The original problem states that: Let $U\subset\mathbb{R}^n$ be a bounded open set. Assume $u\in H^1(U)$ is a bounded weak solution to the uniformly elliptic equation $$-\sum_{i,j=1}^n(a_{ij}u_{x_i})_{x_j}=0\qquad (\text{in}\; U)$$ Let $\phi$ be smooth and convex, and set $w=\phi(u)$, then $$\int_U\sum_{i,j=1}^na_{ij}w_{x_i}v_{x_j}\le0\qquad \forall v\in H_0^1(U),\;\;v\ge 0$$
I have seen two types of solutions in general: one directly claims that $\phi'(u)v\in H_0^1(U)$, and the other attempts to solve the problem through approximation, i.e. $$\int_U\sum_{i,j=1}^na_{ij}w_{x_i}v_{x_j}=\int_U\sum_{i,j=1}^na_{ij}\phi'(u)u_{x_i}v_{x_j}=-\int_U\sum_{i,j=1}^na_{ij}\phi''(u)u_{x_i}u_{x_j}v\le 0$$ for $v\in C_0^\infty(U),v\ge 0$, and an approximation of $H_0^1$ functions concludes the proof.
I have two questions regarding the "solutions" presented above:
(1) Is it correct to claim that $\phi'(u)v\in H_0^1(U)$? I can accept that $\phi'(u)$ is a bounded $H^1$ function with weak derivatives $(\phi'(u))_{x_i}=\phi''(u)u_{x_i}$, but why it remains to be $H^1$ when multiplied by another $H_0^1$ function?
(2) Is it possible to approximate non-negative $H_0^1$ functions with $\textbf{non-negative}$ smooth functions of compact support?
And, if the answer to both questions are negative, is there another way to solve the original problem?
Best Answer
I think both approaches, at the technical level, reduce to proving (2). Indeed, if in (1) you have $v\in C_c^\infty$, then $\phi'(u)v\in H_0^1$ almost trivially. To see that in general you will not have $\phi'(u)v\in H^1$ take $\phi(t)=t^{4}$ so that $v\phi''(u)\nabla u = 12v u^2 \nabla u$, which is not quite good enough to get in $L^2$ (note that the other term $\phi'(u)\nabla v$ is in $L^2$, so the full derivative fails to be in $L^2$).
As for the proof of (2), take $v\in H_0^1$, $v\geq 0$, and consider $w_n\in C_c^\infty$ such that $w_n\to v$ in $H^1$ and a.e. in $U$. Consider $v_n:= \max\{ w_n, 0\}$, then clearly $$|v_n(x)-v(x)|\leq |w_n(x)-v(x)|\qquad \text{a.e. in } U.$$ Moreover $\nabla v_n= \nabla w_n$ a.e. on the set $\{w_n\geq 0\}$ and $\nabla v_n=0$ a.e. on $\{ w_n< 0\}$. In particular $$|\nabla v_n(x)-\nabla v(x)|\leq |\nabla w_n(x)-\nabla v(x)|\mathbf{1}_{\{ w_n\geq 0\}}(x) + |\nabla v(x)|\mathbf{1}_{\{ w_n<0\}}(x), $$ where $\mathbf{1}_A$ denotes the indicator function of the set $A$. By writing $\{ w_n<0\}=\{ w_n<0, v>0\}\cup \{ x_n<0, v=0\}$, and using that $\nabla v=0$ a.e. on the second portion, and that $\mathbf{1}_{\{ w_n<0, v>0\}}\to 0$ a.e. we can conclude that the $L^2$ norm of the second term goes to 0 as $n\to \infty$.
We've shown $v_n\to v$ in $H^1$ and clearly $v_n$ has compact support in $U$, the issue is that $v_n$ is at most Lipschitz, but we can fix this by convolving it with a nice nonnegative mollifier (ensuring that the support is still in $U$) to get a smooth version.