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Let $L$ be a bounded linear operator on a Hilbert space. Without assuming finite dimensions, can we express the operator norm of $L$ in terms of the spectrum of the positive operator $L^{\dagger}L$?

More precisely, does the following hold.

$\sup \big\{ \| L\phi \| \,\big|\, \phi \in \mathcal{H} \land \|\phi\| = 1 \big\} = \sqrt{ \max \sigma(L^{\dagger} L)}$

Marco Tomamichel
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    For $L^L$ we have that $|L^L| = r(L^L)$ - spectral radious of $L^L$, i.e. $r(L^L) = \sup_{\lambda \in \sigma(L^L)}\lambda$. Since $B(H)$ is a C-algebra $|L^L| = |L|^2$. – Frank Tessla Mar 03 '13 at 10:02

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I will convert Frank's comment into the answer.

As it was showed in this answer (theorem 3.4) for every normal element $a$ of a unital $C^*$-algebra $A$ holds $$ \Vert a\Vert=r(a) $$ For your particular case $A=\mathcal{B(H)}$ and $a=L^*L$. Using $C^*$-identity and the fact that $L^*L$ is normal we get $$ \Vert L\Vert^2=\Vert L^*L\Vert=r(L^*L)=\max\sigma(L^*L) $$

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Norbert
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