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

664 S. Albeverio, A. Kosyak / Journal of Functional Analysis 236 (2006) 634–681
Fourier transform for the Gaussian measure μC in the space Rm is:
1
√
(2π) m
Rm
exp i(y,x)dμC(x) = exp − 1
2 (Cy,y)
, y ∈ R m .
Let p = 1. Using (55)–(57) we have
φ1(t11) =
R
exp(it11y1n)dμ(y)= exp − 1
2 c11t 2
11 ;
φ1q(t11; tqq) =
R2
exp i(t11y1n + tqqyqn)dμ(y)= exp − 1
c11t
2
2 11 + 2c1qt11tqq + cqqt 2
qq
;
Mξ 1q
(t11) = iyqn exp(it11y1n)dμ(y)=
R
∂φ1q(t11;
tqq)
∂tqq
=−c1qt11 exp −
tqq=0
1
2 c2 11t2
11 ,
Mξ 1,q (t11) 2 = c 2 1qt2 11 exp−c11t 2
11 ;
Ξ 1q
= maxMξ
1q (t11) 2 = c2 1q exp(−1)
= Ψ 1q ,
we have used the obvious result
t11∈R
c11
1
max f(x)= f =
x∈R a
1
, where f(x)= x exp(−ax), a > 0. (59)
ea
This proves (47) for (p, q) = (1,q).
To prove (44) in the general case we note that
where
p
p
k=1
r=k+1
xkrtrk + tkk
ykn + tqqyqn = a(x) + T,y
R p+1,
y = (y1n,y2n,...,ypn; yqn), T = (t11,t22,...,tpp; tqq) ∈ R p+1 ,
a(x) = a1(x), a2(x), . . . , ap(x); 0 ∈ R p+1 , ak(x) =
x =
1