Lipschitz property of matrix function only depending on singular values

Question

Let $f$ be a function from $\mathbb{R}^{n\times n}$ to $\mathbb{R}$ such that there exists another symmetric function $g$ (invariant under permutation of coordinates) from $\mathbb{R}^{n}$ to $\mathbb{R}$ satisfying: $$f(M)=g(\sigma_1(M),...,\sigma_n(M))$$ where $M\in\mathbb{R}^{n\times n}$ and $\sigma_i(M)\geq 0$ are singular values of $M$ and $f(0)=g(0)=0$.

Now assume $g$ is Lipschitz in every coordinate, i.e. there exists constant $L>0$ such that for any $i=1,...,n$: $$|g(x_1,...,x_i,...,x_n)-g(x_1,...,x_i^\prime,...,x_n)|\leq L|x_i-x_i^\prime|$$

My question is that does $f$ have some sort of Lipschitz property? For example, for any $M,M^\prime\in\mathbb{R}^{n\times n}$, does there exist a constant $C>0$, such that $$|f(M)-f(M^\prime)|\leq C\|M-M^\prime\|_F$$

On the other direction, does the Lipschitz of $f$ imply the Lipschitz of $g$?

Answer 1

Write the block matrix $\hat M:=\left(\begin{array}{ll} 0&M\\M^T&0\end{array}\right)$ whose eigenvalues are $+-$ the singular values of $M$. Now use Hoffman-Wielandt to conclude that the map from $\hat M$ to its eigenvalue is Lipschitz with respect to the Euclidean ($L^2$) norm. Finally use your pointwise Lipschitz assumption and Cauchy-Schwarz to transfer $L^2$ to $L^1$.