Let $H$ be a Hilbert space. let $e$ be a normed vector of $H$. Let $A$ be the application from $H$ in $H$ défined by : [ x \mapsto x-2\langle x, e\rangle e ]
- Show that $A$ is linear continuous, of a norme inferior to 1
- Show by a certain choice of $x$ that $\|A\|_{\mathcal{L}(H)}=1$ It was easy to show that it was linear, but then the continuity and a norm I didn't see a way:
$\begin{array}{l} \|A x\|^{2}=\langle A x, A x\rangle=\langle x-2\langle x, e\rangle e,\langle x-2\langle x, e\rangle e\rangle \\ =\|x\|^{2}-4\langle x, e\rangle\langle e . x\rangle+4\|\langle x, e\rangle\|^{2}\|e\|^{2} \end{array}$
$\langle x, e\rangle=\langle e . x\rangle$ and $e$ is normed vector, so we have
$\|A x\|^{2}=\|x\|^{2}$
but then how to continue?
$\|Ax\|=\|x\|$ implies that $A$ is bounded and $\|A\|=\sup \{\|Ax\|: \|x\|=\sup \{\|x\|: \|x\| \leq 1\}=1$. The norm is attained when $x=e$.