# Lipschitz property of matrix function only depending on singular values

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 $$g$$?

• This question reminds me of math.stackexchange.com/questions/3214226 , although I don't have a precise reduction in either direction. May 26 '19 at 18:26
• Yes, with $C=2L\sqrt{n}$. No, if you want an $n$ independent constant, as the diagonal case (and $f(x)=\sum |x_i|$) shows May 26 '19 at 20:26
• It is OK for $C$ to depend on $n$. How did you get $C=2L\sqrt{n}$? @oferzeitouni May 27 '19 at 2:03

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 $$L^1$$.