Past exam of the calculus of variations course of the University of Oxford / 2009 / 1 / a / ii / Solution

A strictly convex function satisfies the same inequality strictly whenever $x\ne y$ and $0<t<1$.

For the unlabelled Hessian criterion, convexity implies $F(z_0+th)+F(z_0-th)-2F(z_0)\ge 0$. Divide by $t^2$ and let $t\to 0$ to obtain $h^T{D}^{2}F(z_0)h\ge 0$ for every $h$. Thus the Hessian matrix is positive semidefinite.

Conversely, if every Hessian is positive semidefinite, Taylor's integral formula with $h=z-z_0$ givesThe allowed supporting-plane characterization then provesFor the integral inequality put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting tangent at $m$ gives $F(u(x))\ge F(m)+F^{\prime }(m)(u(x)-m)$ almost everywhere. Integrating cancels the linear term, soThis is Jensen's inequality. It remains valid with value $+\infty$ on the right: the affine lower bound has integrable negative part, so the extended integral is well-defined.