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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.

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1 Answer 1

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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 $\partial f(\bar{x}_1,\bar{x}_2) \subsetneq \partial f_1(\bar{x}_1,\bar{x}_2) \times \partial f_2(\bar{x}_1,\bar{x}_2)$.

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