- REDUCED VOLUME OF A STATIC METRIC 383
where we used JV'dJ = 1 a.e. and the Laplacian comparison theorem, i.e.,
d.b..d::::; n - 1. Nc>te that the Laplacian comparison theorem is equivalent to
the Bishop-Gromov volume comparison theorem. D
It is useful to keep the following simple examples in mind when pondering
Lipschitz continuous sub-and super-solutions of the heat equation.
EXAMPLE 8.6 (Heat equation on 51 ). Let M^1 = 51 = ffi./ (2nZ) with
the standard metric g = dB^2. Consider the function
e2
f (B, t) = t + 2' e E (-n, n] and t ER
For each fixed e, f ( e, ·) is a smooth (linear) function of time and, for each
fixed t, we have f (·, t) is Lipschitz on 51 and C^00 except ate= n. Moreover
f is a solution to the heat equation almost everywhere. In particular,
of a^2 f
at ( e, t) = ae2 ( e, t) = i
for all e E 51 - { n} and t E R On the other hand,
d
d r f (e, t) de= 27r > o.
t Js1
This is consistent with the fact that f is not a subsolution of the heat
equation in the weak sense (as we shall now see, f is a supersolution). Note
that
of
ae (e, t) = e
for all e E ( -n, 7r) and t E ffi. (and ~~ is undefined for e = 7r). In particular,
for each t E ffi., ~~ (·, t) has a jump discontinuity at e = n. In the sense of
distributions, we have ~~ (e, t) = e and
IJ2f
(8.5)
8
e 2 (·, t) = 1 - 2n. c57f,
where 67f is the Dirac 6-function centered at e = n. Hence, in the sense of
distributions,
of a^2 f 02 f
at = ae 2 +^2 n ·^6 1f 2:: 3e2 ·
EXERCISE 8.7. Prove (8.5).
SOLUTION TO EXERCISE 8.7. For any C^2 function <p: 51 --+ ffi. we have
f e
2
<p"(B)dB= e
2
tp'(e)\7f - { B<p'(B)de
Js1 2 2 -7[ Js1
= f <p(B)dB-B<p(B)J:_1f
Js1
= f <p(B)dB-2n<p(n).
Js1