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The integral average of an integrable function over a measurable set $E$ of finite positive measure is $⟨v{⟩}_E=|E{|}^{-1}{\int }_{E}v$. Normalization removes the volume factor when comparing estimates on balls of different radii.

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.

Write $E(t)={\int }_{B_t(x_0)}|\nabla u{|}^2$. For $n\ge 2$, choose $c$ in the annular Caccioppoli inequality to be the integral average of $u$ over $A_R$. The supplied Poincare inequality on an annulus givesMoving the $E(R)$ term to the left is the hole-filling argument:Consequently $E(2^{-k}{r}_0)\le {\theta }^{k}E(r_0)$. Put $\beta =-{\log }_2\theta >0$ and choose $\mu =\min \{\beta /2,1/2\}\in (0,1)$. For $2^{-k-1}{r}_0<r\le 2^{-k}{r}_0$, monotonicity givesDyadic endpoints can be assigned to either adjacent interval. Since $E(r_0)\le {\int }_B|\nabla u{|}^2$, the requested dyadic energy decay isBoth constants depend only on the dimension and uniform ellipticity bounds, not on $u,x_0,r,r_0$.

There is a dimensional detail in the printed hint: an annulus is disconnected in dimension one, so that Poincare inequality with a single average is false there. The conclusion still holds. In one dimension the weak equation gives $a(x)u^{\prime }(x)=J$ almost everywhere for a constant flux for a one-dimensional divergence-form equation $J$. Since $\lambda \le a\le \Lambda$,This implies the required estimate with, for example, $\mu =1/2$ and $K=(\Lambda /\lambda {)}^2$. If $J=0$, the estimate is immediate. Thus the proof also covers dimension one without using the inapplicable hint.