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 37: Norms of self adjoint operators

(13.3) Let $H$ be a Hilbert space. Let $T:H\to H$ be a bounded self adjoint. Then $$\Vert T\Vert =\text{sup}\left\{\mid ⟨Tx,x⟩\hspace{0.17em}|\hspace{0.17em}\Vert x\Vert =1\mid \text{.}\right\}$$

		Proof.
	To show: (a) $\Vert T\Vert \ge \text{sup}\left\{\mid ⟨Tx,x⟩\mid \hspace{0.17em}\right|\hspace{0.17em}\Vert x\Vert =1\}\text{.}$ (b) $\Vert T\Vert \le \text{sup}\left\{\mid ⟨Tx,x⟩\mid \hspace{0.17em}\right|\hspace{0.17em}\Vert x\Vert =1\}\text{.}$ (a) Assume $x\in H$ and $\Vert x\Vert =1\text{.}$ To show: $\Vert T\Vert \ge \left|⟨Tx,x⟩\right|\text{.}$ $$\begin{array}{rcl}{\displaystyle \mid ⟨Tx,x⟩\mid }& \le & {\displaystyle \Vert Tx\Vert \Vert x\Vert ,\text{by Cauchy-Schwarz}}\\ & \le & {\displaystyle \Vert T\Vert }\\ & =& {\displaystyle \text{sup}\left\{\frac{\Vert Tx\Vert }{\Vert x\Vert }\hspace{0.17em}\right|\hspace{0.17em}x\in H\},\text{by definition}}\\ & =& {\displaystyle \text{sup}\left\{\Vert Tx\Vert \hspace{0.17em}\right|\hspace{0.17em}\Vert x\Vert =1\}\text{.}}\end{array}$$ (b) Let $x\in H$ with $Tx\ne 0$ and $\Vert x\Vert =1\text{.}$ Let $y=\frac{Tx}{\Vert Tx\Vert }$ and let $\beta =\text{sup}\left\{\mid ⟨Tx,x⟩\mid \hspace{0.17em}\right|\hspace{0.17em}\Vert x\Vert =1\}\text{.}$ Then $$\begin{array}{rcl}{\displaystyle \Vert Tx\Vert }& =& {\displaystyle \frac{⟨Tx,Tx⟩}{\Vert Tx\Vert }}\\ & =& {\displaystyle ⟨Tx,y⟩}\\ & =& {\displaystyle \text{Re}⟨Tx,y⟩}\\ & =& {\displaystyle \frac{1}{4}\left(4\text{Re}⟨Tx,y⟩\right)}\\ & =& {\displaystyle \frac{1}{4}(2⟨Tx,y⟩+2\overline{⟨Tx,y⟩})}\\ & =& {\displaystyle \frac{1}{4}(2⟨Tx,y⟩+2⟨y,Tx⟩)}\\ & =& {\displaystyle \frac{1}{4}(2⟨Tx,y⟩+⟨y,T^{*}x⟩)}\\ & =& {\displaystyle \frac{1}{4}(2⟨Tx,y⟩+2⟨Ty,x⟩),\text{since}\hspace{0.17em}T\hspace{0.17em}\text{is self adjoint,}}\\ & =& {\displaystyle \frac{1}{4}(⟨T(x+y),x+y⟩-⟨T(x-y),x-y⟩)}\\ & \le & {\displaystyle \frac{1}{4}\mid ⟨T(x+y),x+y⟩-⟨T(x-y),x-y⟩\mid }\\ & \le & {\displaystyle \frac{1}{4}(\mid ⟨T(x+y),x+y⟩\mid +\mid ⟨T(x-y),x-y⟩\mid )}\\ & \le & {\displaystyle \frac{1}{4}(\mid ⟨\frac{T(x+y)}{\Vert x+y\Vert },\frac{x+y}{\Vert x+y\Vert }⟩\mid {\Vert x+y\Vert }^2+\mid ⟨\frac{T(x-y)}{\Vert x-y\Vert },\frac{x-y}{\Vert x-y\Vert }⟩\mid {\Vert x-y\Vert }^2)}\\ & \le & {\displaystyle \frac{1}{4}(\beta {\Vert x+y\Vert }^2+\beta {\Vert x-y\Vert }^2)}\\ & =& {\displaystyle \frac{1}{4}\beta (⟨x+y,x+y⟩+⟨x-y,x-y⟩)}\\ & =& {\displaystyle \frac{1}{4}\beta ({\Vert x\Vert }^2+{\Vert x\Vert }^2+{\Vert y\Vert }^2+2\text{Re}⟨x,y⟩-2\text{Re}⟨x,y⟩+{\Vert y\Vert }^2)}\\ & =& {\displaystyle \frac{1}{4}\beta ({\Vert x\Vert }^2+{\Vert y\Vert }^2)·2}\\ & =& {\displaystyle \frac{1}{4}\beta (1+1)·2}\\ & =& {\displaystyle \frac{1}{4}\beta ·2·2}\\ & =& {\displaystyle \beta }\\ & =& {\displaystyle \text{sup}\left\{\mid ⟨Tx,x⟩\mid \hspace{0.17em}\right|\hspace{0.17em}\Vert x\Vert =1\}\text{.}}\end{array}$$ $\square$

Let $H$ be a Hilbert space and let $T:H\to H$ be a nonzero self adjoint compact operator. Then there exists $x\in H$ such that $\Vert x\Vert =1$ and
if $u\in H$ and $\Vert u\Vert =1$ then $\Vert ⟨Tu,u⟩\Vert \le \mid ⟨Tx,x⟩\mid \text{.}$

		Proof.
	By Theorem 13.3 $$\Vert T\Vert =\text{sup}\left\{\mid ⟨Tu,u⟩\mid \hspace{0.17em}\right|\hspace{0.17em}\Vert u\Vert =1\}\text{.}$$ Let $x_1,x_2,\dots \in H$ with $\Vert x_n\Vert =1$ and $$\underset{n\to \infty }{\text{lim}}\mid ⟨T{x}_n,x_n⟩\mid =\Vert T\Vert \text{.}$$ Then $x_{n_1},x_{n_2},\dots$ be a subsequence of $x_1,x_2,\dots$ such that $$\underset{k\to \infty }{\text{lim}}⟨T{x}_{n_k},x_{n_k}⟩$$ exists. Use that $T$ is compact to find a subsequence $x_{n_{k_1}},x_{n_{k_2}},\dots$ of $x_{n_1},x_{n_2},\dots$ such that $w=\underset{j\to \infty }{\text{lim}}T{x}_{n_{k_j}}$ exists. Let $x=\frac{w}{\Vert w\Vert }\text{.}$ To show: (a) $\mid ⟨Tx,x⟩\mid =\Vert T\Vert \text{.}$ (b) $Tx=\lambda x$ with $\Vert T\Vert =\lambda \text{.}$ Let $\lambda =\Vert T\Vert \text{.}$ Since $$\begin{array}{rcl}{\displaystyle 0}& \le & {\displaystyle {\Vert (\lambda I-T)\left(x_{n_{k_j}}\right)\Vert }^2}\\ & =& {\displaystyle {\Vert \lambda x_{n_{k_j}}-T{x}_{n_{k_j}}\Vert }^2}\\ & =& {\displaystyle ⟨\lambda x_{n_{k_j}}-T{x}_{n_{k_j}},\lambda x_{n_{k_j}}-T{x}_{n_{k_j}}⟩}\\ & =& {\displaystyle {\lambda }^2\Vert x_{n_{k_j}}\Vert -\lambda ⟨x_{n_{k_j}},T{x}_{n_{k_j}}⟩-⟨T{x}_{n_{k_j}},\lambda x_{n_{k_j}}⟩+{\Vert T{x}_{n_{k_j}}\Vert }^2}\\ & =& {\displaystyle {\lambda }^2\Vert x_{n_{k_j}}\Vert -2\lambda ⟨x_{n_{k_j}},T{x}_{n_{k_j}}⟩+{\Vert T{x}_{n_{k_j}}\Vert }^2}\\ & \le & {\displaystyle {\lambda }^2-2\lambda ⟨x_{n_{k_j}},T{x}_{n_{k_j}}⟩+{\Vert T\Vert }^2\text{.}}\end{array}$$ Since the right hand side approaches $${\lambda }^2-2{\lambda }^2+{\Vert T\Vert }^2=0\text{as}j\to \infty$$ then $$\underset{j\to \infty }{\text{lim}}{\Vert (\lambda I-T)\left(x_{n_{k_j}}\right)\Vert }^2=0$$ so that $$\underset{j\to \infty }{\text{lim}}(\lambda I-T)\left(x_{n_{k_j}}\right)=0\text{.}$$ So $$\begin{array}{rcl}{\displaystyle \lambda w}& =& {\displaystyle \underset{j\to \infty }{\text{lim}}\lambda T{x}_{n_{k_j}}}\\ & =& {\displaystyle \underset{j\to \infty }{\text{lim}}T\left(\lambda x_{n_{k_j}}\right)}\\ & =& {\displaystyle \underset{j\to \infty }{\text{lim}}T((\lambda I-T)x_{n_{k_j}}+T{x}_{n_{k_j}})}\\ & =& {\displaystyle T(0+w)}\\ & =& {\displaystyle Tw\text{.}}\end{array}$$ $\square$

Notes and References

These are a typed copy of Lecture 37 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on October 7, 2014.

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