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K. Koufany, G. Zhang / Journal of Functional Analysis 236 (2006) 546–580 575<br />

Let f be in M(λs) and suppose f satisfies (25). Then f = Ps(ϕ) is the Poisson transform of a<br />

hyperfunction ϕ on S.<br />

7.2. The necessity of the Hua equation (25)<br />

Consi<strong>de</strong>r the operator<br />

Proposition 7.3. Let s ∈ C. We have<br />

and<br />

In particu<strong>la</strong>r, for σ = 0,(p+ b)/2,<br />

Y = U − −2σ 2 + 2pσ + c<br />

σ(2σ − p − b) W.<br />

WPs,u(z) = Ps,u(z)σ 2 (2σ − p − b)u<br />

UPs,u(z) = Ps,u(z)σ −2σ 2 + 2pσ + c u.<br />

YPs,u(z) = 0,<br />

and the image f = Ps(ϕ) of the Poisson transform of a hyperfunction ϕ on S satisfies<br />

Yf = 0.<br />

Proof. We compute first W on Ps(z, u). By the covariant property of W and transformation<br />

property (10) of the kernel Ps(z, u) we need only to prove that the formu<strong>la</strong> is valid at z = 0. We<br />

use the differentiation h(z, z)¯∂ in p<strong>la</strong>ce of ¯∂ as it will pro<strong>du</strong>ce some more compact formu<strong>la</strong>s; the<br />

eventual result will be the same at z = 0. (The operator h(z, z)¯∂ can be geometrically <strong>de</strong>fined,<br />

see Section 4.) Proceeding as in the proof of Theorem 5.3 by using Lemma 5.2, we have<br />

<br />

h(z, z)¯∂ ∂Ps(z, u) = σ 2 Ps,u(z) b(z, z) z ¯z − u ¯z ⊗ ¯z z −ū z − σPs,u(z)Id.<br />

Its image un<strong>de</strong>r −Ad V ⊗ ¯V →kC is,<br />

−Ad V ⊗ ¯V →kC<br />

since<br />

and<br />

h(z, z)¯∂ ∂Ps(z, u) = σ 2 Ps,u(z)D b(z, z) z ¯z − u ¯z , ¯z z −ū z − σPs,u(z)pZ0,<br />

−AdV ⊗ ¯V →kC<br />

(u ⊗¯v) =−[u, ¯v]=D(u, ¯v)<br />

−AdV ⊗ ¯V →kC<br />

Id = pZ0.<br />

To compute ¯∂Ad V ⊗ ¯V →kC (h(z, z)¯∂)∂Ps(z, u) at z = 0 we observe that for any function f ,<br />

¯∂f(0) is the coefficient of ¯z in the expansion of f near z = 0. By direct computation we find:

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