- MEAN VALUE INEQUALITY FOR SOLUTIONS OF HEAT EQUATIONS 307
be a positive subsolution to (25.6). If (xo, To) E n x (0, T] and r 0 > 0 are
such that(25.8)then^1
(25.9)P9 ( Xo, To, 4ro, - (2ro)^2 ) c n x [O, T]'
CieC2To+C3VKro 1
sup u:=; 2 u(x,T)dμ9(x)dT,
P 9 (xo,To,ro,-r5) ro Vol9 (B9 (xo, ro)) P 9 (xo,To,2ro,-(2r 0 )^2 )where
(1) C1 depends only on Co, n, T, and A,
(2) C2 depends only on SUPnx[O,T] IQI and SUPnx[O,T] IR (x, T)I,
(3) C3 depends only on n.The proof of this theorem shall occupy the rest of this section. The
idea is to use Moser iteration to bound higher and higher LP-norms of the
subsolution u of (25.6) on 'slightly' smaller and smaller parabolic cylinders
based at the same point. Starting with an L^1 bound on a parabolic cylinder,
by applying the Sobolev inequality to a sequence of reverse Poincare-type
inequalities obtained by integration by parts against the equation, we shall
recursively derive an L^00 bound (on the parabolic cylinder based at the same
point with half the radius) as the limit of uniform LP bounds asp --+ oo.
We now proceed with the details.1.2. Proof of the parabolic mean value inequality via Moser
iteration.
To simplify our notation, let~= ~g( 7 ), I · I= I · 19 ( 7 ), and dμ = dμ 9 ( 7 )·
Given our subsolution u : n x [O, T] --+ lR+ to (25.6), define(25.10)
where A 2:: 0. By (25.6), we have
ov
oT-~v+(Q+A)v:::;o.We shall choose A at least as large so that
(25.11) Q+A 2:: 0.
STEP 1. Reverse Poincare-type inequality. We have for any real numberpE[l,oo)
(25.12)! (vP) - ~ (vP) + p (Q +A) vP::; -p (p-l) vP-^2 l~vl^2 ::;^0
(^1) Note that the denominator on the RHS of (25.9) is
r~ Vol9 (B9 (xo, ro)) = Volg+d-rz (P9 (xo, ro, ro, -r~)).