Spectral radius of an operator equals its norm

Question

Let $X$ be a Banach space and $A:X\to X$ a bounded operator. We know that the spectrum of $A$ is always included in the ball $B(0,|A|)$ and the spectral radious $r(A)$ is the smalest radius such that $\sigma(A)\subset B(0,r(A))$. I am just asking for some examples of bounded operators such that $r(A)=|A|$ and other operators such that $r(A)<|A|$.

Answer 1

When $X$ is finite dimensional and Hilbert rather than merely Banach, then self-adjoint operators satisfy $r(A)=\| A \|$. You can prove this with the SVD and the spectral theorem. In the infinite dimensional case you should be able to do the same thing, but the details become more subtle, because not all members of the spectrum are of the same type.

Edit: even in the finite dimensional case, self-adjointness is not necessary. In particular, normality is sufficient. This can again be proven by the SVD and the (normal version of the) spectral theorem; all that changes is that the matrix of signs of eigenvalues that is used to "turn the SVD into the eigenvalue decomposition" may contain complex entries. But as http://ac.els-cdn.com/0024379574900767/1-s2.0-0024379574900767-main.pdf?_tid=5f7ad1b6-f824-11e3-98da-00000aab0f6b&acdnat=1403232265_d4aa8b685fddc47d76be3e0892fc49e3 shows, normality is not necessary either.