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Let $T: X \to Y$ a bounded linear operator. Prove that $$\|T\|=\sup\{\|T(x)\|: x\in X , \|x\|<1 \}.$$

It is $||T||= \sup \{||T(x)||: x\in X ,||x|| \leq1 \}$ so $||T|| \geq \sup\{||T(x)||: x\in X , ||x||<1 \}$.

I can't prove the other inequality.

It isn't so difficult but for some reason I am stuck.

Any help?

Thanks in advance!

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up vote 6 down vote accepted

Define $B_X:=\{x\in X:\Vert x\Vert\leq 1\}$ and $B_X^\circ:=\{x\in X: \Vert x\Vert <1\}$. Since $\Vert T\Vert=\sup\limits_{x\in B_X}\Vert Tx\Vert$, then there exist $\{x_n\}_{n=1}^\infty\subset B_X$ such that $\Vert T\Vert =\lim\limits_{n\to\infty}\Vert Tx_n\Vert$. Consider new sequence $y_n=(1-2^{-n})x_n$. Since $\{x_n\}_{n=1}^\infty\subset B_X$, then $\{y_n\}_{n=1}^\infty\subset B_X^\circ$. Moreover $$ \lim\limits_{n\to\infty}\Vert Ty_n\Vert=\lim\limits_{n\to\infty}\Vert (1-2^{-n})Tx_n\Vert=\lim\limits_{n\to\infty}(1-2^{-n})\cdot\lim\limits_{n\to\infty}\Vert Tx_n\Vert=\Vert T\Vert. $$ Thus we conclude $\sup\limits_{x\in B_X^\circ}\Vert Tx\Vert\geq \Vert T\Vert$, but on the other hand $\sup\limits_{x\in B_X^\circ}\Vert Tx\Vert\leq\sup\limits_{x\in B_X}\Vert Tx\Vert=\Vert T\Vert$. So $$ \Vert T\Vert=\sup\limits_{x\in B_X^\circ}\Vert Tx\Vert=\sup\{\Vert Tx\Vert:x\in X,\Vert x\Vert<1\} $$

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Thank you very much for your time! – passenger Feb 23 '12 at 8:48

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