Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 9 / 1 / i / Solution

The Weak Harnack inequality is an estimate for a nonnegative weak supersolution. Write $F=(f^1,\dots ,f^n)$, let $B_{2R}(x_0)\subset \subset \Omega$, and assume $u\ge 0$ almost everywhere on this ball. There are $p>0$ and $C<\infty$, depending only on the dimension, the uniform ellipticity bounds, $q$, and the scaled norms of the lower-order coefficients, such thatHere $⟨v{⟩}_E=|E{|}^{-1}{\int }_{E}v$ is the integral average. The weak supersolution inequality uses the sign convention $Lu\le \mathrm{div}F+g$. For instance, the coefficient dependence can be expressed usingThus one may use the same $p,C$ on all sufficiently small balls when the global coefficient norms and uniform ellipticity bounds are fixed. The nonnegativity condition is essential to this formulation; a signed weak supersolution may first be shifted, with the resulting change in its forcing included. As usual, supremum and infimum statements for functions in a Sobolev space mean essential suprema and essential infima. The standard multidimensional statement uses $n\ge 2$, so $q/2>1$. In dimension one the analogous formulation requires $q\ge 2$; the printed restriction $q>n$ alone would allow $q/2<1$, which does not suffice. To see the obstruction, smooth the nonnegative capped function $\min \{1,|x|/\epsilon \}$ by a mollifier of width $\tau$. Its positive second derivative is a bump of mass $2/\epsilon$ and has $L^{q/2}$ size of order ${\epsilon }^{-1}{\tau }^{2/q-1}$. For fixed $\epsilon$ and $1<q<2$, this size and the minimum both tend to zero as $\tau \downarrow 0$, whereas the average on a fixed larger interval stays positive. Thus the Weak Harnack inequality cannot have the displayed uniform forcing bound in that range. Shrinking $\epsilon$ and choosing $\tau$ still smaller likewise defeats a uniform Hölder estimate based on that norm alone.