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

470 H. Schultz / Journal of Functional Analysis 236 (2006) 457–489
Proof of Theorem 3.2. eT (∅) = 0, because PT (∅) = 0. If B1,B2 ∈ B(C) with B1 ∩ B2 =∅, then
KT (B1) ∩ KT (B2) ={0}, i.e. range(eT (B1)) ∩ range(eT (B2)) ={0}. According to Lemma 3.6,
[eT (B1), eT (B2)]=0 so that eT (B1)eT (B2) ∈ I( ˜M). Moreover,
range eT (B1)eT (B2) ⊆ range eT (B1) ∩ range eT (B2) ={0},
and we conclude that eT (B1)eT (B2) = eT (B2)eT (B1) = 0.
Now, let (Bn) ∞ n=1 be a sequence of mutually disjoint Borel sets. Then for each N ∈ N we get
from Proposition 3.4 and Lemma 2.7 that
N
N
range eT Bn = KT Bn = KT (B1) +···+KT (BN )
n=1
n=1
= range eT (B1) +···+eT (BN )
and
ker
eT
N
Bn
n=1
N
= KT
n=1
Bn
c
N
= KT
n=1
= ker eT (B1) +···+eT (BN ) .
B c n
N
=
c
KT Bn =
n=1
n=1
N
ker eT (Bn)
Since an element e in I( ˜M) is uniquely determined by its kernel and its range, it follows that eT
is additive, i.e.
N
= eT (B1) +···+eT (BN ) (N ∈ N). (3.12)
n=1
eT
n=1
Bn
Additivity of eT implies that
∞
∞
eT Bn − eT (Bn) = lim
N→∞
n=1
eT
= lim
N→∞ eT
(the limits refer to the measure topology), where
τ supp
∞
= τ
∞
eT
as N →∞.
Bn
n=N+1
PT
∞
Bn
n=1
∞
Bn
n=N+1
n=N+1
N
−
Bn
= μT
eT (Bn)
n=1
∞
n=N+1
Bn
→ 0,
Combining this with (3.13) and Lemma 3.5, we find that eT is σ -additive as well. ✷
(3.13)
Note that in the case where T is a normal operator, B ↦→ PT (B) is just the spectral measure
of T , and eT (B) = PT (B).