Let $A, B$ be some $m\times n$ matrices in $\mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}$ a Lipschitz continuous function, i.e. $\|f(x) - f(y)\|_2 \leq L\|x-y\|_2$ for some $L$.
For the Frobenius norm $\|\cdot\|_F$, we get
$$ \| f(A) - f(B) \|_F \leq L\|A - B\|_F, $$ where $f(A)_{ij} = f(a_{ij})$. Is there some similar statement for the spectral norm?
Would it help if $f$ had a bounded derivative $f'$? Then I could do something like
$$ \| f(A) - f(B) \| = \| f'(\Xi) \odot (A-B)\|$$
where $\Xi = (\xi)_{ij}$ is such that $f(a_{ij}) - f(b_{ij}) = f'(\xi_{ij})(a_{ij} - b_{ij})$ and $\odot$ is the Hadamard product.
I know that there are some cases (I think $f'(\Xi)$ psd could be enough), where one could do $$ f'(\Xi) \text{ has some structure } \implies \| f'(\Xi) \odot (A-B)\| \leq \max_{ij} |f'(\xi_{ij})| \cdot \|(A-B)\|, $$ but I don't think that I can show $f'(\Xi)$ psd in my case.