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

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.

A convex function satisfiesIts value on a line segment never lies above the chord joining its endpoint values.

A strictly convex function satisfies the strict inequalityEndpoint and coincident-point cases are excluded because they necessarily give equality.

A global minimizer is an admissible function $\bar{u}$ for whichEquivalently it attains the infimum over the whole admissible class. Its endpoint traces must equal the prescribed boundary values; comparing only with nearby functions would define a local rather than a global notion.

The integrand is nonnegative, so $\inf I\ge 0$. For each positive integer $n$, setThis is a continuous piecewise affine sawtooth with zero endpoints, $0\le u_n\le 1/n$, and $u_n^{\prime }=\pm 2$ almost everywhere. It belongs to $W^{1,\infty }$, and $(u_n^{\prime }{)}^6=64$ almost everywhere. ThereforeTogether with nonnegativity this proves $\boxed{\underset{\mathcal{A}}{\inf }I=0}$. The functions converge uniformly to zero while their derivatives retain magnitude two; small amplitude does not force small gradient.

If an admissible $u$ attained the infimum zero, each nonnegative term of its integrand would vanish almost everywhere. In particular $|u{|}^{2011}=0$ almost everywhere, so $u=0$ in its Sobolev space and its weak derivative is zero. But then the remaining term is $(-64{)}^2=4096$ almost everywhere, giving $I(u)=4096$, a contradiction. Thus the infimum is zero but no global minimizer exists. The oscillatory sequence has no convergence of derivatives compatible with substitution into this nonconvex energy.