For $A$ bounded operator, $\|A\|\ge |\alpha|$

Question

I want to show that if $A$ is a bounded operator mapping between hilbert spaces $(\mathcal{H_1},\|\cdot\|_X)$ and $(\mathcal{H_2},\|\cdot\|_Y)$, then the set of eigenvalues is bounded above by $\|A\|$.

Let wlog $\lambda\in\mathbb{C}$ be the eigenvalue for a normalized eigenvector $v\in\mathcal{H}$.

Because $A$ is bounded we have that

$\|A\| = \sup_{\|u\|_X=1} \left \|Au\right \|_Y< \infty$

Noting that $\|Au\|_Y=|\lambda|\|u\|_Y\ge |\lambda|$ and taking the supremum on both sides, we get that

$\sup_{\|u\|_X=1} \left \|Au\right \|_Y =\|A\| \ge |\lambda|$ and because eigenvalue and eigenvector were arbitrary, the conclusion follows.

Is my proof correct?

Answer 1

Yes, your proof is correct, though the hypothesis is not exactly: we should have $\mathcal H_1=\mathcal H_2$ in order to speak about eigenvectors.
Note that for a normalized eigenvector $v$, the value $|\lambda|=\|Av\|$ is present in the set that we take the supremum of.