It turns out that every vector field which has its components in L^{p} can be written as a
sum of a gradient and a vector field which has zero divergence. This is a very remarkable
result, especially when applied to vector fields which are only in L^{p}. Recall that for u a
function of n variables, Δu = ∑_{i=1}^{n}
2
∂∂xu2i
.
Definition 31.5.1Define
{ 1-
Φ(y) ≡ − a1 ln1 |y|, if2−nn = 2,
(n−2)an−1 |y| , if n > 2.
where a_{k}denotes the area of the unit sphere, S^{k}.
Then it is routine to verify ΔΦ = 0 away from 0. In fact, if n > 2,
In the above the subscripts following a comma denote partial derivatives.
Lemma 31.5.2For n ≥ 2
Φ,ij(y) = Ωij(ny)
|y |
where
n− 1
Ωijis Lipschitz continuous on S , (31.5.66)
(31.5.66)
Ωij(λy) = Ωij (y) , (31.5.67)
(31.5.67)
for all λ > 0, and
∫
Sn−1 Ωij(y )dσ = 0. (31.5.68)
(31.5.68)
Proof:
Proof:The case n = 2 is left to the reader. 31.5.66 and 31.5.67 are obvious from the
above descriptions. It remains to verify 31.5.68. If n ≥ 3 and i≠j, then this formula
is also clear from 31.5.64. Thus consider the case when n ≥ 3 and i = j. By
symmetry,
∫ ∫
I ≡ 1− ny2dσ = 1− ny2dσ.
Sn−1 i Sn−1 j
Hence
n ∫ ∫ ( )
nI = ∑ 1 − ny2dσ = n− n ∑ y2 dσ
i=1 Sn−1 i Sn−1 i i
∫
= (n − n )dσ = 0.
Sn−1
This proves the lemma.
Let U be a bounded open set locally on one side of its boundary having Lipschitz
boundary so the divergence theorem holds and let B = B
(0,R)
where
B ⊇ U − U ≡ {x− y : x ∈ U,y ∈ U}
Let f ∈ C_{c}^{∞}
(U )
and define for x∈ U,
∫ ∫
u (x) ≡ Φ(y)f (x− y)dy = Φ(x − y )f (y)dy.
B U
Let h
(y)
= f
(x − y)
.Then since Φ is in L^{1}
(B)
,
∫ ∫
Δu (x) = Φ (y )Δf (x − y)dy = Φ (y)Δh (y)dy
B B
→ 0 as ε → 0. The first term in 31.5.69 converges to 0 as
ε → 0 because
||∫ || {
|| Φ (y) ∇h (y) ⋅ndσ||≤ Cnh ε1n−2εn− 1 = Cnhε if n > 2
| ∂B(0,ε) | Ch (ln ε)ε if n = 2
and since ΔΦ
(y)
= 0,
∇Φ (y)⋅∇h (y) = ∇ ⋅(∇Φ (y)h (y)).
Consequently
∫
Δu (x ) = − B∖B(0,ε)∇ ⋅(∇ Φ (y) h(y))dy+ e(ε).
Thus, by the divergence theorem, 31.5.65, and the definition of h above,
∫
Δu (x ) = ∂B(0,ε)f (x− y)∇ Φ (y )⋅ndσ+ e (ε)
∫ ( y ) ( y )
= f (x− y) −-------n ⋅ − --- dσ + e(ε)
∂(B(0,ε) an− 1|y|) |y|
∫ 1
= − ∂B(0,ε)f (x− y )dσ(y) an−1εn−-1 + e(ε).
Letting ε → 0,
− Δu (x) = f (x).
This proves the following lemma.
Lemma 31.5.3Let U be a bounded open set in ℝ^{n}with Lipschitz boundary and letB ⊇ U − U where B = B
(0,R)
. Let f ∈ C_{c}^{∞}
(U )
. Then for x ∈ U,
∫ ∫
Φ (y )f (x− y )dy = Φ(x − y)f (y)dy,
B U
and it follows that if u is given by one of the above formulas, then for all x ∈ U,
− Δu (x) = f (x).
Theorem 31.5.4Let f ∈ L^{p}
(U )
. Then there exists u ∈ L^{p}
(U )
whose weak derivativesare also in L^{p}
(U)
such that in the sense of weak derivatives,
− Δu = f.
It is given by
∫ ∫
u(x) = Φ (y ) ^f (x− y )dy = Φ (x − y)f (y)dy (31.5.70)
B U
(31.5.70)
where
^f
denotes the zero extension of f off of U.
Proof: Let f ∈ L^{p}
(U)
and let f_{k}∈ C_{c}^{∞}
(U)
,
||fk − f||
_{Lp(U)
}→ 0, and let u_{k}
be given by 31.5.70 with f_{k} in place of f. Then by Minkowski’s inequality,
( ∫ ( ∫ ) )
||^ || p 1∕p
||u − uk||Lp(U) = U B Φ(y)|f (x − y)− fk(x− y)|dy dx
( ∫ ( ∫ | |p )1∕p )
≤ |Φ (y)| ||^f (x − y)− fk(x− y )|| dx dy
B U
∫
≤ B |Φ(y)|dy||f − fk||Lp(U) = C (B )||f − fk||Lp(U )
and so u_{k}→ u in L^{p}
(U )
. Also
∫ ∫
uk,i(x) = Φ,i(x − y)fk(y)dy = fk (x − y)Φ,i(y)dy.
U B
Now let
∫
wi ≡ ^f (x− y )Φ,i(y)dy. (31.5.71)
B
(31.5.71)
and since Φ_{,i}∈ L^{1}
(B)
, it follows from Minkowski’s inequality that
||uk,i − wi||Lp(U)
(∫ (∫ | | )p )1∕p
≤ ||f (x− y) −f^(x− y)|||Φ (y)|dy dx
U B k ,i
∫ (∫ | |p )1 ∕p
≤ |Φ,i(y)| ||fk(x − y)− ^f (x − y)|| dx dy
B U
≤ C (B)||fk − f ||Lp(U)
and so u_{k,i}→ w_{i} in L^{p}
(U )
.
Now let ϕ ∈ C_{c}^{∞}
(U)
. Then
∫ ∫ ∫
U wiϕdx = − kli→m∞ U ukϕ,idx = − U u ϕ,idx.
Thus u_{,i} = w_{i}∈ L^{p}
(ℝn )
and so if ϕ ∈ C_{c}^{∞}
(U )
,
∫ ∫ ∫ ∫
fϕdx = lim fϕdx = lim ∇u ⋅∇ ϕdx = ∇u ⋅∇ ϕdx
U k→∞ U k k→ ∞ U k U
and so −Δu = f as claimed. This proves the theorem.
One could also ask whether the second weak partial derivatives of u are in L^{p}
(U)
.
This is where the theory singular integrals is used. Recall from 31.5.70 and 31.5.71 along
with the argument of the above lemma, that if u is given by 31.5.70, then u_{,i} is given by
31.5.71 which equals
∫
U Φ,i(x − y )f (y)dy.
Lemma 31.5.5Let f ∈ L^{p}
(U)
and let
∫
wi(x) ≡ U Φ,i(x− y)f (y)dy.
Then w_{i,j}∈ L^{p}
(U)
for each j = 1
⋅⋅⋅
n and the map f → w_{i,j}is continuous and linear onL^{p}
(U )
.
Proof:First let f ∈ C_{c}^{∞}
(U )
. For such f,
∫ ∫
wi(x) = U Φ,i(x− y)f (y )dy = ℝn Φ,i(x − y)f (y)dy
∫ ∫
= n Φ,i(y)f (x− y)dy = Φ,i(y)f (x− y)dy
ℝ B