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 $g$?



Lipschitz property of matrix function only depending on singular values
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 $L^1$.


