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The functional \begin{equation} F(u) = \int_{\Omega} \langle A(x) \nabla u, \nabla u \rangle \end{equation} where $A$ is a symmetric matrix . You can assume $\Omega$ conviniente such that the expression above make sense. For example, $C^{1}(\Omega, H^{1}_{0}(\Omega))$ or other space such that the functional above be convex. I don't know if the hypothesis that $A$ is simetric is nescessary. Thank you.

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    @DavideGiraudo: Thanks for the suggestion!2012-10-07

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Assuming that $A(x) \geq 0$ (ie, positive semi-definite) for all $x$, then $F$ is convex.

The mapping $h \mapsto \langle A(x_0) h, h \rangle$ on $\mathbb{R}^n$ is convex for each $x_0$. The mapping $u \mapsto \nabla u(x_0)$ is linear. The composition of a convex function with a linear function is convex, hence the functional $\phi_{x_0}(u) = \langle A(x_0) \nabla u (x_0), \nabla u (x_0) \rangle$ is convex.

Convex means that $\phi_{x_0} (\lambda u + (1-\lambda) v) \leq \lambda \phi_{x_0} ( u) + (1-\lambda) \phi_{x_0} ( v)$ for all $\lambda \in [0,1]$. Integrating both sides over $x_0 \in \Omega$ gives $\int_\Omega \phi_{x_0} (\lambda u + (1-\lambda) v) d x_0 \leq \lambda \int_\Omega \phi_{x_0} ( u) d x_0 + (1-\lambda) \int_\Omega \phi_{x_0} ( v) d x_0,$ or equivalently $F(\lambda u + (1-\lambda) v) \leq \lambda F(u) + (1-\lambda) F(v).$ Hence $F$ is convex.