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Consider the Dirichlet kerel:

$f(x) = 1+2\sum_{k=1}^{N}\cos(kx)$.

Now, given a diagonalizable real matrix $A$, one can consider $f(A)$, using the standard notation of matrix functions. Namely, $f(A) = I+2\sum_{k=1}^{N}\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}(kA)^{2j}$.

I am interested in bounding $\|f(A)\|_{\infty}$ from above. I am also assuming:

  • $\|A\|_{\infty} \le c$ for some $c$ that is independent of $N$, $A$ or its dimension.

  • The eigenvalues of $A$ are real and also bounded by $c$.

I have tried following quite a few paths. Among them, are:

1) Trivial bounds, mainly using subadditivity.

2) Using the Cauchy definition, $f(A) = \frac{1}{2\pi i}\int_{\Gamma}f(z)(zI-A)^{-1}dz$.

All of the above led me to upper bounds which are polynomial (or square root) in the dimension of $A$. However, numerical tests for random matrices consistently show very small results (specifically, if we take $N$ to be logarithmic in the dimension of $A$).

Do you have any suggestions?

Thank you.

Clarification: I am using the notation $\|B\|_{\infty} = \max_{i}\sum_{j=1}^{n}|B_{i,j}|$.

Edit: It might be the case, that if $A$ is stochastic and symmetric, $\cos(kA)$ itself has a bounded infinity norm (i.e., independent of both $N$ and $k$). It that is indeed correct, the above question is resolved. I posted this question there, but it might also be appropriate here as well.

I apologize for the duplication.

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  • $\begingroup$ Why don't you just diagonalize and use estimates for the scalar-valued Dirichlet kernel? Any upper bound would have to depend on how much you have to "twist" when you diagonalize, and then the usual $\log(N)$ growth of the Dirichlet kernel. Or have I misunderstood your question? $\endgroup$
    – Yemon Choi
    Commented Mar 12, 2014 at 15:47
  • $\begingroup$ Thank you for your reply, but I am afraid I do not follow. Suppose $A = VDV^{-1}$. Then, $f(A)=Vf(D)V^{-1}$. How do you reach the (so desired) bound of $\log(N)$? $\endgroup$
    – neil
    Commented Mar 12, 2014 at 15:52
  • $\begingroup$ Oops, I was thinking of the wrong norm (namely $L^1({\bf T})$ whereas you seem to want the $L^\infty({\bf T})$ norm). My mistake $\endgroup$
    – Yemon Choi
    Commented Mar 12, 2014 at 17:47

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