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It is fairly easy to show that for a bounded linear operator $T$ on a Hilbert space $H$ $$||T||=\sup_{||x||=1,||y||=1}|\langle y, Tx \rangle |.$$ If $H$ is a complex Hilbert space, can you show that $$||T||=\sup_{||x||=1}|\langle x, Tx \rangle |\;?$$

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Not in general real Hilbert spaces, since there exist, for instance, nonzero operators $T$ such that $(x,Tx)=0$ for all $x$. – 1015 Feb 8 '13 at 22:44
$\langle Qx, x \rangle = 0$ for all $x\in H$ with $H$ complex implies $Q=0$ since $0 = \langle Q(\alpha x + y), \alpha x + y \rangle$. Let $\alpha = 1$ and $\alpha = i$ to get two equations implying $\langle Qx,y\rangle=0$. – CCL Feb 8 '13 at 22:51
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Not when it isn't true, e.g. for the operator $T(x,y)=(y,0)$ on $\mathbb C^2$ with the standard inner product. In that case the $\sup$ is $\frac{1}{2}$ but the operator has norm $1$.

In general, that $\sup$ gives you the numerical radius, which is equivalent but often unequal to the operator norm.

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Yes! There are even cases where $(x,Tx)=0$ for all $x$. – 1015 Feb 8 '13 at 22:43
@julien: Not on complex inner product spaces there aren't. E.g. see… – Jonas Meyer Feb 8 '13 at 22:44
Yes again! Thank you. – 1015 Feb 8 '13 at 22:47
Thanks for the answer. Interesting link. – CCL Feb 8 '13 at 23:26

Yes, if the operator is self-adjoint. Here is a proof from Conway's book, A Course in Functional Analysis:

Let $M = \{\sup |\langle Ax, x\rangle| : \|x\| = 1\}$. If $\|x\| = 1$, then $|\langle Ax,x\rangle | \leq \|A\|$ (since this holds for all pairs of vectors, so certainly works if we use the same in each slot), so $M \leq \|A\|$.

Now let $h,g$ be unit vectors. Self-adjointness gives $$\langle A(h \pm g) , h \pm g\rangle = \langle Ah,h\rangle \pm 2\mbox{Re}\langle Ah,g\rangle + \langle Ag,g \rangle.$$ Subtracting these equations gives $$\langle A(h + g), h + g \rangle - \langle A(h-g), h-g \rangle = 4 \mbox{Re} \langle Ah,g \rangle.$$ Since $|\langle Af, f \rangle| \leq M\|f\|^2$ for any $f$, by Cauchy-Schwarz, we have using the parallelogram law that \begin{align*}&4\mbox{Re}\langle Ah,g\rangle \\\leq &M(\|h + g\|^2 + \|h-g\|^2) \\ =&2M(\|h\|^2 + \|g\|^2) \\ = &4M. \end{align*}

To finish up, choose $e^{i\theta}$ so that $\mbox{Re}(e^{i\theta}\langle Ah, g\rangle) = |\langle Ah,g \rangle|$. Then $$|\langle Ah, g \rangle| = \mbox{Re}(\langle Ae^{i\theta}h, g\rangle ) \leq M.$$ Take the supreumum over all $h,g$ to finish.

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So self-adjointness of $T$ is critical to the argument. – CCL Feb 8 '13 at 23:02
Thanks for the proof. I have seen this proved for self-adjoint operators, but this proof is cleaner than the one I have seen. – CCL Feb 8 '13 at 23:44

Yes, of course. Use the polarization identity: $\newcommand{\z}{\langle}\newcommand{\x}{\rangle}$ $$\z x+y,T(x+y)\x + i \z x+iy,T(x+iy)\x - \\ - \z x-y,T(x-y)\x -i \z x-iy,T(x-iy)\x = \,\dots $$

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