Let $\mathscr{H}$ be a Hilbert space and let $\mathbb{R} \to B(\mathscr{H}), r \mapsto S\_r$ be a continuous (or smooth) family of operators, where $B(\mathscr{H})$ is the space of bounded operators on $\mathscr{H}$.


Denote by $\rho(S\_r)$ the spectral radius of $S\_r$ and assume that $(\sup\_{r \in [0,1]}\rho(S\_r)) < 1$. I want to show for large enough $n$, that $$\left(\sup\_{r \in [0,1]} ||S\_r^n||^{\frac{1}{n}} \right) < 1.$$
Does this hold with the above assumptions? If not, what would one need to assume in addition in order for this to hold? Clearly this would hold if the operator $S\_r$ are all normaloid (i.e. when $\rho(S\_r) = ||S\_r||$), yet I don't want to assume this.



Uniform decay of operator norm for smooth family of operators
This works, essentially because $\|T^k\|^{1/k}$ for a given $k=n$ also controls this quantity for $k\ge n$.


More specifically, suppose that $\|T^n\|^{1/n}\le 1-\delta$. Clearly, $\|T^{kn}\|^{1/kn}\le \|T^n\|^{1/n}$, and for general $N\ge n$, write $N=kn+j$, $0\le j<n$, and estimate
$$
\|T^N\|^{1/N}\le \|T^{kn}\|^{1/N} \|T\|^{j/N} .
$$
For large $k$, we need not worry about the last factor since there is a uniform bound on $\|S\_r\|$, $0\le r\le 1$, so it will be close to $1$. As for the first factor, we further estimate
$$
\|T^{kn}\|^{1/N} = \left( \|T^{kn}\|^{1/kn}\right)^{1-j/N}\le\left( \|T^n\|^{1/n}\right)^{1-j/N}\le (1-\delta)^{1/2} .
$$


Now a standard compactness argument works: Fix a sufficiently small $\delta>0$ so that we can pick, for each $0\le r\le 1$, an $n=n(r)$ such that $\|S\_r^n\|^{1/n}\le 1-\delta$ and then an interval $I=(r-d,r+d)$ such that we still have this inequality (with $1-\delta/2$, say) for this $n$ and all $t\in I$. Finally, cover $[0,1]$ by finitely many of these intervals and take the largest $n$.


