Let $A_1, A_2, \ldots$ be a sequence of finite-dimensional linear transformations. Assume that, for each fixed $x$ and $y$, the sequence $\langle A_kx,y\rangle$ is bounded. Does it follow that $\Vert A_k\Vert_2$ is bounded?
If this doesn't hold, do we have the weaker:
If $A$ is a linear transformation, and if $\langle A^kx,y\rangle$ is bounded over $k$ for each fixed $x,y$, does it follow that $\Vert A^k\Vert_2$ is bounded?
I want to think that at least one of these is true, by some use of continuity or compactness. I basically get:
$$ \sup_k\Vert A_k\Vert = \sup_k\sup_{\Vert x\Vert=\Vert y \Vert=1}\langle A_kx,y\rangle = \sup_{\Vert x\Vert=\Vert y \Vert=1}\sup_k\langle A_kx,y\rangle = \sup_{\Vert x\Vert=\Vert y \Vert=1}K_{x,y}, $$ where each $K_{x,y}$ is finite but depends on $x$ and $y$. I really want to say that $x$ and $y$ are varying over a compact region, and hence the $\sup$ of the $K$'s is finite, but that would require $K$ to vary continuously with $x$ and $y$. Which..... it does? I don't know.
Context:
This is adapted from an exercise of Halmos, where he proves Gelfand's_formula by a somewhat unusual route, without first considering the Neumann series (at least directly). Note that I've read and understand the proof on wikipedia, but I'm interested in understanding the idea behind this exercise.
My problem is specifically how to get from the hint in b) to the main claim of b).
Let $\rho(A)$ denote the spectral radius of $A$, i.e. the largest absolute eigenvalue. The exercise (slightly paraphrased):
a) For fixed $x$ and $y$, $f(\lambda):=\langle(I-\lambda A)^{-1}x,y\rangle$ is analytic in the region $\lambda < \frac1{\rho(A)}$.
b) There exists a constant $K$ such that $|\lambda|^k\Vert A^k\Vert \le K$ whenever $\lambda < \frac1{\rho(A)}$ and $k=0,1,2,\ldots$. (Hint: for each $x$ and $y$ there exists a constant such that $|\lambda^k\langle A^kx,y\rangle| \le K$ for all $k$).
[It’s smooth sailing after b), so I’ll omit the rest of the exercise. b) is really the meat of the theorem].
If they are finite dimensional, then picking a basis $e_1,\dots,e_d$ and applying the condition to $x=e_i$ and $y=e_j$ means that each matrix coefficient $(A_k)_{ij}$ is bounded. Since both $\sup_{i,j}|A_{ij}|$ and $||A||_2$ are norms on the space of $d\times d$ matrices, and all norms on a finite-dimensional space are equivalent, we conclude that $||A_k||_2$ is also bounded.