Functional Analysis, Math 7320 Lecture Notes from September 1, 2016 taken by Wilfredo J. Molina

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Functional Analysis, Math 7320

Lecture Notes from September 1, 2016

taken by Wilfredo J. Molina

The following items are the continuation of a list of examples of the previous set of notes: (e’) If X is a metric space, then (Cb(X,K),k · k∞) is a Banach space, where Cb(X,K) is

the set of continuous, bounded, _K-valued functions on X and k · k∞ is defined by f 7→

sup_x_∈_X|f(x)|.

To see that k · k∞ is a norm, observe that

kfk∞= sup

x∈X

|f(x)|= 0⇐⇒f(x) = 0 for all x∈X.

Moreover, if λ∈_R, then kλfk∞ = supx∈X|λf(x)|=|λ|supx∈X|f(x)|=|λ|kfk∞.

To see that (Cb(X,R),k · k∞) is complete, let (fn)n∈_N be Cauchy in Cb(X,R) and let

ε > 0. Then there is an n0 ∈ N such that kfn−fmk∞ < ε whenever m, n≥ n0, which

implies that for allx∈X,|fm(x)−fn(x)| ≤ kfm−fnk∞< εwhenever m, n≥n0, which

in turn implies that (fn(x))n∈_N is Cauchy in R and thus converges to an element of R.

Definef :X →_Rby x7→limn→∞fn(x).

To see that (fn)n∈_N converges to f and that f is continuous, let ε > 0. Then there is

an n0 ∈ N such that kfm −fnk∞ < ε/2 whenever m, n ≥ n0, which implies that for all

x ∈ X, |fm(x)−fn(x)| < ε/2 whenever m, n ≥ n0, which in turn implies that for all

x∈X,

lim

m→∞|fm(x)−fn(x)|=|f(x)−fn(x)| ≤

ε

2 < ε

whenever n ≥ n0. Therefore, (fn)n∈_N converges uniformly to f, which implies that f is

continuous.

To see that f is bounded, let ε > 0, let n0 ∈ N be such that kf −fn0k < ε, and note that fn0 is bounded, which implies that there is an M ≥0 such that kfn0k∞≤M. Then kfk∞ ≤ kf−fn0k∞+kfn0k∞< ε+M. Therefore,f is bounded.

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(f) If (X,k · kX) is a normed space and (Y,k · kY) is a Banach space, then(B(X, Y),k · kop) is a Banach space, where k · kop is the operator norm defined by

S 7→sup x6=0

kSxkY

kxkX

.

To see that B(X, Y) is a normed space, letS, T ∈ B(X, Y). Then

kS+Tkop= sup x6=0 k(S+T)xkY kxkX ≤sup x6=0 kSxkY kxkX + sup x6=0 kT xkY kxkX =kSkop+kTkop. Letλ ∈_K. Then kλSkop = sup x6=0 kλSxkY kxkX =|λ|sup x6=0 kSxkY kxkX =|λ|kSkop. Observe that 0 =kTkop= sup x6=0 kT xkY kxkX ⇐⇒ kT xkY = 0 for all 06=x∈X ⇐⇒T = 0.

To see thatB(X, Y)is complete, let(Tn)n∈Nbe a Cauchy sequence inB(X, Y), let ε >0,

and let x ∈ X. Then there is an n0 ∈ N such that kTm −Tnkop < ε/kxkX whenever

m, n ≥ n0, which implies that kTmx−TnxkY ≤ kTm −TnkopkxkX < ε, which in turn

implies that (Tnx)n∈N is a Cauchy sequence in Y and thus converges to an element in Y

since Y is complete. Therefore, define T :X →Y byx7→limn→∞Tnx.

To see that T is linear, letx, y ∈X and let λ∈_K. Then T(λx+y) = lim

n→∞Tn(λx+y) = λnlim→∞Tnx+ limn→∞Tny=λT x+T y.

To see that T is bounded, letε= 1 and letx∈X such that kxkX ≤1. Then there is an

n0 ∈N such that for all n≥n0,

kT xkY ≤ kTnxkY +kT x−TnxkY <kTnkop+ 1.

To see that limn→∞Tn = T, let ε > 0 and let x ∈ X such that kxkX ≤ 1. Then

there is an n0 ∈ N such that kT x−TnxkY < ε/2 whenever n ≥ n0. Since (Tn)n∈_N is a

Cauchy sequence in B(X, Y), there is an N0 ∈N such that kTm−Tnkop< ε/2 whenever m, n≥N0. Therefore,

kT x−TnxkY ≤ kT x−TmxkY +kTmx−TnxkY < ε

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1.3 Completion

1.3.1 Definition. Let (X, dX) and (Y, dY)be metric spaces and f :X →Y.

(1) If dY(f(x), f(y)) =dX(x, y) for all x, y ∈X, then f is called an isometry.

(2) If f(X) = Y and f is an isometry, then f is called an isometric isomorphism, and we write X ∼=Y.

(3) If f is an isometry, f(X) =Y, and Y is complete, then f is called a completion.

1.3.2 Theorem. Let (X, dX)and (Y, dY) be metric spaces.

(1) X has a completion.

(2) If f : X → Y is uniformly continuous, then there is a unique uniformly continuous map

ˆ

f : ˆX → Yˆ such that fˆ|X = f, where η : X → ( ˆX, d_Xˆ) and ϕ : Y → ( ˆY , d_Yˆ) are

completions.

(3) If X has completions Xˆ and η :X →Y, thenXˆ and Y are isometrically isomorphic. Proof. (1) Fix x0 ∈ X and define η : X → Cb(X,R) by x 7→ fx, where fx : X → R is

defined by y 7→ dX(x, y)−dX(y, x0). Recall that Cb(X,R) is a metric space with the metric

induced by k · k∞. Since fx(y) = dX(x, y)− dX(y, x0) ≤ dX(x, x0), fx is bounded. Since

dX is continuous, fx is continuous. Therefore, fx ∈ Cb(X,R). Let x1, x2, y ∈ X. Then

fx1(y)−fx2(y) = dX(x1, y)−dX(y, x2), which implies that |fx1(y)−fx2(y)| ≤ dX(x1, x2). Observe that equality is attained if y=x2. Therefore,kfx1−fx2k∞=dX(x1, x2), which implies that η is an isometry. Let Xˆ = η(X). Since (Cb(X,R),k · k∞) is a Banach space and Xˆ is

closed, Xˆ is complete, which implies thatXˆ is a completion of X.

(2) Identify X with η(X), identify Y with ϕ(Y), let xˆ∈X, and letˆ ε > 0. Since X is dense in

ˆ

X, there is a sequence(xn)n∈N in X that converges tox. Sinceˆ f is uniformly continuous, there

is a δ >0 such that for all x, x0 ∈X with d_Xˆ(x, x0)< δ, d_Yˆ(f(x), f(x0))< ε. Since (xn)n∈N is

Cauchy, there is an n0 ∈N such that d_Xˆ(xm, xn) < δ whenever m, n≥ n0, which implies that

d_Yˆ(f(xm), f(xn)) < ε whenever m, n ≥ n0, which in turn implies that (f(xn))n∈N is Cauchy

and thus converges to an element of Yˆ. Extend f to fˆby definingfˆ(ˆx) = limn→∞f(xn).

To see that fˆis well defined, let(x0_n)n∈N be a sequence inX that converges to x. Then, by theˆ

previous argument, ( ˆf(x0_n))n∈N is Cauchy, which implies that ( ˆf(x1),fˆ(x

0

1),fˆ(x2),fˆ(x02), . . .)

is Cauchy and converges to xˆ since its subsequence ( ˆf(xn))n∈_N converges to x, which in turnˆ

implies that ( ˆf(x0_n))n∈N converges to x.ˆ

To see that fˆ|X = f, let x ∈ X and let (xn)n∈_N be such that xn = x for all n ∈ N. Then ˆ

f(x) = limn→∞f(xn) =f(x).

To see that fˆis uniformly continuous, let ε > 0. Since fˆ|X is uniformly continuous, there is

a δ > 0 such that for all x, y ∈ X with d_Xˆ(x, y) < δ, d_Yˆ( ˆf(x),fˆ(y)) < ε/3. Let x,ˆ yˆ ∈ Xˆ

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and y, respectively, which implies that there is anˆ n0 ∈ N such that d_Xˆ(xn,xˆ) < δ/3 and

d_Xˆ(yn,yˆ) < δ/3 whenever n ≥ n0. Since d_Xˆ(xn, yn) ≤ d_Xˆ(xn,xˆ) +d_Xˆ(ˆx,yˆ) +d_Xˆ(ˆy, yn) < δ

whenever n ≥n0, d_Yˆ( ˆf(xn),fˆ(yn))< ε/3 whenever n ≥n0. Since( ˆf(xn))n∈_N and ( ˆf(yn))n∈_N

converge to fˆ(ˆx) and fˆ(ˆy), respectively, there is an N0 ∈ N such that d_Yˆ( ˆf(xn),fˆ(ˆx)) <

ε/3 and d_Yˆ( ˆf(yn),fˆ(ˆy)) < ε/3 whenever n ≥ N0. Letting N = max{n0, N0} yields that

d_Yˆ( ˆf(ˆx),fˆ(ˆy))≤d_Yˆ( ˆf(ˆx),fˆ(xn)) +d_Yˆ( ˆf(xn),fˆ(yn)) +d_Yˆ( ˆf(yn),fˆ(ˆy))< ε.

To see thatfˆis unique, leth: ˆX →Yˆ be a uniformly continuous extension off and let(xn)n∈_Nbe

a sequence in X converging toxˆ∈X. Sinceˆ h is continuous,limn→∞h(xn) = h(limn→∞xn) =

h(ˆx). Since h(xn) = f(xn) for all n ∈N, h(ˆx) = limn→∞h(xn) = limn→∞f(xn) = ˆf(ˆx). (3) Identify X with η(X). To see thatη is uniformly continuous, letε =δ >0 and observe that if x, y ∈ X are such that dX(x, y) < δ, then dY(η(x), η(y)) = dX(x, y) < δ = ε since η is an

isometry. Moreover, observe that the completion of a complete metric space is its identity map. Therefore, η extends toηˆ: ˆX →Y.

To see that ηˆis an isometry, let x,ˆ yˆ∈X. Then there are sequencesˆ (xn)n∈_N and (yn)n∈_N in X

that converge to xˆ and y, respectively. Therefore,ˆ

d_Xˆ(xn, yn) = dX(xn, yn) = dY(η(xn), η(yn)) =d_Yˆ(η(xn), η(yn)),

and taking the limit as n approaches infinity yields that d_Xˆ(ˆx,yˆ) =d_Yˆ(η(ˆx), η(ˆy)).

As above, η−1 :η(X)→X extends to an isometry dη−1 :Y →X. Therefore,ˆ dη−1◦ηˆ: ˆX →Xˆ

is an extension of the identity map on X, which implies that dη−1◦ηˆ : ˆX → Xˆ is the identity

map on X. Similarly,ˆ ηˆ◦dη−1 : Y → Y is the identity map on Y. Therefore, ηˆis an isometric

isomorphism.

To visualize the functions fx constructed in the proof of (1) above, let X = Q and let x0 = 0.

Then the graphs off−3 and f2 are as follows:

-4 _-2 2 4 Rational Numbers -3 -2 -1 1 2 3 Real Numbers f-3 f2

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Moreover, the graph of |f−3−f2| is as follows: -4 -2 2 4 Rational Numbers 1 2 3 4 5 Real Numbers |f-3-f2| Plot of |f-3-f2| with x0=0.