Basic Properties of Sobolev Spaces
1.3 Some Inequalities for Functions of One Variable
1.3.1 Basic Facts on Hardy-type Inequalities
Most of this section is concerned with variants and extensions of the following Hardy inequality (cf. Hardy, Littlewood, and P´olya [351], Sect. 9.9).
Iff(x)≥0, then ∞
0
x−rF(x)pdx≤ p
|r−1| p ∞
0
x−r xf(x)p
dx, (1.3.1)
where p >1,r= 1 and F(x) =
x 0
f(t) dt forr >1, F(x) =
∞
x
f(t) dt forr <1.
In this section we make some remarks concerning (1.3.1) and related in- equalities.
(i) First of all, (1.3.1) fails ifr= 1 for each of the above definitions ofF.
One can see it by choosing eitherf or 1−f as the characteristic function of the interval [0,1].
(ii) The constant factor in front of the integral on the right-hand side of (1.3.1) is sharp. Let, for example,r >1, and let
∞
0
x−r|F|pdx≤C ∞
0
x−r+p|f|pdx (1.3.2) hold. We take
fε(x) =
0 forx >1, xr−pp−1−ε forx >1, where εis a positive sufficiently small number. Then
Fε(x) = x
0
fε(t) dt=
0 forx <1,
r−11
p −ε(xr−1p −ε−1) forx >1.
We have ∞
0
x−r|Fε|pdx=
p r−1−pε
p ∞
1
1−x1−rp +εp dx x1+pε. Since (1−x1−rp +ε)p= 1 +O(x1−rp +ε) forx >1, the last integral is equal to
∞
1
dx
x1+pε +O(1) = (pε)−1+O(1).
The right-hand side of (1.3.2) is ∞
0
xr+p|fε|pdx= ∞
1
x−1−pεdx= (pε)−1. Hence (1.3.2) becomes
pp (r−1−pε)p
1
pε+O(1)
≤C(pε)−1, which implies
C≥ pp (r−1)p. The caser <1 is quite similar.
(iii) A multidimensional variant of Hardy’s inequality is
Rm
u(y)p|y|−sdy≤ p
|s−m| p
Rm
∇u(y)p|y|p−sdy, (1.3.3) where u∈C0∞(Rm) ands=m. In the cases > mwe require u(0) = 0. This inequality follows from the one-dimensional inequality (1.3.3) written as
∞
0
|v(r)|p rs−m+1dr≤
p
|s−m| p ∞
0
v(r)prm+p−s−1dr,
if one passes to spherical coordinates r = |x|, ω = x/r and remarks that
|∂u∂r| ≤ |∇u|. One should use, of course, that
Rm· · ·dx=
Sm−1
dsω
∞
0
· · ·rm−1dr.
The constant in (1.3.3) is sharp, which can be shown by radial functions.
(iv) A more general Hardy’s inequality
Rm+n
u(z)p|y|−sdz≤ p
|s−m| p
Rm+n
∇u(z)p|y|p−sdz,
where u∈C0∞(Rm+n),z = (x, y),x∈Rn, y∈Rm, s=m, and additionally u(x,0) = 0 for allx∈Rn ifs > m, is obtained by integration of (1.3.3), with u(z) instead of u(y), overRn.
(v) Finally, a few words about the critical values=mexcluded in (1.3.3).
The one-dimensional inequality holds ∞
1
|u(t)|p t(logt)pdt≤
p p−1
p ∞
1
|u(t)|ptp−1dt, (1.3.4) where u(1) = 0 andp >1. In fact, by introducing the new variablex= logt, we rewrite (1.3.4) as the Hardy inequality
∞
0
|u(x)˜ |p xp dx≤
p p−1
p ∞
0
|u˜(x)|pdx with ˜u(x) =u(ex).
As a consequence of (1.3.4) we obtain
Rn+m
|u(z)|pdz
|y|m(log|y|)p ≤ p
p−1 p
Rn+m
∇u(z)p|y|p−mdz
for allu∈C0∞(Rn+m) such thatu(z) = 0 for|y| ≤1.
1.3.2 Two-weight Extensions of Hardy’s Type Inequality in the Case p≤q
Theorem 1. Let μ and ν be nonnegative Borel measures on (0,∞) and let ν∗ be the absolutely continuous part ofν. The inequality
∞
0
x 0
f(t) dt qdμ(x)
1/q
≤C ∞
0
f(x)pdν(x) 1/p
, (1.3.5)
holds for all Borel functionsf and1≤p≤q≤ ∞if and only if B := sup
r>0
μ
[r,∞)1/q r 0
dν∗ dx
−1/(p−1)
dx (p−1)/p
<∞. (1.3.6)
Moreover, if C is the best constant in (1.3.5), then B≤C≤B
q q−1
(p−1)/p
q1/q. (1.3.7)
If p= 1or q=∞, thenB=C.
In the caseq=∞ the condition (1.3.6) means that B= sup
r >0 :μ([r,∞))>0
<∞, and dνdx∗ >0for almost allx∈[0, B].
We begin with the proof of the following less general theorem on absolutely continuous measuresμand ν.
Theorem 2.Let 1≤p≤q≤ ∞. In order that there exists a constantC, independent of f, such that
∞
0
w(x) x
0
f(t) dt qdx
1/q
≤C ∞
0
v(x)f(x)pdx 1/p
, (1.3.8) it is necessary and sufficient that
B:= sup
r>0
∞
r
w(x)qdx
1/q r 0
v(x)−pdx 1/p
<∞, (1.3.9) where p=p/(p−1). Moreover, if C is the best constant in (1.3.8)and B is defined by (1.3.9), then (1.3.7)holds. If p= 1 orp=∞, then B=C.
Proof. The case 1< p≤q <∞.Necessity. If f ≥0 and supp f ⊂[0, r], then from (1.3.8) it follows that
∞
r
w(x)qdx
1/q r 0
f(t) dt≤C r
0
v(x)f(x)pdx 1/p
.
Let r
0
v(x)−p/(p−1)dx <∞.
We setf(x) =|v(x)|−p forx < r andf(x) = 0 forx > r. Then ∞
r
w(x)qdx
1/q r 0
v(x)−pdx 1/p
≤C. (1.3.10)
If r
0
v(x)−pdx=∞,
then we arrive at the same result, replacing v(x) by v(x) +εsgnv(x) with ε >0 in (1.3.8) and passing to the limit asε→0.
Sufficiency.We put h(x) =
x 0
v(t)−pdt 1/qp
. By H¨older’s inequality,
∞
0
w(x) x
0
f(t) dt qdx
p/q
≤ ∞
0
w(x)q x 0
f(t)h(t)v(t)pdt q/p
× x
0
h(t)v(t)−pdt q/p
dx p/q
. (1.3.11)
Now we prove that ∞
0
ϕ(x) x
0
f(y) dy r
dx 1/r
≤ ∞
0
f(y) ∞
y
ϕ(x) dx 1/r
dy, (1.3.12) provided that ϕ≥0,f ≥0 and r≥1. In fact, the left-hand side in (1.3.12) is equal to
∞
0
∞
0
ϕ(x)1/rf(y)χ[y,∞)(x) dy r
dx 1/r
,
whereχ[y,∞)is the characteristic function of the halfaxis [y,∞). By Minkows- ki’s inequality the last expression does not exceed
∞
0
∞
0
ϕ(x)1/rf(y)χ[y,∞)(x)r
dx 1/r
dy
= ∞
0
f(y) ∞
y
ϕ(x) dx 1/r
dy.
By (1.3.12), the right-hand side in (1.3.11) is majorized by ∞
0
f(t)h(t)v(t)p ∞
t
w(x)q x 0
h(y)v(y)−pdy q/p
dx p/q
dt.
(1.3.13) Using here the expression forh, we rewrite the integral inxas
∞
t
w(x)q x 0
v(y)−p y 0
v(z)−pdz −1/q
dy q/p
dx. (1.3.14) Since
x 0
v(y)−p y 0
v(z)−pdz −1/q
dy=q x
0
v(x)−pdy −1/q
,
(1.3.14) is equal to (q)q/p
∞
t
w(x)q x 0
v(y)−pdy
q/(pq)
dx.
By the definition ofB this expression is majorized by Bq/q(q)q/p
∞
t
w(x)q ∞
x
w(y)qdy −1/q
dx
=Bq−1(q)q/pq ∞
t
w(x)qdx 1/q
≤Bq(q)q/pq t
0
v(x)−pdx −1/p
=Bq(q)q/pqh(t)−q. (1.3.15) Therefore, (1.3.11) has the following majorant
∞
0
f(t)v(t)h(t)p
Bq(q)q/pqh(t)−qp/q
dt
=Bq(q)p/pqp/q ∞
0
v(t)f(t)pdt.
Hence, (1.3.8) holds with the constantB(q)(p−1)/pq1/q. Now we consider the limit cases.
Ifp=∞thenq=∞and (1.3.8) follows from the obvious estimate ess sup
0<x<∞
w(x) x
0
f(t) dt
≤ess sup
0<x<∞
w(x) x 0
dt
|v(t)|ess sup
0<t<x
v(t)f(t). Ifp= 1, q <∞, then from (1.3.12) it follows that
∞
0
w(x) x
0
f(t) dt qdx
1/q
≤ ∞
0
f(t) ∞
t
w(x)qdx 1/q
1
|v(t)|v(t)dt
≤B ∞
0
v(t)f(t)dt.
Letq=∞,p= 1. Then
ess sup
0<x<∞
w(x) x
0
f(t) dt
≤ess sup
0<x<∞
w(x)ess sup
0<t<x
1
|w(t)| x
0
v(t)f(t)dt
≤B ∞
0
v(t)f(t)dt.
Ifp >1, then ess sup
0<x<∞
w(x) x
0
f(t) dt
≤ess sup
0<x<∞
w(x) x 0
v(t)−pdt
1/p x 0
v(t)f(t)pdt 1/p
≤B x
0
v(t)f(t)pdt 1/p
.
This completes the proof of Theorem 2.
Proof of Theorem 1.Setting f = 0 on the support of the singular part of the measureν, we obtain that (1.3.5) is the equivalent to
∞
0
x 0
f(t) dt qdμ(x)
1/q
≤C x
0
f(x)pdν∗ dx dx
1/p
.
The estimateB ≤ C can be derived in the same way as in the proof of Theorem 2, if |v(x)|p is replaced by dν∗/dxand
∞
r
w(x)qdx byμ([r,∞)).
Now we establish the upper bound for C. We may assume f ≥ 0. Let {gn} be a sequence of decreasing absolutely continuous functions on [0,∞) satisfying
0≤gn(x)≤gn+1(x)≤μ [x,∞)
,
nlim→∞gn(x) =μ [x,∞)
, for almost allx. We have
∞
0
x 0
f(t) dt q
dμ(x) 1/q
= ∞
0
μ [x,∞)
d x
0
f(t) dt q1/q
.
By the monotone convergence theorem the right-hand side is equal to sup
n
∞
0
gn(x) d x
0
f(t) dt q1/q
= sup
n
∞
0
x 0
f(t) dt q
−gn(x) dx
1/q
. (1.3.16)
The definition of the constant B and the sequence{gn} imply ∞
r
−gn(x) dx
1/q r 0
dν∗ dx
−p/p
dx 1/p
≤B.
From this and Theorem 2 we conclude that the right-hand side in (1.3.16) is not greater than
B(q)(p−1)/pq1/q ∞
0
f(x)pdν∗ dx dx
1/p
,
which completes the proof.
Replacingxbyx−1 we derive the following assertion from Theorem 1.
Theorem 3.Let 1≤p≤q≤ ∞. In order that there exist a constantC, independent off and such that
∞
0
∞
x
f(t) dt qdμ(x)
1/q
≤C ∞
0
f(x)pdν(x) 1/p
, (1.3.17) it is necessary and sufficient that the value
B := sup
r>0
μ
(0, r)1/q ∞
r
dν∗ dx
−1/(p−1)
dx
(p−1)/p
,
be finite. The best constant in (1.3.17) satisfies the same inequalities as in Theorem 1.
Analogously, by the change of variable
(0,∞)x→y =x−x−1∈(−∞,+∞), from Theorem 1 we obtain the next assertion.
Theorem 4.Let 1≤p≤q≤ ∞. In order that there exist a constantC, independent of f and such that
+∞
−∞
∞
x
f(t) dt qdμ(x)
1/q
≤C +∞
−∞
f(x)pdν(x) 1/p
, (1.3.18) it is necessary and sufficient that
B:= sup
r∈(−∞,+∞)
μ
(−∞, r)1/q ∞
r
dν∗ dx
−1/(p−1)
dx (p−1)/p
<∞. ConstantsB andC are related in the same way as in Theorem 1.
1.3.3 Two-Weight Extensions of Hardy’s Inequality in the Case p > q
Lemma. Let 1 ≤q < p ≤ ∞ and let ω be a nonnegative Borel function on (0, b), where b ∈ (0,∞]. In order that there exist a constant C, independent of ψ and such that
b 0
ω(t) t
0
ψ(τ) dτ qdt
1/q
≤C b
0
ψ(t)pdt 1/p
, (1.3.19)
it is necessary and sufficient that B:=
b 0
b t
ω(τ) dτ
p/(p−q)
t(q−1)p/(p−q)dt
(p−q)/pq
<∞. (1.3.20) IfC is the best constant in (1.3.19), then
p−q p−1
(q−1)/q
q1/qB≤C≤ p
p−1
(q−1)/q
q1/qB forq >1 andB=C forq= 1.
Proof. Sufficiency. First consider the case q > 1. We may assume ψ(t) ≥ 0. Integrating by parts on the left-hand side of (1.3.19) and using H¨older’s inequality with exponentsp/(p−q),p/(q−1), andp, we obtain
b 0
ω(t) t
0
ψ(τ) dτ q
dt 1/q
=q1/q b
0
b t
ω(τ) dτ ψ(t) t
0
ψ(τ) dτ q−1
dt 1/q
≤q1/q b
0
ψ(t)pdt
1/p b 0
t 0
ψ(τ) dτ p
t−pdt
(q−1)/p
× b
0
t(q−1)p/(p−q) b
t
ω(τ) dτ
p/(p−q)
dt
(p−q)/p1/q
. (1.3.21) From (1.3.20) and Hardy’s inequality (1.3.2), it follows that (1.3.21) is ma- jorized by
B p
p−1
(q−1)/q
q1/q b
0
ψ(t)pdt 1/p
.
Necessity.Consider, for example, the case b=∞. The proof is similar for b <∞. If (1.3.19) holds for the weightω with the constantC, then it holds
for the weightωn=ωχ[0,N], whereχ[0,N] is the characteristic function of the segment [0, N], with the same constant. We put
fN(x) = ∞
x
ωN(t) dt
1/(p−q)
x(q−1)/(p−q), BN =
∞
0
∞
t
ωN(τ) dτ
p/(p−q)
t(q−1)p/(p−q)dt
(p−q)/p
. From (1.3.19) we have
CBNq/(p−q)=C ∞
0
fN(x)pdx 1/p
≥ ∞
0
ωN(t) t
0
fN(τ) dτ q
dt 1/q
. (1.3.22) Integrating by parts, we find that the right-hand side in (1.3.22) is equal to
q
∞
0
fN(t) ∞
t
ωN(τ) dτ t
0
fN(τ) dτ q−1
dt 1/q
. (1.3.23)
Since
t 0
fN(τ) dτ q−1
= t
0
x(q−1)/(p−q) ∞
x
ωN(τ) dτ
1/(p−q)
dx q−1
≥ ∞
t
ωN(τ) dτ
(q−1)/(p−q)
t(p−1)(q−1)/(p−q) p−1
p−q 1−q
, we see that (1.3.23) is not less than
q1/q p−1
p−q
(1−q)/q ∞
0
∞
t
ωN(τ) dτ
p/(p−q)
t(q−1)p/(p−q)dt 1/q
=q1/q p−1
p−q
(1−q)/q
BNp/(p−q). Therefore,
BN ≤q−1/q p−q
p−1
(1−q)/q
C, and the same estimate is valid forB.
In the caseq= 1 the condition (1.3.6) becomes especially simple:
B = b
0
b t
ω(τ) dτ p
dt 1/p
<∞.
To prove that in this caseC≤Bwe integrate by parts on the left-hand side of (1.3.19) and apply H¨older’s inequality with exponentspandp(cf. (1.3.21)).
Then the right-hand side of (1.3.19) has the upper bound b
0
∞
t
ω(τ) dτ p
dt
1/p b 0
ψ(t)pdt 1/p
. Thus, we proved thatC≤B.
To derive the inequalityB≤C we substitute fN(x) =
∞
x
ωN(t) dt
1/(p−1)
,
into (1.3.23). This yieldsBN ≤C and henceB≤C. The lemma is proved.
Theorem 1.Let 1≤q < p≤ ∞. Then(1.3.8)holds if and only if ∞
0
x 0
dy
|v(y)|p
q−1 ∞
x
w(y)qdy
p/(p−q)
dx
|v(x)|p
(p−q)/pq
<∞. (1.3.24) If C is the best constant in (1.3.8)and B stands for the left-hand side of (1.3.24), then
p−q p−1
(q−1)/q
q1/qB≤C≤ p
p−1
(q−1)/q
q1/qB for1< q < p≤ ∞ andB=C forq= 1,1< p≤ ∞.
Proof.We may assume thatf ≥0, since the right-hand side in (1.3.8) does not change and the left-hand side increases iff is replaced by|f|. We may as well assumef(x) = 0 for sufficiently large values ofx. Let us put
t(x) = x
0
v(y)−pdy.
Then (1.3.8) becomes b
0
w(t)˜ qv(t)˜ pϕ(t)qdt 1/q
≤C b
0
ϕ(t)pdt 1/p
, where ˜w(t(x)) =w(x), ˜v(t(x)) =v(x),
ϕ t(x)
= x
0
f(y) dy, b= ∞
0
v(y)−pdy.
Now, in the case 1≤q < p <∞the result follows from the Lemma.
Letp=∞. Then B =
∞
0
x 0
dy
|v(y)|
q−1 ∞
x
w(y)qdy dx
|v(x)| 1/q
=q−1/q ∞
0
w(x)q x 0
dy
|v(y)| q
dx 1/q
.
Hence ∞
0
w(x) x
0
f(t) dt qdx
1/q
≤Bq1/qess sup
0<x<∞|vf|.
To prove the necessity we note thatvdoes not vanish on the set of positive
measure and putf = 1/v. The theorem is proved.
The following more general assertion can be derived from Theorem 1 in the same way as Theorem1.3.2/1 was derived from Theorem 1.3.2/2.
Theorem 2. Let μ and ν be nonnegative Borel measures on (0,∞) and letν∗ be the absolutely continuous part ofν. Inequality (1.3.8) with1 ≤q <
p≤ ∞holds for all Borel functionsf if and only if B=
∞
0
μ
[x,∞) x
0
dν∗ dy
−p
dy
q−1p/(p−q) dν∗
dx −p
dx
(p−q)/pq
<∞.
The best constantC in (1.3.17)is related withB in the same manner as in Theorem1.
The change of variable (0,∞)x→y =x−x−1∈(−∞,+∞) leads to the following necessary and sufficient condition for the validity of (1.3.18):
+∞
−∞
μ
(−∞, x] ∞
x
dν∗ dy
−p
dy
q−1p/(p−q) dν∗
dx −p
dx <∞, where 1≤q < p≤ ∞.