- LINEAR AND INTERPOLATED DIFFERENTIAL HARNACK ESTIMATES 119
(2.128)A bound for reasonable solutions is given by the following result (see
Lemma 1.2 and Proposition 1.1 in Ni and Tam [290]) ..PROPOSITION 2.97 (Exponential bound for ha.$). Suppose a solution ha.$
of (2.128) satisfies, for some constants A and B, the following inequalities:
(2.129) [ha.$ (x, 0) [ :=:; eA(l+ro(x))
and(2.130)Then there exists a constant C < oo such that
[ha.$ (x, t) [ :'S eC(l+ro(x)).Furthermore, if (ha.$ (x, 0)) ;:: 0, then (ha.$ (x, t)) ;:: 0 for all t > 0.
The analogue of the linear trace differential Harnack estimate for the
Riemannian Ricci flow, Theorem A.57, is as follows (see Theorem 1.2 on
p. 633 of Ni and Tam [290]). The Kahler linear trace differential
Harnack quadratic is defined by (compare with (A.27))Z (h, V) ~ ~ga.$ ( \7 $div (h)a. + \7 a. div (h)$) + Ra.$hf3&.+ ga.$ (div (h)a. V$ +div (h)$ Va.)+ ha.$Vf3Va + ~,
where Vis a vector field of type (1, 0), H ~ ga.$ha.$' and
(2.131) div (h)a. ~ g"f$\7 "Iha.$·THEOREM 2.98 (Kahler linear trace differential Harnack estimate). Sup-
pose that (Mn, g (t)), t E [O, T), is a complete solution of the Kahler-Ricci
flow with bounded nonnegative bisectional curvature and (ha.$) ;:: 0 is a so-
lution of the Kahler-Lichnerowicz Laplacian heat equation (2.128) satisfying
(2.129) and (2.130). Then
Z (h, V) ;:: 0
on M x [O, T) for any vector field V of type (1, 0).
The proof of this theorem requires a number of calculations which we
state and prove. In these calculations the theme is to derive a heat-type
equation for each of the quantities under consideration. We then need to
combine terms fa a good way so that we obtain a supersolution to the heat
equation. The way this is accomplished, as in the Riemannian case, is to
look for terms which vanish on gradient Kahler-Ricci solitons.