Gateaux differentiable and

(Dvf)(x0) = Df(x0)v

6 3 Function Spaces

n A multiindex α = (α1, . . . , αn) is an element of Z with nonnegative entries. n Its length is |α| = α1 + . . . αn. The Euclidean norm of an element x ∈ R is p 2 2 n denoted the same way, |x| = x1 + ··· + xn. An open ball in R of radius Q r is B(x, r) = {y; |x − y| < r}. The factorial of a multiindex is α! = i αi!. The derivatives are denoted ∂α: ∂|α|f ∂αf = α1 αn ∂x1 . . . ∂xn ∂f The vector ∇f = (∂1f, . . . , ∂nf) is the gradient, and ∂if = . ∂xi If U is an open subset of Rn and m is nonnegative, we say that s function f : U → R is in Cm(U) if f is m times differentiable and ∂αf are continuous for |α| ≤ m. When m = 0 we sometimes write C(U). The support of a continuous function is the complement of the largest open set where the function is identically zero. The subset of Cm(U) formed with compactly supported m functions is denoted C0 (U). If a function f has continuous derivatives of any order we say that f is infinitely differentiable, f ∈ C∞(U). If the function ∞ has compact support, we write f ∈ C0 (U). If 1 ≤ p < ∞ we define

1 Z p p kfkp = |f| dx U for any measurable function in U. If the integral is ﬁnite we say that f ∈ Lp(U). Here dx = λ is Lebesgue measure. For p = ∞ we deﬁne

kfk∞ = inf{c ≥ 0 ; |f| ≤ c λ − a.e. in U} Sometimes we write Lp short for Lp(U). The Lp spaces can be deﬁned starting from any complete measure µ. Theorem 5 Let 1 ≤ p < ∞. Let f be a measurable function in U. Assume that there exists a constant C such that Z

fφdx ≤ Ckφkp U

∞ p0 holds for all φ ∈ C0 (U). Then f ∈ L (U) and

kfkp0 ≤ C.

7 Remark 1 Note carefully that p = ∞ is not allowed in the theorem. The theorem together with H¨older’sinequality imply Z

kfkp0 = sup φfdx . kφkp≤1 U

∞ p where the supremum is taken over all C0 (U) functions with L norm less or equal to 1. The range of p0 is (1, ∞].

p n ∞ n Deﬁnition 7 If f ∈ L (R ) and φ ∈ C0 (R ) we deﬁne the convolution φ∗f by Z (φ ∗ f)(x) = φ(y)f(x − y)dy

∞ n Proposition 6 For φ, ψ, χ ∈ C0 (R ): (i) φ ∗ ψ = ψ ∗ φ (ii)(φ ∗ ψ) ∗ χ = φ ∗ (ψ ∗ χ) (iii) ∂α(φ ∗ ψ) = (∂αφ) ∗ ψ = φ ∗ (∂αψ) (iv.) supp(φ ∗ ψ) ⊂ suppφ + suppψ.

Here suppφ means the support of the function φ and A + B = {x; x = a + b, a ∈ A, b ∈ B}.

p ∞ n Proposition 7 (Young) Let f ∈ L , 1 ≤ p ≤ ∞ and φ ∈ C0 (R ). Then

kφ ∗ fkLp ≤ kφk1kfkp holds

p Proposition 8 Let 1 ≤ p < ∞. Then C0(U) is dense in L (U), that is for p any f ∈ L (U) there exists a sequence of functions φj ∈ C0(U) that converges in Lp to f.

p Remark 2 Note that p = ∞ is not allowed. Note also that C0(U) ⊂ L (U).

Exercise 7 Give an example of a function f ∈ L∞(R) such that there is no sequence of compactly supported continuous functions that converges to it in L∞. (Hint: what can you say about the uniform limit of a sequence of continuous functions?)

8 n Deﬁnition 8 Let h ∈ R . We denote the translate of f by h by τh(f)

τh(f)(x) = f(x − h).

We denote the ﬁnite diﬀerence by δh

δh(f) = f(x − h) − f(x)

Exercise 8 Let f ∈ Lp(Rn), 1 ≤ p < ∞. Let > 0. Then there exists δ > 0 so that kδhfkp ≤ holds for all h ∈ Rn with |h| ≤ δ. (Hint: approximate ﬁrst f in Lp by a continuous function with compact support. Then use the uniform continuity of that function.)

Exercise 9 Let f ∈ Lp and g ∈ Lp0 with p, p0 conjugate exponents. Prove that f ∗ g is uniformly continuous.

Proposition 9 Let φ ∈ L1(Rn) and assume that Z φdx = 1

Consider for > 0 x φ (x) = −nφ Let f ∈ Lp, 1 ≤ p < ∞. Then

lim kf − (φ ∗ f) kp = 0 →0

Idea of proof. Because R φdx = 1 we have Z φ ∗ f − f = φ(z)δz(f)dz

Then Z kφ ∗ f − fkp ≤ |φ(z)|kδzfkpdz holds. The proof is completed by using Lebesgue dominated convergence.

9 ∞ n R Deﬁnition 9 Take φ ∈ C0 (R ), φ ≥ 0, φdx = 1, suppφ ⊂ {x; |x| < 1}. The family x φ = −nφ is called a standard molliﬁer.

∞ n p n Proposition 10 Assume that φ ∈ C0 (R ), f ∈ L (R ), 1 ≤ p < ∞. Then φ ∗ f ∈ C∞(Rn) and ∂α (φ ∗ f) = (∂αφ) ∗ f

∞ p Proposition 11 C0 (U) is dense in L (U), 1 ≤ p < ∞.

Idea of proof. By Proposition 13 it is enough to approximate continuous functions with compact support. Convolution with a standard molliﬁer will do the trick.

Proposition 12 Let K ⊂ U be a compact included in an open set in Rn. ∞ Then there exists ψ ∈ C0 (U) such that 0 ≤ ψ ≤ 1 and ψ(x) = 1 holds for all x ∈ K.

Idea of proof. Take > 0 small enough so that dist(K, ∂U) > 3. (Any point in K has a ball centered at x and of radius 3 included in U). K = {x : ∃k ∈ K, |k − x| ≤ }. Take φ a standard molliﬁer and put

ψ = φ ∗ 1K .

Note that if x ∈ K and z ∈ suppφ then x − z ∈ K and therefore ψ(x) = R φ(z)dz = 1.

Proposition 13 (Partition of unity). Let U = ∪a∈AUa be a union of open n ∞ n sets in R . There exists a countable family ψj ∈ C0 (R ), j ∈ N, such that the following properties hold.

(i) ∀j ∈ N, ∃a ∈ A, supp ψj ⊂ Ua. (ii) ∀K ⊂ U, K compact, {j; suppψj ∩ K 6= ∅} is ﬁnite. P (iii) ψj(x) = 1. j∈N

10 (1) (1) (2) Idea of proof. First construct a family Bj ⊂ Bj ⊂ Bj of open balls (1) (2) such that U = ∪jBj and ∀j ∈ N, ∃a ∈ A, Bj ⊂ Ua.(B is the closure of (2) the ball B.) This is done, for instance, by choosing as the family Bj the collection of all open balls with rational centers (all coordinates) and rational (1) radii, so that their closure is contained in some Ua, and for Bj the family (2) of balls with same centers as the balls in Bj but half the radius. If x ∈ U, n ∃a ∈ A, ∃r > 0,B(x, 3r) ⊂ Ua. Then ∃ y ∈ Q , ∃q ∈ Q so that x ∈ B(y, q) (1) and 0 < q < r. Then B(y, 2q) ⊂ Ua which puts B(y, q) in the list Bj . (1) ∞ (2) Thus ∪Bj ⊃ U. By the previous proposition there exists φj ∈ C0 (Bj ) (1) Qj so that φj ≡ 1 on Bj . Deﬁne now ψ1 = φ1 and ψj+1 = φj+1 i=1(1 − φi) (2) for j ≥ 1. Note that supp ψj ⊂ Bj and also, by induction, ψ1 + ··· + ψj = Qj 1 − i=1(1 − φi). If K ⊂⊂ U ( a notation for K compact, included in (1) U), then ﬁnitely many Bj cover K. Let J be the largest index of in the J (1) cover, .i.e K ⊂ ∪j=1Bj . If j ≥ J + 1 then ψj(x) = 0 at x ∈ K and ψ1(x) + ··· + ψJ (x) = 1.

4 Fourier series

Definition 10 Let f : [0, 2π] → C, f ∈ L1([0, 2π]). For j ∈ Z we define the jth Fourier coefficient of f by 1 Z 2π fˆ(j) = f(θ)e−ijθdθ. 2π 0 Definition 11 If f, g ∈ L2([0, 2π]) we define their scalar product and norm as Z 2π 1 2 hf, gi = f(θ)g(θ)dθ, kfk 2 = hf, fi. L (T) 2π 0 Let n ≥ 0 and set n X ˆ ijθ (Pnf)(θ) = f(j)e . j=−n

Note that Pnf is a trigonometric polynomial. Using the deﬁnition of the Fourier coeﬃcients we see that 1 Z 2π (Pnf)(θ) = Dn(θ − t)f(t)dt 2π 0

11 with Dn(θ) the Dirichlet kernel

n 1 X sin n + θ D (θ) = eijθ = 2 . n sin θ j=−n 2 Exercise 10 Verify the last equality above using trigonometric identities.

Theorem 6 Let f ∈ L2([0, 2π]). Then

2 2 (i) kPnfkL (T) ≤ kfkL (T), 2 (ii) limn→∞ kf − PnfkL (T) = 0, 2 2 P (iii) kfk 2 = f(j) . L (T) j∈Z b

Consider now the average of n + 1 operators Pj, 1 S = (P + ··· + P ) n n + 1 0 n and observe that 1 Z 2π (Snf)(θ) = Fn(θ − t)f(t)dt 2π 0 where Fn is the Fej´erkernel

n !2 X |j| 1 sin n+1 θ F (θ) = 1 − eijθ = 2 . n n + 1 n + 1 sin θ j=−n 2

1 Pn Exercise 11 Verify that the average n+1 j=0 Pj is given by the integral above and verify the identities defining the Fejérkernel. The following are properties of the Fejér kernel:

(i) Fn ≥ 0,Fn is continuous, 2π − periodic, 1 R 2π (ii) 2π 0 Fn(θ)dθ = 1, Cδ (iii) ∀δ > 0, ∃Cδ > 0, so that supθ∈[δ,2π−δ] |Fn(θ)| ≤ n+1 . Theorem 7 Let f be a continuous, complex valued function so that f(0) = f(2π). Then Snf converge uniformly to f, that is

lim sup |f(θ) − (Snf)(θ)| = 0. n→∞ θ∈[0,2π]

12 Remark 3 This theorem implies that any continuous, periodic function can be approximated uniformly by trigonometric polynomials.

Remark 4 The same statement is not true for Pn. Functional analysis ar- guments show that there exist many continuous functions f for which the uniform convergence of Pnf to f fails. In fact, even pointwise convergence fails: one can specify any point θ0, and ﬁnd (many) functions such that, for each such f, the sequence of numbers |Pnf(θ0)| is not even bounded, as n → ∞.

Idea of proof. First, in view of periodicity, we have

1 Z 2π 1 Z 2π (Snf)(θ) = Fn(θ − t)f(t)dt = Fn(t)f(θ − t)dt. 2π 0 2π 0 Then we write 1 Z 2π (Snf − f)(θ) = Fn(t)(δtf(θ)) dt 2π 0 with (δtf)(θ) = f(θ − t) − f(θ), the ﬁnite diﬀerence. This is possible because of property (ii) of Fn. Now pick > 0. Because f is (uniformly) continuous and 2π-periodic, there exists δ > 0 so that, if t ∈ Jδ = [0, δ] ∪ [2π − δ, 2π] then |δtf(θ)| ≤ 2 holds for all θ ∈ [0, 2π]. Because Fn is nonnegative then

Z 2π−δ 1 2CδM Fn(t) | δt(f)(θ)| dt ≤ ≤ 2π δ n + 1 2 holds for all θ. (Above, M = supt∈[0,2π] |f(t)|).

13 Definition 12 Let f : U ⊂ Rn → C be a continuous bounded function defined in the open set U. Let 0 < α < 1. We say that is Höldercontinuous of order α, f ∈ C0, α(U) if

|f(x) − f(y)| [f]0,α = sup α < ∞. x6=y |x − y|

The same deﬁnition is used if U is any set. When the function f is periodic in each direction with period 2π we write f ∈ C0,α(Tn). The quantity

kfk0,α = kfk∞ + [f]0,α is the H¨oldernorm.

Proposition 14 Let f : [0, 2π] → C. Then

df dk−1f (i) If f, dθ ,... dθk−1 , are continuous and periodic, k and if d f ∈ L1([0, 2π]), then f(j) ≤ C|j|−k holds. dθk b (ii) If f ∈ C0,α([0, 2π])is periodic, then |fb(j)| ≤ C|j|−α holds.

1 Proposition 15 (Bernstein) Let α > 2 . There exists a constant Cα > 0 so that, for any f ∈ C0,α(T),

X fb(j) ≤ Cαkfk0,α j∈Z holds.

Idea of proof. Let m be a nonnegative integer, and consider the annulus

m m+1 Am = {j ∈ Z; 2 ≤ |j| < 2 }.

2π −m 2π 4π Take h = 3 2 and note that, for any j ∈ Am we have |jh| ∈ 3 , 3 , and consequently √ −ijh e − 1 ≥ 3. Therefore 2 2 X 1 X −ijh 2 fb(j) ≤ e − 1 fb(j) 3 j∈Am j∈Am

14 But −ijh (e − 1)fb(j) = δdhf(j) and consequently, from Parseval,

2 X 1 2 fb(j) ≤ kδhfk 2 . 3 L (T) j∈Am

Now, because f ∈ C0,α(T) we have

2 2α 2 kδ fk 2 ≤ C|h| kfk h L (T) 0,α Consequently, 2 X 2 −2αm fb(j) ≤ Ckfk0,α2 j∈Am holds (in many proofs from now on C might change harmlessly from line to line).Using Schwartz:

X p −αm ( 1 −α)m 2 fb(j) ≤ C #{j ∈ Am}2 kfk0,α = Ckfk0,α2 . j∈Am

1 Because α > 2 the right hand side is summable in m, and that implies that the left hand side is summable.

Deﬁnition 13 A sequence of numbers {λn} is said to be lacunary if there exists a constant q > 1 such that λn+1 > qλn. A trigonometric series is lacunary if all the frequencies appearing in it are of the form ±λn, with {λn} lacunary.

Lemma 1 Let f ∈ L1(T) and assume that fb(j) = 0 for all j such that 1 ≤ |n − j| ≤ 2N. Assume also that f(t) = O(t) ast t → 0. Then,   4 −1 −1 −2 1 |fb(n)| ≤ 2π N sup t f(t) + N kfkL (T) 1 |t|≤N − 4

Proof.. We use the condition fb(j) = 0 as follows. If gN is a trigonometric polynomial of degree 2N that satisﬁes gcN (0) = 1, then Z 2π 1 −int fb(n) = e f(t)gN (t)dt. 2π 0

15 −2 2 We choose gN = kFN k 2 F (t) where FN is the Fej´erkernel of order N. L (T) N We note that π2 0 ≤ F (t) ≤ N (n + 1)t2 follows from the formula for the kernel, and also 2 2 X |j| N kF k 2 = 1 − ≥ N L (T) N + 1 2 |j|≤N so we deduce 4 −3 −4 0 ≤ gN (t) ≤ 2π N t . Now we break the integral into three pieces:

fb(n) ≤ I1 + I2 + I3.

1 R I1 = 2π |t|≤N −1 |f(t)|gN (t)dt ≤ −1 −1 1 R 2π N sup|t|≤N −1 |t f(t)| 2π 0 gN (t)dt −1 −1 = N sup|t|≤N −1 |t f(t)| The second piece is

1 R I2 = − 1 |f(t)|gN (t)dt ≤ 2π N −1≤|t|≤N 4 − 1 3 −3 −1 R N 4 −3 π N sup 1 |t f(t)| −1 t dt |t|≤N − 4 N 3 −1 −1 ≤ π N sup 1 |t f(t)| |t|≤N − 4 Finally, the third piece is

1 R I3 = − 1 |f(t)|gN (t)dt ≤ 2π |t|≥N 4 3 −2 1 π N kfkL (T) and this concludes the proof. A corollary is

Corollary 1 Let {λn} be a a lacunary sequence and assume that X f ∼ an cos(λnt)

1 −1 is in L (T). Assume that f is diﬀerentiable at one point. Then an = o(λn ).

16 Proof. WLOG we assume that f is differentiable at t = 0. Replacing f by f − f(0) cos(t) − f 0(0) sin(t), we may assume that f(t) = o(t) as t → 0. The lacunarity condition implies that there exists a positive constant c such that for all integers j such that 1 ≤ |λn − j| ≤ cλn we have fb(j) = 0. We may apply then the lemma with λn in place of n in the lemma, and 2N = cλn. −1 We obtain that an = 2fb(λn) = o(λn ). Corollary 2 The Weierstrass function P 2−n cos(2nt) is continuous and nowhere differentiable. P Exercise 12 Let f ∼ an cos(λnt) is Höldercontinuous with exponent 0 < −α α < 1 at t = t0 and assume that {λn} is lacunary. Deduce that an = o(λn ). Prove that this implies that f is everywhere Höldercontinuous with exponent α.

5 Fourier Transform

Deﬁnition 14 Let f ∈ L1(Rn) and ξ ∈ Rn. The Fourier transform of f at ξ is Z F(f)(ξ) = fˆ(ξ) = e−iξ·xf(x)dx.

Pn n Here ξ · x = j=1 ξjxj is scalar product in R . The function f might as well be taken complex valued and f ∈ L1 means that |f| is integrable, or, equivalently, both the real part Re(f) and the imaginary part Im(f) are in L1.

∞ n Proposition 16 Let φ, ψ ∈ C0 (R ).

(i) F(φ ∗ ψ) = φˆψˆ −ih·ξ ˆ (ii) F(τhφ)(ξ) = e φ(ξ) ih·x ˆ (iii) F(e φ) = τhφ ˆ (iv) F(∂jφ)(ξ) = iξjφ ˆ (v) F(ixjφ) = −∂j(φ) (vi) R φψdxˆ = R φψdxˆ

17 Proposition 17 Let f ∈ L1(Rn). Then

ˆ ∞ n ˆ (i) f ∈ L (R ) and kfk∞ ≤ kfk1

ˆ (ii) lim|ξ|→∞ f(ξ) = 0 (iii) fˆ is uniformly continuous.

Point (ii) is the Riemann-Lebesgue lemma. ∞ n Idea of proof. For (iii) let > 0. Take φ ∈ C0 (R ) so that kf − φk1 ≤ 3 . ˆ ˆ Then use |f(ξ) − φ(ξ)| ≤ 3 together with the fact that Z φˆ(ξ + h) − φˆ(ξ) = e−ih·x − 1 e−iξ·xφ(x)dx which implies that

ˆ ˆ −ih·x sup φ(ξ + h) − φ(ξ) ≤ sup e − 1 kφk1. ξ x∈supp φ

Choosing |h| ≤ δ with δ small enough we have sup φˆ(ξ + h) − φˆ(ξ) ≤ . ξ 3

For (ii) Take > 0 use the same φ. Note that, in view of the previous proposition 1 |φˆ(ξ)| ≤ k∇φk |ξ| 1 ˆ Thus, for |ξ| large enough |φ(ξ)| ≤ 2 .

Exercise 13 Let Φ(x) = e−|x|2 . Then φ = Φˆ is

n 2 π 2 − |ξ| φ(ξ) = e 4 and φ, Φ ≥ 0, φ, Φ ∈ L1 ∩ L∞ and Φ(0) = 1, R φdξ = (2π)n.

Lemma 2 Assume φ ≥ 0, φ = Φˆ, Φ(0) = 1 φ, Φ ∈ L1 ∩ L∞ and Z (2π)−n φdξ = 1.

18 Then, for any f ∈ L1 Z Z eix·ξfˆ(ξ)Φ(ξ)dξ = f(y)φ(y − x)dy.

Moreover, the function Z −n ix·ξ ˆ f(x) = (2π) e f(ξ)Φ(ξ)dξ converges in L1 to f.

Idea of proof. After verifying the identity and rescaling we note that −n −x f(x) = f ∗ ψ where ψ(x) = (2π) φ . Because ψ has integral equal to one and is positive, the convergence follows from Proposition 14.

1 1 Exercise 14 Let fj be a sequence of L functions that converges to f in L . Then there exists a subsequence of fj that converges almost everywhere to f.

Theorem 8 (Fourier Inversion). Assume that f ∈ L1(Rn) and fˆ ∈ L1(Rn). Then Z f(x) = (2π)−n eix·ξfˆ(ξ)dξ holds dx-a.e.

Idea of proof. Pick Φ, φ like in the lemma. By the Lemma, f converges to f in L1, and hence, by the exercise, a subsequence converges to f(x) almost everywhere. On the other hand, by Lebesgue dominated convergence f(x) converges (for each ﬁxed x) to (2π)−n R eix·ξfˆ(ξ)dξ.

Theorem 9 Let f ∈ L1 ∩ L2. Then Z Z |f(x)|2dx = (2π)−n |fˆ(ξ)|2dξ.

Idea of proof. WLOG f is smooth, compactly supported. Consider f˜(x) = f(−x) where z is the complex conjugate. Then let g = f ∗f˜. By construction, gˆ(ξ) = |fˆ(ξ)|2 ≥ 0 and g(0) = R |f(x)|2dx. We may apply the Fourier inversion formula.