1
$\begingroup$

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!

$\endgroup$
1

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Can you please explain how you got $|\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle |\leq P\|x+y\|^2+P\|x-y\|^2$? $\endgroup$
    – Teodorism
    Nov 17, 2019 at 20:42
  • 1
    $\begingroup$ For the future readers, the answer is here: math.stackexchange.com/a/3440029/371990 $\endgroup$
    – Teodorism
    Nov 18, 2019 at 0:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .