# Lipschitz constant for $\|AB - C\|^2_F$ with Lipschitz continuous gradient.

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?

• 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. – Pantelis Sopasakis May 14 at 16:32
• Dear Pantelis f is a function of A. While B and C are constant. – Chandan Pradhan May 14 at 19:54
• Hint: This function is not Lipschitz continuous. It is locally Lipschitz. – Pantelis Sopasakis May 14 at 23:28
• 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. – Chandan Pradhan May 15 at 5:34
• 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. – Chandan Pradhan May 15 at 5:59

## 1 Answer

$$\newcommand{\R}{\mathbb{R}}$$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).