the spectrum is contained in the interval ${\cal I} = [-1,1]$. We also assume that $f(z)$ is real for real $z$, so that $f$ maps Hermitian matrices to Hermitian matrices.

Let ${\mathbb P}_k$ denote the set of all complex polynomials of degree at most $k$ on $\cal I$. Given $p\in{\mathbb P}_k$, the matrix $p(A)\in \mathcal{A}$ is well defined and it is banded with bandwidth at most $km$. So for any polynomial $p\in{\mathbb P}_k$ and any pair of indices $i,j$ such that $|i-j|>km$ we have $$\begin{aligned}
    \|[f(A)]_{ij}\|&=&  \|[f(A)-p(A)]_{ij}\|\label{step1}\\
%    &\leq&\|f(A)-p(A)\|_{op}\label{step2}\\
    &\leq&\|f(A)-p(A)\|\label{step2}\\
    &=&\rho(f(A)-p(A))\label{step3}\\
    &=&\sup(\sigma(f(A)-p(A)))=\sup(\sigma((f-p)(A)))\label{step4}\\
    &=&\sup((f-p)(\sigma(A)))\leq E_k(f),\label{step5}
    \end{aligned}$$ where $E_k(f)$ is the best uniform approximation error for the function $f$ on the interval $\cal I$ using polynomials of degree at most $k$: $$E_k(f):= \min_{p\in {\mathbb P}_k} \max_{t\in {\cal I}} |f(t) - p(t)|\,.$$ In the above computation:

-   (\[step2\]) follows from (\[step1\]) as a consequence of the definition of operator norm,

-   (\[step3\]) follows from (\[step2\]) because $A$ is Hermitian, so $f(A)-p(A)$ is also Hermitian,

-   the spectral theorem allows us to obtain (\[step5\]) from (\[step4\]).

Next, we recall the classical Bernstein’s Theorem concerning the asymptotic behavior of $ E_k(f) $ for $k\to \infty$; see, e.g., [@meinardus page 91]. This theorem states that there 