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A function matrix $f : X \to \mathbb R$ is a convex Lipschitz continuous matrix function with Lipschitz constant $\mathrm L$ with respect to a fixed given norm $\|\cdot\|$, i.e., $|f(A)-f(B)| \leq \mathrm L \cdot \|A-B\|$, $\forall A,~B \in X$.

Considering $X$ as positive definite cone of $n$-by-$n$ matrices, are the following maps Lipschitz continuous with respect to $\|\cdot\|_1$? (I was able to prove the convexity.)

  1. $\mathrm{Trace}$ of inverse: $\mathrm{Trace}((\cdot)^{-1})$?
  2. Inverse of the minimum eigenvalue: $1/\lambda_{\min}(\cdot)$?

If yes, what would be the Lipschitz constants for them?

Any help would be greatly appreciated.

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    $\begingroup$ No, because the map $1/t$ is not Lipschitz continuous as a function $(0,\infty) \to (0, \infty)$. Take, for instance, $A_t$ to be the diagonal matrix with all diagonal entries $1$ except for the top-left, which is $1/t$. Then the 1-distance to the identity matrix is $1-1/t$, but $|f(A_t) - f(A)| = t-1$ for both of your functions. This implies $L \ge t$ for all $t$, which is impossible. $\endgroup$
    – mme
    Commented Sep 7, 2022 at 18:37
  • $\begingroup$ Great! Thank you, @mme. Is the map $1/t$ Lipschitz continuous as a function $(c,\infty) \to \mathbb R$? I mean being PD prevents to be Lipschitz, yes? $\endgroup$
    – Reza
    Commented Sep 7, 2022 at 19:18
  • $\begingroup$ Yes, it is. (You should check this as an exercise for yourself). I believe you can prove that on the set $P_c$ of positive-definite matrices with all eigenvalues $\lambda \ge c > 0$, both of your functions are Lipschitz (but I didn't think about this long enough to say confidently). $\endgroup$
    – mme
    Commented Sep 7, 2022 at 20:00

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