TST ANA2 Quiz 4

Problem 1

Exercise 353.1(Exercise).

Let $A\subseteq {\mathbb{R}}^2$ be a countable set. Then prove that ${\mathbb{R}}^2\setminus A$ is path connected.

Let $x,y\in {\mathbb{R}}^2\setminus A$. For each $r\in [\Vert x-y\Vert /2,\infty )$, there exists a circle passing through $x$ and $y$ with radius $r$. All circles in this family intersect at $x$ and $y$, and thus do not intersect at any other point. Each circle in this family can be used to prescribe a path from $x$ to $y$. Since this is an uncountable family, and $A$ is countable, it follows that there must exist some $r$ such that the path along the circle with radius $r$ does not intersect $A$.

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Problem 2

$V$ is a real Hilbert space, not necessarily separable.

Part a

Exercise 353.2(Exercise).

Prove that if $V=W\oplus W^{\perp }$ then $W$ and $W^{\perp }$ are closed.

Let $\{w_n{\}}_{n=1}^{\infty }\subseteq W$ converge to $v=w+w^{\perp }$. We have $⟨w_n,w^{\perp }⟩=0$ for all $n$.

$$\begin{array}{rl}⟨w+w^{\perp },w^{\perp }⟩& =⟨\underset{n\to \infty }{\lim }{w}_n,w^{\perp }⟩\\ & =\underset{n\to \infty }{\lim }⟨w_n,w^{\perp }⟩\\ & =0.\end{array}$$

Therefore, $w^{\perp }=0$, and $W$ is closed. A symmetric argument shows $W^{\perp }$ is closed.

Exercise 353.3(Exercise).

Provide an example of a subspace of a Hilbert space which is not closed.

Let $H$ be a separable Hilbert space with a countable orthonormal basis $\mathcal{B}$. Then, $\text{span }\mathcal{B}$ is a subspace of $H$ which is not closed. We have seen that $\text{span }\mathcal{B}$ is dense in $H$; if $\text{span }\mathcal{B}$ were equal to $H$, we would be able to write $H$ as a countable union of proper, closed, nowhere dense subspaces, contradicting Baire’s theorem.

Part b

Exercise 353.4(Exercise).

Every closed convex subset of a Hilbert space has a unique element on minimal norm.

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Let $S$ be a closed convex subset of $V$. Let $\alpha ={\inf }_{x\in S}\Vert x\Vert$. Let $\{x_n{\}}_{n=1}^{\infty }\subseteq S$ be any sequence such that $\Vert x_n\Vert \to \alpha$. Since $S$ is convex, we have

$$\Vert \frac{x_n+x_m}{2}\Vert ⩾\alpha$$

for all $n,m$. Now,

$$\begin{array}{rl}\Vert x_n-x_m{\Vert }^2+\Vert x_n+x_m{\Vert }^2& =2\Vert x_n{\Vert }^2+2\Vert x_m{\Vert }^2\\ \Vert x_n-x_m{\Vert }^2& ⩽2\Vert x_n{\Vert }^2+2\Vert x_m{\Vert }^2-4\alpha .\end{array}$$

It follows that $\Vert x_n-x_m\Vert \to 0$ as $n,m\to 0$. Since $V$ is complete, $\{x_n\}$ must converge, say to $x$. $\Vert x\Vert =\Vert {\lim }_{n\to \infty }{x}_n\Vert ={\lim }_{n\to \infty }\Vert x_n\Vert =\alpha$. If $\{y_n\}$ is another sequence such that $\Vert y_n\Vert \to \alpha$ and $\{y_n\}\to y$, the interlaced sequence $x_1,y_1,x_2,y_2,\dots$ also has the property of the norms of its elements converging to $\alpha$, hence $x=y$. Thus, $x$ is the unique element of minimal norm in $S$.

Part c

Let $W$ be a closed subspace of $V$.

Exercise 353.5(Exercise).

Given any $v\in V$, show that the set $v+W$ is closed and convex.

Suppose $\{v+w_n{\}}_{n=1}^{\infty }\subseteq v+W$ is Cauchy. It follows immediately that $\{w_n\}$ is Cauchy. Since $W$ is closed, $\{w_n\}$ converges, say to $w$. $\Vert (v+w_n)-(v+w)\Vert =\Vert w_n-w\Vert \to 0$, so $\{v+w_n\}\to v+w$. Thus, $v+W$ is closed.

Suppose $v+w_1,v+w_2\in v+W$.

$$\begin{array}{rlr}(v+w_1)t+(v+w_2)(1-t)=v+w_{1}t+w_2(1-t)\in v+W& & (t\in [0,1]).\end{array}$$

Thus, $v+W$ is convex.

Let $v+w\in v+W$ be the unique element with minimum norm, as in Exr 4. Then, $-w$ is unique element such that $\Vert v-(-w)\Vert$ is minimum, that is, $w^{\prime }:=-w$ is the element of $W$ that is closest to $v$.

Exercise 353.6(Exercise).

Prove that $v-w^{\prime }\in W^{\perp }$.

Let $\overline{w}\in W$. For all $ϵ>0$,

$$\begin{array}{rl}\Vert v-w^{\prime }{\Vert }^2& ⩽\Vert v-w^{\prime }-ϵ\overline{w}{\Vert }^2\\ ⟨v-w^{\prime },v-w^{\prime }⟩& ⩽⟨v-w^{\prime },v-w^{\prime }⟩-2ϵ⟨v-w^{\prime },\overline{w}⟩+{ϵ}^2⟨\overline{w},\overline{w}⟩\\ & ⟨v-w^{\prime },\overline{w}⟩⩽ϵ\Vert \overline{w}{\Vert }^2.\end{array}$$

It follows that $⟨v-w^{\prime },\overline{w}⟩=0$ for all $\overline{w}\in W$.

Therefore, any $v\in V$ can be expressed as

$$v=w^{\prime }+(v-w^{\prime }),$$

where $w^{\prime }\in W$ and $v-w^{\prime }\in W^{\perp }$. Since $W\cap W^{\perp }=\{0\}$, It follows that this decomposition is unique, and $V=W\oplus W^{\perp }$.

Exercise 353.7(Exercise).

$W=W^{\perp \perp }$.

We have $V=W\oplus W^{\perp }=W^{\perp }\oplus W^{\perp \perp }$. Let $w\in W$. There exist $x\in W^{\perp }$, $y\in W^{\perp \perp }$ such that $w=x+y$.

$$\begin{array}{rl}⟨w,x⟩& =⟨x,x⟩+⟨y,x⟩\\ 0& =\Vert x{\Vert }^2+0,\end{array}$$

so $x=0$ and $w=y$. It follows that $W\subseteq W^{\perp \perp }$; a symmetric argument lets us conclude $W=W^{\perp \perp }$.

Exercise 353.8(Exercise).

For a (not necessarily closed) subspace $W$ of $V$, its closure $\overline{W}$ equals $W^{\perp \perp }$.

We will first prove that $W^{\perp }=(\overline{W}{)}^{\perp }$. Let $\tilde{w}\in W^{\perp }$. Let $\{w_n{\}}_{n=1}^{\infty }\subseteq W$ converge to $\overline{w}\in \overline{W}$. Then,

$$⟨\tilde{w},\overline{w}⟩=⟨\tilde{w},\underset{n\to \infty }{\lim }{w}_n⟩=\underset{n\to \infty }{\lim }⟨\tilde{w},w_n⟩=0.$$

Thus, $\tilde{w}\in (\overline{W}{)}^{\perp }$, and $W^{\perp }=(\overline{W}{)}^{\perp }$.

Finally, $\overline{W}=((\overline{W}{)}^{\perp }{)}^{\perp }=W^{\perp \perp }$.