Let $E$ be a $\mathbb R$-Banach space and $A$ be a bounded linear operator on $E$. Let $$r(A):=\lim_{n\to\infty}\left\|A^n\right\|^{\frac1n}$$ denote the spectral radius of $A$. I know that $|\lambda|\le r(A)$ for all $\lambda\in\mathbb R$ belonging to the sepctrum $\sigma(A)$ of $A$. Moreover, I know that $\sigma(A)$ is compact.
Are we able to show that there is a $r(A)=\max_{\lambda\in\sigma(A)}|\lambda|$?
I know that this is true in the complex setting. However, as noted in the comments, in the real case $\sigma(A)$ might be empty. However, assuming that $E$ is a $\mathbb R$-Hilbert space and $A$ is self-adjoint, we know that $$\left\|A\right\|_{\mathfrak L(E)}=\max_{z\in\sigma(A)}|\lambda|.$$ On the other hand, if $E$ is a $\mathbb R$-Hilbert space and $A$ is normal, $$r(A)=\left\|A\right\|_{\mathfrak L(E)}.$$
So, assuming that $E$ is a $\mathbb R$-Hilbert space and $A$ is self-adjoint, we should obtain the desired claim. But since this particularly implies that $\sigma(A)$ is nonempty, I'm unsure whether I miss something crucial or this is indeed true.