20.07.2013 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

D. Kucerovsky / Journal of Functional Analysis 236 (2006) 395–408 399<br />

subalgebra, we are to show the same for (1 + c)B(1 + c). Because the image of C in the corona is<br />

not unital, there exists c ′ ∈ C such that c ′ (1 + c) is in C but not in B, so that (1 + c)c ′ Bc ′ (1 + c)<br />

contains a full stab<strong>le</strong> subalgebra. But this then gives a full stab<strong>le</strong> subalgebra of the <strong>la</strong>rger algebra<br />

(1 + c)B(1 + c). This, in fact, is exactly how the nonunital version of the main result of [6] is<br />

<strong>de</strong>rived from the easier unital case.<br />

The second major theorem that we need is the Kasparov stabilization theorem [10], one of<br />

the fundamental tools of Kasparov’s KK-theory [11], which will here be used to “diagonalize” a<br />

singly generated hereditary subalgebra of a stab<strong>le</strong> algebra. One form of the Kasparov stabilization<br />

theorem is as follows.<br />

Theorem 4.3. [19] Let E be a Hilbert B-mo<strong>du</strong><strong>le</strong> that is countably generated in M(E). Ifwe<br />

<strong>de</strong>note the standard Hilbert B-mo<strong>du</strong><strong>le</strong> by HB, then E ⊕ HB is unitarily equiva<strong>le</strong>nt to HB.<br />

In the statement of the theorem, a Hilbert mo<strong>du</strong><strong>le</strong> E is said to be countably generated in<br />

M(E) if there is a sequence of multipliers (mi) ⊆ M(E) such that the e<strong>le</strong>ments {mib: b ∈ B}<br />

span a <strong>de</strong>nse submo<strong>du</strong><strong>le</strong> of E. This <strong>de</strong>finition generalizes Kasparov’s, since the generators mi do<br />

not need to be in the mo<strong>du</strong><strong>le</strong> E. It is likely that the assumption of σ -uniticity can be removed in<br />

the next theorem.<br />

Theorem 4.4. Let B be stab<strong>le</strong> and σ -unital. A hereditary subalgebra, ℓBℓ ∗ , generated by an<br />

e<strong>le</strong>ment ℓ of the multipliers M(B) is isomorphic to a hereditary subalgebra generated by a<br />

multiplier projection, P .Ifℓ is not contained in a proper i<strong>de</strong>al of the multipliers, then neither<br />

is P .<br />

Proof. The closed right i<strong>de</strong>al E := ℓB is, if we take B to act in the natural way from the right,<br />

a Hilbert B-mo<strong>du</strong><strong>le</strong> with inner pro<strong>du</strong>ct 〈a,b〉:=a∗b, and is countably generated (in the generalized<br />

sense <strong>de</strong>scribed above) with generators (ℓ1/n ) ∞ n=1 . Thus, by the above form of the Kasparov<br />

stabilization theorem (Theorem 4.3), there is a unitary U in L(E ⊕ HB, HB) imp<strong>le</strong>menting an<br />

isomorphism of E ⊕ HB and HB.<br />

Let P be the projection of E ⊕ HB onto the first factor, E. The projection T := UPU∗ ∈<br />

L(HB) has image isomorphic (by a unitary equiva<strong>le</strong>nce) to E, and thus by the <strong>de</strong>finition of the<br />

compact operators on a Hilbert mo<strong>du</strong><strong>le</strong>,<br />

K(T HB) ∼ = K(ℓB).<br />

Now, however, recalling the <strong>de</strong>finition of the compact operators on a Hilbert mo<strong>du</strong><strong>le</strong>, we see that<br />

K(ℓB) is generated by e<strong>le</strong>ments of the form ℓb1b ∗ 2 ℓ∗ , where the bi are in B. Hence (up to unitary<br />

equiva<strong>le</strong>nce), T K(HB)T ∼ = ℓBℓ ∗ . However, T is in L(HB) = M(B ⊗K), and K(HB) = B ⊗K,<br />

which thus comp<strong>le</strong>tes the proof of the first part of the <strong>le</strong>mma since B is stab<strong>le</strong>. For the fullness<br />

c<strong>la</strong>imed in the <strong>la</strong>st part of the theorem, notice that if we write L(E ⊕ HB, HB)P L(HB,E⊕ HB)<br />

as a formal two-by-two matrix, and consi<strong>de</strong>r the lower right-hand corner, we are to show that<br />

L(ℓB, HB)L(HB,ℓB)is <strong>de</strong>nse in L(HB). However,<br />

L(ℓB, HB)L(HB,ℓB)= L(B, HB)ℓ ∗ ℓL(HB,B)<br />

= L(B, HB)L(B)ℓ ∗ ℓL(B)L(HB,B)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!