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I see the following paragraph in a paper

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I think there is a wrong in the paper because, I think we have only $$\|T\|_A=\sup_{\substack{\xi,\eta\in \overline{R(A)},\\\|\xi\|_A \leq 1, \|\eta\|_A \leq 1}}|\langle T\xi\;|\;\eta\rangle _A| \,.$$

Thank you for your help.

Student
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2 Answers2

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As you say, you need to restrict the $x$ and $y$ to the range of $A$.

Take $H=\mathbb C^2$, with $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$. Take $T=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. Then for $\omega=(\omega_1,\omega_2)$, $$ \|\omega\|_A=|\omega_1| $$ and $$R(A)=\{(t,0):\ t\in\mathbb C\}.$$ Thus $T\omega=0$ for all $\omega\in R(A)$ and \begin{align} \|T\|_A&=0. \end{align} On the other hand, since $AT=T$, \begin{align} \sup\{|\langle Tx,y\rangle_A|:\ \|x\|_A=\|y\|_A=1\} &=\sup\{|\langle Tx,y\rangle|:\ \|x\|_A=\|y\|_A=1\}\\ \ \\ &=\sup\{|x_2\overline{y_1}|:\ |x_1|=|y_1|=1\}\\ \ \\ &=\infty \end{align} (on the other hand, if we force $x,y\in R(A)$, then $x_2=0$ and the two expressions agree).

Martin Argerami
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Because $|\langle Tx,y\rangle| \le \|Tx\|\|y\| \le \|T\|\|x\|\|y\|$, then $$ \sup_{\|x\|=1,\|y\|=1}|\langle Tx,y\rangle| \le \|T\|. $$ Without losss of generality, assume $\|T\|\ne 0$. Let $0 < \epsilon < \|T\|$ be given. There exists $x'$ such that $\|x'\|=1$ and $\|Tx'\| \ge \|T\|-\epsilon$. Therefore, $$ \sup_{\|x\|=1,\|y\|=1}|\langle Tx,y\rangle| \ge |\langle Tx',\frac{1}{\|Tx'\|}Tx'\rangle| = \|Tx'\| \ge \|T\|-\epsilon. $$ Because this is true of all $\epsilon > 0$, then $\sup_{\|x\|=1,\|y\|=1}|\langle Tx,y\rangle| \ge \|T\|$.

Disintegrating By Parts
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    Why we have supposed that $A=I$? Thank you – Student Dec 16 '17 at 14:28
  • @Student : Where can you not add $A$ to the argument without a problem? – Disintegrating By Parts Dec 16 '17 at 14:32
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    I think that $$|T|A=\sup{\substack{x,y\in \overline{R(A)},\|x|_A \leq 1, |y|_A \leq 1}}|\langle Tx;|;y\rangle _A| .$$ – Student Dec 16 '17 at 14:34
  • My problem is why $x,y\in H$ – Student Dec 16 '17 at 14:35
  • @Student : Treat $\langle \cdot,\cdot\rangle_A$ as an inner product with induced norm $|\cdot|_A$. Then $T$ is assumed to be a bounded operator with respect to this norm. The arguments go through as for the usual norm. $A$ is defined on all of $\mathcal{H}$, and there is no completion process going on. – Disintegrating By Parts Dec 16 '17 at 17:41
  • $$\langle\cdot;|;\cdot\rangle_{A}:\mathcal{H}\times \mathcal{H}\longrightarrow \mathbb{C},;(x,y)\longmapsto\langle x;| ;y\rangle_{A} =\langle Ax;| ;y\rangle.$$ Note that $\langle .;|;.\rangle_A$ defines a semi-inner product on $\mathcal{H}$ – Student Dec 16 '17 at 18:23
  • I think there is a wrong in the paper. please see this question https://math.stackexchange.com/questions/2567495/generalisation-of-the-norm-of-bounded-linear-operators?noredirect=1&lq=1 – Student Dec 16 '17 at 18:24
  • The inequality in the second line is not necesarily true if you keep the $A$. The answer is reasoning as if $T$ leaves $\overline{R(A)}$ invariant, which is not the case in general. – Martin Argerami Dec 17 '17 at 02:36
  • $A$ is defined on all of $\mathcal{H}$ and $T : \mathcal{H}\rightarrow\mathcal{H}$. – Disintegrating By Parts Dec 17 '17 at 05:44
  • but $A$ hasn't a dense range. – Student Dec 17 '17 at 07:58
  • @Student : It's the definition of the norm of the operator. Put $A$ on the norm. But $A$ on the inner product. Check how the norm was defined. – Disintegrating By Parts Dec 17 '17 at 14:36
  • @Student : I think the confusion is that you're thinking they're renorming the range of $A$. They're renorming $\mathcal{H}$ using $A$. The new norm is defined on all on $\mathcal{H}$. – Disintegrating By Parts Dec 17 '17 at 14:44
  • @DisintegratingByParts: I fail to see what's your point. The two expressions in the paper quoted by the OP are different, as shown in my example. You seem to insist in that they are equal; that means that you claim my example is somehow wrong, so I would like you to point out where/how. – Martin Argerami Dec 17 '17 at 16:42