Lemma 12.9.12F7130F1BDC30

Proof. We first estimate positive and negative powers, then compare their small moments by an estimate for $\log u$.

For $p\ne 0,1$, test the weak equation by $\eta ^2u^{p-1}$, where $\eta $ is a smooth cutoff. The principal term is

The other terms contain respectively $2\eta u^{p-1}A\nabla u\cdot \nabla \eta $, $(p-1)\eta ^2u^{p-1}B\cdot \nabla u$, and $2\eta u^ pB\cdot \nabla \eta $. Taking absolute values after moving the principal term to one side and applying Young's inequality absorbs half of its energy. Thus, for either sign of $p$,

with $C_1$ depending only on the ellipticity bounds. Each test is legitimate since this individual smooth positive solution is bounded above and away from zero on the cutoff support.

Put $\beta =m/(m-2)>1$ and choose $p_0=\beta ^{-k-1/2}$, with $k$ to be fixed later. Apply the Euclidean Sobolev inequality to $\eta u^{p/2}$ on nested cubes shrinking from $Q$ to $Q/2$. Choose the cutoff gaps proportional to $2^{-j}$ and the powers $p_ j=p_0\beta ^ j$. The positive powers never equal one: the closest two are $\beta ^{-1/2}$ and $\beta ^{1/2}$. Consequently the constants in (12.9.13), together with the cutoff terms from $\nabla (\eta u^{p/2})$, grow at most polynomially in $p_ j$ and exponentially in $j$. Since $\sum _ j(j+1)\beta ^{-j}<\infty $, their iteration product converges. Negative powers give the same calculation for $u^{-1}$. We obtain

Here $C_2$ also depends on $p_0$ and the fixed cube size.

Testing instead by $\eta ^2/u$ gives

For every subcube $S\subset Q$ of side length $\ell $, a cutoff supported in $2S\subset 3Q$, equal to one on $S$, has derivative bounded by $C/\ell $. Poincaré's inequality and Cauchy–Schwarz therefore give

where $K_0\geq 1$ is uniform and $(\log u)_ S$ denotes the mean. We spell out the exponential consequence of this mean oscillation bound.

Write $f=\log u$ and fix $a>2$. Select the maximal dyadic subcubes $S$ of $Q$ on which $\fint _ S|f-f_ Q|>aK_0$. Their total measure is at most $|Q|/a$. The parent of each selected cube has average at most $aK_0$, hence $|f_ S-f_ Q|\leq 2^ maK_0$. Outside their union, differentiation of dyadic averages gives $|f-f_ Q|\leq aK_0$ almost everywhere. Repeat this selection inside each chosen cube, subtracting its own mean. The union at generation $j$ has measure at most $a^{-j}|Q|$, and outside generation $j+1$ the accumulated mean changes and last oscillation give

The intersection of all generations is null. Choosing $c>0$ so that $\exp (c2^ ma)<a$ and summing over the generations yields

with dimensional constants. Now fix $k$ large enough that $p_0<c/K_0$. Multiplying the estimates for the positive and negative exponentials cancels $f_ Q$ and gives

Combining this with (12.9.14) proves the result. Every constant comes from the stated coefficient bounds and the fixed cube. $\square$