# Is this combination of convex functional is still convex?

Let $u$, $v\in C_c^\infty$ and $\Omega\subset \mathbb R^N$ is open bounded, smooth boundary. We also assume that $0\leq v\leq 1$.

Define $$F(u,v):=\int_\Omega |\nabla u|^2v^2dx.$$ Do we have $F(u,v)$ is convex? i.e., for any $0<t<1$, do we have $$F(tu_1+(1-t)u_2, tv_1+(1-t)v_2)\leq tF(u_1,v_1)+(1-t)F(u_2,v_2)$$

I know for instance that the product of two nonnegative, nondecreasing convex function is convex, but it does not apply to this case since $\nabla u$ is not always non-negative.

• $(x,y)\mapsto x^2y^2$ is not convex. $(1,0)\mapsto 0$, $(0,1)\mapsto 0$, but $(\frac12,\frac12)\mapsto \frac1{16}>0$.
– user856
Dec 12, 2015 at 16:07
• @Rahul hmm, you are right. Dec 12, 2015 at 16:08
• The same logic applies to your problem: Choose $u_1,v_1$ and $u_2,v_2$ such that such that $\nabla u_1\ne0$, $v_1=0$, and $\nabla u_2=0$, $v_2\ne0$.
– user856
Dec 12, 2015 at 18:16

## 1 Answer

As noted in a comment by Rahul, this functional is not convex. Indeed, let $B_1,B_2$ be two disjoint closed balls in $\Omega$ and let $u,v$ be bump functions with support $B_1,B_2$ respectively. Then $F(u,v)=F(v,u)=0$, but $$F\left(\frac{u+v}{2},\frac{u+v}{2}\right) = F(u/2,u/2)+F(v/2,v/2)>0$$ showing that $F$ is not convex.

$F$ is separately convex in each variable, though.