Brandon Behring

Real Variables # 9: Hilbert Spaces

Exercise 3

||f + g||2 + ||f − g||2 = 2(||f||2 + ||g||2)

Proof: We have

||f + g||2 = (f + g, f + g) = (f, f) + (g, f) + (f, g) + (g, g) = ||f||2 + (f, g) + (f, g) + ||g||2 = ||f||2 + ||g||2 + 2<(f, g)

and switching g with −g we have

||f − g||2 = ||f||2 + ||g||2 − 2<(f, g)

thus

||f + g||2 + ||f − g||2 = 2(||f||2 + ||g||2).

Exercise 4a l2 is complete. Proof: Let us assume that C is complete. Take a Cauchy sequence

(k) ∞ (k) (k) 2 xk = (xi )i=1 = (x1 , x2 , ···) ∈ l ,

then for any > 0 there exists N such that for all m, n > N

∞ X (m) (n) 2 2 ||xm − xn||l2 = |xi − xi | < . i=0

(m) (n) 2 so for all i ∈ N we must have |xi − xi | < . Choosing a speciﬁc i, we (k) have that the sequence of the i-th entries xi is a Cauchy sequence in C and thus must converge to some xi ∈ C by the completeness of C. Doing this for all entries we construct a new sequence

x = (x1, x2, ··· ...)

1 We must now verify x ∈ l2. We have

2 ||xm − x||l2 = lim ||xm − xn||l2 < . n→∞

2 2 thus xm − x ∈ l . Since l is a vector space (Minkowski) we have x = 2 (x − xm) + xm ∈ l . Exercise 4b l2 is separable. Proof: Just as we assumed C is complete, we assume the complex ratio- nals are dense in C. We must show that any element of l2 is arbitrarily close to an element of the form

α = (α1, α2, ··· , αn, 0, 0,...)

where αi are of the form αk = ak + ibk with ak, bk ∈ Q. 2 Take an arbitrary element x = (x1, x2, x3,...) ∈ l . Since this converges in the norm, the tail end of must converge. That is, for suﬃciently large N we must have for all > 0 that

∞ X |x |2 < i 2 k=N+1

Additionally, since the Q is dense in R, we must have that for any xi there is an αi arbitrary close to it. Therefore, we can ﬁnd > 0 such that

N X |x − α |2 < . i i 2 k=1

Finally,

2 2 ||x − α||l2 = ||(x1 − α1, x2 − α2, ··· , xn − αn, xn+1, ···)|| N ∞ X 2 X 2 = |xi − αi| + |xi| k=1 k=N+1 = + = . 2 2

Exercise 5 Establish the following relations L2(Rd) and L1(Rd):

(a) Neither the inclusion L2(Rd) ⊂ L1(Rd) nor the inclusion L1(Rd) ⊂ L2(Rd) is valid. Proof: Let us take the case d = 1. The general dimensions work exactly the same but obscure what is actually happening. For big p,

2 Lp behaves badly for large values of f. While for small p, Lp behaves badly for small values of f. Let us consider

( √1 for |x| ≤ 1 f(x) = |x| 0 for |x| > 1.

Then

Z ||f||1 = |f|dx

Z +1 1 = p dx −1 |x| +1p = 2 · |−1 |x|dx = 4 < ∞ so f ∈ L1(R); however, Z 2 2 ||f||2 = |f| dx Z +1 1 = dx −1 |x| 1 = 2 · |0 log |x|dx = ∞ so f∈ / L2(R). Next, take 0 for |x| ≤ 1 g(x) = |x|−3/4 for |x| > 1. Then

Z ||g||1 = |g|dx Z ∞ = 2 |x|−3/4dx 1 ∞ 1/4 = 6 · |1 |x| dx = ∞ so f∈ / L1(R); however, Z 2 2 ||g||2 = |g| dx

3 Z ∞ = 2 |x|−3/2dx 1 ∞ −1/2 = −4 · |1 |x| dx = 4 < ∞

so f ∈ L2(R). (b) Let f ∈ L2 be supported on a set of ﬁnite measure E. Then

1/2 ||f||1 ≤ m(E) ||f||2.

Proof: Applying the Cauchy-Schwarz inequality we have

Z ||f|1 = |f| E Z = |f|χEdx R = (|f|, χE)

≤ ||f||2||χE||2 Z 1/2 2 = ||f||2 |χE| dx R 1/2 = m(E) ||f||2

(c) If f is bounded and f ∈ L1 then f ∈ L2 with

1/2 1/2 ||f||2 ≤ M ||f||1 .

Proof: Using |f(x)| < M we have

Z 2 2 ||f|2 = |f| dx ZR ≤ M|f|dx R = M||f||1

Thus

1/2 1/2 ||f||2 ≤ M ||f||1 .

4 Exercise 6 The following are dense subspaces of L2(Rd). (a) The simple functions Proof: It is suﬃcient to consider f ≥ 0. The simple approxima- tion theorem (Chapter 1, Theorem 4.1) gives us a sequence of non- ∞ negative simple functions {φk}k=1 that converges to f. Because we have 2 2 2 2 3 2 |φn − f | ≤ 2 (|φn| + |f| ) ≤ 2 |f| 2 and by deﬁnition of L2 that |f| is integrable, the Dominated Con- vergence theorem tells us that

2 ||φn − f||L (R) → 0. It should be clear how to apply this to Lp(R). (b) The step functions. Proof: A simple function is a linear combination of characteristic Pn functions of sets on a ﬁnite measure, φ(x) = i=1 aiχEi . A step function is nothing more than a ﬁnite combination of characteristic function of sets over cubes. However, any measurable set of ﬁnite measure can be approximated as close as we would like by a union of cubes.This line of reasoning is worked out in Chapter 1 Theorem 4.3 which gives us an almost disjoint family of rectangles {Rj} with M 2 m(E∆∪j=1 Rj) < . We now have that a simple function χE can be P approximated by ψ = j xRj and diﬀer at most on a set of measure 2 and thus

M 1/2 ||χE − ψ||L2 = m(E∆ ∪j=1 Rj) = . (c) The continuous functions of compact support. Proof: Any step function can be approximated in LP by a continu- ous function. This follows by the same argument in the proof of the case in L1 ( Theorem 2.2.4).

∞ 2 d Exercise 7 Let {φk}k=1 is an orthonormal basis for L (R ). Prove that the collection {φk,j}1≤,k,j<∞ with φk,j(x, y) = φk(x)φj(y) is an orthonormal basis of L2(R2 × Rd). Proof: Assuming (φi, φj) = δij, we have using Fubini’s theorem

Z Z (φij, φmn) = φij(x, y)φmn(x, y)dxdy ZR ZR = φi(x)φj(y)φm(x)φn(y)dxdy ZR R Z = φi(x)φm(x)dx φj(y)φn(y)dxdy R R = (φi, φm)(φj, φn) = δimδjn

5 so {φk,j}1≤,k,j<∞ is an orthonormal series. By Theorem 2.3 we have 4 equivalent ways to verify an orthornormal set is in fact a basis We will choose option (iv), which our context amounts to {φk,j}1≤,k,j<∞ is a basis 2 2 d 2 2 d for L (R × R ) if F (x, y) ∈ L (R × R ) and (F, φk,j) = 0 for all k, j, then F = 0.

Let us assume we have (F, ψk,j) = 0 for all k, j

Z Z (F, φk,j) = F (x, y)φk,k(x, y)dxdy R R Z Z = F (x, y)φk(y)dy φk(x)dx R R Z = Fj(x)φk(x) R = (Fj, φk) = 0

where Z Fj(x) = (F (x, y), φj) = F (x, y)φj(y)dy. R ∞ 2 d Because {φk}k=1 is an orthonormal basis for L (R ), we have that Fj(x) ∈ 2 d L (R ) and (Fj, φk) = 0 for all k implies Fj(x) = 0 for all j. However, we 2 2 d now have that F (x, y) ∈ L (R × R ) and (F (x, y), φj) = 0. Since φj is a basis, this implies F (x, y) = 0. Exercise 8 Let η(t) be a ﬁxed strictly positive continuous function on [a, b]. 2 Deﬁne Hη = L ([a, b], η) to be the space of all measurable functions f on [a, b] such that

Z b |f(t)|2η(t)dt < η. a

Deﬁne the inner product on Hη by Z b (f, g)η = f(t)g(t)η(t)dt. a

(a1) Show that Hη is a Hilbert space. Let us verify the conditions of a Hilbert space.

(i) Hη is a vector space over C. Take f, g ∈ Hη, then

Z 2 ||af + bg||Hη = |af + bg| η(t)dt Z Z ≤ 4|a| |f|2η(t)dt + 4|b| |g|2η(t)dt

= 4|a|||f||Hη + 4|b|||g||Hη < ∞.

6 so af + bg ∈ Hη.

(ii) Hη is equipped with an inner product (f, g)η, so that f 7→ (f, g) is a linear on Hη for every ﬁxed g ∈ Hη,

Z b (cf1 + df2, g)η = [cf1(t) + df2(t)] g(t)η(t)dt a Z b Z b = c f1(t)g(t)η(t)dt + d f2(t)g(t)η(t)dt a a = c(f1, g)η + d(f2, g)η

and that (f, g)η = (g, f)η because η(t) is real

Z b (g, f)η(f, g)η = g(t)f(t)η(t)dt a Z b = f(t)g(t)η(t)dt = (f, g)η. a

Finally,

Z b (f, f) = |f|2η(t)dt ≥ 0 a because η(t) (and |f|) is positive. (iii) We have

Z b 2 2 ||f||η = (f, f) = |f| η(t)dt = 0 a if and only if |f|2η(t) = 0. Since η(t) > 0, we must have f = 0. (iv) This a redundant condition as it follows from (i) and (ii). (v) We will examine this in part (b) as we examine the limitations on function η. (vi) For separability follow the same strategy as for L2. (a2) Show that the mapping U : f 7→ η1/2f gives a unitary correspon- 2 dence between Hη 7→ L ([a, b]) (i) U is linear, that is

U(αf + βg) = αfη1/2 + βgη1/2 = αU(f) + βU(g).

(ii) U is a bijection as we have the unique inverse map U −1 : f 7→ −1/2 2 η f from L ([a, b]) 7→ Hη as η is strictly positive.

7 (iii) We have

2 ||Uf||L2([a,b]) = (Uf, Uf) Z b = η(t)1/2f(t)η(t)1/2f(t)dt a Z b = f(t)f(t)η(t)dt a = (f, f) = ||f||2 η Hη

(b) Let η > 0 be a measurable function deﬁned on [a, b], then Hη is complete. Proof: We assume L2[a, b] is complete as it follows from Theorem 1.2. Take a Cauchy sequence fn in Hη then for any > 0 for m, n > N we have

Z b 2 ||fm − fn||η = |fm(t) − fn(t)| η(t)dt a Z b √ √ 2 = | ηfm(t) − ηfn(t)| < a

√ 2 This gives a cauchy sequence gn = ηfn ∈ L [a, b]. Since this a complete vector space we have g ∈ L2[a, b] such that if we deﬁne √ 2 f = g/ η we have gn → g in the L [a, b] norm, thus Z b Z b 2 2 2 ||gn − g||L2([a,b])| = |gn(t) − g(t)| dt = |fn(t) − f(t)| ηdt → 0, a a However, from this relation we also see that

2 ||gn − g||L2([a,b])| = ||fn − f||Hη . g √ So fn → f = η in the Hη norm. Further Hη is complete because g ∈ L2[a, b] tells us

g ||f|| = ||√ || Hη η Hη Z b |g(t)|2 = p η(t)dt a η(t) Z b 2 = |g(t)| dt = ||g||L2[a,b] < ∞ a

so f ∈ Hη.

8 2 iθ Exercise 9 Let H1 = L ([−π, π)] be the Hilbert space of functions F (e ) on 1 R π iθ iθ the unit circle with inner product (F,G) = 2π −π F (e )G(e )dθ. Let 2 H2 be the space L (R). Using the mapping

i − x x 7→ i + x

of R to the unit circle, show that: (a) The correspondence U : F → f, with

1 i − x f(x) = √ F (i + x) π i + x

gives a unitary mapping of H1 to H2. Proof: Let us verify the three requirements of a unitary mapping. (i) U is linear,

1 i − x i − x U(αf + βg) = √ αF + βG (i + x) π i + x i + x 1 i − x 1 i − x = α √ F + β √ G (i + x) π i + x (i + x) π i + x = αU(f) + βU(g)

(ii)/(iii) We have

||UF ||H1 = (UF,UF )H1 Z = f(x)f(x)dx R Z 1 i − x 1 i − x = √ F √ F dx (i + x) π i + x (−i + x) π i + x R Z 1 i − x i − x = F F dx (x2 + 1)π i + x i + x R

iθ i−x Now, let e = i+x . As x varies from −∞ to ∞, θ varies from +π to −π. Thus, the map is bijective. Then, taking the derivative with respect to x,

dθ 2i ieiθ = dx (x + i)2

9 or dθ 2 1 = − dx eiθ (x + i)2 i + x 1 1 = −2 = −2 i − x (x + i)2 x2 + 1

Thus Z 1 i − x i − x ||UF || = F F dx H2 (x2 + 1)π i + x i + x R Z +π 1 = F eiθ F (eiθ)dθ −π 2π

= ||F ||H1

(b) As a result

1 i − xn∞ √ (i + x) π i + x n=−∞

is an orthonormal basis of L2(R). Proof: A unitary map is a linear bijection that preserves norms. Thus following the remarks after Corollary 4.2.5 we have that it must map a basis to a basis. Take {eınθ} as the usual basis of L2([−π, π]). n iθ ınθ This corresponds to the basis of functions Fn(z) = z so F (e = e with n ∈ Z. Then U : Fn 7→ fn or 1 i − xn∞ f(x) = U(Fn) = √ . (i + x) π i + x n=−∞

Problem 2 (not ﬁnished) Consider the collection of exponentials {eiλx} on R, with λ ∈ R. Let H0 denote the space of ﬁnite linear combinations of these exponentials. For f, g ∈ H0, we deﬁne the inner product as

1 Z +T (f, g) = lim f(x)g(x)dx. T →∞ 2T −T (a) Show that this limit exists and

N X (f, g) = aλk bλk . k=1

Proof: Take two f, g ∈ H0. Then by deﬁnition of H0 they are ﬁnite linear combinations of exponentials

10 N X iλmx f(x) = aλm e m=1 N X iλnx g(x) = bλn e . n=1

The inner product is then

1 Z +T X i(λm−λn)x (f, g) = lim aλ bλ e dx T →∞ 2T m n m,n −T 1 X = lim aλ bλ 2T δij T →∞ 2T m n m,n N X = aλk bλk . k=1

which converges

(b) Take any f ∈ H0, then by the preceding problem

N 2 X 2 ||f|| = (f, f) = |aλk | k=1

and X sup |f(x)| ≥ |aλk | x k=0 so

||f|| ≤ sup |f(x)|. x

Take H to be the completion of H created by the Cantor diagonal- ization of Cauchy sequences in H0. Then H is not separable because we have an uncountable set of orthogonal vectors eiλkx. (c) Take a continuous function f ∈ AP . By deﬁnition it is a uniform N (n) limit on of elements in h ∈ H where h = P a eiλkx. Each R n 0 n i=k λk (n) a eiλkx is periodic with period τ = 2π/λ . λk k k I need to show that the sum of two periodic functions satisﬁes this condition and that the uniform limit of almost periodic functions also satisﬁes this condition.....

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