Helms L.L. Potential Theory
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298 7 Interpolation and Monotonicity
Therefore,
δ
b
|u(x)|≤
2
a
δ
a
0
|u|
(b)
a,Ω
+ K
a
∞
j=0
2
−bj
|u|
(b)
a,Ω
≤
2
a
δ
a
0
|u|
(b)
a,Ω
+(2
b
K
a
/(2
b
− 1))|u|
(b)
a,Ω
(7.29)
for all x ∈ Ω
δ
,δ ≤ δ
0
/2, and it follows that |u|
(b)
0,Ω
≤ C|u|
(b)
a,Ω
where
C =2
a
/δ
a
0
+(2
b
K
a
/(2
b
−1)). This completes the proof of Inequality (7.27).
Turning to Inequality (7.28), assume that b<0. Consider x, y ∈
*
Ω
δ
and let
t be a fixed positive number. Two cases will be considered.
Case (i) (δ<t). In this case, choose k ≥ 1 such that t<2
k
δ ≤ 2t.There
are then points x
1
∈
*
Ω
2δ
,x
2
∈
*
Ω
4δ
,...,x
k
∈
*
Ω
2
k
δ
such that |x − x
1
|≤
Kδ,|x
1
− x
2
|≤K2δ,...,|x
k−1
− x
k
|≤K2
k−1
δ. Letting x
0
= x and x
t
=
x
k
∈
*
Ω
2
k
δ
⊂
*
Ω
t
,
|x − x
t
|≤
k
i=1
|x
i−1
− x
i
|≤Kδ
k
i=1
2
i−1
≤ Kδ2
k
.
Since
a+b
|u(x
) − u(y
)|
|x
− y
|
a
≤|u|
(b)
a,Ω
for all x
,y
∈
*
Ω
, 0 <<δ
0
,
and x
t
∈
*
Ω
t
⊂
*
Ω
δ
,
|u(x) − u(x
t
)|≤
k
i=1
|u(x
i−1
) − u(x
i
)|
≤|u|
(b)
a,Ω
k
i=1
|x
i−1
− x
i
|
a
(2
i−1
δ)
−a−b
≤|u|
(b)
a,Ω
k
i=1
(K2
i−1
δ)
a
(2
i−1
δ)
−a−b
≤ K
a
t
−b
k
i=1
1
2
k−i
−b
≤ Ct
−b
|u|
(b)
a,Ω
.
Inthesameway,fory ∈
*
Ω
δ
there is a y
t
∈
*
Ω
t
with |y −y
t
|≤Kδ2
k
such that
|u(y) − u(y
t
)|≤Ct
−b
|u|
(b)
a,Ω
.
7.6 Monotonicity 299
It follows that
|u(x) − u(y)|≤|u(x) − u(x
t
)| + |u(x
t
) − u(y
t
)|+ |u(y
t
) − u(y)|
≤ 2Ct
−b
|u|
(b)
a,Ω
+ |x
t
− y
t
|
a
(2
k
δ)
−a−b
|u|
(b)
a,Ω
≤ 2Ct
−b
|u|
(b)
a,Ω
+ |x
t
− y
t
|
a
t
−a−b
|u|
(b)
a,Ω
.
Since |x
t
−y
t
|≤2Ct+|x−y|, putting t = |x−y|, |x
t
−y
t
|≤(2C +1)|x−y| and
|u(x) − u(y)|
|x − y|
−b
≤ (2C +(2C +1)
a
)|u|
(b)
a,Ω
, |x − y| >δ. (7.30)
Case (ii) (δ ≥ t)Inthiscase,
|u(x) − u(y)|≤|x − y|
a
δ
−a−b
|u|
(b)
a,Ω
so that
|u(x) − u(y)|≤|u|
(b)
a,Ω
|x − y|
a
t
δ
a+b
t
−a−b
≤|u|
(b)
a,Ω
|x − y|
a
t
−a−b
.
Taking t = |x − y|,
|u(x) − u(y)|
|x − y|
−b
≤|u|
(b)
a,Ω
, |x − y|≤δ. (7.31)
Combining this inequality with Inequality (7.30) and taking the supremum
over x, y ∈
*
Ω
δ
,
|u|
(b)
−b,Ω
≤ C|u|
(b)
a,Ω
.
This completes the proof in the a ≤ 1case.Thea>1 case will be dealt with
by an induction argument. Assume that the lemma is true for a − 1andb is
neither a negative integer or 0. By Lemma 7.6.4, if a
> 1, then
|u|
(b)
a
,Ω
≤ C
|α|≤1
|D
α
u|
(b+1)
a
−1,Ω
≤ C
|α|≤1
|D
α
u|
(b+1)
a−1,Ω
≤ C|u|
(b)
a,Ω
. (7.32)
The 0 ≤ a
≤ 1 <acase can be proved using the preceding results as follows.
Note that
|u|
(b)
a
,Ω
≤ C|u|
(b)
1,Ω
by the “a ≤ 1” case. Suppose first that 1 <a≤ 2sothata −1 ≤ 1. Applying
the mean value theorem,
300 7 Interpolation and Monotonicity
δ
1+b
|u|
1,
*
Ω
δ
= δ
1+b
|u|
0+1,
*
Ω
δ
= δ
1+b
(u
0,
*
Ω
δ
+[u]
0+1,
*
Ω
δ
)
≤ δ
1+b
u
0,
*
Ω
δ
+
|α|=1
D
α
u
0,
*
Ω
δ
≤|u|
(1+b)
0,Ω
+
|α|=1
|D
α
u|
(1+b)
0,Ω
.
Since a − 1 ≤ 1, by the above “a ≤ 1” case,
|D
α
u|
(1+b)
0,Ω
≤ C|D
α
u|
(1+b)
a−1,Ω
≤ C sup
δ>0
δ
a+b
|u|
a,
*
Ω
δ
= C|u|
(b)
a,Ω
;
and by Theorem 7.3.2,
|u|
(1+b)
0,Ω
≤|u|
(1+b)
a−1,Ω
=sup
δ>0
δ
a+b
|u|
a−1,
*
Ω
δ
≤ sup
δ>0
δ
a+b
|u|
a,
*
Ω
δ
= |u|
(b)
a,Ω
.
Therefore,
|u|
(b)
1,Ω
≤ C|u|
(b)
a,Ω
, 1 <a≤ 2
so that
|u|
(b)
a
,Ω
≤ C|u|
(b)
1,Ω
≤ C|u|
(b)
a,Ω
, 0 ≤ a
≤ 1 <a≤ 2.
The remaining cases are proved by induction using Inequality (7.32).
Remark 7.6.8 The above proof does not apply when b = 0, as can be seen in
the derivation of Inequality (7.29). The requirement that b not be a negative
integer is exhibited in Inequality (7.32); for example, if b = −1,
|u|
(−1)
a
,Ω
=
|α|≤1
|D
α
u|
(0)
a
−1,Ω
and the induction process cannot be applied, since b = 0 in the terms in the
sum.
If a + b ≥ 0,a
+ b
≥ 0,b
≤ b, a
≥ a, by combining Inequalities (7.25)
and (7.26)
|u|
(b)
a,Ω
≤ C|u|
(b
)
a
,Ω
(7.33)
where C = C(a, b, a
b
,d(Ω)). We now will consider compactness properties
of the H
(−b)
a
(Ω) spaces. Recall that b = m + σ with 0 <σ≤ 1.
Lemma 7.6.9 (Gilbarg and H¨ormander [24]) If 0 <a
<a,0 <b
<
b, b ≤ a, b
≤ a
, and neither b nor b
is a nonnegative integer, then ev-
ery bounded sequence (u
j
) in H
(−b)
a
(Ω) has a subsequence that converges in
H
(−b
)
a
(Ω).
7.6 Monotonicity 301
Proof: Suppose |u
j
|
(−b)
a,Ω
≤ M for all j ≥ 1. By Inequality (7.26), |u
j
|
b,Ω
=
|u
j
|
(−b)
b,Ω
≤ C|u
j
|
(−b)
a,Ω
≤ M,j ≥ 1, from which it follows that each u
j
∈
C
m
(Ω
−
). It follows that the sequences {D
α
u
j
}, 0 ≤|α|≤m are bounded
and equicontinuous on Ω
−
. By Theorems 0.2.2 and 0.2.3, there is a subse-
quence {u
j
k
} and a function u such that D
α
u
j
k
− D
α
u
0,Ω
→ 0ask →
∞, |α| =0,...,m. By letting k →∞in the definition of |u
j
k
|
b,Ω
, |u|
b,Ω
≤ M ,
and |u
j
k
−u|
b,Ω
≤ 2M, k ≥ 1. Consider any c ∈ (0,b). Then c = tb +(1−t)0
for some 0 <t<1. By Lemma 7.6.3,
|u
j
k
− u|
c,Ω
= |u
j
k
− u|
(−c)
c,Ω
≤ C(|u
j
k
− u|
(−b)
b,Ω
)
t
(|u
j
k
− u|
0,Ω
)
1−t
≤ C(2M)
t
(u
j
k
− u
0,Ω
)
1−t
and lim
k→∞
|u
j
k
− u|
c,Ω
=0.Ifa
= b
<b, this result is applicable with
c = a
so that |u
j
k
−u|
(−b
)
a
,Ω
→ 0ask →∞. It therefore can be assumed that
a
>b
. By comparing the slopes of the lines in R
2
through (a, −b)andthe
points (0, 0), (a
, −b
), the rest of the proof can be split into three cases.
Case (i) a/b = a
/b
. In this case, the point (a
, −b
) lies on the line joining
(0, 0) to (a, −b)sothatthereisa0<t<1 such that
|u
j
k
− u|
(−b
)
a
,Ω
≤ C
|u
j
k
− u|
(−b)
a,Ω
t
|u
j
k
− u|
(0)
0,Ω
1−t
≤ C(2M)
t
( u
j
k
− u
0,Ω
)
1−t
.
Therefore, lim
k→∞
|u
j
k
− u|
(−b
)
a
,Ω
=0.
Case (ii) a/b > a
/b
. In this case, the line through (a
, −b
)and(a, −b) will
intersect the line y = −x in a point (c, −c)with0<c<b.Thus,thereisa
0 <t<1 such that
|u
j
k
− u|
(−b
)
a
,Ω
≤ C(2M)
t
|u
j
k
− u|
(−c)
c,Ω
1−t
and lim
k→∞
|u
j
k
− u|
(−b
)
a
,Ω
=0.
Case (iii) a/b < a
/b
.Inthiscase,choosec so that b
<c<min (a
,b).
This is possible since b
<a
and b
<b.Let(c, −d) be the point on the line
through (a
, −b
)and(a, −b). Since −d>−c,forsome0<t<1,
|u
j
k
− u|
(−b
)
a
,Ω
≤ C(2M )
t
|u
j
k
− u|
(−d)
c,Ω
1−t
≤ C(2M )
t
|u
j
k
− u|
(−c)
c,Ω
1−t
by Inequality (7.6.3). Since 0 <c<b,lim
k→∞
|u
j
k
− u|
(−b
)
a
,Ω
=0.
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Chapter 8
Newtonian Potential
8.1 Introduction
If Ω is an open subset of R
n
, a function k : Ω × Ω → R is called a kernel
and can be used to transform a measurable function g : Ω → R into a new
function ˆg : Ω → R by putting
ˆg(x)=
Ω
k(x, y)g(y) dy, x ∈ Ω,
provided the integral is defined. Properties of k and g will be examined now
that will ensure that ˆg has some desirable property; e.g., that ˆg is continuous
on Ω. The most obvious criterion for continuity of ˆg is that g be integrable
on Ω,thatk be bounded on Ω × Ω,andthatk(·,y) is continuous on Ω for
each y ∈ Ω;forif{x
n
} is an arbitrary sequence in Ω with limit x,then
lim
n→∞
ˆg(x
n
) = lim
n→∞
Ω
k(x
n
,y)g(y) dy =
Ω
k(x, y)g(y) dy =ˆg(x)
by the Lebesgue dominated convergence theorem. If the integrability condi-
tion on g is dropped, additional conditions on k must be imposed. Rather
than stating some preliminary results specific to the Newtonian, the results
will be formulated in a more general context for use in later chapters.
8.2 Subnewtonian Kernels
Throughout this section, Ω will be an open subset of R
n
,Σwill be a relatively
open subset of ∂Ω,andk will be a kernel on (Ω ∪ Σ) × (Ω ∪ Σ)thatis
continuous in the extended sense on (Ω ∪Σ)×(Ω ∪Σ) and continuous off the
diagonal of (Ω ∪Σ) × (Ω ∪Σ); in addition, it will be assumed that for each
L.L. Helms, Potential Theory, Universitext, 303
c
Springer-Verlag London Limited 2009
304 8 Newtonian Potential
y ∈ Ω the partial derivatives D
x
i
k(x, y),D
x
i
x
j
k(x, y), and D
x
i
x
j
x
k
k(x, y)are
continuous on (Ω∪Σ) ∼{y} and continuous in the extended sense on (Ω∪Σ),
and have the following properties:
|k(x, y)|≤C|x − y|
−n+2
, (8.1)
|D
x
i
k(x, y)|≤C|x − y|
−n+1
, 1 ≤ i ≤ n, (8.2)
|D
x
i
x
j
k(x, y)|≤C|x − y|
−n
, 1 ≤ i, j ≤ n, (8.3)
|D
x
i
x
j
x
k
k(x, y)|≤C|x − y|
−n−1
, 1 ≤ i, j, k ≤ n (8.4)
for all x ∈ Ω ∪ Σ,y ∈ Ω,whereC is a constant depending upon parameters
n, d(Ω), etc. Such a kernel will be called a subnewtonian kernel.IfΩ
is a bounded open subset of R
2
, the logarithmic potential −log |x − y| is
a subnewtonian potential. See Section 8.3 for details. Throughout this
section k(x, y) will denote a subnewtonian kernel.Mostoftheresults
of this section are adapted from [25].
Example 8.2.1 Consider α satisfying 2 <α<nand a bounded open region
Ω. The function k(x, y)=|x−y|
−n+α
is called a Riesz potential of order α.
It is easily seen that such potential functions are subnewtonian. An extensive
treatment of Riesz potentials can be found in [36].
Lemma 8.2.2 If >0 and |β| =0, 1,then
Ω∩B
x,
|D
β
(x)
k(x, y)|dy ≤
Cσ
n
2−|β|
2 −|β|
,x∈ Ω ∪ Σ;
moreover,
lim
→0
Ω∩B
x,
|D
β
(x)
k(x, y)|dy =0 uniformly for x ∈ Ω ∪ Σ.
Proof: The second assertion follows from the first. By Inequalities (8.1) and
(8.2),
Ω∩B
x,
|D
β
(x)
k(x, y)|dy ≤ C
B
x,
1
|x − y|
n−2+|β|
dy =
Cσ
n
2−|β|
2 −|β|
.
Lemma 8.2.3 If Γ is a compact subset of Ω and |β| =0, 1 ,then
Γ
|D
β
(x)
k(x, y)|dy < +∞
for all x ∈ Ω ∪ Σ.
Proof: Fix x ∈ Ω ∪ Σ.Forδ>0,
Γ
|D
β
(x)
k(x, y)|dy =
Γ ∩B
x,δ
|D
β
(x)
k(x, y)|dy +
Γ ∼B
x,δ
|D
β
(x)
k(x, y)|dy.
8.2 Subnewtonian Kernels 305
By the preceding lemma, δ>0 can be chosen so that the first term is less
than 1. Since D
β
(x)
k(x, ·) is continuous and therefore bounded on the compact
set {y; y ∈ Γ, |x − y|≥δ}, the second term is finite.
Definition 8.2.4 An operator K corresponding to the kernel k is defined as
follows. If f is a measurable function on Ω and x ∈ Ω ∪ Σ,let
Kf(x)=
Ω
k(x, y)f(y) dy
if finite.
If Σ is a subset of the boundary of the open set Ω, C
k
(Ω ∪Σ) will denote
the set of functions on Ω whose derivatives of order less than or equal to k
are continuous on Ω and have continuous extensions to Ω ∪ Σ.
Theorem 8.2.5 If f is a bounded and integrable function on Ω,thenKf ∈
C
0
b
(Ω ∪ Σ).
Proof: It will be shown first that Kf (x) is defined for all x ∈ Ω ∪ Σ.Fix
x ∈ Ω ∪ Σ and >0. Then
|Kf(x)|≤
Ω∩B
x,
|k(x, y)||f(y)|dy + C
Ω∼B
x,
1
|x − y|
n−2
|f(y)|dy
≤ (sup
Ω
|f|)
Ω∩B
x,
|k(x, y)|dy +
C
n−2
Ω∼B
x,
|f(y)|dy.
The first integral is finite by Lemma 8.2.3 and the second is finite by the
integrability of f. Now consider a function ψ ∈ C
2
(R) satisfying 0 ≤ ψ ≤
1, 0 ≤ ψ
≤ 2,ψ(t)=0fort ≤ 1, and ψ(t)=1fort ≥ 2. For each >0and
x ∈ Ω ∪ Σ,let
K
f(x)=
Ω
k(x, y)ψ(|x − y|/)f(y) dy.
Since f is integrable on Ω, by Inequality (8.1)
|K
f(x)|≤
Ω∼B
x,
|k(x, y)|ψ(|x − y|/)|f(y)|dy
≤ C
Ω∼B
x,
1
|x − y|
n−2
|f(y)|dy
≤
C
n−2
Ω
|f(y)|dy < +∞
(8.5)
and K
f(x) is defined for all x ∈ Ω ∪ Σ.Let{x
j
} be a sequence in Ω ∪ Σ
such that x
j
→ x ∈ Ω ∪ Σ.Since
|k(x, y)ψ(|x − y|/)f(y)|≤
C
n−2
|f(y)|,y∈ Ω ∼ B
x,
,
306 8 Newtonian Potential
and the latter function is integrable on Ω, lim
j→∞
K
f(x
j
)=K
f(x)bythe
Lebesgue dominated convergence theorem. This shows that K
f ∈ C
0
b
(Ω∪Σ).
By Lemma 8.2.2,
|Kf(x) − K
f(x)|≤
Ω
|k(x, y)(1 − ψ(|x − y|/))f(y)|dy
≤ (sup
Ω
|f|)
Ω∩B
x,2
|k(x, y)|dy → 0
uniformly for x ∈ Ω ∪ Σ as → 0. Thus, Kf is continuous on Ω ∪ Σ.Asto
the second assertion, by what has just been proved there is an
0
> 0such
that |Kf (x)|≤1+|K
0
f(x)| and the latter term is bounded uniformly for
x ∈ Ω ∪ Σ by Inequality (8.5).
Remark 8.2.6 Only Inequalities (8.1) and (8.2) are used in the proof of the
following theorem.
Theorem 8.2.7 If f is bounded and integrable on Ω,thenKf ∈ C
1
(Ω ∪Σ),
D
x
i
Kf(x)=
Ω
D
x
i
k(x, y)f(y) dy for all x ∈ Ω ∪ Σ,
and [Kf ]
1+0,Ω
< +∞.
Proof: Fix 1 ≤ i ≤ n. Define g : Ω ∪ Σ → R by the equation
g(x)=
Ω
D
x
i
k(x, y)f(y) dy x ∈ Ω ∪ Σ.
Let ψ be the function described in the preceding proof. For each >0and
x ∈ Ω ∪ Σ,let
g
(x)=
Ω
D
x
i
k(x, y)
ψ(|x − y|/)f(y) dy.
Since
|
D
x
i
k(x, y)
ψ(|x − y|/)f(y)|≤
C
|x − y|
n−1
ψ(|x − y|/)|f(y)|≤
C
n−1
|f(y)|
and the latter function is integrable on Ω, g
is continuous on Ω ∪ Σ by the
Lebesgue dominated convergence theorem. By Lemma 8.2.2,
|g(x) − g
(x)| =
Ω
D
x
i
k(x, y)
1 − ψ(|x − y|/)
f(y) dy
≤
sup
Ω
|f|
Ω∩B
x,2
|D
x
i
k(x, y)|dy → 0
8.2 Subnewtonian Kernels 307
as → 0 uniformly for x ∈ Ω ∪ Σ. It follows that g is continuous on Ω ∪ Σ.
Letting
K
f(x)=
Ω
k(x, y)ψ(|x − y|/)f(y) dy
and using the fact that f is integrable on Ω,
|K
f(x)| =
Ω∼B
x,
k(x, y)ψ(|x − y|/)f(y) dy
≤
Ω∼B
x,
C
|x − y|
n−2
|f(y)|dy
≤
C
n−2
Ω
|f(y)|dy < +∞
and K
f is defined for all x ∈ Ω ∪ Σ. It follows from Lemma 8.2.2 that
|Kf(x) − K
f(x)|≤
Ω
|k(x, y)(1 − ψ(|x − y|/))f(y) dy
≤
sup
Ω
|f|
Ω∩B
x,2
|k(x, y)|dy → 0
uniformly for x ∈ Ω ∪ Σ as → 0. Thus, lim
→0
K
f = Kf uniformly on
Ω ∪ Σ.Forfixed>0andx, y ∈ Ω,
D
x
i
k(x, y)ψ(|x − y|/)
= k(x, y)ψ
(|x − y|/)
1
x
i
− y
i
|x − y|
+
D
x
i
k(x, y)
ψ(|x − y|/). (8.6)
Since ψ
(|x − y|/)=0inB
x,
and |k(x, y)|≤C/
n−2
outside B
x,
, the first
term on the right is dominated by 2C/
n−1
; the second term on the right is
dominated by C/
n−1
.Thus
|D
x
i
k(x, y)ψ(|x − y|/)
|≤
2C
n−1
(8.7)
uniformly for x, y ∈ Ω. Letting
i
denote a unit vector having 1 as its ith
coordinate,
K
f(x + λ
i
) − Kf (x)
λ
=
Ω
k(x + λ
i
,y)ψ(|x + λ
i
− y|/) − k(x, y)ψ(|x − y|/)
λ
f(y) dy
=
Ω
D
x
i
k(ξ
λ
,y)ψ(|ξ
λ
− y|/)
f(y) dy