Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Assignment 2 Solutions

1. Let $$\begin{array}{rcl}{\displaystyle A}& =& {\displaystyle \{(x,y)\in {ℝ}^2\hspace{0.17em}|\hspace{0.17em}{x}^2+y^2<1\}\text{and}}\\ {\displaystyle B}& =& {\displaystyle \{(x,y)\in {ℝ}^2\hspace{0.17em}|\hspace{0.17em}{(x-2)}^2+y^2<1\}\text{.}}\end{array}$$ Determine, with proof, whether $X=A\cup B,$ $Y=\bar{A}\cup \bar{B}$ and $Z=\bar{A}\cup B$ are connected subsets of ${ℝ}^2$ with the usual topology.

   		Solution.
   	The picture is $$\begin{array}{c} A p B (rcosθ,rsinθ) \end{array}$$ with $$\begin{array}{rcl}{\displaystyle \bar{A}}& =& {\displaystyle \left\{(x,y)\hspace{0.17em}\right|\hspace{0.17em}{x}^2+y^2\le 1\}\text{and}}\\ {\displaystyle \bar{B}}& =& {\displaystyle \left\{(x,y)\hspace{0.17em}\right|\hspace{0.17em}{(x-2)}^2+y^2\le 1\}\text{.}}\end{array}$$ The sets $A,B,\bar{A}$ and $\bar{B}$ are all path connected since $$\bar{A}=\left\{(r\text{cos}\hspace{0.17em}\theta ,r\text{sin}\hspace{0.17em}\theta )\hspace{0.17em}\right|\hspace{0.17em}r\in [0,1]\hspace{0.17em}\text{and}\hspace{0.17em}\theta \in [0,2\pi )\}$$ and $p:[0,1]\to \bar{A}$ given by $p\left(t\right)=(rt\text{cos}\hspace{0.17em}\theta ,rt\text{sin}\hspace{0.17em}\theta )$ is a path from $(0,0)$ to $(r\text{cos}\hspace{0.17em}\theta ,r\text{sin}\hspace{0.17em}\theta )\text{.}$ $\bar{B}$ is a translate of $\bar{A}$ by $2$ so it is also path connected. Namely, $$\bar{B}=(2,0)+\bar{A}=\left\{(r\text{cos}\hspace{0.17em}\theta +2,r\text{sin}\hspace{0.17em}\theta )\hspace{0.17em}\right|\hspace{0.17em}r\in [0,1],\theta \in [0,2\pi )\}$$ and $p\prime :[0,1]\to \bar{B}$ given by $p\left(t\right)=(rt\text{cos}\hspace{0.17em}\theta +2,rt\text{sin}\hspace{0.17em}\theta )$ is a path from $(2,0)$ to $(r\text{cos}\hspace{0.17em}\theta +2,r\text{sin}\hspace{0.17em}\theta )\text{.}$ (a) Let $X=A\cup B\text{.}$ Then $$X=A\cup B,A\cap B=\varnothing ,A\ne \varnothing \text{and}B\ne \varnothing$$ and so $X$ is not connected. (b) Let $Y=\bar{A}\cup \bar{B}\text{.}$ Then $\bar{A}$ is path connected and $\bar{B}$ is path connected and $(1,0)\in \bar{A}\cap \bar{B}\text{.}$ It follows that $Y=\bar{A}\cup \bar{B}$ is path connected and connected. (c) Let $Z=\bar{A}\cup B=\bar{A}\cup ((1,0)\cup B)\text{.}$ Then every point of $\bar{A}$ is path connected to $(1,0)\text{.}$ Every point of $(1,0)\cup B$ is path connected to $(1,0)\text{.}$ So $Z$ is path connected and connected (since path connected implies connected, see the Rubinstein notes, the sentence between Definition 8.13 and Example 8.14). $\square$

2. Let $X$ and $Y$ be topological spaces and assume that $Y$ is Hausdorff. Let $f:X\to Y$ and $g:X\to Y$ be continuous functions.
   1. Show that the set $\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)=g\left(x\right)\}$ is a closed subset of $X\text{.}$
   2. Show that if $f:X\to ℝ$ and $g:X\to ℝ$ are continuous then $$f-g\text{is continuous.}$$
   3. Show that if $f:X\to ℝ$ and $g:X\to ℝ$ are continuous then $$\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)<g\left(x\right)\}\text{is open.}$$

   		Solution.
   	(a) To show: ${\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)=g\left(x\right)\}}^c$ is open. To show: $\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)\ne g\left(x\right)\}$ is open. Let $A=\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)\ne g\left(x\right)\}\text{.}$ Let $x\in A\text{.}$ To show: $x$ is an interior point of $A\text{.}$ Since $Y$ is Hausdorff and $f\left(x\right)\ne g\left(x\right)$ there exist open sets $U$ and $V$ of $Y$ with $$f\left(x\right)\in U,g\left(x\right)\in V\text{and}U\cap V\ne \varnothing \text{.}$$ $$\begin{array}{c} X x U V Y f(x) g(x) ⟶f ⟶g \end{array}$$ To show: There exists an open set $B$ of $X$ such that $x\in B$ and $B\subseteq A\text{.}$ Let $N=f^{-1}\left(U\right)$ and $P=g^{-1}\left(V\right)$ and let $B=N\cap P\text{.}$ To show: (aa) $B$ is open. (ab) $x\in B\text{.}$ (ac) $B\subseteq A\text{.}$ (aa) Since $f$ is continuous, $N=f^{-1}\left(U\right)$ is open. Since $g$ is continuous, $P=g^{-1}\left(V\right)$ is open. So $B=N\cap P$ is open. (ab) Since $f\left(x\right)\in U,$ then $x\in f^{-1}\left(U\right)=N\text{.}$ Since $g\left(x\right)\in V,$ then $x\in g^{-1}\left(V\right)=P\text{.}$ So $x\in N\cap P=B\text{.}$ (ac) To show: $B\subseteq A\text{.}$ To show: If $x\prime \in B$ then $x\prime \in A\text{.}$ Assume $x\prime \in B\text{.}$ To show: $f\left(x\prime \right)=g\left(x\prime \right)\text{.}$ Since $x\prime \in B,$ then $f\left(x\prime \right)\subseteq f\left(N\right)\subseteq U$ and $g\left(x\prime \right)\subseteq g\left(P\right)\subseteq V\text{.}$ Since $U\cap V=\varnothing$ then $f\left(x\prime \right)\ne g\left(x\prime \right)\text{.}$ So $x\prime \in A\text{.}$ So $B\subseteq A\text{.}$ So $x$ is an interior point of $A\text{.}$ So $A$ is open and $\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)=g\left(x\right)\}$ is closed. (b) Assume $f:X\to ℝ$ and $g:X\to ℝ$ are continuous. To show: $f-g$ is continuous. The function $f-g$ is the composition of $$\begin{array}{ccc}X& \stackrel{f\times g}{⟶}& ℝ\times ℝ\\ x& ⟼& (f\left(x\right),g\left(x\right))\end{array}\text{and}\begin{array}{ccc}ℝ\times ℝ& ⟶& ℝ\\ (x,y)& ⟼& x-y\end{array}$$ and the composition of continuous functions is continuous. To show: (ba) $\hspace{0.17em}$ $$\begin{array}{cccc}f\times g:& X& ⟶& ℝ\times ℝ\\ & x& ⟼& (f\left(x\right),g\left(x\right))\end{array}$$ is continuous. (bb) $\hspace{0.17em}$ $$\begin{array}{ccc}ℝ\times ℝ& ⟶& ℝ\\ (x,y)& ⟼& x-y\end{array}$$ is continuous. (ba) This is part 1 of theorem 3.6 in the Rubinstein notes. No proof is given there, so it might be desirable to provide a proof. (bb) To show: If $(x_1,y_1),(x_2,y_2),\dots$ is a sequence in ${ℝ}^2$ and $\underset{n\to \infty }{\text{lim}}(x_n,y_n)=(x,y)$ then $$\underset{n\to \infty }{\text{lim}}(x_n-y_n)=x-y\text{.}$$ To show: $\underset{n\to \infty }{\text{lim}}(x_n-y_n)=\left(\underset{n\to \infty }{\text{lim}}{x}_n\right)-\left(\underset{n\to \infty }{\text{lim}}{y}_n\right)\text{.}$ This is a fact usually proved in $2^{\text{nd}}$ year Real analysis. Again, it might be desirable to provide a proof. (c) Let $B=\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)<g\left(x\right)\}\text{.}$ Then $B=\{x\in X\hspace{0.17em}|\hspace{0.17em}(f-g)\left(x\right)<0\}={(f-g)}^{-1}\left({ℝ}_{<0}\right)\text{.}$ Since ${ℝ}_{<0}$ is open in $ℝ$ and, by (b), $f-g$ is continuous, $B={(f-g)}^{-1}\left({ℝ}_{<0}\right)$ is open. $\square$

3. Let $X$ be a complete normed vector space over $ℝ\text{.}$ A sphere in $X$ is a set $$S(a,r)=\{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,a)=\Vert x-a\Vert =r\},\text{for}\hspace{0.17em}a\in X\hspace{0.17em}\text{and}\hspace{0.17em}r\in {ℝ}_{>0}\text{.}$$
   1. Show that each sphere in $X$ is nowhere dense.
   2. Show that there is no sequence of spheres $\left\{S_n\right\}$ in $X$ whose union is $X\text{.}$
   3. Give a geometric interpretation of the result in (b) when $X={ℝ}^2$ with the Euclidean norm.
   4. Show that the result of (b) does not hold in every complete metric space $X\text{.}$

   		Solution.
   	(a) Let $a\in X$ and $r\in {ℝ}_{>0}\text{.}$ To show: ${\left(\overline{S(a,r)}\right)}^{\circ }=\varnothing \text{.}$ To show: (aa) $\overline{S(a,r)}=S(a,r)\text{.}$ (ab) ${S(a,r)}^{\circ }=\varnothing \text{.}$ (aa) To show: $S(a,r)$ is closed. $$S(a,r)=f^{-1}\left(\left\{r\right\}\right),$$ where $$\begin{array}{cccc}f:& X& ⟶& ℝ\\ & x& ⟼& d(x,a)\end{array}$$ Since $f$ is continuous (see Theorem 3.6 of the Rubinstein notes) and $\left\{r\right\}$ is closed, then $S(a,r)=f^{-1}\left(\left\{r\right\}\right)$ is closed. (ab) To show: ${S(a,r)}^{\circ }=\varnothing \text{.}$ To show: If $x\in S(a,r)$ then $x$ is not an interior point of $S(a,r)\text{.}$ Assume $x\in S(a,r)\text{.}$ To show: If $\epsilon \in {ℝ}_{>0}$ then $B(x,\epsilon )⊈S(a,r)\text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ For $t\in {ℝ}_{>0}$ let $$c\left(t\right)=a+t(x-a)\begin{array}{c} a r c(t) x=c(1) S(a,r) ⏟ \end{array}$$ Since $$\begin{array}{rcl}{\displaystyle d(c\left(t\right),x)}& =& {\displaystyle \Vert a+t(x-a)-x\Vert }\\ & =& {\displaystyle \Vert (t-1)(x-a)\Vert }\\ & =& {\displaystyle \mid t-1\mid \Vert x-a\Vert }\\ & =& {\displaystyle \mid t-1\mid ·r,}\end{array}$$ then $c\left(t\right)\in B(x,\epsilon )$ for $r·\mid t-1\mid <\epsilon \text{.}$ So $c\left(t\right)\in B(x,\epsilon )$ for $1-\frac{\epsilon }{r}<t<1+\frac{\epsilon }{2}\text{.}$ Since $$d(c\left(t\right),a)=\Vert a+t(x-a)-a\Vert =t\Vert x-a\Vert =t·r,$$ then $c\left(t\right)\in S(a,r)$ only if $t=1\text{.}$ So $B(x,\epsilon )⊈S(a,r)\text{.}$ So $x$ is not an interior point of $S(a,r)\text{.}$ So ${S(a,r)}^{\circ }=\varnothing \text{.}$ So ${\left(\overline{S(a,r)}\right)}^{\circ }={S(a,r)}^{\circ }=\varnothing$ and $S(a,r)$ is nowhere dense in $X\text{.}$ (b) To show: If $a_1,a_2,\dots \in X$ and $r_1,r_2,\dots \in {ℝ}_{>0}$ then $X\ne \bigcup _{n\in {\mathbb{Z}}_{>0}}S(a_n,r_n)\text{.}$ Assume $a_1,a_2,\dots \in X$ and $r_1,r_2,\dots \in {ℝ}_{>0}\text{.}$ By part (a), each $S(a_n,r_n)$ is nowhere dense in $X\text{.}$ By the Baire theorem, $X$ is not a countable union of nowhere dense sets. So, $X\ne \bigcup _{n\in {\mathbb{Z}}_{>0}}S(a_n,r_n)\text{.}$ (c) If $X={ℝ}^2$ then $S(a,r)$ is a circle with centre $a$ and radius $r\text{.}$ So (b) says that the plane ${ℝ}^2$ cannot be completely covered by a sequence of circles. (d) To show: There exists a complete metric space $X$ and spheres $S(a_1,r_1),S(a_2,r_2),\dots$ in $X$ with $X=\bigcup _{n\in {\mathbb{Z}}_{>0}}S(a_n,r_n)\text{.}$ Let $X=\mathbb{Z}$ (so $X$ is an infinite set with the discrete topology). Then $\mathbb{Z}$ is a complete metric space (since it is a closed subset of the complete metric space $ℝ\text{).}$ Let $0=a_1=a_2=\cdots$ and $r_n=n$ for $n\in {\mathbb{Z}}_{>0}\text{.}$ Then $$S(a_n,r_n)=S(0,n)=\{-n,n\}$$ in $\mathbb{Z}\text{.}$ So $$\mathbb{Z}=\bigcup _{n\in {\mathbb{Z}}_{>0}}\{-n,n\}=\bigcup _{n\in {\mathbb{Z}}_{>0}}S(0,n)\text{.}$$ $\square$

4. Prove that if $X$ and $Y$ are path connected, then $X\times Y$ is also path connected.

   		Solution.
   	Assume $X$ and $Y$ are path connected. To show: $X\times Y$ is path connected. To show: If $(x_1,y_1),(x_2,y_2)\in X\times Y$ then there exists a path $p:[0,1]\to X\times Y$ connecting $(x_1,y_1)$ and $(x_2,y_2)\text{.}$ Assume $(x_1,y_1)\in X\times Y$ and $(x_2,y_2)\in X\times Y\text{.}$ Since $X$ and $Y$ are path connected we know that there exist continuous functions $p_1:[0,1]\to X$ and $p_2:[0,1]\to Y$ with $$\begin{array}{rcl}{\displaystyle p_1\left(0\right)}& =& {\displaystyle x_1,}\\ {\displaystyle p_1\left(1\right)}& =& {\displaystyle x_2,}\end{array}\text{and}\begin{array}{rcl}{\displaystyle p_2\left(0\right)}& =& {\displaystyle y_1,}\\ {\displaystyle p_2\left(1\right)}& =& {\displaystyle y_2\text{.}}\end{array}$$ To show: There exists a continuous function $p:[0,1]\to X\times Y$ with $p\left(0\right)=(x_1,y_1)$ and $p\left(1\right)=(x_2,y_2)\text{.}$ Let $p:[0,1]\to X\times Y$ be given by $p\left(t\right)=(p_1\left(t\right),p_2\left(t\right))\text{.}$ Then $$\begin{array}{rcl}{\displaystyle p\left(0\right)}& =& {\displaystyle (p_1\left(0\right),p_2\left(0\right))=(x_1,y_1),}\\ {\displaystyle p\left(1\right)}& =& {\displaystyle (p_1\left(1\right),p_2\left(1\right))=(x_2,y_2),}\end{array}$$ and $p$ is continuous by Theorem 3.6 in the Rubinstein notes. More specifically Theorem 3.6 part (a) states: Let $f$ be a function from $X$ to $Y_1$ and let $g$ be a function from $X$ to Y2. Define the function $h$ from $X$ to the product $Y_1\times Y_2$ by $$h\left(x\right)=(f\left(x\right),g\left(x\right)),$$ for $x\in X\text{.}$ Then $h$ is continuous if and only if both functions $f$ and $g$ are continuous. This statement is not proved in the Rubinstein notes so it might be desirable to provide a proof. $\square$

5. Let $p\in {ℝ}_{>1}$ and define $q\in {ℝ}_{>1}$ by $\frac{1}{p}+\frac{1}{q}=1\text{.}$
   1. Define the normed vector space ${\ell }^p\text{.}$
   2. Show that ${\ell }^p$ is a Banach space.
   3. Prove that the dual of ${\ell }^p$ is ${\ell }^q\text{.}$

   		Solution.
   	(a) $\hspace{0.17em}$ $${\ell }^p=\left\{(x_1,x_2,\dots )\hspace{0.17em}\right|\hspace{0.17em}{x}_i\in ℝ\hspace{0.17em}\text{and}\hspace{0.17em}{\Vert (x_1,x_2,\dots )\Vert }_p<\infty \}$$ where $$\Vert (x_1,x_2,\dots )\Vert ={\left(\sum _{i\in {\mathbb{Z}}_{>0}}{\mid x_i\mid }^p\right)}^{\frac{1}{p}}\text{.}$$ (b) If $V$ and $W$ are normed vector spaces let $$B(V,W)=\{T:V\to W\hspace{0.17em}|\hspace{0.17em}T\hspace{0.17em}\text{is a linear operator and}\hspace{0.17em}\Vert T\Vert \hspace{0.17em}\text{is bounded}\}$$ where $$\Vert T\Vert =\text{sup}\left\{\frac{\Vert Tx\Vert }{\Vert x\Vert }\hspace{0.17em}\right|\hspace{0.17em}x\in V\}\text{.}$$ Recall the following from the Rubinstein notes: Theorem 11.8 If $W$ is a Banach space then $B(V,W)$ is a Banach space. Theorem 4.6 The space $ℝ$ with the usual metric is complete. Together these imply that $B(V,ℝ)$ is complete. In part (c) we will show that $${\ell }^p=B({\ell }^2,ℝ)\text{.}$$ Thus ${\ell }^p$ is a Banach space. (c) To show: ${\ell }^q$ is the dual of ${\ell }^p\text{.}$ To show: ${\ell }^q=B({\ell }^p,ℝ)\text{.}$ Define $$\begin{array}{cccc}\phi :& {\ell }^q& ⟶& B({\ell }^p,ℝ)\\ & y& ⟼& {\phi }_y:& {\ell }^p& ⟶& ℝ\\ & & & & x& ⟼& ⟨y,x⟩\end{array}$$ where $$⟨y,x⟩=\sum _{i\in {\mathbb{Z}}_{>0}}{y}_i{x}_i,$$ if $y=(y_1,y_2,\dots )$ and $x=(x_1,x_2,\dots )\text{.}$ To show: (ca) $\phi$ is a linear transformation. (cb) $\phi$ is invertible. (cc) If $y\in {\ell }^q$ then $\Vert {\phi }_y\Vert =\Vert y\Vert \text{.}$ (ca) To show: (caa) If $y_1,y_2\in {\ell }^q$ then $\phi (y_1+y_2)=\phi \left(y_1\right)+\phi \left(y_2\right)\text{.}$ (cab) If $y\in {\ell }^q$ and $c\in ℝ$ then $\phi \left(cy\right)=c\phi \left(y\right)\text{.}$ (caa) Assume $y_1,y_2\in {\ell }^q\text{.}$ To show: $\phi (y_1+y_2)=\phi \left(y_1\right)+\phi \left(y_2\right)\text{.}$ To show: If $x\in {\ell }^p$ then ${\phi }_{y_1+y_2}\left(x\right)=({\phi }_{y_1}+{\phi }_{y_2})\left(x\right)\text{.}$ Assume $x\in {\ell }^p\text{.}$ To show: ${\phi }_{y_1+y_2}\left(x\right)=({\phi }_{y_1}+{\phi }_{y_2})\left(x\right)\text{.}$ $$\begin{array}{rcl}{\displaystyle {\phi }_{y_1+y_2}\left(x\right)}& =& {\displaystyle ⟨y_1+y_2,x⟩}\\ & =& {\displaystyle ⟨y_1,x⟩+⟨y_2,x⟩}\\ & =& {\displaystyle {\phi }_{y_1}\left(x\right)+{\phi }_{y_2}\left(x\right)}\\ & =& {\displaystyle ({\phi }_{y_1}+{\phi }_{y_2})\left(x\right)\text{.}}\end{array}$$ (cab) Assume $y\in {\ell }^q$ and $c\in ℝ\text{.}$ To show: $\phi \left(cy\right)=c\phi \left(y\right)\text{.}$ To show: If $x\in {\ell }^p$ then ${\phi }_{cy}\left(x\right)=\left(c{\phi }_y\right)\left(x\right)\text{.}$ Assume $x\in {\ell }^p\text{.}$ To show: ${\phi }_{cy}\left(x\right)=\left(c{\phi }_y\right)\left(x\right)\text{.}$ $${\phi }_{cy}\left(x\right)=⟨cy,x⟩=c⟨y,x⟩=c\left({\phi }_y\left(x\right)\right)=\left(c{\phi }_y\right)\left(x\right)\text{.}$$ So $\phi :{\ell }^q\to B({\ell }^p,ℝ)$ is a linear transformation. (cb) To show: $\phi :{\ell }^2\to B({\ell }^p,ℝ)$ is invertible. To show: There exists $\psi :B({\ell }^p,ℝ)\to {\ell }^q$ such that $\phi \circ \psi =\text{id}$ and $\psi \circ \phi =\text{id.}$ Let $\psi :B({\ell }^p,ℝ)\to {\ell }^q$ be given by $$\psi \left(\gamma \right)=(\gamma \left(e_1\right),\gamma \left(e_2\right),\dots )$$ where $e_i=(0,0,\dots ,0,1,0,0,\dots )$ with $1$ in the $i^{\text{th}}$ spot. To show: (cba) $\phi \circ \psi =\text{id.}$ (cbb) $\psi \circ \phi =\text{id.}$ (cba) To show: If $\gamma \in B({\ell }^p,ℝ)$ then $\phi \left(\psi \left(\gamma \right)\right)=\gamma \text{.}$ Assume $\gamma \in B({\ell }^p,ℝ)\text{.}$ To show: $\phi \left(\psi \left(\gamma \right)\right)=\gamma \text{.}$ To show: If $x\in {\ell }^p$ then $\phi \left(\psi \left(\gamma \right)\right)\left(x\right)=\gamma \left(x\right)\text{.}$ Assume $x\in {\ell }^p\text{.}$ Let $x=(x_1,x_2,\dots )\text{.}$ To show: $\phi \left(\psi \left(\gamma \right)\right)\left(x\right)=\gamma \left(x\right)\text{.}$ $$\begin{array}{rcl}{\displaystyle \phi \left(\psi \left(\gamma \right)\right)\left(x\right)}& =& {\displaystyle \phi (\gamma \left(e_1\right),\gamma \left(e_2\right),\dots )\left(x\right)}\\ & =& {\displaystyle ⟨(\gamma \left(e_1\right),\gamma \left(e_2\right),\dots ),(x_1,x_2,\dots )⟩}\\ & =& {\displaystyle \sum _{i\in {\mathbb{Z}}_{>0}}\gamma \left(e_i\right)x_i}\\ & =& {\displaystyle \gamma \left(\sum _{i\in {\mathbb{Z}}_{>0}}{x}_i{e}_i\right)}\\ & =& {\displaystyle \gamma \left(x\right)\text{.}}\end{array}$$ (cbb) To show: $\psi \circ \phi =\text{id.}$ To show: If $y\in {\ell }^q$ then $\psi \left(\phi \left(y\right)\right)=y\text{.}$ Assume $y\in {\ell }^q\text{.}$ Let $y=(y_1,y_2,\dots )\text{.}$ To show: $\psi \left(\phi \left(y\right)\right)=y\text{.}$ $$\psi \left(\phi \left(y\right)\right)=\psi \left({\phi }_y\right)=({\phi }_y\left(e_1\right),{\phi }_y\left(e_2\right),\dots )=(y_1,y_2,\dots ),$$ since $${\phi }_y\left(e_i\right)=⟨y,e_i⟩=⟨(y_1,y_2,\dots ),(0,\dots ,0,1,0,\dots ,0)⟩=y_i\text{.}$$ So $\psi \left(\phi \left(y\right)\right)=y\text{.}$ (cc) To show: If $y\in {\ell }^q$ then $\Vert {\phi }_y\Vert ={\Vert y\Vert }_q\text{.}$ Assume $y\in {\ell }^q\text{.}$ Let $y=(y_1,y_2,\dots )\text{.}$ To show: (cca) $\Vert {\phi }_y\Vert \le {\Vert y\Vert }_q\text{.}$ (ccb) $\Vert {\phi }_y\Vert \ge {\Vert y\Vert }_q\text{.}$ (cca) To show: If $x\in {\ell }^p$ then $\mid {\phi }_y\left(x\right)\mid \le {\Vert x\Vert }_p{\Vert y\Vert }_q\text{.}$ Assume $x\in {\ell }^p\text{.}$ Let $x=(x_1,x_2,\dots )\text{.}$ Then $$\mid {\phi }_y\left(x\right)\mid =\mid \sum _{n\in {\mathbb{Z}}_{>0}}{x}_n{y}_n\mid \le {\Vert x\Vert }_p{\Vert y\Vert }_q$$ by Hölder's inequality. So $\Vert {\phi }_y\Vert \le {\Vert y\Vert }_q\text{.}$ (ccb) To show: $\Vert {\phi }_y\Vert \ge {\Vert y\Vert }_q\text{.}$ To show: There exists $x\in {\ell }^p$ with $\mid {\phi }_y\left(x\right)\mid \ge {\Vert x\Vert }_p{\Vert y\Vert }_q\text{.}$ Let $$x=(\text{sgn}\left(y_1\right){\mid y_1\mid }^{q-1},\text{sgn}\left(y_2\right){\mid y_2\mid }^{q-1},\dots )\text{.}$$ Then $$\begin{array}{rcl}{\displaystyle {\Vert x\Vert }_p}& =& {\displaystyle {\left(\sum _{n\in {\mathbb{Z}}_{>0}}{\mid x_n\mid }^p\right)}^{\frac{1}{p}}}\\ & =& {\displaystyle {\left(\sum _{n\in {\mathbb{Z}}_{>0}}{\mid \text{sgn}\left(y_n\right){\mid y_n\mid }^{q-1}\mid }^p\right)}^{\frac{1}{p}}}\\ & =& {\displaystyle {\left(\sum _{n\in {\mathbb{Z}}_{>0}}{\mid y_n\mid }^{pq-p}\right)}^{\frac{1}{p}}}\\ & =& {\displaystyle {\left(\sum _{n\in {\mathbb{Z}}_{>0}}{\mid y_n\mid }^{pq(1-\frac{1}{q})}\right)}^{\frac{1}{p}}}\\ & =& {\displaystyle {\left(\sum _{n\in {\mathbb{Z}}_{>0}}{\mid y_n\mid }^{pq\frac{1}{p}}\right)}^{\frac{1}{p}}}\\ & =& {\displaystyle {\left({\left(\sum _{n\in {\mathbb{Z}}_{>0}}{\mid y_n\mid }^q\right)}^{\frac{1}{q}}\right)}^{q\frac{1}{p}}}\\ & =& {\displaystyle {\Vert y\Vert }_q^{q\frac{1}{p}}}\\ & =& {\displaystyle {\Vert y\Vert }_q^{q(1-\frac{1}{q})}}\\ & =& {\displaystyle {\Vert y\Vert }_q^{q-1}\text{.}}\end{array}$$ So $$\begin{array}{rcl}{\displaystyle \mid {\phi }_y\left(x\right)\mid }& =& {\displaystyle \mid \sum _{n\in {\mathbb{Z}}_{>0}}{x}_n{y}_n\mid }\\ & =& {\displaystyle \mid \sum _{n\in {\mathbb{Z}}_{>0}}\left(\text{sgn}\left(y_n\right)\mid y_n\mid \right)\left(\text{sgn}\left(y_n\right){\mid y_n\mid }^{q-1}\right)\mid }\\ & =& {\displaystyle \sum _{n\in {\mathbb{Z}}_{>0}}{\mid y_n\mid }^q}\\ & =& {\displaystyle {\Vert y\Vert }_q^q}\\ & =& {\displaystyle {\Vert y\Vert }_q{\Vert y\Vert }_q^{q-1}}\\ & =& {\displaystyle {\Vert y\Vert }_q{\Vert x\Vert }_p\text{.}}\end{array}$$ So $$\Vert {\phi }_y\Vert \ge {\Vert y\Vert }_q\text{.}$$ $\square$

6. Let $X=C^1[0,1],$ $Y=C[0,1]$ so that functions in $X$ are continuously differentiable and functions in $Y$ are continuous. $$\begin{array}{rcl}{\displaystyle Y}& =& {\displaystyle C[0,1],\text{with norm given by}\Vert f\Vert =\text{sup}\left\{\mid f\left(t\right)\mid \hspace{0.17em}\right|\hspace{0.17em}t\in [0,1]\},\hspace{0.17em}\text{and}}\\ {\displaystyle X}& =& {\displaystyle C^1[0,1],\text{with norm given by}{\Vert f\Vert }_0=\Vert f\Vert +\Vert f\prime \Vert ,}\end{array}$$ where $f\prime =\frac{df}{dt}\text{.}$ Let $D:X\to Y$ be the differentiation operator $Df=\frac{df}{dt}\text{.}$
   1. Show that $D:(X,{\Vert ·\Vert }_0)\to (Y,\Vert ·\Vert )$ is a bounded linear operator with $\Vert D\Vert =1\text{.}$
   2. Show that $D:(X,\Vert ·\Vert )\to (Y,\Vert ·\Vert )$ is an unbounded linear operator. (Hint: Consider the sequence of elements $t^n$ in $X\text{).}$

   		Solution.
   	(a) To show: $\Vert D\Vert =1\text{.}$ To show: (aa) $\Vert D\Vert \ge 1\text{.}$ (ab) $\Vert D\Vert \le 1\text{.}$ (aa) If $n\in {\mathbb{Z}}_{>0}$ then $$\frac{\Vert D{t}^n\Vert }{{\Vert t^n\Vert }_0}=\frac{\Vert n{t}^{n-1}\Vert }{\Vert n{t}^{n-1}\Vert +\Vert t^n\Vert }=\frac{n}{n+1}=\frac{1}{1+\frac{1}{n}}\text{.}$$ So $\Vert D\Vert \ge \frac{1}{1+\frac{1}{n}}$ for $x\in {\mathbb{Z}}_{\ge 0}\text{.}$ So $\Vert D\Vert \ge 1\text{.}$ (ab) Let $f\in C^1[0,1]\text{.}$ Since $$\Vert Df\Vert =\Vert \frac{df}{dt}\Vert \le \Vert f\Vert +\Vert \frac{df}{dt}\Vert ={\Vert f\Vert }_0$$ then $\frac{\Vert Df\Vert }{{\Vert f\Vert }_0}\le 1\text{.}$ So $\Vert D\Vert =1\text{.}$ (b) To show: If $n\in {\mathbb{Z}}_{\ge 0}$ then $\Vert D\Vert \ge n\text{.}$ Assume $n\in {\mathbb{Z}}_{\ge 0}\text{.}$ Since $$\frac{\Vert D{t}^n\Vert }{\Vert t^n\Vert }=\frac{\Vert n{t}^n\Vert }{\Vert t^n\Vert }=\frac{n}{1}=n,$$ then $$\Vert D\Vert \ge n\text{.}$$ So $D$ is unbounded. $\square$

7. Let $\{a_1,a_2,\dots \}$ be a bounded sequence of complex numbers. Define an operator $T:l^2\to l^2$ by; $$T(b_1,b_2,\dots )=(0,a_1{b}_1,a_2{b}_2,\dots )\text{.}$$
   1. Show that $T$ is a bounded linear operator and find $\Vert T\Vert \text{.}$
   2. Compute the adjoint operator $T^{*}\text{.}$
   3. Show that if $T\ne 0$ then $T^{*}T\ne T{T}^{*}\text{.}$
   4. Find the eigenvalues of $T^{*}\text{.}$

   		Solution.
   	(a) Let $e_i=(0,0,\dots ,\stackrel{i\text{th}}{0},1,0,\dots )$ so that $\Vert e_i\Vert =1\text{.}$ Then $${\Vert T{e}_i\Vert }^2={\mid a_i\mid }^2={\mid a_i\mid }^2{\Vert e_i\Vert }^2,$$ so $\Vert T\Vert \ge \mid a_i\mid \text{.}$ So $$\Vert T\Vert \ge \text{sup}\{\mid a_1\mid ,\mid a_2\mid ,\dots \}\text{.}$$ To show: $\Vert T\Vert =\text{sup}\{\mid a_1\mid ,\mid a_2\mid ,\dots \}\text{.}$ To show: $\Vert T\Vert \le \text{sup}\{\mid a_1\mid ,\mid a_2\mid ,\dots \}\text{.}$ Let $b=(b_1,b_2,\dots )\in {\ell }^2\text{.}$ Then $$\begin{array}{rcl}{\displaystyle {\Vert Tb\Vert }^2}& =& {\displaystyle \sum _{i\in {\mathbb{Z}}_{>0}}{\mid a_i{b}_i\mid }^2}\\ & =& {\displaystyle \sum _{i\in {\mathbb{Z}}_{>0}}{\mid a_i\mid }^2{\mid b_i\mid }^2}\\ & \le & {\displaystyle \sum _{i\in {\mathbb{Z}}_{>0}}{k}^2{\mid b_i\mid }^2}\\ & =& {\displaystyle k^2{\Vert b\Vert }^2,}\end{array}$$ where $k=\text{sup}\{\mid a_1\mid ,\mid a_2\mid ,\dots \}\text{.}$ So $\Vert T\Vert \le k=\text{sup}\{\mid a_1\mid ,\mid a_2\mid ,\dots \}\text{.}$ So $\Vert T\Vert =\text{sup}\{\mid a_1\mid ,\mid a_2\mid ,\dots \}\text{.}$ (b) Since ${\ell }^2$ is a Hilbert space, $T^{*}$ is determined by $$⟨b,T^{*}c⟩=⟨Tb,c⟩=\sum _{i\in {\mathbb{Z}}_{>0}}{a}_i{b}_i\overline{c_{i+1}}=\sum _{i\in {\mathbb{Z}}_{>0}}{b}_i\overline{\left(\stackrel{‾}{a_i}{c}_{i+1}\right)}\text{.}$$ So $T^{*}c=(\overline{a_1}{c}_2,\overline{a_2}{c}_3,\overline{a_3}{c}_4,\dots )\text{.}$ (c) To show: If $T^{*}T=T{T}^{*}$ then $T=0\text{.}$ Assume $T^{*}T=T{T}^{*}\text{.}$ Let $c=(c_1,c_2,\dots )\in {\ell }^2\text{.}$ Then $$\begin{array}{rcl}{\displaystyle T^{*}T\left(c\right)}& =& {\displaystyle T^{*}(0,a_1{c}_1,a_2{c}_2,\dots )}\\ & =& {\displaystyle (\overline{a_1}{a}_1{c}_1,\overline{a_2}{a}_2{c}_3,\dots )}\\ & =& {\displaystyle ({\mid a_1\mid }^2{c}_1,{\mid a_2\mid }^2{c}_2,\dots )}\end{array}$$ and $$\begin{array}{rcl}{\displaystyle T{T}^{*}\left(c\right)}& =& {\displaystyle T(\overline{a_1}{c}_2,\overline{a_2}{c}_3,\dots )}\\ & =& {\displaystyle (0,a_1\overline{a_1}{c}_2,a_2\overline{a_2}{c}_3,\dots )}\\ & =& {\displaystyle (0,{\mid a_1\mid }^2{c}_2,{\mid a_2\mid }^2{c}_3,\dots )\text{.}}\end{array}$$ Thus $T^{*}T=T{T}^{*}$ implies $${\mid a_1\mid }^2{c}_1=0,{\mid a_2\mid }^2{c}_2={\mid a_1\mid }^2{c}_2,{\mid a_3\mid }^2{c}_3={\mid a_2\mid }^2{c}_3,\dots$$ for all $c=(c_1,c_2,\dots )\in {\ell }^2\text{.}$ So $T^{*}T=T{T}^{*}$ implies ${\mid a_1\mid }^2=0,$ ${\mid a_2\mid }^2={\mid a_1\mid }^2,$ ${\mid a_3\mid }^2={\mid a_2\mid }^2,\dots \text{.}$ So $T^{*}T=T{T}^{*}$ implies $0=a_1=a_2=\dots \text{.}$ So $T=0\text{.}$ (d) Assume $c=(c_1,c_2,\dots )\in {\ell }^2$ is an eigenvector of $T^{*}$ with eigenvalue λ. Then $$\lambda (c_1,c_2,\dots )=T^{*}c=(\overline{a_1}{c}_2,\overline{a_2}{c}_3,\dots )$$ so that $$\overline{a_1}{c}_2=\lambda c_1,\overline{a_2}{c}_3=\lambda c_2,\dots$$ and $$c_2=\frac{\lambda }{\overline{a_1}}{c}_1,c_3=\frac{\lambda }{\overline{a_2}}{c}_2=\frac{\lambda }{\overline{a_1{a}_2}}{c}_1,\dots$$ So $$c=(c_1,\frac{\lambda }{\overline{a_1}}{c}_1,\frac{{\lambda }^2}{\overline{a_1{a}_2}}{c}_1,\dots )=c_1(1,\frac{\lambda }{\overline{a_1}},\frac{{\lambda }^2}{\overline{a_1{a}_2}},\dots )\text{.}$$ Since $c\in {\ell }^2,$ $$1+\mid \frac{1}{\overline{a_1}}\mid \mid \lambda \mid +\mid \frac{1}{\overline{a_1{a}_2}}\mid {\mid \lambda \mid }^2+\cdots$$ converges. By the root test, this series converges if $$\underset{n\to \infty }{\text{lim}}{\left(\frac{{\mid \lambda \mid }^n}{\mid a_1{a}_2\cdots a_n\mid }\right)}^{\frac{1}{n}}=\underset{n\in {\mathbb{Z}}_{>0}}{\text{limsup}}{\left(\frac{{\mid \lambda \mid }^n}{\mid a_1{a}_2\cdots a_n\mid }\right)}^n<1\text{.}$$ So the series converges if $$\mid \lambda \mid <\underset{n\in {\mathbb{Z}}_{>0}}{\text{limsup}}{\mid a_1{a}_2\cdots a_n\mid }^{\frac{1}{n}}$$ and the series diverges if $$\mid \lambda \mid >\underset{n\in {\mathbb{Z}}_{>0}}{\text{limsup}}{\mid a_1{a}_2\cdots a_n\mid }^{\frac{1}{n}}$$ So $\lambda \in ℂ$ with $\mid \lambda \mid <L$ are eigenvalues of $T^{*},$ where $L=\underset{n\in {\mathbb{Z}}_{>0}}{\text{limsup}}{\mid a_1{a}_2\cdots a_n\mid }^{\frac{1}{n}}\text{.}$ $\square$

8. Let $\left[a_{ij}\right]$ be an infinite complex matrix, $i,j=1,2,\dots ,$ such that if $j\in {\mathbb{Z}}_{>0}$ then $$c_j=\sum _i\mid a_{ij}\mid \text{converges,}\text{and}c=\text{sup}\{c_1,c_2,\dots \}<\infty \text{.}$$ Show that the operator $T:{\ell }^1\to {\ell }^1$ defined by $$T(b_1,b_2,\dots )=(\sum _j{a}_{1j}{b}_j,\sum _j{a}_{2j}{b}_j,\dots )$$ is a bounded linear operator and that $\Vert T\Vert =c\text{.}$

   		Solution.
   	To show: (a) $T$ is a linear operator. (b) $T:{\ell }^1\to {\ell }^1$ is well defined. (c) $\Vert T\Vert \le c\text{.}$ (d) $\Vert T\Vert \ge c\text{.}$ (a) To show: (aa) If $b=(b_1,b_2,\dots )$ and $b\prime =(b_1^{\prime },b_2^{\prime },\dots )$ then $T(b+b\prime )=T\left(b\right)+T\left(b\prime \right)\text{.}$ (ab) If $b=(b_1,b_2,\dots )$ and $\lambda \in ℂ$ then $T\left(\lambda b\right)=\lambda T\left(b\right)\text{.}$ (aa) Let $b=(b_1,b_2,\dots )$ and $b\prime =(b_1^{\prime },b_2^{\prime },\dots )\text{.}$ To show: $T(b+b\prime )=T\left(b\right)+T\left(b\prime \right)\text{.}$ To show: ${T(b+b\prime )}_i={(T\left(b\right)+T\left(b\prime \right))}_i\text{.}$ $$\begin{array}{rcl}{\displaystyle {T(b+b\prime )}_i}& =& {\displaystyle T{(b_1+b_1^{\prime },b_2+b_2^{\prime },\dots )}_i}\\ & =& {\displaystyle \sum _j{a}_{ij}(b_j+b_j^{\prime })}\\ & =& {\displaystyle \sum _j{a}_{ij}{b}_j+\sum _j{a}_{ij}{b}_j^{\prime }}\end{array}$$ and $${(T\left(b\right)+T\left(b\prime \right))}_i=\sum _j{a}_{ij}{b}_j+\sum _j{a}_{ij}{b}_j^{\prime }\text{.}$$ So $T(b+b\prime )=T\left(b\right)+T\left(b\prime \right)\text{.}$ (ab) Let $b=(b_1,b_2,\dots )$ and $\lambda \in ℂ\text{.}$ To show: $T\left(\lambda b\right)=\lambda T\left(b\right)\text{.}$ To show: $T{\left(\lambda b\right)}_i={\left(\lambda T\left(b\right)\right)}_i\text{.}$ $$\begin{array}{rcl}{\displaystyle T{\left(\lambda b\right)}_i}& =& {\displaystyle \sum _j{a}_{ij}{\left(\lambda b\right)}_j}\\ & =& {\displaystyle \sum _j{a}_{ij}\left(\lambda b_j\right)}\\ & =& {\displaystyle \lambda \sum _j{a}_{ij}{b}_j}\\ & =& {\displaystyle \lambda T{\left(b\right)}_i}\\ & =& {\displaystyle {\left(\lambda T\left(b\right)\right)}_i\text{.}}\end{array}$$ So $T\left(\lambda b\right)=\lambda T\left(b\right)\text{.}$ So $T$ is a linear operator if it is a function. (b) To show: $T:{\ell }^1\to {\ell }^1$ is well defined. To show: (ba) If $b=(b_1,b_2,\dots )\in {\ell }^1$ then $Tb$ is defined. (bb) If $b=(b_1,b_2,\dots )\in {\ell }^1$ then $Tb\in {\ell }^1\text{.}$ (ba) Let $b=(b_1,b_2,\dots )\in {\ell }^1\text{.}$ To show: $Tb$ is defined. To show: If $i\in {\mathbb{Z}}_{>0}$ then ${\left(Tb\right)}_i$ is defined. Assume $i\in {\mathbb{Z}}_{>0}\text{.}$ Then $${\left(Tb\right)}_i=\sum _j{a}_{ij}{b}_j$$ which converges since $\sum _j\mid a_{ij}\mid$ converges and $\sum _j\mid b_j\mid$ converges. (More details: By the Cauchy-Schwarz inequality $$\begin{array}{rcl}{\displaystyle \sum _i\mid a_{ij}\mid \mid b_j\mid }& =& {\displaystyle ⟨(\mid a_{i1}\mid \mid a_{i2}\mid ,\dots ),(\mid b_1\mid ,\mid b_2\mid ,\dots )⟩}\\ & \le & {\displaystyle \left(\sum _j\mid a_{ij}\mid \right)\left(\sum _j\mid b_j\mid \right)}\\ & =& {\displaystyle \Vert {\left(a_{ij}\right)}_{j\in {\mathbb{Z}}_{>0}}\Vert \Vert {\left(b_j\right)}_{j\in {\mathbb{Z}}_{>0}}\Vert }\end{array}$$ and since $\sum _j\mid a_{ij}\mid \mid b_j\mid$ converges, $\sum _j{a}_{ij}{b}_j$ converges.) (bb) To show: If $b=(b_1,b_2,\dots )\in {\ell }^1$ then $Tb\in {\ell }^1\text{.}$ Assume $(b_1,b_2,\dots )\in {\ell }^1\text{.}$ To show: $Tb\in {\ell }^1\text{.}$ To show: $\sum _i\mid T{\left(b\right)}_i\mid$ converges. $$\begin{array}{rcl}{\displaystyle \sum _i\mid T{\left(b\right)}_i\mid }& =& {\displaystyle \sum _i\mid \sum _j{a}_{ij}{b}_j\mid }\\ & \le & {\displaystyle \sum _i\sum _j\mid a_{ij}{b}_j\mid }\\ & =& {\displaystyle \sum _j\sum _i\mid a_{ij}\mid \mid b_j\mid }\\ & \le & {\displaystyle \sum _j\sum _i{c}_j\mid b_j\mid }\\ & \le & {\displaystyle \sum _{j}c\mid b_j\mid }\\ & =& {\displaystyle c\Vert b\Vert \text{.}(*)}\end{array}$$ So $Tb\in {\ell }^1\text{.}$ (c) To show: $\Vert T\Vert =c\text{.}$ Since $T{e}_j=(a_{1j},a_{2j}),\dots$ then $$\Vert T{e}_j\Vert =\sum _{i\in {\mathbb{Z}}_{>0}}\mid a_{ij}\mid =c_j=c_j\Vert e_j\Vert \text{.}$$ so $\Vert T\Vert \ge c_j\text{.}$ So $\Vert T\Vert \ge c,$ since $c=\text{sup}\{\mid c_1\mid ,\mid c_2\mid ,\dots \}\text{.}$ To show: $\Vert T\Vert \le c\text{.}$ Let $b=(b_1,b_2,\dots )\in {\ell }^1\text{.}$ By the mysterious computation in $(*),$ $$\mid Tb\mid =\sum _{i\in {\mathbb{Z}}_{>0}}\mid T{\left(b\right)}_i\mid \le c\Vert b\Vert \text{.}$$ So $\Vert T\Vert \le c\text{.}$ So $\Vert T\Vert =c\text{.}$ $\square$

Notes and References

These are a typed copy of Assignment 2 Solutions from a series of handwritten lecture notes for the class Metric and Hilbert Spaces.

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