Does bounded derivative imply bounded Frechet derivative for real analytic functions?

Question

Let $f$ be a real analytic function where the convergence radius of the Taylor expansion on any $x\in\mathbb{R}$ is $\infty$. Moreover, $\|f'\|_{\infty}\leq O(1)$. Let $A, B$ be two Hermitian matrices. Is it true that

$\lim_{\delta\rightarrow 0}\frac{f(A+\delta B)-f(A)}{\delta}\preceq O(1) |B|$, where $|B|=\sqrt{B^2}$?

Answer 1

I don't think. I would use the Cauchy integral formula for complex analytic functions $$h(z) = \frac{1}{2i\pi}\int_{|s| =r} \frac{h(s)}{s-z} ds$$ And since $f(A+zB)$ is complex analytic in $z$ $$\lim_{z \to 0} \frac{f(A+zB)-f(A)}{z} =\frac{1}{2i\pi} \int_{|s| = r} \frac{f(A+sB)}{s^2} ds$$ Thus, it is enough to assume $f(A+sB)-f(A)$ is bounded for $|s| \le r/\|B\|$ when $A$ is hermitian, which is the case if $f'(z)$ is bounded for $|\Im(z)| \le r$.