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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?!

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  • $\begingroup$ No, that's not true; you ignored the supremum. Your last equality only holds for those $y$ at which the supremum is attained. $\endgroup$ – joriki Apr 20 '11 at 5:18
  • $\begingroup$ 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 $Tx$). $\endgroup$ – t.b. Apr 20 '11 at 5:25
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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.

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  • $\begingroup$ What exactly do you mean by the homogeneity of the norm? Thanks! $\endgroup$ – r.g. Apr 20 '11 at 5:17
  • $\begingroup$ @Rohan: $\|\lambda x\| = |\lambda| \|x\|$ $\endgroup$ – t.b. Apr 20 '11 at 5:17
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    $\begingroup$ @Rohan: By the way what you're asking about is a special instance of the Riesz representation theorem $\endgroup$ – t.b. Apr 20 '11 at 5:49
  • $\begingroup$ By what I'm asking for, do you mean the Theorem VI.1. In that case, that is true, I'm just developing analogues to operators! $\endgroup$ – r.g. Apr 20 '11 at 8:34
  • $\begingroup$ @Rohan: I meant the fact that the linear functional $x \mapsto \langle \cdot, y\rangle$ has norm $\|y\|$ follows from (part of) the Riesz representation theorem. Could you clarify what exactly you mean by "developing analogues to operators"? $\endgroup$ – t.b. Apr 20 '11 at 8:41

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