Can the spectral radius be checked on finite-dimensional subspaces?

Question

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.

Ben Grossmann · Answer

The answer is yes.

Let $0 < epsilon < 1$. There exists a $k$ for which $|A^k|^{1/k} > rho(A) - epsilon$, i.e. $|A^k| > [rho(A) - epsilon]^k geq rho(A)^k - kepsilon^k$. 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.

Can the spectral radius be checked on finite-dimensional subspaces?

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