1
$\begingroup$

I'm a little surprised that I'm stuck on this point, but I have my bad days. I am trying to understand a statement made in the proof of Theorem VI.1 in p.184 of Reed and Simon Volume I Functional Analysis. Let $L(H,\mathbb{C})$ denote the space of linear functionals (operators) from the Hilbert space $H$ to another Hilbert space of complex numbers $\mathbb{C}$. Now, Let $T$ be an operator in $L(H,H)$, the space of operators from $H$ into $H$. Then consider $Tx$ in $L(H,\mathbb{C})$.

It says the operator norm of $Tx$ in $L(H,\mathbb{C})$ is the same as the norm of $Tx$ in $H$.

I know what an operator norm is. To define the operator norm of $Tx$ in $L(H,\mathbb{C})$:

$\sup_{y\neq 0} \frac{|\langle Tx,y \rangle|}{\langle y,y \rangle^{1/2}}$

and the norm of $Tx$ in H is just $\langle Tx,Tx \rangle^{1/2}$.

The conclusion of the two being equal tells me that assuming $\langle y,y \rangle=1$, we have:

$|\langle Tx,y \rangle|= \langle Tx,Tx \rangle^{1/2}$

Is that even true? Where did I go wrong?!

  • 0
    of course, @joriki is right about that, but I figured this was simply a typo because it is so obviously wrong (take $y$ orthogonal to $Tx$).2011-04-20

1 Answers 1

4

In the definition of the operator norm we have $\|T\| = \sup_{\|x\| = 1} \|Tx\| = \sup_{x \neq 0} \frac{\|Tx\|}{\|x\|}$ simply by homogeneity of the norm.

Then note that by Cauchy-Schwarz we have $|\langle Tx,y\rangle| \leq \|Tx\|\,\|y\|$, so for $\|y\| = 1$ this gives $|\langle Tx,y\rangle| \leq \|Tx\|$ and hence $\sup_{\|y\| = 1} |\langle Tx, y \rangle| \leq \|Tx\|$. Of course, we may assume $Tx \neq 0$. Taking $y = \frac{Tx}{\|Tx\|} = \frac{Tx}{\langle Tx,Tx \rangle^{1/2}}$ then yields $|\langle Tx,y\rangle| = \|Tx\|$, so $\sup_{\|y\| = 1} |\langle Tx, y \rangle| = \|Tx\|$ as desired.

  • 0
    @Jonas: Agreed, I can't argue with that, of course. I just thought mentioning Riesz might be helpful, but certainly the isometry part is the easiest bit of the proof of that result.2011-04-22