Estimation optimale du gradient du semi-groupe de la chaleur sur le ...

S. Albeverio, A. Kosyak / Journal of Functional Analysis 236 (2006) 634–681 643
det C
= exp −
(2π) m 1
exp(tEpq)
2
∗ C exp(tEpq)x, x
dx = dμBpq(t)(x),
where (Bpq(t)) −1 = Cpq(t) = exp(tEpq) ∗ C exp(tEpq) (we note that det C = det Cpq(t)).
Hence, using (2) we get
We shall prove that
H μ LI+tEpq
B ,μB =
det Cpq(t) + C
2
det Cpq(t) det C
det 2 Cpq(t)+C
2
1/4
=
det C
det Cpq(t)+C
2
1/2
. (3)
= det C + t2
4 cppA q q(C), (4)
where A p q (C), 1 p,q m, denote the cofactors of the matrix C corresponding to the row p
and the column q. Wehave
hence
det Cpq(t)+C
2
det C
and finally, using (3) we get
where
H μ mLI+tEpq m
B ,μB =
B (n) =
1r,sm
= det C + t2
det C
det Cpq(t)+C
2
∞
n=q+1
4 cppA q q(C)
= 1 +
det C
t2
4 cppbqq,
1/2
= 1 + t2
4 cppbqq
−1/2 H μ LI+tEpq
B (n)
,μB (n) =
∞
n=q+1
b (n)
rs Ers and C (n) := B (n) −1 =
1 + t2
4 c(n)
1r,sm
pp b(n) qq
c (n)
rs Ers.
−1/2
,
So using the properties of the Hellinger integral for two Gaussian measures we conclude that
m LI+tEpq m
μB ⊥ μB ∀t ∈ R \{0} ⇔
∞
n=q+1
1 + t2
4 c(n)
⇔ S L m
pq μB =∞.
pp b(n) qq
−1/2
= 0