Past exam of the calculus of variations course of the University of Oxford / 2011 / 1 / a / Solution

For the supporting inequality, assume first that $F$ is convex and fix $z,z_0$. For $0<t\le 1$,Subtract $F(z_0)$, divide by $t$ and let $t\downarrow 0$. Differentiability givesConversely assume this inequality for every pair. Put $w=(1-t)z_0+tz$ and apply it with base point $w$ to both $z_0$ and $z$. Multiply those two inequalities by $1-t$ and $t$ and add. The gradient terms cancel, proving $F(w)\le (1-t)F(z_0)+tF(z)$. Thus the supporting inequality characterizes a differentiable convex function.

For the integral inequality let $\bar{z}=|\Omega {|}^{-1}{\int }_{\Omega }u$. Apply the supporting inequality at $\bar{z}$ to $u(x)$ and integrate. Its affine term has integral $DF(\bar{z})\cdot ({\int }_{\Omega }u-|\Omega |\bar{z})=0$, yieldingThis is Jensen's inequality. Since $u$ is integrable, the supporting affine lower bound has integrable negative part, so the right side is well-defined as an extended real integral. If its positive part is infinite, the inequality remains valid.