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 29: Norms of linear operators

Let $V$ and $W$ be vector spaces over $𝔽\text{.}$

A linear operator is a function $T:V\to W$ such that

(a)	if $v_1,v_2\in V$ then $T(v_1+v_2)=T\left(v_1\right)+T\left(v_2\right),$
(b)	if $v\in V$ and $\lambda \in 𝔽$ then $T\left(\lambda v\right)=\lambda T\left(v\right)\text{.}$

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

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 $x\in V$ then $\Vert Tx\Vert \le c\Vert x\Vert \text{.}$

The operator norm of a bounded linear operator $T:V\to W$ is $$\begin{array}{cc}\Vert T\Vert =\text{sup}\left\{\frac{\Vert Tx\Vert }{\Vert x\Vert }\hspace{0.17em}\right|\hspace{0.17em}x\in V\hspace{0.17em}\text{and}\hspace{0.17em}x\ne 0\}\text{.}& (*)\end{array}$$

HW: Let $V$ and $W$ be normed vector spaces. Show that $$B(V,W)=\{\text{bounded linear operators}\hspace{0.17em}T:V\to W\}$$ with norm $\Vert ·\Vert :B(V,W)\to {ℝ}_{\ge 0}$ given by $(*)$ is a normed vector space.

Let $V$ and $W$ be normed vector spaces. If $W$ is a Banach space then $B(V,W)$ is a Banach space.

		Proof.
	Assume $W$ is a Banach space. To show: $B(V,W)$ is complete. To show: If $T_1,T_2,\dots$ is a Cauchy sequence in $B(V,W)$ then $T_1,T_2,\dots$ converges to $T\in B(V,W)\text{.}$ Assume $T_1,T_2,\dots$ is a Cauchy sequence in $B(V,W)\text{.}$ To show: There exists $T\in B(V,W)$ such that $\underset{n\to \infty }{\text{lim}}\Vert T_n-T\Vert =0\text{.}$ Let $T:V\to W$ be given by $$T\left(x\right)=\underset{n\to \infty }{\text{lim}}{T}_n\left(x\right)\text{.}$$ To show: (a) $\underset{n\to \infty }{\text{lim}}{T}_n\left(x\right)$ exists in $W\text{.}$ (b) $T\in B(V,W)\text{.}$ (c) $\underset{n\to \infty }{\text{lim}}\Vert T_n-T\Vert =0\text{.}$ From here it is possible to follow the proof in the Rubinstein notes. $\square$

Let $V$ and $W$ be normed vector spaces and let $T:V\to W$ be a linear operator. Then

(a)	$T\in B(V,W)$ if and only if $T$ is continuous.
(b)	$T$ is continuous if and only if $T$ is uniformly continuous.

		Proof.
	To show: (a) If $T\in B(V,W)$ then $T$ is uniformly continuous. (b) If $T$ is uniformly continuous then $T$ is continuous. (c) If $T$ is continuous then $T\in B(V,W)\text{.}$ (b) was a direct consequence of the way that we set up the definitions of 'uniformly continuous' and 'continuous'. (a) Assume $T\in B(V,W)\text{.}$ To show: $T$ is uniformly continuous. To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $x,y\in V$ and $d(x,y)<\delta$ then $d(Tx,Ty)<\epsilon \text{.}$ This is a consequence of the computation $$d(Tx,Ty)=\Vert Tx-Ty\Vert =\Vert T(x-y)\Vert \le \Vert T\Vert \Vert x-y\Vert =\Vert T\Vert d(x,y)\text{.}$$ (c) To show: If $T$ is continuous then $T\in B(V,W)\text{.}$ Assume $T$ is continuous. To show: $T$ is bounded. To show: There exists $C\in {ℝ}_{>0}$ such that if $u\in V$ then $\Vert Tu\Vert \le C\Vert u\Vert \text{.}$ Since $T$ is continuous, $T$ is continuous at $0\text{.}$ So, there exists $\delta \in {ℝ}_{>0}$ such that if $x\in V$ and $\Vert x\Vert \le \delta$ then $\Vert Tx\Vert <1\text{.}$ Let $C=\frac{2}{\delta }\text{.}$ To show: If $u\in V$ then $\Vert Tu\Vert \le C\Vert u\Vert \text{.}$ Assume $u\in V\text{.}$ Let $x=\frac{\delta }{2}\frac{u}{\Vert u\Vert }$ so that $\Vert x\Vert <\frac{\delta }{2}\text{.}$ Then $$1>\Vert Tx\Vert =\Vert T\left(\frac{\delta }{2}\frac{u}{\Vert u\Vert }\right)\Vert =\frac{\delta }{2\Vert u\Vert }\Vert Tu\Vert \text{.}$$ So $$\Vert Tu\Vert <\frac{2}{\delta }\Vert u\Vert =C\Vert u\Vert \text{.}$$ So $T$ is bounded. $\square$

Notes and References

These are a typed copy of Lecture 29 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on September 16, 2014.

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