Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 107 / 5

Adopt the sign convention that a weak supersolution for $Lu=D_i(a^{ij}{D}_{j}u)$ satisfies $Lu\le 0$, or $\int a^{ij}{D}_{j}u{D}_i\phi \ge 0$ for every nonnegative compactly supported test function. For a nonnegative supersolution, the Weak Harnack inequality gives some $q>0$ and $C_H\ge 1$, depending only on $n,\lambda ,\Lambda$, such thatHere $|E{|}^{-1}{\int }_E$ denotes the average over $E$; equivalently the right side gives a lower bound for the essential infimum. One may take a sufficiently small positive exponent, so no assumption $q\ge 1$ is needed. The inequality is unchanged under translations and dilations of the Euclidean balls.

Since a Sobolev space element is defined up to changes on a null set, the infimum is initially essential. After choosing a continuous representative it is the ordinary infimum. Without choosing such a representative, an arbitrary pointwise infimum cannot be bounded by an integral.

Work first with essential bounds. On a Euclidean ball $B_R(x_0)⋐B_1$, put $M={\mathrm{esssup}}_{B_R}u$, $m={\mathrm{essinf}}_{B_R}u$ and $\omega =M-m$. The functions $M-u$ and $u-m$ are nonnegative weak solutions, hence weak supersolutions. In $B_{R/2}$, at least one of the sets $\{u\le (M+m)/2\}$ and $\{u\ge (M+m)/2\}$ has at least half the measure.

In the first case, the Weak Harnack inequality applied to $M-u$ givesIn the second case, applying it to $u-m$ gives the same lower bound for the improvement of the minimum. Thus the oscillation decay estimate isThis measure argument works even when $q<1$, avoiding a false use of the triangle inequality in that range.

Iteration yields ${\mathrm{osc}}_{B_r(x_0)}u\le C(r/R{)}^{\mu }{\mathrm{osc}}_{B_R(x_0)}u$, with, for example, $\mu =\min \{1/2,-\log \tau /\log 2\}\in (0,1)$. At Lebesgue points the shrinking oscillations define a unique continuous representative, and the estimate extends to every point by continuity. For $x,y\in B_{1/4}$ at distance less than $1/4$, apply the estimate to a Euclidean ball centred at $x$ with initial radius $1/2$ and final radius comparable to $|x-y|$. This Euclidean ball stays in $B_1$. Larger distances use the trivial bound $2\Vert u{\Vert }_{\infty }$. We obtainThe constants depend only on $n,\lambda ,\Lambda$. This derives the relevant conclusion of the De Giorgi-Nash-Moser theorem directly from part (a).

First prove the hinted Caccioppoli inequality. Testing the weak formulation with $u_k{\phi }^2$ is legitimate by approximation in the zero-boundary Sobolev space. Let ${\Lambda }_{\ast }=\mathrm{esssup}\Vert a(x){\Vert }_{\mathrm{op}}\le n\Lambda$. Thenby Young inequality. HenceNo derivative of the measurable coefficient matrix is taken.

Set $w_k=u_k/{\sup }_{B_1}|u_k|$. The denominator is finite and positive because $u_k$ is continuous on the compact closure and nonzero. Each $w_k$ solves the same linear equation and $|w_k|\le 1$. For every $\theta <1$, choose a cutoff function equal to one on $B_{\theta }$ and supported in a slightly larger interior Euclidean ball. The energy bound gives a uniform $W^{1,2}(B_{\theta })$ bound.

Take radii ${\theta }_j\uparrow 1$. Weak compactness and a diagonal subsequence argument produce one subsequence converging weakly in $W^{1,2}(B_{{\theta }_j})$ for every $j$, with consistent restrictions defining $v\in W_{\mathrm{loc}}^{1,2}(B_1)$. Part (b) and the Arzelà-Ascoli theorem allow a further subsequence converging uniformly on $\overline{B_{1/4}}$. Its uniform limit agrees almost everywhere with the weak limit and supplies its continuous representative there.

For every smooth compactly supported test function, choose $j$ containing its support. Since $a^{ij}{D}_i\phi \in L^2$ and $D{w}_k$ converges weakly there,Thus the compactness of normalized weak elliptic solutions givesThe diagonal step is needed to retain the equation beyond the small Euclidean ball of uniform convergence. The limit need not be nonzero: for the Laplacian in dimension at least two, the normalized harmonic functions $\mathrm{Re}(x_1+i{x}_2{)}^k$ tend uniformly to zero on each smaller Euclidean ball.