# Clarke Subdifferential for a function of two variables.

I am looking of an example where equality don't hold in the below relation:

let $$f: \mathbb{R} \times \mathbb{R} \mapsto \mathbb{R}$$ be locally lipschitz and regular at $$x = (x_1 , x_2)^T \in \mathbb{R} \times \mathbb{R}$$

Denote: $$\partial^C f_1 (x_1 , x_2)$$ Clarke Subdifferential of $$f(. , x_2)$$ at $$x_1$$ and $$\partial^C f_2 (x_1 , x_2)$$ Clarke Subdifferential of $$f(x_1 , . )$$ at $$x_2$$, then the following holds: $$\partial^C f (x_1 , x_2) \subseteq \partial^C f_1 (x_1 , x_2) \times \partial^C f_2 (x_1 , x_2)$$

I am looking for a counter example where the above relation holds with a strict subset (Equality don't hold).

The way I thought about it was:

since regular I know that: $$f^\circ ((x_1,x_2);(v_1,v_2)) = f^\prime ((x_1,x_2);(v_1,v_2))$$ Also, $$f_1^\circ ((x_1,x_2);v_1) = f^\circ ((x_1,x_2);(v_1,0)) = f^\prime ((x_1,x_2);(v_1,0))$$ similarly for $$f_2^\circ ((x_1,x_2);v_2) = f^\circ ((x_1,x_2);(0,v_2)) = f^\prime ((x_1,x_2);(0,v_2))$$ So I need a convex function of two variables (to ensure regularity) where differentiability fails at a point and at that point the clarke subdifferential for each variable will be an interval and for the clarke subdifferential for the whole function there is a point where it does not belong to the cartesian product of the clarke subdifferential of each variable alone and hence equality don't hold, but I can't find an example for this.

## 1 Answer

How about $$f(x_1, x_2) = \lvert x_1 + x_2 \rvert = \max\{(x_1 + x_2), -(x_1 + x_2)\}$$ at $$(\bar{x}_1,\bar{x}_2) = (0,0)$$?

Let $$\text{conv}$$ denote the convex hull.

From Proposition 2.3.12 of Clarke, we have $$\partial f(\bar{x}_1,\bar{x}_2) = \text{conv}([1, 1], [-1, -1])$$.

Moreover, $$\partial f_1(\bar{x}_1,\bar{x}_2) = \partial f_2(\bar{x}_1,\bar{x}_2) = [-1,1]$$, which means that $$\partial f(\bar{x}_1,\bar{x}_2) \subsetneq \partial f_1(\bar{x}_1,\bar{x}_2) \times \partial f_2(\bar{x}_1,\bar{x}_2)$$.