Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2008 / B / 5

A convex function satisfies $F((1-t)x+ty)\le (1-t)F(x)+tF(y)$ for every $0\le t\le 1$ and every $x,y$.

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

For necessity of the strict supporting-tangent property, convexity gives a nonnegative gap along a segment from $z_0$ to $z$. If the gap at its other endpoint were zero, convexity of the gap and its zero initial value would force it to be zero all along the segment. That would make $F$ affine on this nontrivial segment, contrary to strict convexity. Conversely let $z=(1-t)x+ty$ with distinct endpoints. Apply the strict supporting inequality at $z$ to both endpoints, multiply by $1-t$ and $t$, and add. The linear terms cancel, yielding strict convexity. ThusFor strict Jensen's inequality, put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting gap $F(u)-F(m)-DF(m)\cdot (u-m)$ is nonnegative everywhere and strictly positive where $u\ne m$. That set has positive measure because $u$ is not almost everywhere constant. Its integral is therefore strictly positive. The linear term integrates to zero, giving the requested strict integral Jensen inequality.

For the unlabelled Euler-Lagrange assertion, use its integrated form $f^{\prime }(u^{\prime })=c$ almost everywhere. This follows from the first variation $\int f^{\prime }(u^{\prime }){\varphi }^{\prime }=0$ for all compactly supported tests. If $u\in C^1$, the left side is continuous and so equals $c$ everywhere. Its derivative values lie in the finite fiber $(f^{\prime }{)}^{-1}(c)$, while the continuous image of the interval under $u^{\prime }$ is connected. Thus $u^{\prime }$ is constant. Endpoint values fix it to $m=(\beta -\alpha )/(b-a)$, giving exactly the affine $\bar{u}$. Conversely that affine function has constant $f^{\prime }({\bar{u}}^{\prime })$, so is a solution. This proves uniqueness among $C^1$ solutions; the scalar integrand is the one-dimensional case of the notation in the source.

If $f$ is strictly convex, Jensen's inequality applied to $u^{\prime }$ givesEquality in the strict inequality can occur only when $u^{\prime }$ is almost everywhere constant. Absolute continuity and the endpoints then give $u=\bar{u}$. Hence the affine function is the unique global minimizer.

In this special case the endpoints are both zero, overriding the previous strict ordering. The affine function is zero and $I(0)=1$. Since $0<f(v)\le 1$, every admissible energy is at most one, proving zero is a global maximizer.

For $u_N(x)=N\min \{x,1-x\}$, the endpoint values vanish and its slopes are $\pm N$ almost everywhere. HenceEvery admissible $u$ has essentially bounded derivative, so its integrand is bounded below by the positive constant $1/(1+\Vert u^{\prime }{\Vert }_{\infty }^{2008})$. Thus no admissible function has zero energy and there is no global minimizer.