NONLINEAR POTENTIAL THEORY-THE BESSEL POTENTIAL APPROACH
JAN MALÝ
1. Energy and potentials
Let 0 < α ≤ n and 1 < p < ∞. We consider the fractional Sobolev space Wα,p (Rn ) of functions
u = Gα ∗ f , where Gα is the Bessel kernel of order α and f ∈ Lp (Rn ). Let u ∈ Wα,p (Rn ).
The function f is uniquely determined by u and is labelled as G−α ∗ u. Indeed, if Gα ∗ f = 0, then
for each test function ϕ ∈ S(Rn ) we have
Z
Z
0=
(Gα ∗ f ) ϕ dx =
f Gα ∗ ϕ dx
Rn
Rn
and the functions of the form Gα ∗ ϕ are dense in S.
The (α, p)-norm of u is defined as the Lp -norm of f . Let µ be a Radon measure on Rn . If the
functional
Z
(1.1)
u 7→
u dµ,
u ∈ S(Rn )
Rn
is continuous with respect to the Wα,p -norm, we say that µ has a finite (α, p) energy and we extend
the functional (1.1) to all of Wα,p (Rn ). We denote the action of this functional on u ∈ Wα,p (Rn ) as
hµ, ui. This way we identify the family of all measures of finite (α, p)-energy with the positive cone of
Wα,p (Rn )∗ .
The energy of a measure µ ∈ Wα,p (Rn )∗ is defined as
1
1
p
hµ,
G
∗
f
i
−
E
(µ)
=
sup
kf
k
.
(1.2)
α
α,p
p
p0
p
f ∈Lp (Rn )
Notice that this is a special application of the so called Fenchel duality in convex analysis. Consider a
measure µ ∈ Wα,p (Rn )∗ . By direct methods of the calculus of variations we obtain that there exists a
minimizer f = fµ of the functional (1.2) and that it satisfies the Euler-Lagrange equation
Z
(1.3)
hµ, Gα ∗ gi −
|f |p−2 f g dx = 0,
g ∈ Lp (Rn ).
Rn
Since hµ, Gα ∗ gi = hGα ∗ µ, gi, we deduce that
|f |p−2 f = Gα ∗ µ.
Now we use the fact that µ ≥ 0 and the kernel Gα is positive as well, so that f ≥ 0 and thus Gα ∗µ = f p−1 ,
this means
(1.4)
0
f = (Gα ∗ µ)p −1 .
We define
(1.5)
0
Vα,p µ = Gα ∗ fµ = Gα ∗ (Gα ∗ µ)p −1 .
This object is called the Vα,p -potential of µ, it has been introduced by Maz’ya and Havin. Notice that
the Vα,p -potential of µ can make a sense even if µ is of infinite (α, p)-energy. Consider the linear case
p = 2. Then
Vα,2 µ = G2α ∗ µ,
which makes sense for all Radon measures and defines an a.e. finite function e.g. whenever µ has a
compact support. On the other hand, the Dirac measure has a finite energy only if Wα,p (Rn ) ,→ C(Rn ),
this means for αp > n, in the linear case for 2α > n.
Unfortunately, there is still a case when Vα,p µ has a bad sense, namely, if p ≤ 2 − α
n and µ is the
0
Dirac measure at the origin, then fµ ∈ Gαp −1 ∈
/ L1 (Rn ) and the convolution with Gα cannot help.
1
1.1. Theorem. Let µ ∈ Wα,p (Rn )∗ be a measure and u be the Vα,p -potential of µ. Then
0
0
E(µ) = kGα ∗ µkpp = kµkpWα,p (Rn )∗ = kukpα,p = hµ, ui.
Proof. Let f = fµ . Denote the norm of f by a and the dual norm of µ by b, so that
a = kf kp = kukα,p ,
b = kµkWα,p (Rn )∗ .
Choose v = Gα ∗ g ∈ Wα,p (Rn ) with kgkp ≤ 1. Then
1
ap
1
1
u
ahµ, vi = hµ, avi − kagkpp + kagkpp ≤ hµ, ui − kf kpp +
= hµ, ui = ahµ, i.
p
p
p
p
a
Therefore the supremum in the definition of the dual norm b is attained and it is equal to
have
1
1
ap
(1.6)
hµ, ui = ab,
Eα,p µ = hµ, ui − kf kpp = ab − .
0
p
p
p
1
a hµ,
ui. We
Substitute g = f in (1.3), then we obtain
(1.7)
ab − ap = hµ, Gα ∗ f i −
Z
|f |p dx = 0.
Rn
0
Then we deduce that ab = ap , b = ap−1 and a = bp /p . From (1.6) and (1.7) it follows
ap
ab
ab
1
Eα,p (µ) = ab −
= ab −
= 0
0
p
p
p
p
Therefore
0
Eα,p (µ) = ab = ap = bp .
Finally, by (1.4),
0
|Gα ∗ µ|p = f p
and thus
0
kGα ∗ µkp = kf kpp = ap .
2. Capacity
2.1. Definition (Capacity). The (Bessel) capacity is a set function which associates with each E ⊂ Rn
the number
o
n
o
n
Capα,p (E) = inf kf kpp : f ∈ F(E) = inf kf kpp : f ∈ F(E) ,
where F(E) = {f ∈ Lp (Rn ) : Gα ∗ f ≥ 0}
and F(E) is the closure of F(E) in Lp (Rn ).
The capacity of E can be ∞, but if E is bounded, then Capα,p (E) < ∞.
2.2. Theorem. Let E ⊂ Rn . If Capα,p (E) < ∞, then there exists a unique minimizer fE of kf kpp in
0
F(E) and a unique Radon measure µE such that fE = (Gα ∗ µE )p −1 .
Proof. We obtain the existence of f = fE by direct methods of calculus of variations. It is obvious
that f ≥ 0, otherwise |f | would be another minimizer. Let v ∈ S + and g = G−α ∗ v. Then obviously
f + tg ∈ F(E) and thus kf + tgkpp ≥ kf kpp . By differentiation of k · kpp at f along the direction g we obtain
Z
f p−1 g dx ≥ 0.
Rn
Hence
Z
v 7→
f p−1 G−α ∗ v dx
Rn
is a positive linear functional on S + (Rn ) and therefore by the Riesz representation theorem, there exists
a Radon measure µ = µE such that
Z
hµ, vi =
f p−1 G−α ∗ v dx,
v ∈ S(Rn ),
Rn
2
which implies
Z
f p−1 g dx,
hµ, Gα ∗ gi =
(2.1)
g ∈ Lp (Rn ).
Rn
But by (1.3) this means that u is the Vα,p -potential of µ, so that
0
f = (Gα ∗ µ)p −1 .
2.3. Definition. Let E ⊂ Rn be a set of finite α, p-capacity. Then Gα ∗ fE is called the capacitary
potential for Capα,p (E) and µE is the capacitary distribution for Capα,p (E).
2.4. Corollary. Let E ⊂ Rn be a set of finite α, p-capacity, f = fE , u be the capacitary potential for
Capα,p (E) and µ be the capacitary distribution for Capα,p (E). Then u = Vα,p µ and
Z
p
p
Capα,p (E) = Eα,p (µ) = kukα,p = kf kp =
u dµ.
Rn
Proof. This follows from the results of Section 1.
2.5. Lemma. There exists a constant C such that
(ψδ ∗ Gα )(x) ≤ CGα (x/2),
0 < δ < 1.
Proof. If |x| ≥ 2δ, then
ψδ ∗ Gα (x) ≤
Gα (y) ≤ Gα (x/2).
sup
y∈B(x,δ)
If |x| < 2δ and α < n, then
Z
Z
C
C
Gα (y) dy ≤ n
|y|α−n dy ≤ Cδ α−n ≤ C|x|α−n ≤ CGα (x/2).
ψδ ∗ Gα (x) ≤ n
δ B(x,δ)
δ B(0,3δ)
The proof for α = n uses the estimate of Gn in terms of the logarithmic function, we skip the details. 2.6. Theorem (Properties of capacitary potentials and distributions). Let E ⊂ Rn be a set of finite
α, p-capacity, f = fE , µ = µE and u = Gα ∗ f . Then
(a)
(b)
(c)
(d)
u is lower semicontinuous.
u ≥ 1 on the interior of E.
µ is supported by E ∩ {u ≤ 1}.
u ≤ 1 holds µ-a.e.
Proof. (a) This is obvious, as the kernel Gα is lower semicontinuous.
(b) If g ∈ F(E), then Gα ∗ g ≥ 1 on E. The operator g 7→ Gα ∗ g is continuous from Lp to Lp , and
thus Gα ∗ f ≥ 1 a.e. on E. Let x be an interior point of E. For δ > 0 small enough we have Gα ∗ f ≥ 1
a.e. on B(x, δ) and thus (ψδ ∗ Gα ∗ f )(x) ≥ 1. For the passage to limit as δ → 0 we use the Lebesgue
theorem: Let h(x) = Gα (x/2). If Gα ∗ f (x) = ∞, there is nothing to prove. Otherwise (h ∗ f )(x) is also
finite and by Lemma 2.5,
0 ≤ (ψδ ∗ Gα )(x − y) f (y) ≤ Ch(x − y) f (y),
0 < δ < 1,
which yields a majorant for the passage to limit. It follows that u(x) = Gα ∗ f (x) ≥ 1.
(c) Let v ∈ S(Rn ) and g = G−α v. and suppose that the support of v is disjoint with E. Then
f + tg ∈ F(E). Differentiating kf kpp at f along the direction g yields
Z
hµ, vi =
f p−1 g dx = 0.
Rn
It follows that spt µ ⊂ E. Similarly, if the support of v is contained in {u > 1}, then by lower semicontinuity, the minimum of u on the support of v is attained and bigger than 1. It follows that for |t| small
enough, f + tg is in F(E) and we proceed as above.
(d) is an obvious consequence of (c).
3
2.7. Theorem. (a) If E1 , E2 ⊂ Rn , E1 ⊂ E2 , then Capα,pS
(E1 ) ≤ Capα,p (E2 ).
(b) Let Ej be a nondecreasing sequence of sets and E = j Ej . Then
Capα,p (E) = lim Capα,p (Ej )
j
S
P
n
(c) If E1 , E2 , · · · ⊂ R , then Capα,p ( j Ej ) ≤ j Capα,p (Ej ).
Proof. (a) is obvious.
(b): It is enough to consider the ≤ inequality in case that the limit on the right is finite. We find
test functions gj ∈ F(Ej ) such that kgj kpp − Capα,p (Ej ) → 0. Now, the sequence gj is bounded in Lp ,
so passing to a subsequence we may assume that it converges weakly in Lp . Then some their convex
combinations hj converge strongly in Lp and still we may choose them so
Pthat Gα ∗ hj ≥ 1 on Ej and
khj kpp − Capα,p (Ej ) → 0. Passing to a subsequence, we may assume that j khj − hj+1 kp < ∞. Choose
ε > 0 and find k ∈ N such that
khk kp − Capα,p (Ek )1/p < ε
and
X
khj − hj+1 kp < ε.
j≥k
Set
g = hk +
X
|hj − hj+1 |.
j≥k
Then
kgkp ≤ Capα,p (Ek )1/p + 2ε
and
g ≥ sup hj ,
j≥k
so that
Gα ∗ g ≥ 1 on E.
It follows that
Capα,p (E) ≤ kgkpp ≤ (Capα,p (Ek )1/p + 2ε)p ≤ (lim Capα,p (Ej )1/p + 2ε)p
j
and letting ε → 0 we conclude the proof.
(c) In view of (b), it is enough to prove just the finite subadditivity. If f1 , f2 : Rn → [0, ∞] are
measurable and Gα ∗ fi ≥ 1 on Ei , i = 1, 2, then Gα ∗ (max{f1 , f1 } ≥ 1 on E1 ∪ E2 , so that
Z
Z
Z
Z
Capα,p (E1 ∪ E1 ) ≤
max{f1p , f2p } dx ≤
(f1p + f2p ) dx =
f1p dx +
f2p dx,
Rn
Rn
Rn
Rn
and passing to the infimum over f1 and f2 we obtain
Capα,p (E1 ∪ E1 ) ≤ Capα,p (E1 ) + Capα,p (E2 )
n
2.8. Theorem. Let E ⊂ R be a set of finite α, p-capacity. Then
n
o
n
o
Capα,p (E) = inf Capα,p (G) : G ⊃ E, G open = inf Eα,p (ν) : ν a measure, Vα,p ν ≥ 1 on E .
Proof. Choose ε > 0. Find a test function g ∈ F(E) such that kgkpp < Capα,p (E) + ε. Then there is
t > 1 such that tp kgkpp < Capα,p (E) + ε. Since the Bessel potentials are lower semicontinuous, there
exists an open set G ⊃ E such that Gα ∗ tg > 1 on G. This proves the first equality. Now, if ν is a
capacitary distribution for Capα,p (G), then Vα,p ν ≥ 1 on E and Capα,p (E) ≤ Eα,p (ν) = Capα,p (G). 2.9. Theorem. The capacity Capα,p satisfies the Choquet axioms, thus,
Capα,p (A) = sup{Capα,p (K) : K ⊂ A}
whenever A is an analytic set.
Proof. It follows from Theorems 2.7 and 2.8.
2.10. Theorem (Dual definition of capacity). Let E ⊂ Rn be a Capα,p -capacitable set. Then
Capα,p (E) = sup{ν(Rn ) : spt ν ⊂ E, Vα,p ν ≤ 1 ν-a.e.}.
4
Proof. If K ⊂ E is compact, then spt µK ⊂ K ⊂ E and Vα,p µK ≤ 1 µK -a.e. by Theorem 2.6 (d). Also,
Z
Z
(2.2)
Capα,p (K) =
Vα,p µK dµK ≤
1 dµK = µK (Rn ).
Rn
Rn
If E is capacitable,
Capα,p (E) = sup Capα,p (K) ≤ sup µK (Rn ).
K⊂E
K⊂E
This proves the ≤ inequality. Conversely, consider a Radon measure ν such that spt ν ⊂ E and Vα,p ν ≤
1 ν-a.e. Let g ∈ F(E). Then
Z
ν(Rn ) ≤ ν(E) ≤
Gα ∗ g dν ≤ kgkp kGα ∗ νkp0 .
Rn
Now, by Theorem 1.1,
kGα ∗
0
νkpp0
Z
Z
1 dν = ν(Rn ).
Vα,p ν dν ≤
=
Rn
Rn
We have obtained
0
ν(Rn ) ≤ kgkp kGα ∗ νkp0 ≤ kgkp ν(Rn )1/p ,
so that
ν(Rn ) ≤ kgkpp .
Infimum over g ∈ F(E) gives the ≥ inequality.
2.11. Corollary. If K ⊂ Rn is compact, then Capα,p (K) = µK (K) = µK (Rn ).
Proof. The ≥ inequality follows from the preceding theorem, the ≤ inequality is (2.2).
3. Metric properties of capacity
In this section we suppose 0 < α < n, p > 1.
3.1. Convention. We use the symbols C, c for a generic strictly positive constants which can change at
each occurence, so that, for example, the inequality
2C ≤ C
is legal.
3.2. Theorem. There exist a, b > 0 depending on α, n, p such that for each ball B(r) of radius r ∈ (0, 1)
we have:
(a) if αp < n, then arn−αp ≤ Capα,p (B(r)) ≤ brn−αp ,
(b) if αp = n, then a(log(1/r))1−p ≤ Capα,p (B(r)) ≤ b(log(1/r))1−p ,
(c) if αp > n, then a ≤ Capα,p (B(r)) ≤ b.
Proof. We first estimate Gα ∗ χB(r) Let |x| < 2r. Then
Z
Z
(Gα ∗ χB(r) )(x) =
Gα (x − y) dy ≤ C
B(r)
|x − y|α−n dy ≤ Crα .
B(r)
If |x| ≥ 2r, then
Z
Gα (x − y) dy ≤ Crn Gα (x/2) ≤ Crn |x|α−n e−|x|/6
(Gα ∗ χB(r) )(x) =
B(r)
It follows that
Z
p0
(Gα ∗ χB(r) ) dx ≤ Cr
n+p0 α
+ Cr
np0
Z
Rn
Now, if αp < n, then
Z
∞
0
0
tn−1+αp −np e−t/6 dt.
2r
∞
0
0
tn−1+αp −np e−t/6 dt ≤
2r
Z
∞
0
0
0
2r
If αp = n, then
Z
∞
t
n−1+αp0 −np0 −t/6
e
Z
dt ≤
2r
If αp > n, then
Z
∞
2r
2r<t<1
0
0
tn−1+αp −np e−t/6 dt ≤
0
tn−1+αp −np dt = Crn+αp −np .
Z
0
5
∞
1
dt
+ 6 ≤ C log .
t
r
0
0
tn−1+αp −np e−t/6 dt = C
Let g : Rn → [0, ∞] be a measurable function such that Gα ∗ g ≥ 1 on B(r). Then
Z
Z
rn ≤ C
Gα ∗ g dx = C
(Gα ∗ χB(r) )g dx ≤ kGα ∗ χB(r) kp0 kgkp .
Rn
B(r)
Taking infimum over g we obtain
Capα,p (B(r)) ≥ crpn kGα ∗ χB(r) k−p
p0
Now, let us distingush the cases again. If pα < n, then
0
pn (1−p)(n+p α)
rpn kGα ∗ χB(r) kp−p
r
= crn−pα .
0 ≥ cr
If pα = n, then
0
0
pn n+p α
rpn kGα ∗ χB(r) k−p
+ rnp log 1r )1−p ≥ c(log 1r )1−p .
p0 ≥ cr (r
If pα > n, then n + pα0 > np0 and
0
0
0
pn n+p α
rpn kGα ∗ χB(r) k−p
+ rnp )1−p ≥ crnp+np (1−p) = c.
p0 ≥ cr (r
For the converse inequality, notice that
Capα,p (B(r)) ≤ Capα,p (B(1))
if αp > n. For the remaining cases we construct suitable competitors. If pα < n, set f = r−α χB(r) .
Then for |x| < r we have
Z
−α
Gα ∗ f (x) ≥ cr
|x − y|α−n dy ≥ c.
B(r)
In this case
Z
Capα,p (B(r)) ≤ C
f p dx ≤ Crn−αp .
Rn
If pα = n, set f (y) = (log 1r )−1 |y|−α χB(1)\B(r) (y). If |x| < r < |y| < 1, then |x − y|α−n ≥ |2y|α−n and
thus
Z
Z
|x − y|α−n |y|−α dy ≥ c(log 1r )−1
|y|−n dy ≥ c.
Gα ∗ f (x) ≥ c(log 1r )−1
B(1)\B(r)
B(1)\B(r)
It follows that (for αp = n)
Z
Capα,p (B(r)) ≤ C
Rn
f p dx ≤ C(log 1r )−p
Z
B(1)\B(r)
|x|−n dx = C(log 1r )1−p .
3.3. Theorem. Suppose that αp < n. There exists b > 0 depending on α, n, p such that for each E ⊂ Rn
we have
Capα,p (E) ≤ bH1n−αp (E)
(3.1)
Proof. Let
E⊂
[
B(xj , rj )
j
with rj < 1. Then by Theorem 2.7(c),
Capα,p (E) ≤
X
Capα,p (B(xj , rj )) ≤ C
j
X
rjn−αp
j
and passing to infimum over such coverings we obtain (3.2).
3.4. Theorem. Suppose that αp < n and 1 < q < p. There exists b > 0 depending on α, n, p, q such that
for each E ⊂ Rn we have
(3.2)
n−αq
H∞
(E) ≤ b Capα,p (E).
6
Proof. We may assume that Capα,p (E) < ∞. Consider a function g : Rn → [0, ∞] such that g ∈ Lp (Rn )
and
Gα ∗ g ≥ 1 on E.
Choose x ∈ E. Denote h(t) = tα−n e−t/3 .
Z
Z
1≤
g(y)Gα (x − y) dy ≤ C
Rn
Z
g(y)h(|x − y|) dy ≤ C
Rn
Z
=C
0
∞ Z
Z
Rn
∞
|h0 (t)|g(y) dt dy
|x−y|
g(y) dy |h0 (t)| dt.
B(x,t)
On the other hand, |h0 (t)| ≤ Ctα−n−1 for t < 1 and thus
Z ∞
q
|h0 (t)|tn−α p dr ≤ C.
0
It follows that there exists r = rx > 0 such that
0
|h (r)|r
q
n−α p
Z
0
≤ C|h (r)|
g(y) dy,
B(x,r)
so that
r
q
−α p
Z
≤C−
Z
g(y) dy ≤ C −
B(x,r)
g p (y) dy
1/p
,
B(x,r)
which yields
rn−αq ≤ C
Z
g p (y) dy.
B(x,r)
The integrability of g p gives a uniform bound for rx . Now, using the Vitali theorem, we find a finite or
infinite sequence (B(xj , rj ))j of disjointed balls such that
[
E⊂
B(xj , 5rj )
j
and
rjn−αq ≤ C
Z
g p (y) dy.
B(xj ,rj )
It follows that
n−αq
H∞
(E)
≤C
XZ
j
p
Z
g (y) dy ≤ C
B(xj ,rj )
g p (y) dy.
Rn
Taking infimum over g we obtain the desired inequality.
4. Frostman measure
4.1. Definition (Fractional maximal function). Let µ be a Radon measure on Rn and α > 0. The
fractional maximal function of µ is defined as
Mα µ(x) = sup rα
r>0
µ(B(x, r))
µ(B(x, r))
= sup rα−n
|B(x, r)|
αn
r>0
If f is a measurable function, we define the fractional maximal function of f by through identifying f
with the measure with density |f |.
4.2. Definition. A sequence (B(xj , rj ))j is called a Whitney covering of an open set G if B(xj , 2rj ) ⊂ G,
B(xj , 2rj ) 6⊂ G, the balls B(xj , rj ) cover G and the balls B(xj , rj /5) are disjointed.
Sk
4.3. Remark. Any Whitney covering must be an infinite sequence, as j=1 B(xj , rj ) is a compact subset
of G from any k ∈ N.
4.4. Theorem. Let G be an open set of finite measure. Then there exists a Whitney covering of G such
that r1 ≥ r2 ≥ r3 ≥ . . . .
Proof. The existence of a Whitney covering is a standard consequence of the Vitali covering argument. If
we choose δ > 0, the set of all j with rj > δ is finite as the balls B(xj , rj /5) are disjointed and contained
in G. Therefore we can reorder rj in descending way.
7
4.5. Theorem (Frostman). There exists a > 0 depending on n and α with the following property: If
G ⊂ Rn is an open set of finite measure, then there exists an absolutely continuous Radon measure µ
such that Mα µ ≤ 1 everywhere and Mα µ ≥ a on G.
Proof. Let (B(xj , rj ))j be a Whitney covering such that r1 ≥ r2 ≥ . . . . We construct a sequence (aj )j
of real numbers and a sequence fj of functions recursively. We write ρj = rj /5. We start with a1 = 1
and f1 = ρ−α
1 χB(x1 ,ρ1 ) . If j > 1, we set
fj = fj−1 + aj ρ−α
j χB(x1 ,ρj ) ,
where aj is the greatest constant preserving the inequality Mα fj ≤ 1. Obviously aj ≤ 1, this is obtained
testing at xj by r = ρj . Set f = lim fj . Then the inequality Mα f ≤ 1 is obvious and it remains to prove
the lower estimate. Choose z ∈ G and find j such that z ∈ B(xj , rj ). By the maximality of aj , there
exists x ∈ G and r > 0 such that B(xj , ρj ) ∩ B(x, r) 6= ∅ and
Z
(2r)−α ≤ −
fj .
B(x,r)
ρ−α
j
Since fj ≤
in G, it follows that 2r ≥ ρj and 14r ≥ rj + ρj + 2r ≥ |x − z| + r. We have B(z, 14r) ⊃
B(x, r) and thus
Z
Z
α
α−n α
Mα f (z) ≥ (14r) −
fj ≥ 14
r −
fj ≥ 7α 14−n .
B(z,14r)
B(x,r)
4.6. Definition. If we agree on a fixed constant a (e.g. as it the proof of Theorem 4.5), we call a Frostman
measure for G any measure satisfying the properties obtained in Theorem 4.5.
5. Capacity and Hausdorff content: p = 1
It can be shown that the Bessel kernel can be replaced by the Riesz kernel if we are interested in
(α, p)-capacity and αp < n. The Riesz and Bessel capacities are the comparable on sets with unifomly
bounded diameter. The Riesz capacity can be also studied when p = 1 and α < n. However, if p = 1
and α ∈ N, then the Wα,1 space is not equivalent to the usual Sobolev space W α,1 . To obtain better
correspondence of capacities, we consider another capacities related to the fractional maximal functions.
The fractional maximal functions play both roles of Riesz and Vα,p -potentials.
5.1. Definition. Let 0 < α < n. We define the (α, 1)-capacity of E ⊂ Rn as
capα,1 (E) = inf{kf k1 : Mα f ≥ 1 on E}.
If µ is a Radon measure on Rn , we define the (α, 1)-energy of µ as
Z
Eα,1 (µ) = Mα µ dµ.
Frostman measures have some properties similar to those of capacitary distributions.
5.2. Theorem. There exist a, b > 0 depending on n and α with the following property: If G ⊂ Rn is an
open set of finite measure and µ is a Frostman measure for G, then
a2 capα,1 (G) ≤ a µ(Rn ) ≤ Eα,1 (µ) ≤ b capα,1 (G).
Proof. Let f be the density of µ. There is a > 0 such that Mα f ≥ a on G, so that
Z
a capα,1 (G) ≤
f dx = µ(Rn )
Rn
Further,
Z
Eα,1 (µ) =
Mα f dµ.
G
Since a ≤ Mα f ≤ 1 on G, we obtain
aµ(Rn ) ≤ Eα,1 (µ) ≤ µ(Rn ).
Finally, if g is an L1 -function such that Mα g ≥ 1 on G, then for each x ∈ G we find r > 0 such that
Z
(2r)−α ≤ −
g dy.
B(x,r)
8
Using a Vitali type argument, we find a (finite or infinite) sequence B(xj , rj ) of pairwise disjoint balls
such that
Z
−α
(2rj ) ≤ −
g dy.
B(xj ,rj )
and
G⊂
[
B(xj , 5rj ).
j
Then
n
µ(R ) ≤
X
µ(B(xj , 5rj )) ≤ C
X
j
rjn−α
≤C
XZ
j
Z
g≤C
B(xj ,rj )
j
g.
Rn
Passing to infimum over g we obtain
µ(Rn ) ≤ C capα,1 (G).
n
5.3. Theorem. There exists b > 0 depending on n and α with the following property: If G ⊂ R is an
open set and ν is a measure in G such that Mα ν ≤ 1 holds ν-a.e., then
ν(Rn ) ≤ b capα,1 (G).
Proof. Let A = {x ∈ G : Mα ν(x) ≤ 1}. Consider an integrable function g ≥ 0 on Rn such that Mα g ≥ 1
on G (if it does not exist, the inequality is trivial). For each x ∈ A we find a radius r = rx > 0 such that
Z
rn−α ≤ 2
g.
B(x,r)
The integrability of g gives a uniform bound for rx . By a VitaliStype argument we find a finite or infinite
sequence (B(xj , rj ))j of pairwise disjoint balls such that A ⊂ j B(xj , 5rj ) and
Z
n−α
ν(B(xj , 5rj )) ≤ Crj
≤C
g.
B(xj ,rj )
Then summing up we obtain
Z
n
ν(R ) = ν(A) ≤ C
g.
Rn
Passing to infimum over g we obtain the desired inequality.
5.4. Theorem. There exist a, b > 0 depending on n and α with the following property: If G ⊂ Rn is an
open set, then
n−α
a capα,1 (G) ≤ H∞
(G) ≤ b capα,1 (G).
n−α
Proof.
Assume (as we can) that
(G) < ∞ and consider a covering (B(xj , rj ))j of G such that
P n−α
P H∞
< ∞. Then |G| < αn j rn < ∞. Let µ be a Frostman measure for G. Then
jr
X
X
µ(G) ≤
µ(B(xj , rj )) ≤ C
rjn−α ,
j
j
therefore
n−α
capα,1 (G) ≤ CH∞
(G).
n
Let capα,1 (G) < ∞ and g : R → [0, ∞] is an integrable function satisfying Mα g ≥ 1 on G, Then for
each x ∈ G we find a radius r = rx > 0 such that
Z
n−α
r
≤2
g.
B(x,r)
The integrability of g gives a uniform bound for rx . By a VitaliStype argument we find a finite or infinite
sequence (B(xj , rj ))j of pairwise disjoint balls such that G ⊂ j B(xj , 5rj ) and
Z
rjn−α ≤ C
g.
B(xj ,rj )
Then summing up we obtain
n−p
H∞
(G) ≤ C
X
rjn−α ≤ C
j
Passing to infimum over g we obtain the desired inequality.
9
Z
g.
Rn