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In this question $\mathcal{H}$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{B}(\mathcal{H})$ the algebra of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}$. Let $T\in \mathcal{B}(\mathcal{H})$, is this inequality true $$\displaystyle \sup_{\|x\|=\|y\|=1}|\langle Tx\;| \;y\rangle|\leq \displaystyle \sup_{\|x\|=1}|\langle Tx\;| \;x\rangle|\;?$$ And thank you for your help.

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  • $\begingroup$ As 5xum argued below, this is not true. However, if $T$ is a selfadjoint and positive semi-definite contraction, the above holds with taking the square root on the RHS. $\endgroup$ – amsmath Oct 16 '17 at 13:16
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Certainly not.

Take $\mathcal H = \mathbb R^2$ and take $A$ the linear operator mapping $(x,y)$ to $(-y, x)$ which is a bounded linear operator.

Then, $$\langle A\vec x, \vec x\rangle =\langle((-y, x), (x,y)\rangle = -xy+xy= 0$$ for all $x$, however $\langle A(1,0), (0,1)\rangle =\langle(0, 1), (0,1)\rangle = 0+1= 1$.


The inequality is, however, true if you reverse it, because the set

$$\{\langle Tx, x\rangle| x\in\mathcal H, \|x\|=1\}$$

is a subset of the set

$$\{\langle Tx, y\rangle | x,y\in\mathcal H, \|x\|=\|y\|=1\}$$

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A nice counterexample, because it addresses also the complex case, is to take $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix}. $$ Then $$ \sup\{\langle Ax,y\rangle:\ \|x\|=\|y\|=1\}=1, $$ while $$ \sup\{\langle Ax,x\rangle:\ \|x\|=1\}=\frac12. $$

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