# Is the functional $F(u) = \int_{\Omega} \langle A(x) \nabla u, \nabla u \rangle$ convex?

The functional $$F(u) = \int_{\Omega} \langle A(x) \nabla u, \nabla u \rangle$$ 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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I would presume you need $A(x)$ to be positive semi-definite a.e.? – copper.hat Jun 6 '12 at 22:03
Yes. You can assume for exmaple that $A$ is elipitic. – user29999 Jun 6 '12 at 22:15
If $A(x) \geq 0$, then each functional $\phi_{x_0}(u) = \langle A(x_0) \nabla u (x_0), \nabla u (x_0) \rangle$ is convex. Integration preserves order, hence the integral of these functionals (over $x_0$) will be convex. – copper.hat Jun 6 '12 at 22:25
How could I see the details of these statements? – user29999 Jun 6 '12 at 22:29
Sure; the functional $x \mapsto \langle A(x_0) x, x \rangle$ (on $\mathbb{R}^n$) is convex because the second derivative $A(x_0)+A(x_0)^T$ is positive semi-definite. The mapping $u \mapsto \nabla u(x_0)$ is linear. The composition of a convex function with a linear function is convex. – copper.hat Jun 7 '12 at 0:02

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.