# Operator Norm $\| T\|$

Would somebody mind explaining why if $T$ is a continuous and bounded operator on a Hilbert space $H$, we have

$$\|T\| = 1 \;\;\;\Rightarrow \;\;\;\|Te_n \| = \|e_n\|\;\;\mbox{for all }\;\;x\in H$$

where $(e_n)_{n\in \mathbb{N}}$ is an ON Basis of $H$?

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This is not true. We only have $\|Tx\| \leq \|x\|$, for all $x \in X$. For example, if $X = \ell_2$ with orthonormal basis $\{e_n\}$, define the map $T$ by $T(\sum a_n e_n) =a_1 e_1$. Then $\|T\|=1$, but $\|Te_2\|=\|0\|=0$. Asking that $\|Tx\|=\|x\|$ for all $x$ is a much stronger condition; a map $T$ satisfying this condition is called an isometry. – Tom Cooney May 4 '12 at 17:07
What about now, if its just for an ON Basis/family? – rk101 May 4 '12 at 17:10
The exact same counterexample applies. In the example above, $\|T\|=1$, but $\|Te_n\|=0$ for $n \neq 1$. – Tom Cooney May 4 '12 at 17:11
Ok Thanks for your help – rk101 May 4 '12 at 17:12
Indeed, $T$ satisfies $\|T e_n\| = \|e_n\|$ for all $e_n$ in an orthonormal basis if and only if $T$ is an isometry. – Nate Eldredge May 4 '12 at 21:17

Let $H=\ell^2$ the set of real (or complex) suare-summable sequences with the natural inner-product. Let $e^{(n)}$ the sequence such that the only non-vanishing term is the $n$-th, which is $1$, and define $T(x)=\sum_{n=0}^{+\infty}\frac n{n+1}\langle e^{(n)},x\rangle e^{(n)}$. Then $T$ is well defined, linear, of norm $1$ since $\lVert Te^{(n)}\rVert=\frac n{n+1}$ and for all $x$, $\lVert Tx\rVert\leq 1$. But $Te^{(k)}=\frac k{k+1}e^{(k)}$ for each $k$ so $\lVert Te^{(k)}\rVert=\frac k{k+1}<1$. In fact, in this case, for all $x$ we have $$\lVert Tx\rVert^2=\sum_{n=0}^{+\infty}\frac{n^2}{(n+1)^2}|\langle x,e^{(n)}\rangle|^2<\lVert x\rVert,$$ except if $x=0$. So the supremum in the definition of the norm of $T$ is not a maximum.