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5.3. Theorem. Let V be an n-dimensional totally real subspace of�n . For every p ∈]1, +∞[,
there exists a constant c = c(p, n) such that
�Du�Lp ≤ c�∂u�Lp for every u ∈ C ∞ c (�,�n ), and for every u ∈ C ∞ c (Cl(À),�n ) such that u(s) ∈ V for every s ∈Ê.
We shall also need the following regularity result for weak solutions of ∂. Denoting by ∂ :=
∂s − i∂t the anti-Cauchy-Riemann operator, we have:
5.4. Theorem. (Regularity of weak solutions of ∂) Let V be an n-dimensional totally real subspace
of�n , and let 1 < p < ∞, k ∈Æ.
(i) Let u ∈ L p
loc (�,�n k,p
), f ∈ Wloc (�,�n ) be such that
��〈u, Re ∂ϕ〉dsdt = −Re
��〈f, ϕ〉dsdt,
for every ϕ ∈ C ∞ c (�,�n ). Then u ∈ W k+1,p
loc (�,�n ) and ∂u = f.
(ii) Let u ∈ Lp k,p
(À,�n ), f ∈ W (À,�n ) be such that
�À〈u,
�À〈f, Re ∂ϕ〉dsdt = −Re ϕ〉dsdt,
for every ϕ ∈ C ∞ c (�,�n ) such that ϕ(Ê) ⊂ V . Then u ∈ W k+1,p (À,�n ), ∂u = f, and the
trace of u onÊtakes values into the ω0-orthogonal complement of V ,
V ⊥ω 0 := {ξ ∈�n | ω0(ξ, η) = 0 ∀η ∈ V }.
5.5. Remark. If we replace the upper half-planeÀin (ii) by the right half-plane {Re z > 0} and
the test mappings ϕ ∈ C ∞ c (�,�n ) satisfy ϕ(iÊ) ⊂ V , then the trace of u on iÊtakes value into
V ⊥ , the Euclidean orthogonal complement of V inÊ2n .
Two linear subspaces V, W ofÊn are said to be partially orthogonal if the linear subspaces
V ∩ (V ∩ W) ⊥ and W ∩ (V ∩ W) ⊥ are orthogonal, that is if their projections into the quotient
Ên /V ∩ W are orthogonal.
5.6. Lemma. Let V and W be partially orthogonal linear subspaces ofÊn . For every p ∈]1, +∞[,
there exists a constant c = c(p, n) such that
for every u ∈ C ∞ c (Cl(À+ ),�n ) such that
�Du�Lp ≤ c�∂u�Lp (81)
u(s) ∈ N ∗ V ∀s ∈ [0, +∞[, u(it) ∈ N ∗ W ∀t ∈ [0, +∞[. (82)
Proof. Since V and W are partially orthogonal,Ên has an orthogonal splittingÊn = X1 ⊕ X2 ⊕
X3 ⊕ X4 such that
Therefore,
V = X1 ⊕ X2, W = X1 ⊕ X3.
N ∗ V = X1 ⊕ X2 ⊕ iX3 ⊕ iX4, N ∗ W = X1 ⊕ X3 ⊕ iX2 ⊕ iX4.
Let U ∈ U(n) be the identity on (X1 ⊕ X2) ⊗�, and the multiplication by i on (X3 ⊕ X4) ⊗�.
Then
UN ∗ V =Ên , UN ∗ W = X1 ⊕ X4 ⊕ iX2 ⊕ iX3 = N ∗ (X1 ⊕ X4).
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