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 $g$?

  • $\begingroup$ This question reminds me of math.stackexchange.com/questions/3214226 , although I don't have a precise reduction in either direction. $\endgroup$ May 26 '19 at 18:26
  • $\begingroup$ 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 $\endgroup$ May 26 '19 at 20:26
  • $\begingroup$ It is OK for $C$ to depend on $n$. How did you get $C=2L\sqrt{n}$? @oferzeitouni $\endgroup$ 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 $L^1$.

  • $\begingroup$ Thank you! That's a really elegant argument! $\endgroup$ May 28 '19 at 17:16

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