Assume we have a (bounded) linear operator $A \colon X \to X$ on a (complex) Hilbert space. Then $\rho(A) = \inf_{k > 0} \|A^k\|^{1/k}$, where $\rho(A)$ denotes the spectral radius. Furthermore, we have
$\inf_{k > 0} \|A^k\|^{1/k} = \inf_{k > 0} \sup_{h \in A: \|h\| = 1} \|A^k h\|^{1/k} \geq \sup_{h \in A: \|h\| = 1} \inf_{k > 0} \|A^k h\|^{1/k}$.
Question. Is the inequality above an equality?
It is clear, that this is true if $A$ is selfadjoint or has a discrete spectrum.
One can also consider the sequence space $c_{00} = \{ (a_0, a_1, \dots) \colon \text{ all } a_i = 0 \text{ except for finitely many.} \}$ together with some norm you like. This space is not complete. For the shift operator $A = (a_0, a_1, a_2, \dots) \mapsto (a_1, a_2, a_3, \dots)$ one has $$ \inf_{k > 0} \|A^k\|^{1/k} = 1 > 0 = \sup_{h \in A: \|h\| = 1} \inf_{k > 0} \|A^k h\|^{1/k}. $$ Thus, the assumption of completeness is quint-essential.
The answer is yes.
Let $0 < \epsilon < 1$. There exists a $k$ for which $\|A^k\|^{1/k} > \rho(A) - \epsilon$, i.e. $\|A^k\| > [\rho(A) - \epsilon]^k \geq \rho(A)^k - k\epsilon^k$. Thus, there exists an $x \in H$ with $\|x\| = 1$ for which $\|A^kx\| > \|A^k\| - \epsilon^k = \rho(A) - (k+1)\epsilon^k$. We note that $$ \|A^kx\|^{1/k} > (\rho(A)^k - (k+1)\epsilon^k)^{1/k} \geq \rho(A) - \frac{k}{k+1} \epsilon. $$ It follows that $$ \sup_{h \in A: \|h\| = 1} \inf_{k > 0} \|A^k h\|^{1/k} \geq \inf_{k > 0} \|A^k x\|^{1/k} \geq \rho(A) - \epsilon. $$ Because this holds for arbitrary $\epsilon \in (0,1)$, we conclude that $$ \sup_{h \in A: \|h\| = 1} \inf_{k > 0} \|A^k h\|^{1/k} \geq \rho(A), $$ which was what we wanted.