For a normal operator $A$ on a Hilbert space $H$, I know that spectral radius $r(A)=\|A\|$. The question says is it true that either $\|A\|$ or $-\|A\|$ belongs to spectrum $\sigma(A)$?
I have started this way. We can obtain $\lambda_n \in \sigma(A)$ such that $\lvert\lambda_n\rvert \to \|A\|$. For a moment I thought I am through as this will imply $\lambda_n \to \|A\|$ and since $\sigma(A)$ is closed, $\|A\| \in \sigma(A)$. But soon I realised my mistake. But now I am stuck on how to proceed.
Hint: The spectrum of a normal operator need not be real. Try to look for a counter-example among (low dimensional) matrices.