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

K. Koufany, G. Zhang / Journal of Functional Analysis 236 (2006) 546–580 571
Now let k = j, then [Eβ,vα]=D(cj , ¯w)vα = D(vα, ¯w)cj =〈vα,w〉cj , which vanishes except
when vα = w and in that case,
and
1
[Eβ,vα]¯vα = cj ¯vα =
2 ξj − 1
2 ζj
− D(cj , ¯cj ) ¯w,
1
[Eβ,vα]¯vα ≡
2 ξj − 1
2 ζj
(η ¯w + η ′ ¯w ) + η ¯w + η ′ ¯w .
1
Case III. α ∈ Ψ3, with α| −
t =
C 2γj−1, and the root vector vα ∈ Vj−1,0. In this case, we have
[Eβ,vα]¯vα = D(cj , ¯w)vα ¯vα ≡ ζD(cj , ¯w)vα − D
D(cj , ¯w)vα,cj η ¯vα
≡ ζD(cj , ¯w)vαη ¯vα + D
cj , D(cj ,w)vα vα.
However, by the commutator relation (JP15) in [9] we have
D(w, ¯vα), D(cj , ¯cj ) = D
D(w, ¯vα)cj , ¯cj − D cj , D(vα, ¯w)cj =−D cj , D(vα, ¯w)cj
since D(w, ¯vα)cj = 0 by the Peirce rule that D(w, ¯vα)cj ∈{Vj,j−1 ¯Vj−1,0Vjj }={0}, thus
D
cj , D(cj , ¯w)vα vα = D(cj ,cj ), D(w, ¯vα) vα = D(cj ,cj )D(w, ¯vα)vα = D(vα, ¯vα)w
since D(cj ,cj )vα = 0.
It is easy to see, by direct matrix computation, that,
Hence, modulo U(gC)kC
Consequently,
vα∈Vj−1,0
vα∈Vj−1,0
[Eβ,vα]v
vα∈Vj−1,0
∗ α
and this finishes the proof. ✷
b ¯w if w = ej−1,j ,
D(vα, ¯vα)w =
0 ifw = ej,j−1.
b(η ¯w + η
D(vα, ¯vα)w ≡
′ ¯w ) if w = ej−1,j ,
0 if w = ej,j−1.
≡
vα
ζD(cj , ¯w)vαη ¯vα +
b(η ¯w + η ′ ¯w ) if w = ej−1,j ,
0 if w = ej,j−1