Let there are complex matrices given by $A \in C^{m×n}$, $B \in C^{n \times s}$ and $C \in \mathbb{C}^{m \times s}$ and I have $f(A)= \|AB - C\|_F^2$ which has a gradient given by $g(A) = (AB - C)B^H$ w.r.t $A$, then as the $f$ has a Lipschitz continous gradient, I have obatined the Lipschitz constant as

$L = \|BB^H\|_F$.

Can someone suggest if it is the correct answer or I am making a mistake?

  • $\begingroup$ What is the variable of $f$? Are $B$ and $C$ constant or free variables? You could write for example $f(A)$ or $f(A, B, C)$ to clarify this. $\endgroup$ – Pantelis Sopasakis May 14 at 16:32
  • $\begingroup$ Dear Pantelis f is a function of A. While B and C are constant. $\endgroup$ – Chandan Pradhan May 14 at 19:54
  • $\begingroup$ Hint: This function is not Lipschitz continuous. It is locally Lipschitz. $\endgroup$ – Pantelis Sopasakis May 14 at 23:28
  • $\begingroup$ Is the function $f(A)$ is not Lipschitz continuous? My understanding is that the $f(A)$ has Lipschitz continuous gradient $g(A)$ w.r.t $A$ under the frobenious norm. Pardon me if I am wrong as I have just started with these concepts. $\endgroup$ – Chandan Pradhan May 15 at 5:34
  • $\begingroup$ Specifically I got the result by following calculation $||g(A_1) - g(A_2)||_F = ||(A_1 - A_2)(BB^H)||_F \leq || BB^H||_F||(A_1 - A_2)||_F $, giving me $L = ||BB^H||_F$: Pardon me if I am wrong as I have just started with these concepts. $\endgroup$ – Chandan Pradhan May 15 at 5:59

$\newcommand{\R}{\mathbb{R}}$We say that a function $f:D\to\R^m$, where $D\subseteq \R^n$, is Lipschitz continuous on $D$ with Lipschitz constant $L\geq 0$ if $$ \|f(x) - f(y)\| \leq L \|x-y\|, $$ for all $x, y \in D$.

What is important to note here is that $L$ does not depend on $x$ or $y$.

If $f$ is a $C^1$ function (your function is, indeed, $C^1$), then the best Lipschitz constant on a set $X\subseteq D$ is given by

$$ L = \sup_{x\in X}\|\nabla f(x)\|, $$

but in your case $\nabla f(x)$ is not bounded on $\R^n$, therefore, $f$ is not Lipschitz on $\R^n$, but it is Lipschitz on any compact set $K\subset\subset\R^n$. This property is known as local Lipschitz continuity.

In order to understand this intuitively, think of $f(x) = x^2$ with $x\in\R$. This is locally Lipschitz, but bot globally (see this thread for reference).


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