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

Lecture 30: Examples of linear operators

Let $V$ and $W$ be normed vector spaces over $𝔽,$ where $𝔽=ℝ$ or $𝔽=ℂ\text{.}$

A bounded linear operator from $V$ to $W$ is a linear operator $T:V\to W$ such that there exists $C\in {ℝ}_{>0}$ such that if $u\in V$ then $\Vert Tu\Vert \le C\Vert u\Vert \text{.}$

The norm of $T$ is the minimal $C$ that works.

Let $V$ be a normed vector space and let $$\begin{array}{cccc}1:& V& ⟶& V\\ & x& ⟼& x\end{array}\text{and}\begin{array}{cccc}0:& V& ⟶& V\\ & x& ⟼& 0\end{array}\text{.}$$ Then $\Vert 1\Vert =1$ and $\Vert 0\Vert =0\text{.}$

Let $V={ℝ}^n$ or $V={ℂ}^n$ and let $T:V\to W$ be a linear operator. Then the function $$\begin{array}{cccc}\mathrm{\Phi }:& V& ⟶& {ℝ}_{\ge 0}\\ & x& ⟼& \Vert Tx\Vert \end{array}$$ is continuous (Really? Why?). Let $$S^{n-1}=\{x\in V\hspace{0.17em}|\hspace{0.17em}\Vert x\Vert =1\}\text{.}$$ Since $S^{n-1}$ is closed and bounded, $S^{n-1}$ is compact. Since $\mathrm{\Phi }$ is continuous, $\mathrm{\Phi }\left(S^{n-1}\right)$ is compact. So $\left\{\Vert Tx\Vert \hspace{0.17em}\right|\hspace{0.17em}\Vert x\Vert =1\}$ is closed and bounded in ${ℝ}_{\ge 0}\text{.}$ So $\Vert T\Vert =\text{sup}\left(\mathrm{\Phi }\left(S^{n-1}\right)\right)\text{.}$

Let $C[0,1]=\{f:[0,1]\to ℝ\hspace{0.17em}|\hspace{0.17em}f\hspace{0.17em}\text{is continuous}\},$ with norm given by $$\Vert f\Vert ={\int }_0^1\mid f\left(t\right)\mid dt\text{.}$$ Let $T:C[0,1]\to ℝ$ be given by $$Tf=e{v}_0\left(f\right)=f\left(0\right)\text{.}$$ Let $f_1,f_2,\dots$ be the sequence in $C[0,1]$ given by $$f_n\left(t\right)=\{\begin{array}{ll}-n^{2}t+n,& \text{for}\hspace{0.17em}t\in [0,\frac{1}{n}],\\ 0,& \text{for}\hspace{0.17em}t\in [\frac{1}{n},1]\text{.}\end{array}\begin{array}{c} 1 n 1 t n y \end{array}$$ Then $$\Vert f_n\Vert =\frac{n\left(\frac{1}{n}\right)}{2}=\frac{1}{2}$$ (the area under $y=f_n\left(t\right)\text{),}$ and $$\underset{n\to \infty }{\text{lim}}T{f}_n=\infty$$ in ${ℝ}_{\ge 0}\cup \left\{\infty \right\}$ since $$T{f}_1=1,T{f}_2=2,T{f}_3=3,\dots$$ So there does not exist $C\in {ℝ}_{\ge 0}$ such that if $f\in C[0,1]$ then $\Vert Tf\Vert \le C\Vert f\Vert \text{.}$

Let ${\lambda }_1,{\lambda }_2,\dots$ be a bounded sequence in $ℝ\text{.}$ Define $T:{\ell }^{\infty }\to {\ell }^{\infty }$ by $$T(a_1,a_2,\dots )=({\lambda }_1{a}_1,{\lambda }_2{a}_2,\dots )\text{.}$$ The norm on ${\ell }^{\infty }$ is $$\Vert (a_1,a_2,\dots )\Vert =\text{sup}\{\left|a_1\right|,\mid a_2\mid ,\dots \}\text{.}$$ A basis of ${\ell }^{\infty }$ is $\{e_1,e_2,e_3,\dots \}$ where $$e_i=(0,0,\dots ,0,\stackrel{i\text{th spot}}{1},0,0,\dots )\text{.}$$ Then $$T{e}_i=(0,\dots ,0,{\lambda }_i,0,0,\dots )\text{and}\Vert T{e}_i\Vert =\mid {\lambda }_i\mid =\mid {\lambda }_i\mid \Vert e_i\Vert \text{.}$$ So $$\Vert T\Vert \ge \text{sup}\{\mid {\lambda }_1\mid ,\mid {\lambda }_2\mid ,\dots \}\text{.}$$ Let $x=(x_1,x_2,\dots )$ in ${\ell }^{\infty }\text{.}$ Then $$\begin{array}{rcl}{\displaystyle \Vert Tx\Vert }& =& {\displaystyle \Vert ({\lambda }_1{x}_1,{\lambda }_2{x}_2,\dots )\Vert }\\ & =& {\displaystyle \text{sup}\{\mid {\lambda }_1\mid \mid x_1\mid ,\mid {\lambda }_2\mid \mid x_2\mid ,\dots \}}\\ & \le & {\displaystyle \text{sup}\{\mid {\lambda }_1\mid ,\mid {\lambda }_2\mid ,\dots \}·\text{sup}\{\mid x_1\mid ,\mid x_2\mid ,\dots \}}\\ & =& {\displaystyle C\Vert x\Vert }\end{array}$$ where $C=\text{sup}\{\mid {\lambda }_1\mid ,\mid {\lambda }_2\mid ,\dots \}\text{.}$ So $\Vert T\Vert \le C\text{.}$ So $$\Vert T\Vert =\text{sup}\{\mid {\lambda }_1\mid ,\mid {\lambda }_2\mid ,\dots \}\text{.}$$

Let $C[a,b]=\{f:[a,b]\to ℝ\hspace{0.17em}|\hspace{0.17em}f\hspace{0.17em}\text{is continuous}\}$ with the sup norm $$\Vert f\Vert =\text{sup}\left\{\mid f\left(t\right)\mid \hspace{0.17em}\right|\hspace{0.17em}t\in [a,b]\}\text{.}$$ Let $T:C[a,b]\to ℝ$ be given by $$Tf={\int }_a^{b}f\left(t\right)dt\text{.}$$ If $X:[a,b]\to ℝ$ is the function given by $x\left(t\right)=1$ then $$Tx={\int }_a^{b}dt=b-a\text{and}\Vert Tx\Vert =(b-a)=(b-a)\Vert x\Vert$$ since $\Vert x\Vert =1\text{.}$ So $\Vert T\Vert \ge b-a\text{.}$ If $f\in C[a,b]$ then $$\Vert Tf\Vert =\mid {\int }_a^{b}f\left(t\right)dt\mid \le {\int }_a^b\mid f\left(t\right)\mid dt\le \Vert f\Vert (b-a)\text{.}$$ So $\Vert T\Vert \le b-a\text{.}$ Thus $\Vert T\Vert =b-a\text{.}$

Integral operators. Let $$C[a,b]=\{f:[a,b]\to ℂ\hspace{0.17em}|\hspace{0.17em}f\hspace{0.17em}\text{is continuous}\}$$ with norm given by $$\Vert f\Vert =\text{sup}\left\{\mid f\left(t\right)\mid \hspace{0.17em}\right|\hspace{0.17em}t\in [a,b]\}\text{.}$$ Let $K:[a,b]\times [a,b]\to ℂ$ be a continuous function. Define $T:C[a,b]\to C[a,b]$ by $$\left(Tf\right)\left(t\right)={\int }_a^{b}K(t,s)f\left(s\right)ds$$ (generalised matrix multiplication!).

To show:

(a) If $f\in C[a,b]$ then $Tf\in C[a,b]\text{.}$ (b) $T$ is a bounded linear operator.

(a) Assume $f\in C[a,b]\text{.}$ To show: $Tf$ is continuous. In fact we will show: $Tf$ is uniformly continuous. To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $t,t\prime \in [a,b]$ and $\mid t-t\prime \mid <\delta$ then $$\mid Tf\left(t\right)-Tf\left(t\prime \right)\mid <\epsilon \text{.}$$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ Since $[a,b]\times [a,b]$ is compact and $K$ is continuous then $K$ is uniformly continuous. Let $\delta \in {ℝ}_{>0}$ be such that if $s,s\prime ,t,t\prime \in [a,b]$ and $d((s,t),(s\prime ,t\prime ))<\delta$ then $(K(s,t)-K(s\prime ,t\prime ))<\frac{\epsilon }{(b-a)\Vert f\Vert }\text{.}$ To show: If $t,t\prime \in [a,b]$ and $\mid t-t\prime \mid <\delta$ then $\mid Tf\left(t\right)-Tf\left(t\prime \right)\mid <\epsilon \text{.}$ Assume $t,t\prime \in [a,b]$ and $\mid t-t\prime \mid <\delta \text{.}$ To show: $\mid Tf\left(t\right)-Tf\left(t\prime \right)\mid <\epsilon \text{.}$ $$\begin{array}{rcl}{\displaystyle \mid Tf\left(t\right)-Tf\left(t\prime \right)\mid }& =& {\displaystyle \mid {\int }_a^b(K(s,t)-K(s,t\prime ))f\left(s\right)ds\mid }\\ & \le & {\displaystyle {\int }_a^b\mid K(s,t)-K(s,t\prime )\mid ·\mid f\left(s\right)\mid ds}\\ & <& {\displaystyle \frac{\epsilon }{(b-a)\Vert f\Vert }·(b-a)\Vert f\Vert }\\ & =& {\displaystyle \epsilon \text{.}}\end{array}$$ So $Tf$ is uniformly continuous. So $Tf$ is continuous and so $Tf\in C[a,b]\text{.}$ (b) To show: $T$ is a bounded linear operator. To show: There exists $C\in {ℝ}_{>0}$ such that if $f\in C[a,b]$ then $\Vert Tf\Vert \le C\Vert f\Vert \text{.}$ Since $[a,b]\times [a,b]$ is compact and $K$ is continuous $K([a,b]\times [a,b])$ is compact and $K([a,b]\times [a,b])$ is bounded. Let $$C=\text{sup}\left(\left\{\mid K(s,t)\mid \hspace{0.17em}\right|\hspace{0.17em}s,t\in [a,b]\}\right)\text{.}$$ To show: If $f\in C[a,b]$ then $\Vert Tf\Vert \le C|\left|f\right|\text{.}$ Assume $f\in C[a,b]\text{.}$ Then $$\mid Tf\left(t\right)\mid \le {\int }_a^b\mid K(s,t)\mid ·\mid f\left(s\right)\mid ds\le (b-a)C\Vert f\Vert \text{.}$$ So $$\Vert Tf\Vert =\text{sup}\left\{\mid Tf\left(t\right)\mid \hspace{0.17em}\right|\hspace{0.17em}t\in a,b\}\le (b-a)C\Vert f\Vert \text{.}$$ So $\Vert T\Vert \le (b-a)C\text{.}$ So $\Vert T\Vert$ is bounded.

Notes and References

These are a typed copy of Lecture 30 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on September 17, 2014.

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