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

A convex function satisfiesThus its value on every line segment lies below the chord joining the endpoint values.

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.

An admissible $\bar{u}$ is a global minimizer ifIn particular it attains the infimum; no closeness condition on competing admissible functions is imposed.

The integrand is nonnegative. Define the Lipschitz triangular wavesThey have zero endpoint values, slope $\pm 2$ almost everywhere, and height $1/N$. Hence $(u_N^{\prime 4}-16{)}^4=0$ almost everywhere. On each of the $N$ equal cells, direct integration of the fourth power of the two affine pieces gives $1/(5N^5)$. ThereforeThis is an oscillating minimizing sequence with no minimizer.

If an admissible $u$ attained the infimum, nonnegativity and $I(u)=0$ would force $u^4=0$ almost everywhere, so its continuous Sobolev representative would be identically zero. Its derivative would then vanish almost everywhere, but the other integrand term would equal $16^4>0$. This is impossible. Equivalently, zero energy would simultaneously require $u=0$ and $|u^{\prime }|=2$ almost everywhere. Thus