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

S. Albeverio, A. Kosyak / Journal of Functional Analysis 236 (2006) 634–681 667
Let e1(t) = c11t11 + c12t22 = 0sot11 =−c12t22/c11. In this case
Finally
(CT , T ) = c11t 2 11 + 2c12t11t22 + c22t 2 22 =
c2 12
− 2
c11
c2
12
+ c22 t
c11
2 22
c12t11 + c22t22 = − c12c12
+ c22 t22 =
c11
c11c22 − c2 12
c11
M12
12
=
c11
= M12
12
c11
Mξ 22 (t) 2 = Miy2nexp it11 + it22(x12y1n + y2n) 2
=
∂φ2(t)
=
M 12
12
c11 t22
We have used the inequality
2 M
exp − 12
12
c11 t2
22
1 + c11t 2 22
M12
12
c11
Hence if we denote t = (t11,t22) ∈ R 2 we have using (43)
∂t22
2
.
t 2 22 ,
e1(t)=0,t21=t22
t 2 22 exp
M12
12
− + c11 t
c11
2
22 .
1 + x exp x, x ∈ R. (62)
Ξ 22 = max
t∈R2
22
Mξ (t) 2 22 (M
>Ψ := 12
12 )2 exp(−1)
c11(M12 12 + c2 .
11 )
This proves (47) for (p, q) = (2, 2). For (2,q),2