# Equivalent operator norm on a Hilbert space.

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

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

• Be careful:If the vector space is equal to $\{0\}$ then there will be no element with norm equal to one. – daw Feb 13 '19 at 10:41

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