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Let $T$ be any bounded linear operator on Hilbert space $H$ then we know that the operator norm of $T$ can be defined by $\|T\| = \sup\{ |\langle Tx,y\rangle| : \|x\|=\|y\|=1\}$. Now how I can prove the following formula :

$\|T\| = \sup\{ |\langle Tx,y\rangle| : \|x\| < 1 , \|y\| < 1 \}$.

I just find the page Equivalent definition operator norm but I think it's not exactly true for my question. Actually I can prove that

$\|T\| = \sup\{ |\langle Tx,y\rangle| : \|x\| \le 1 , \|y\| \le 1 \}$

, but I can not conclude that

$\|T\| = \sup\{ |\langle Tx,y\rangle| : \|x\| < 1 , \|y\| < 1 \}$.

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  • $\begingroup$ Be careful:If the vector space is equal to $\{0\}$ then there will be no element with norm equal to one. $\endgroup$ – daw Feb 13 '19 at 10:41
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By your definition $$ \|T\| = \sup\{ |\langle Tx,y\rangle| : \|x\|=\|y\|=1\}, $$ we can find sequences of $x_n,y_n\in H$ such that $\|x_n\|=\|y_n\|=1$ and $|\langle Tx_n,y_n\rangle| \to \|T\|$. More specifically, we can choose $x_n,y_n$ so that $$ |\langle Tx_n,y_n\rangle| \ge \|T\| -\frac 1n. $$ Now, consider the sequence $x'_n=\frac n{n+1}x_n, y'_n=\frac n{n+1}y_n$. It is obvious that $\|x'_n\|=\|y'_n\| <1$. Direct computation shows that $$ |\langle Tx'_n,y'_n\rangle|=\frac {n^2}{(n+1)^2}|\langle Tx_n,y_n\rangle| \ge \frac {n^2}{(n+1)^2}\|T\| - \frac {n}{(n+1)^2}. $$ By taking $n\to\infty$, we can see that $$ \sup_{n\in\Bbb N} |\langle Tx'_n,y'_n\rangle| \ge \|T\|, $$ and this is exactly what you need.

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  • $\begingroup$ This works only if $H\ne\{0\}$ $\endgroup$ – daw Feb 13 '19 at 10:45

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