Given $A \in \mathbb{R}^{n \times n}$ that is symmetric, stochastic and diagonalizable, and $k \in \mathbb{N}$, I am interested in bounding $\|\cos(kA)\|_{\infty}$ from above.
$\| \|_{\infty}$ is the induced $\ell_{\infty}$ norm, i.e. $\|A\|_{\infty} = \max_{i}\sum_j|A_{i,j}|$, and the cosine of a matrix can be defined, for instance, via its Taylor expansion, $\cos(A) = \sum_t \frac{(-1)^{t}}{(2t)!}A^{2t}$.
The trivial bound is of course $\cosh(k)$. However, it seems highly untight for large $k$. For instance, we also have $\|\cos(kA)\|_{\infty} \le \sqrt{n}\|\cos(kA)\|_{2} = \sqrt{n}$ (since $\cos(kA)$ is normal and its largest eigenvalue is at most $1$).
Can you see a tighter bound in the special case where $A$ is doubly stochastic?
Thank you very much.
Denote $f(x) = \cos(kx)$, and let $g(a) = f\left(1-2\frac{|a|}{d}\right)$ for $a \in \{0,1\}^{d}$.
If we take $A$ to be the adjacency matrix of the hypercube graph with $2^{d}$ vertices, the question is equivalent to finding the so-called spectral norm of $g$ as a Boolean function, i.e. $$\|f(A)\|_{\infty} = \| \hat{g}\|_{1}.$$
I could not bound it analytically, however numerical experiments suggest that the above behaves like $d$ and unfortunately is not bounded by a constant.
However, I do not have any further insight as to which matrices the above expression is bounded by a constant (maybe regular graphs of constant degree?).