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

hence
C = C3 =
S. Albeverio, A. Kosyak / Journal of Functional Analysis 236 (2006) 634–681 669
c11 c12 c13
c12 c22 c23
c13 c23 c33
⎛
C1(t) = I + C(t) = ⎝
= diag(t21,t31,t32)
, C(t) (61)
⎛
= ⎝
1 + c11t 2 21 c11t21t31 c12t21t32
c11t21t31 1 + c11t 2 31 c12t31t32
c11t 2 21 c11t21t31 c12t21t32
c11t21t31 c11t 2 31 c12t31t32
c12t21t32 c12t31t32 c22t 2 32
⎞
⎠
c12t21t32 c12t31t32 1 + c22t 2 32
⎛
c11 + t
⎝
−2
21
c11 c12
c11 c11 + t −2
31
c12
c12 c12 c22 + t −2
32
⎞
⎠ ,
⎞
⎠ diag(t21,t31,t32).
We prove the following inequality for an operator C of order n such that I + C>0:
det(I + C) exp tr C. (63)
Indeed by Hadamard inequality (see [7] or [13, Section 2.5.4]) we have for positive operator C
of order n
det C
Using the Hadamard inequality and (62) we have for an operator C such that I + C>0
i=1
n
i=1
cii.
n
det(I + C) (1 + cii) (62) n
n
exp cii = exp
i=1
i=1
cii
= exp(tr C),
where we denote by tr C the trace of an operator C in the space C n . Using (63) and (61) we
conclude that
det I + C(t) tr C(t) = exp
p−1
k=1
where α2 k = p r=k+1 t2 rk since by (61) we have
Using (26) we get
tr C(t) =
1k