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

Since ϕ = ϕ0 + ϕ1 ∧ ωN, one has
UΦdz=
and therefore,
R N
J. Van Schaftingen / Journal of Functional Analysis 236 (2006) 490–516 503
R N
R n
uϕ dx =
R n
uϕ0 dx +
R n
uϕ1 dx ∧ ωN
UΦdz
uDk(Rn )ϕ0L1 (Rn ) +uDk−1(Rn )ϕ1L1 (Rn ) |ωN |.
When k>1, the conclusion comes from Theorem 3.1 and from the inequality
ϕ0 L 1 (R n ) +ϕ1 L 1 (R n ) Φ0 L 1 (R n ) +Φ1 ∧ ωN L 1 (R n ) Φ0 + Φ1 ∧ ωN L 1 (R n ) .
The last inequality comes from the fact that Φ0(x) and Φ1(x) ∧ ωN are orthogonal for every
x ∈ RN . When k = 1, one has ϕ1 = 0, and the conclusion comes similarly.
Conversely, let us now estimate uDk(Rn ) by Proposition 2.6. Let ψ ∈ D(Rn ; Λk−1Rn ). Consider
a family (ηλ)λ>0 in D(R) such that ηλ 0,
R ηλ dt = 1 and
R |η′ λ | dt → 0asλ→∞, and let Ψλ(x, t) = ηλ(t)ψ(x). For every λ>0,
R N
Therefore,
udψdx
=
R n
UdΨλdz =
R N
R N
Letting λ →∞yields the conclusion. ✷
U(dηλ∧ ψ + ηλdψ)dz = 0 +
R n
udψ dx.
UdΨλdz UDk(R N
) ψL1 (Rn ) +dψL1 (Rn ) η ′ λL1
(R) .
Remark 3.3. When kk(see Proposition 4.6). On the
other hand, the extension of a function in Dn(R n ) lies in DN(R N ) by Proposition 2.9. In view of
the trace theory of the next section, one could wonder whether when 1 kk.
3.3. Trace theory
The restriction of continuous functions from R N to R n can be extended to a continuous operator
from VK(R n ) to Vk(R n ) when N − K = n − k.
Theorem 3.4. Let n N, 1 K N and k = K − (N − n). Let U ∈ Vk(R N ) be continuous.
Define for x ∈ R n ,
u(x) = U(x,0).