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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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    Please define $B$.2012-12-30
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    @Tomás I'm going to guess that $B[w,v]=\int_U a^{ij} w_{x_i} v_{x_j}$. And yes, user54688 should add something to the post, like the assumptions on $a^{ij}$ in addition to what tomasz said.2012-12-31
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    $\phi$ convex implies that $\phi''\geq 0$. Try to write $B[w,v]$ in terms of $\phi''$ and see what happens.2013-01-01
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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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    The real pros of Einstein summation use sub-superscript pairing...2013-01-02
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    Please @PavelM, explain me the difference.2013-01-02
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    I was mostly kidding, but anyway: it is useful and often necessary to distinguish between a vector space $V$ and its dual $V^*$, even though in finite dimensions they are isomorphic. So one has a basis $(e^i)$ for $V$ and a basis $(f_i)$ for $V^*$. The coordinates are set in opposite way: $x=x_i e^i$ and $f=f^i x_i$. This helps one visually identify the expressions that are invariant under change of basis: $f^i x_i$ is invariant while $x_i x_i$ is not. // As Wikipedia says, when working with fixed basis on $\mathbb R^n$ one may identify $V$ and $V^*$, but on manifolds this would make a mess.2013-01-02
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    Also, http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors2013-01-02
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    Ok thank you @PavelM . Im really learning a lot from you and I really apreciate your efforts to help people improve their skills.2013-01-02
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    $\phi'(u)$ might not even be $H^{1}$ eg. $u=loglog(1-1/|x|)$ and $\phi'(x)=e^{e^{x}}-e$.2014-11-25
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    I don't see the point of your comment @TKM2014-11-25
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    we have the condition that u is bounded @TKM2015-05-02
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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