Operator Norm of a Self-adjoint Operator

functional-analysislinear algebra

I am trying to prove $\|A\|=\sup_{||x||\leq 1}|\langle x,Ax\rangle|$ for a self adjoint bounded linear operator $A$ on a complex Hilbert space. I know how to prove $\|A\|\geq\sup_{||x||\leq 1}|\langle x,Ax\rangle|$. For the converse, however,
$$ \|A\| = \sup_{||x||\leq 1} \| Ax \|$$
$$ = \sup_{||x||\leq 1} \langle Ax,Ax\rangle^{1/2}$$
$$ = \sup_{||x||\leq 1} \langle x,A^*Ax\rangle^{1/2} $$
$$ = \sup_{||x||\leq 1} \langle x,A^2x\rangle^{1/2} $$
And then I'm stuck!

Best Answer

One can easily verify that $\|A\|=\sup\{|\langle Ax,y\rangle| : x,y \in \mathcal H,\ \|x\|=\|y\|=1\}$ in a complex Hilbert space $\mathcal H$.

Notice that $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle = 2\langle Ax,y\rangle+2\langle Ay,x\rangle.$$ Since $A$ is self adjoint, $\langle Ay,x\rangle=\langle y,Ax\rangle=\overline{\langle Ax,y\rangle}.$ So $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle=4\newcommand{\re}{\operatorname{Re}}\re\langle Ax,y\rangle.$$ Let $P:=\sup_{||x||\leq 1}|\langle x,Ax\rangle|.$ Then $$\begin{align*} |4\re\langle Ax,y\rangle| &=|\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle |\\ &\leq P\|x+y\|^2+P\|x-y\|^2\\ &=2P(\|x\|^2+\|y\|^2). \end{align*}$$ So, whenever $\|x\|=\|y\|=1$, we have $$|\re\langle Ax,y\rangle|\leq P\tag{$\color{red}{1}$}\label1.$$

Suppose $\langle Ax,y\rangle=re^{i\theta}$ with $\|x\|=\|y\|=1$. I will construct an element $z$ with $\|z\|=1$ such that $|\langle Ax,y\rangle|=|\re\langle Az,y\rangle|$. Then we can apply $(\ref 1)$ to $|\re\langle Az,y\rangle|$ and we will get $|\langle Ax,y\rangle|\leq P $.

Consider $z=e^{-i\theta} x$. Then $\|z\|=1$. Also, note that $$\langle Az,y\rangle=e^{-i\theta}\langle Ax,y\rangle=r=\re\langle Az,y\rangle,$$ and $|\langle Ax,y\rangle|=r$. So $$|\langle Ax,y\rangle|=r=|\re\langle Az,y\rangle|\leq P.$$ Hence $\|A\|\leq P$.

I learnt this proof from Functional Analysis by B.V. Limaye.

Related Question