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Assume $u\in H^1(U)$ is a bounded weak solution of $-\sum_{i,j=1}^n(a^{ij}u_{x_i})_{x_j}=0 ~~~in~ U$

Let $\phi:R\rightarrow R$ be convex and smooth,and set $w=\phi(u)$ Show $w$ is a weak subsolution; that is $B[w,v]\leq 0$

for all $v\in H^1_0(U),~v\geq0$ $B=\int_U \sum_{i,j=1}^na^{ij}v_{x_i}w_{x_j}$

I used integration by part and eliptic property, actually my problem is that I don't know when $\phi$ is convex $\phi'(u)$ is positive or not?or

$\int_U \sum_{i,j=1}^na^{ij}u_{x_jx_j}v\phi'(u)dx$ is positive?

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    If this is taken from an exercise of the Evans textbook, then $B$ is defined, according to the author, as $B[w,v]=\int_U \sum_{i,j=1}^n a^{ij}w_{x_i} v_{x_j} \, dx.$ (Just saying.)2015-02-15

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Im gonna use here the Einstein summation convention and $\frac{\partial v}{\partial x_i}=v_i$. Also, Im gonna assume ellipticity on $(a_{ij})$ and that $w\in H^1$, because that only with your hypothesis this is not always true. Note that $(v\in C_0^1(U),\ v\geq 0)$\begin{eqnarray} B[w,v] &=& \int_U a_{ij}v_iw_j \nonumber \\ &=& \int_U a_{ij}\phi'(u)u_iv_j \nonumber \\ &=& \int_U a_{ij}u_i(\phi'(u)v)_j-\int_U(a_{ij}u_iu_j)v\phi''(u) \end{eqnarray}

To conclude, you have to show that $\phi'(u)v\in H_0^1$ and $-(a_{ij}u_iu_j)v\phi''(u)\leq 0$, then you use the fact that $C_0^1$ is dense in $H_0^1$. Can you do this?

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    @Tomás Could you give some details about the density argument? What convergence theorem should we use to pass to the limit?2015-10-25