n. we also arxiv:2109.08506v1 [math.ap] 17 sep 2021

arX

iv:2

109.

0850

6v1

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17

Sep

2021

Adams inequalities with exact growth conditionfor Riesz-like potentials on Rn

Liuyu Qin

Abstract

We derive sharp Adams inequalities with exact growth condition for the

Riesz potential as well as more general Riesz-like potentials on Rn. We also

obtain Moser-Trudinger inequalities with exact growth condition for the frac-

tional Laplacian, and for general homogeneous elliptic differential operators

with constant coefficients.

1. Introduction and main results

The Moser-Trudinger inequality with exact growth condition on Rn takes the form

∫

Rn

exp⌈nα−2⌉

[β(α, n)|u|

nn−α

]

1 + |u|n

n−α

dx ≤ C||u||n/αn/α for all u ∈ W α,n

α (Rn), ||∇αu||n/α ≤ 1

(1.1)where ⌈x⌉ denotes the ceiling of x, i.e. the smallest integer greater than or equal x,and where expN is the regularized exponential, that is

expN(t) = et −N∑

k=0

tk

k!

and where for α ∈ (0, n) an integer the higher order gradient ∇α is defined as

∇αu =

(−∆)

α2 u if α is even

∇(−∆)α−12 u if α is odd.

Such inequality was proved first by Ibrahim, Masmoudi and Nakanishi [IMN] forn = 2 and α = 1, followed by Masmoudi and Sani who dealt with the cases n = 4,α = 2 in [MS1], any n and α = 1 in [MS2], and any n and any integer α in [MS3]. In[LT] Lu and Tang dealt with the case α = 2, for any n. In all these results the explicitsharp exponential constant β(α, n) (see (1.21)) that was found was the same as thesharp exponential constant in the classical Moser-Trudinger inequality on boundeddomains due to Adams [A1]:

∫

Ω

exp[β(α, n)|u|

nn−α

]dx ≤ C|Ω| for all u ∈ W

α,nα

0 (Ω), ‖∇αu‖n/α ≤ 1 (1.2)

where |Ω| denotes the Lebesgue measure of Ω. Recall that the exponential constantis sharp in the sense that it cannot be replaced by a larger constant.

1

http://arxiv.org/abs/2109.08506v1

The main new result behind the proof of (1.1) in [IMN], [MS1] (α = 1) is whatthe authors call “optimal descending growth condition” (ODGC). In essence, suchresult gives an optimally adjusted exponential growth of radial functions outside ballsof radius R, given the Ln norms of their gradients. In [MS2] and [LT] the sameresult is proven for radial functions under Ln/2 norm conditions on their Laplacians,and in [MS3] under Lorentz Ln/2,q norm conditions on their Laplacians. The keyinitial step that allowed the authors to only consider the radial Sobolev functions wasthe application of well-known, powerful symmetrization inequalities, specifically thePolya-Szego and Talenti’s inequalities.

There are a few other types of sharp Moser-Trudinger inequalities in the wholeRn. The most common one states that for all u ∈ W α,n

α (Rn) satisfying the under theRuf condition

‖u‖n/αn/α + ‖∇αu‖

n/αn/α ≤ 1 (1.3)

the following estimate holds

∫

Rn

exp⌈nα−2⌉

[β(α, n)|u|

nn−α

]dx ≤ C. (1.4)

This result was first derived by Ruf, [R] for α = 1 in dimension n = 2 and laterextended to all dimensions by Li-Ruf [LR]. The general case was settled by Fontana-Morpurgo in [FM2], where the authors prove (1.4) under (1.3) for arbitrary n andinteger α, but also for fractional powers of ∆, and for homogeneous elliptic operatorswith constant coefficients.

Under norm conditions weaker than (1.3) estimate (1.4) is in general false, but itbecomes true if one lowers the exponential constant. For example under the condition

max‖u‖n/α, ‖∇

αu‖n/α≤ 1 (1.5)

inequality (1.4) holds with exponential constant θβ(α, n), for any θ ∈ (0, 1). Thisresult was originally derived for α = 1 by Cao [C] and Panda [P] in dimension 2 andby Do O [D] in any dimension. Later Adachi-Tanaka [AT] re-proved the result andcast it in a dilation invariant form. In [FM2] the authors derived the general case asa corollary of (1.4) under (1.3), and showed that under either (1.5) or under

‖u‖rn/αn/α + ‖∇αu‖

rn/αn/α ≤ 1, r > 1 (1.6)

for any θ ∈ (0, 1)

∫

Rn

exp⌈nα−2⌉

[θβ(α, n)|u|

nn−α

]dx ≤ C(1− θ)−1+1/r, (1.7)

where r = ∞ under (1.5).It is important to point out that the Masmoudi-Sani result is the strongest one to

date, in the sense that it directly implies (1.4) under the Ruf condition (see [MS1],[MS2], [MS3]).

2

Our initial goal was to derive the sharp Adams inequality with exact growthcondition for the Riesz potential

Iαf(x) =

∫

Rn

|x− y|α−nf(y)dy,

that is

∫

Rn

exp⌈nα−2⌉

[1

|B1||Iαf |

nn−α

]

1 + |Iαf |n

n−α

dx ≤ C||Iαf ||n/αn/α, ||f ||n

α≤ 1, (1.8)

where |B1| is the volume of the unit ball of Rn and where the exponential constantis sharp. Note that the exponential constant |B1|

−1 is the same constant as in theoriginal inequality due to Adams [A1]:

∫

Ω

exp

[1

|B1||Iαf(x)|

nn−α

]dx ≤ C|Ω| for all f ∈ L

nα (Ω), ‖f‖n/α ≤ 1. (1.9)

Clearly (1.8) implies (1.1), in the same way that (1.9) implies (1.2) due to the factthat Iα is the inverse of (−∆)α/2 on smooth, compactly supported functions.

In this paper we prove that (1.8) is true, and not only for the Riesz kernel but fora subclass of the Riesz-like kernels introduced by Fontana-Morpurgo, which we callstrictly Riesz-like kernels.

To describe our result let us recall the definition given in [FM2]:

Definition 1. A measurable function K : Rn \ 0 → R is a Riesz-like kernel oforder α ∈ (0, n) if it satisfies the following properties:

K(x) = g(x∗)|x|α−n +O(|x|α−n+δ1) , x∗ =x

|x|, 0 < |x| ≤ B (A1)

|K(x)| ≤ H1|x|α−n (A2)

|K(z1)−K(z2)| ≤ H2|z1 − z2|max|z1|α−n−1, |z2|

α−n−1, z1, z2 6= 0 (A3)

where g : Sn−1 → R is a measurable function and δ,H1, H2, B are positive constants.

If we add an additional condition (A4) as below, we have more restrictive controlof the kernel K when |x| is large:

Definition 2. A measurable function K : Rn \ 0 → R is a strictly Riesz-like

kernel of order α ∈ (0, n) if it is Riesz-like and satisfies the following property:

|K(x)| ≤ |g(x∗)||x|α−n +O(|x|α−n−δ2) , |x| > B (A4)

where g : Sn−1 → R is a measurable function and B, δ1, δ2, H1, H2 are positiveconstants.

3

Here the “big O” notation in (A1) means that |O(|x|α−n+δ1)| ≤ C|x|α−n+δ1 for allx such that 0 < |x| ≤ B. And the same notation in (A4) means that |O(|x|α−n−δ2)| ≤C|x|α−n−δ2 . It is clear that (A3) implies that g is Lipschitz. Also, (A1),(A3) and (A4)imply (A2).

Clearly, any kernel of type g(x∗)|x|α−n with g Lipschitz on the sphere, providesan example of strictly Riesz-like kernel.

For m ∈ N, a kernel K is called m-regular if K ∈ Cm(Rn \ 0) and

|DhxK(x)| ≤ C|x|α−n−|h|, x 6= 0, |h| ≤ m

where h = (h1, ..., hn) is a multi-index with |h| = h1+ ...+hn and where DhxK denotes

the h-th derivative of K with respect to x. Clearly the Riesz kernel is m-regular forall m, and any 1-regular K satisfies condition (A3).

Let us denote T the convolution operator with kernel K:

Tf(x) = K ∗ f(x) =

∫

Rn

K(x− y)f(y)dy.

For vector valued functions K = (K1, ..., Km), ; f = (f1, ..., fm) we define Tf in thesame way with

Kf = K1f1 + ... +Kmfm, |f | = (f 21 + ...+ f 2

m)1/2.

The results and proofs in this paper apply to both scalar and vector cases, so wewill not distinguish between these two cases, except in the proof of sharpness.

The main result of this paper is the following:

Theorem 1. Let 0 < α < n, and K is strictly Riesz-like. There exists C =C(n, α,K) > 0 such that for all compactly supported f with

||f ||nα≤ 1

we have

∫

Rn

exp⌈nα−2⌉

[1

Ag|Tf |

nn−α

]

1 + |Tf |n

n−α

dx ≤ C||Tf ||n/αn/α, (1.10)

where

Ag =1

n

∫

Sn−1

|g(x∗)|n

n−αdx∗. (1.11)

If K is n-regular, then the exponential constant A−1g in (1.10) cannot be replaced by a

larger number. Furthermore, if K is n-regular, then (1.10) cannot hold if the powern

n−αin the denominator is replaced by any smaller power.

Here dx∗ is the surface measure of the unit sphere Sn−1, induced by the Lebesguemeasure.

4

As pointed out in [FM2] Adams type estimates involving an integral of the regular-ized exponential over the whole space, have equivalent “local” formulations in termsof the standard exponential. Via the exponential regularization lemma (Lemma A insection 3) estimate (1.10) is equivalent to the following local version

∫

E

exp

[1

Ag|Tf |

nn−α

]

1 + |Tf |n

n−α

dx ≤ C(|E|+ ||Tf ||

n/αn/α

)(1.12)

valid for all measurable E with finite measure, and under ||f ||nα≤ 1.

We mention that inequality (1.10) still holds if we have “≤” instead of “=” incondition (A1), that is, if (A1) is replaced by

|K(x)| ≤ |g(x∗)||x|α−n + |O(|x|α−n+δ1)|, 0 < |x| ≤ B. (A1’)

But in order to have sharpness in the exponential constant A−1g we need to assume

condition (A1).We point out that for Theorem 1 to hold it is not enough to assume that K be

only Riesz-like. It is relatively easy to find an example of a Riesz-like kernel such thatthe inequality in Theorem 1 cannot hold, but the one in [FM2, Theorem 5] holds. Insection 6 remark 2, we will address this example, which indicates that it is necessaryfor us to strengthen our assumption for large |x|, i.e. (A4), so that K has samebehavior near the origin and at infinity.

One of the main difficulties we had to overcome toward a proof of Theorem 1,even for the Riesz potential as in (1.8), was to find a suitable replacement of theoptimal growth condition result for the potential Tf , under norm conditions on f .Clearly, in this context no tools such as the Polya-Szego or Talenti’s inequalities areavailable, which makes an initial reduction to radial functions impossible, even in thecase of the Riesz potential. The way we bypass this problem is by carefully splittingthe function f and by making use of an improved O’Neil inequality. Very looselyspeaking, we will consider a suitable 1-parameter family of sets Fτ depending on f ,and with measure τ , and we will split f as f = fτ + f ′

τ , with fτ = fχFτ. By use of an

improved O’Neil inequality we will prove an estimate of type

(Tf)∗(t) ≤ Ufτ (t) + U ′f ′τ (τ), 0 < t ≤ τ (1.13)

where (Tf)∗ is the symmetric decreasing rearrangement of Tf , and where U, U ′ aretwo suitable, real-valued (nonlinear) functionals stemming from the O’Neil inequality(see estimate (3.19)). The first term in (1.13) is handled by an Adams inequality forsets of finite measure due to Fontana-Morpurgo (see Theorem A). The challengingpart is the proof of an optimal descending growth condition for the function U ′f ′

τ (τ)(see Proposition 1). In [IMN], [MS1], [MS2], [LT] a version ODGC was first proved forsequences, followed by a suitable discretization of radial Sobolev functions. We willalso make use of the discrete ODGC for sequences (See Lemma 5, Section 4), howeverthe discretization of U ′f ′

τ (τ) turns out to be rather involved (see Proposition 1 andits proof, given in Section 5).

5

As a consequence of Theorem 1 we derive the following general Adachi-Tanakatype inequality:

Corollary 1. If K is a strictly Riesz-like kernel, then for any θ ∈ (0, 1) there existsC independent of θ such that for all compactly supported f with

||f ||n/α ≤ 1, (1.14)

we have ∫

Rn

exp⌈nα−2⌉

[θ

Ag|Tf(x)|

nn−α

]dx ≤

C

1− θ||Tf ||

n/αn/α (1.15)

where Ag is the same as in (1.11). If K is n-regular and K /∈ Ln

n−α (|x| ≥ 1) then in-equality (1.15) is sharp, in the sense that the exponential integrals cannot be uniformlybounded if θ = 1.

Estimate (1.15) improves the one obtained in [FM2, Theorem 6], which does nothave ‖Tf‖n/α on the right hand side, and which has (1 − θ)−1 only in the case Khomogeneous.

At the level of Moser-Trudinger inequalities, Theorem 1 implies almost immedi-ately the Masmoudi-Sani result (1.1), for integer powers α. Similarly, as a consequenceof Theorem 1, we will obtain a Moser-Trudinger inequality with exact growth con-dition for the fractional Laplacian (−∆)α/2, for any α ∈ (0, n) , and also for generalhomogeneous elliptic operators.

To describe such result, recall that the Sobolev space W α,p(Rn) consists of alllocally summable functions u : Rn → R such that for each multiindex h with|h| ≤ α, the h-th weak partial derivative of u exists and belongs to Lp(Rn). For noninteger α, the space W α,n

α (Rn) will denote the Bessel potential space

W α,p(Rn) = u ∈ S ′ : (I −∆)α/2u ∈ Lp(Rn) = Gα ∗ f, f ∈ Lp(Rn), (1.16)

where Gα is the kernel of the Bessel potential (I −∆)−α/2 and its Fourier transformis (1 + 4π2|ξ|2)−α/2.

We also recall that a homogeneous elliptic differential operator of even order α < nwith real constant coefficients has the form

Pu =∑

|k|=α

akDku (1.17)

for u ∈ C∞c (Rn), with symbol

pα(ξ) = P (2πiξ) = (2π)α(−1)α/2∑

|k|=α

akξk, |pα(ξ)| ≥ c0|ξ|

α, ξ ∈ Rn

for some c0 > 0. The fundamental solution of P is given by a convolution operatorwith kernel gP :

gP (x) =

∫

Rn

e−2πix·ξ

pα(ξ)dξ (1.18)

in the sense of distributions. Since pα is homogeneous of order α, the kernel gP isalso homogeneous with order α− n.

6

Theorem 2. For 0 < α < n, let P be either (−∆)α2 , ∇(−∆)

α−12 for α odd, or a

homogeneous elliptic operator of even order α < n with constant coefficients. Thenthere exists C = C(α, n, P ) > 0 such that for every u ∈ W α,n

α (Rn) with

||Pu||n/αn/α ≤ 1 (1.19)

we have ∫

Rn

exp⌈nα−2⌉

[γ(P )|u(x)|

nn−α

]

1 + |u(x)|n

n−α

dx ≤ C||u||n/αn/α (1.20)

where

γ(P ) =

c− n

n−αα

|B1|, if P = (−∆)

α2

((n− α− 1)cα+1)− n

n−α

|B1|, if P = ∇(−∆)

α−12 and α odd,

(1.21)

with

cα =Γ(n−α

2)

2απn/2Γ(α2)

(1.22)

and whereγ(P ) =

n∫Sn−1 |gP (x

∗)|n

n−αdx∗

if P elliptic and α even. The exponential constant γ(P ) is sharp. Moreover, the aboveinequality (1.20) cannot hold if the power n

n−αin the denominator is replaced by any

smaller power.

As an immediate consequences of Theorem 2, we have the following Corollary:

Corollary 2. Let Ω be a bounded and open set in Rn, 0 < α < n an integer. There

exists C > 0 such that for all u ∈ Wα,n

α0 (Ω) with ||∇αu||n

α≤ 1 we have

∫

Ω

exp⌈nα−2⌉

[γ(P )|u(x)|

nn−α

]

1 + |u(x)|n

n−α

dx ≤ C||u||n/αn/α. (1.23)

The exponential constant γ(P ) is sharp. Furthermore, the above inequality (1.23)cannot hold if the power n

n−αin the denominator is replaced by any smaller power.

Although the proof of (1.23) uses Adams inequality on Ω, it is still not an easydirect consequence from Adams [A1]. We also mention that the above inequality(1.23) is different from the inequalities in [A1] because of the norm of u in the RHS.For example, take α = 1, n = 2, then by Corollary 2 we have

∫

Ω

e4πu2− 1

1 + u2dx ≤ C||u||22 (1.24)

7

and [A1] gave ∫

Ω

e4πu2

dx ≤ C|Ω| (1.25)

So for fixed Ω, we can see that, if ||u||22 becomes very small, then so is the LHS of(1.24), but this point may not be reflected by the second inequality (1.25).

As we mentioned earlier, Riesz-like kernels were introduced in [FM2], where theauthors proved, among other things, that if K is a Riesz-like kernel, then under theRuf condition

||f ||n/αn/α + ||Tf ||

n/αn/α ≤ 1 (1.26)

the following Adams inequality holds:

∫

Rn

exp⌈nα−2⌉

[1

Ag|Tf(x)|

nn−α

]dx ≤ C (1.27)

where Ag is as in (1.11). and where the exponential constant A−1g is sharp if the kernel

is n−regular. In section 8 we will prove that such result is implied by Theorem 1 ifK is strictly Riesz-like.

2. Improved O’Neil Lemma, O’Neil functional and

Adams inequality

Suppose that (M,µ) and (N, ν) are measure spaces. Given a measurable functionf : M → [−∞,∞] its distribution function will be denoted by

mf (s) = µ(x ∈M : |f(x)| > s), s ≥ 0.

Assume that the distribution function of f is finite for s ≥ 0. The decreasing rear-rangement of f will be denoted by

f ∗(t) = infs ≥ 0 : mf (s) ≤ t, t > 0

and we define

f ∗∗(t) =1

t

∫ t

0

f ∗(u)du, t > 0

which is sometimes called the maximal function of f ∗.Given a ν × µ-measurable function k : N ×M → [−∞,∞], assume that the level

sets of k(x, ·) and k(·, y) have finite measure for all x ∈ N and all y ∈M . Let k∗1(x, t)and k∗2(y, t) be the decreasing rearrangement of k(x, y) with respect to the variabley(resp. x) for fixed x(resp. y), and define

k∗1(t) = supx∈N

k∗1(x, t)

k∗2(t) = supy∈M

k∗2(y, t).

8

Lastly, let T be an integral operator defined as

Tf(x) =

∫

M

k(x, y)f(y)dµ(y). (2.1)

One of the main tools used in the proof is the following version of O’Neil lemma,whose proof was given in ([Q, Section 1.2.3]). Its proof is based on the proof oflemma 9 in [FM3] with some small modifications.

Lemma 1. (Improved O’Neil lemma)Let k : N ×M → [−∞,∞] be measurable and

k∗1(t) ≤ Dt− 1

q′ , k∗2(t) ≤ Bt− 1

q′ , t > 0 (2.2)

with q′ > 1. Let f(x, y) : N ×M → R be a measurable function on N ×M . For eachx ∈ N , let fx : M → R be defined as fx(y) = f(x, y) on M and assume fx ∈ Lq(M).(q′)−1 + q−1 = 1. Suppose there is a measurable function f : M → [−∞,∞] ,f ∈ Lq(M) such that

For all x ∈ N, f(y) ≥ |fx(y)| µ− a.e. y ∈M (2.3)

Let

T ′f(x) = Tfx(x) =

∫

M

k(x, y)fx(y)dµ(y), (2.4)

then there is a constant C0 = C0(D,B, q) such that

(T ′f)∗∗(t) ≤ C0maxτ− 1

q′ , t− 1

q′

∫ τ

0

f∗(u)du+ess sup

x∈N

∫ ∞

τ

f ∗x(u)k

∗1(x, u)du, ∀t, τ > 0.

(2.5)

Note that in this Lemma we consider the rearrangement of Tfx(x), where thefunction fx depends on x. This makes it different from the other improved O’Neillemma in [FM3], which estimated the rearrangement of Tf for a fixed function f .In order to apply the above Lemma 1 to the proof of our main theorem, we also needthe following lemma regarding the rearrangement of the sum of two functions whosesupports are mutually disjoint.

Lemma 2. Let f1, f2 : M → [−∞,∞] be measurable functions. Suppose that thesupports of f1 and f2 are mutually disjoint, µ(suppf1) = z and

|f1| ≥ ||f2||∞ µ− a.e. x ∈ x : f1(x) 6= 0

Then we have

(f1 + f2)∗(u) =

f ∗1 (u) if 0 < u < z

f ∗2 (u− z) if u > z

, (2.6)

and(f1 + f2)

∗∗(u) = f ∗∗1 (u) for 0 < u ≤ z. (2.7)

9

Proof of lemma 2: Given the assumptions on f1, f2 we have

mf1+f2(s) = mf1(s) +mf2(s) (2.8)

and mf2(s) = 0 if mf1(s) < z

mf1(s) = z if mf2(s) > 0.(2.9)

For u < z, by (2.8), (2.9) we have mf1+f2(s) ≤ u if mf1(s) ≤ u. It is also clearthat mf1(s) ≤ u whenever mf1+f2(s) ≤ u. So mf1(s) ≤ u if and only if mf1+f2(s) ≤ u.We get

(f1 + f2)∗(u) = infs ≥ 0 : mf1+f2(s) ≤ u = infs ≥ 0 : mf1(s) ≤ u = f ∗

1 (u).

Let u > z. If mf2(s) ≤ u−z then mf1+f2(s) ≤ u. We will show that mf1+f2(s) ≤ uimplies mf2(s) ≤ u − z. Suppose there exists s ≥ 0 such that mf1+f2(s) ≤ u andmf2(s) > 0. Then by (2.9) we have mf1(s) = z and hence

mf2(s) = mf1+f2(s)−mf1(s) ≤ u− z.

Therefore,

(f1 + f2)∗(u) = infs ≥ 0 : mf1+f2(s) ≤ u = infs ≥ 0 : mf2(s) ≤ u− z = f ∗

2 (u− z).

Lastly, (2.7) holds by (2.6) and the definition of the maximal function.

It is clear that as a consequence of Lemma 1, if fx = f for all x ∈ N , the O’Neilestimate takes the following form:

(Tf)∗∗(t) ≤ C0t− 1

q′

∫ t

0

f ∗(u)du+

∫ ∞

t

k∗1(u)f∗(u)du.

Under the hypothesis that m(f, s) < ∞ for s ≥ 0, we are able to define the O’Neilfunctional U as follows:

Uf(t) = C0t− 1

q′

∫ t

0

f ∗(u)du+

∫ ∞

t

k∗1(u)f∗(u)du

where C0 is the constant in the improved O’Neil lemma 1.

We now state an Adams inequality due to Fontana and Morpurgo ([FM3, Corollary2]) in terms of the O’Neil functional. Although the original theorem in their paper isnot stated in this form, it is clear from the proof in [FM3] that everything also worksfor the O’Neil functional instead of the original operator. This result plays a crucialrole in the proof of our main result.

Theorem A ([FM3, Corollary 2]) Suppose ν(N) <∞, µ(M) <∞, and that

k∗1(t) ≤ A1q′ t

− 1q′(1 +H(1 + | log t|)−γ

), 0 < t ≤ µ(M) (2.10)

10

k∗2(t) ≤ Bt− 1

q′ . 0 < t ≤ ν(N) (2.11)

Then there exists a constant C = C(q, γ, A,B,H) such that for each f ∈ Lq(M) with||f ||q ≤ 1 ∫ ν(N)

0

exp

[1

A

(Uf(t)

)q′]dt ≤ C

(ν(N) + µ(M)

). (2.12)

3. Proof of the inequalities in Theorem 1

Let us start with following lemma from [FM2], in order to clarify the equivalencebetween exponential inequalities on sets x ∈ Rn : |Tf(x)| ≥ 1 and regularizedexponential inequalities over Rn.

Lemma A ([FM2, Lemma 9]) Let (N, ν) be a measure space and 1 < p < ∞,a > 0. Then for every u ∈ Lp(N) we have

∫

|u|≥1

ea|u|p′

dx−ea||u||pp ≤

∫

N

(ea|u|

p′

−

⌈p−2⌉∑

k=0

ak|u|kp′

k!

)dx ≤

∫

|u|≥1

ea|u|p′

dx+ea||u||pp

(3.1)and also

∫

|u|≥1

ea|u|p′

1 + |u|p′dx−ea||u||pp ≤

∫

N

ea|u|p′

−∑⌈p−2⌉

k=0ak |u|kp

′

k!

1 + |u|p′dx ≤

∫

u≥1

ea|u|p′

1 + |u|p′dx+ea||u||pp.

(3.2)In particular, the following three inequalities are equivalent:

∫

N

exp⌈p−2⌉ [a|u|p′]

1 + |u|p′dx ≤ C||u||pp, (3.3)

∫

|u|≥1

ea|u|p′

1 + |u|p′dx ≤ C||u||pp, (3.4)

∫

E

ea|u|p′

1 + |u|p′dx ≤ C(||u||pp + |E|) (3.5)

for every measurable set E with finite measure.

In order to prove (1.10), it is enough to show that

∫

|Tf |≥1

exp

[1

Ag|Tf |

nn−α

]

1 + |Tf |n

n−α

dx ≤ C||Tf ||n/αn/α. (3.6)

Lett0 =

∣∣x : |Tf | ≥ 1∣∣.

11

Note that by this definition we have (Tf)∗(t) ≥ 1 for 0 < t < t0 and (Tf)∗(t) < 1 fort > t0.

Now we will show that (3.6) is equivalent to

∫ t0

0

exp

[1

Ag

((Tf)∗(t)

) nn−α

]

1 +((Tf)∗(t)

) nn−α

dt ≤ C||Tf ||n/αn/α. (3.7)

Let us denote the rearrangement of Tf with respect to a measurable set E as (Tf)∗Eand its corresponding maximal function as (Tf)∗∗E . Clearly

(Tf)∗E(t) =((Tf)χE

)∗(t), 0 < t ≤ |E|. (3.8)

Let

F (z) =e

1Agzn/(n−α)

1 + zn/(n−α)(3.9)

and E =x : |Tf | ≥ 1

, then the LHS of (3.6) can be written as

∫

E

F (|Tf |)dx =

∫ t0

0

F ((Tf)∗E(t))dt =

∫ t0

0

F ((Tf)∗(t))dt, (3.10)

The first equality holds since for F non-negative and measurable on [0,∞), for g

measurable on Rn and if E is a level set of g, we have∫EF |g|dx =

∫ |E|

0F g∗Edt

(see for example [K, Theorem 1.1.1]). To prove the second equality, note that

|(Tf)χE(x)| ≥ ||(Tf)χEc||∞, for a.e. x ∈ E,

hence by Lemma 2 we get

(Tf)∗E(t) = (Tf)∗(t), for 0 < t < t0.

To estimate (Tf)∗, we first define 1-parameter families of sets Eτ , Fτ (dependingon f) as follows.

For τ > 0, let Eτ be the set such that

|Eτ | = τ

x : |Tf(x)| > (Tf)∗(τ) ⊆ Eτ ⊆ x : |Tf(x)| ≥ (Tf)∗(τ).(3.11)

In order to show that such Eτ exists, we denote V1 = x : |Tf(x)| > (Tf)∗(τ)and V2 = x : |Tf(x)| ≥ (Tf)∗(τ). By definition of rearrangement, we have thatµ(V1) ≤ τ and µ(V2) ≥ τ. If µ(V2) = τ , we take Eτ = V2. Otherwise, consider thecontinuous function g(r) = µ(V1) + µ(Br ∩ (V2 \ V1)) for r ≥ 0, where Br = B(0, r)is the ball centered at 0 with radius r. It is clear that g(0) = µ(V1) ≤ τ , andg(r) → µ(V2) as r → ∞. Since µ(V2) > τ , there exists a r such that g(r) = τ , andEτ = V1 ∪ (Br ∩ V2 \ V1) is a measurable set that satisfies the condition (3.11).

12

Similarly, let Fτ be the set such that

|Fτ | = τ

x : |f(x)| > f ∗(τ) ⊆ Fτ ⊆ x : |f(x)| ≥ f ∗(τ).(3.12)

Letfτ = fχFτ

, f ′τ = fχF c

τ

and r(τ) = (τ/|B1|)1/n so that

|B(0, r(τ))| = τ. (3.13)

Remark 1. If f and K are radially decreasing, then both Eτ and Fτ are either openor closed balls of volume τ .

Next, for all x ∈ Eτ define

W (τ, x) =

∫ 2τ

τ

k∗1(u)(f′τχB(x,r(τ)))

∗(u− τ)du (3.14)

M(τ, x) = |T (f ′τχBc(x,r(τ)))(x)|. (3.15)

Lastly, for fixed τ > 0, take the essential supremum in (3.14) and (3.15), and let

Wτ = ess supx∈Eτ

W (τ, x) (3.16)

Mτ = ess supx∈Eτ

M(τ, x). (3.17)

We want to point out that for all τ > 0 we have Wτ < ∞ and Mτ < ∞, by the factthat f is compactly supported and ||f ||n/α ≤ 1. Also, note that for each x and τ ,

f = fτ + f ′τ = fτ + f ′

τχB(x,r(τ)) + f ′τχBc(x,r(τ))

andTf(x) = Tfτ (x) + T (f ′

τχB(x,r(τ)))(x) + T (f ′τχBc(x,r(τ)))(x). (3.18)

From now on we will use the following notation:

q =n

α, q′ =

n

n− α.

Recall that the O’Neil functional is defined as

Uf(t) = C0t− 1

q′

∫ t

0

f ∗(u)du+

∫ ∞

t

k∗1(u)f∗(u)du.

Our first step toward a proof of (3.7) is to establish the following estimate:

(Tf)∗(t) ≤ (Tf)∗∗(t) ≤ Ufτ (t) +Wτ +Mτ for 0 < t ≤ τ. (3.19)

13

Recall the definition of (Tf)∗E in (3.8), for any measurable set E . The definition ofEτ implies that

|(Tf)χEτ(x)| ≥ ||(Tf)χEc

τ||∞, for a.e. x ∈ Eτ ,

hence we can apply Lemma 2 to get

(Tf)∗∗Eτ(t) = (Tf)∗∗(t), for 0 < t ≤ τ. (3.20)

Letfx,τ = fτ + f ′

τχB(x,r(τ)),

note that by the definition of Mτ in (3.17) and the decomposition of Tf in (3.18),

|(Tf)χEτ(x)| ≤ |(Tfx,τ)χEτ

(x)|+Mτ , for x ∈ Eτ .

Due to subadditivity of (·)∗∗ (see [BS, Chapter 2 inequality (3.12)]) and (3.20), wehave

(Tf)∗(t) ≤ (Tf)∗∗(t) = (Tf)∗∗Eτ(t) ≤ (Tfx,τ)

∗∗Eτ(t) +Mτ , 0 < t ≤ τ. (3.21)

Therefore in order to prove (3.19) it is enough to show the following:

(Tfx,τ)∗∗Eτ(t) ≤ Ufτ (t) +Wτ for 0 < t ≤ τ. (3.22)

In other words, we only need to show the rearrangement of Tfx,τ (x) over the setEτ satisfies (3.22). Let us apply the improved O’Neil Lemma (Lemma 1) with N =Eτ , M = Rn, q = n/α, fx = fx,τ , and

f = |fτ + f ′τ | = |f | (3.23)

so that (2.3) holds. For x ∈ Eτ , let

T ′f(x) = Tfx,τ (x) =

∫

Rn

K(x− y)(fτ + f ′τχB(x,r(τ)))(y)dy. (3.24)

By Lemma 9 in [FM1], we have that (A1), (A3) implies the condition (2.2) in Lemma1. From now on we will use k∗ to denote k∗1 since k(x, y) = K(x− y) is a convolutionkernel, and k∗(t) = k∗1(t). We obtain

(T ′f)∗∗Eτ(t) ≤ C0t

− 1q′

∫ t

0

f ∗(u)du+ ess supx∈Eτ

∫ ∞

t

k∗1(x, u)f∗x,τ(u)du

= C0t− 1

q′

∫ t

0

f ∗(u)du+ ess supx∈Eτ

∫ ∞

t

k∗(u)f ∗x,τ(u)du.

(3.25)

By definition of Fτ , fτ and (3.23), we apply Lemma 2 to get

f ∗(u) = (fτ + f ′τ )

∗(u) = f ∗τ (u) if 0 < u < τ (3.26)

14

and

f ∗x,τ (u) = (fτ + f ′

τχB(x,r(τ)))∗(u) =

f ∗τ (u) if 0 < u < τ(f ′τχB(x,r(τ))

)∗(u− τ) if u > τ.

(3.27)

Therefore, (3.25) can rewritten as

(T ′f)∗∗Eτ(t) ≤ C0t

− 1q′

∫ t

0

f ∗τ (u)du+

∫ τ

t

k∗(u)f ∗τ (u)du

+ ess supx∈Eτ

∫ 2τ

τ

k∗(u)(f ′τχB(x,r(τ))

)∗(u− τ)du

= Ufτ (t) +Wτ .

(3.28)

Hence (3.22) is proved and (3.19) follows.

Next we consider the following inequality (also in [MS2, the inequality below(4.7)], with slightly different form)

(a+ b)p ≤ λ1−pap + (1− λ)1−pbp a, b ≥ 0, 0 < λ < 1, p > 1, (3.29)

which can be proved by writing a + b as (aλ−1/p′)λ1/p′

+ (b(1 − λ)−1/p′)(1 − λ)1/p′

and apply Holder inequality. Then we use estimation (3.19) and apply the aboveinequality to the integrand in (3.7), with

p = q′, a = (Ufτ )∗(t), and b =Wτ +Mτ .

We get

exp

[1

Ag

((Tf)∗(t)

)q′]

1 +((Tf)∗(t)

)q′ ≤ C

exp

[1

Ag

(Ufτ (t) +Wτ +Mτ

)q′]

1 +(Ufτ (t) +Wτ +Mτ

)q′

≤ C

exp

[(1− λ)1−q

′

Ag

(Wτ +Mτ

)q′]

1 +(Wτ +Mτ

)q′ · exp

[λ1−q

′

Ag

(Ufτ (t)

)q′].

(3.30)

To get the first inequality in (3.30), let F (z) be defined as in (3.9). Note that F (z) ≥

C > 0 on [0,∞) and is increasing in z for z ≥ 1 + A(n−a)/ng . Also recall that for

0 < t < t0 we have (Tf)∗(t) ≥ 1. We consider two cases. If 1 ≤ (Tf)∗(t) ≤

1 + A(n−a)/ng , then we have F ((Tf)∗(t)) ≤ C and F (Ufτ (t) +Wτ +Mτ ) ≥ C > 0. If

(Tf)∗(t) ≥ 1 +A(n−a)/ng , then by (3.19) and the fact that F (z) is increasing, the first

inequality follows.Let t1 > 0 be the number such that

∫ t10f ∗(u)qdu

||f ||qq=

1

4

15

and

ǫτ = min

1

4,

∫ τ0f ∗(u)qdu

||f ||qq

.

We estimate (3.30) using the following two lemmas. The first one is an integralestimate (essentially the Adams inequality):

Lemma 3. If we define

I2(τ, t, λ) = exp

[λ1−q

′

Ag

(Ufτ (t)

)q′], τ > 0, t > 0, λ > 0,

then ∫ τ

0

I2(τ, t, ǫτ )dt ≤ Cτ, 0 < τ ≤ t1. (3.31)

Proof of Lemma 3: First note that when τ ≤ t1, we have

ǫτ =

∫ τ0f ∗(u)qdu

||f ||qq.

If we let

f :=fτ

ǫ1/qτ

,

then we have that f has measure of support µ(suppf) ≤ τ with

||f ||q ≤ 1,

also assumption (A1) implies the estimate (2.2) on k∗ (See [FM1, Lemma 9]). Thatis, conditions (2.10) and (2.11) are satisfied. Therefore by Theorem A, the Adamsinequality for the O’Neil functional, we obtain (3.31).

The estimation for I1(τ, λ) which is stated in the following lemma is essential forthe rest of the proof. Let us assume the lemma for now, and its proof will be givenin sections 4-5.

Lemma 4. Let 0 < τ ≤ t0, 0 ≤ λ < 1. Define

I1(τ, λ) =

exp

[(1− λ)1−q

′

Ag

(Wτ +Mτ

)q′]

1 +(Wτ +Mτ

)q′ .

Then there exists constant C > 0 such that for all x1, x2 ∈ Eτ

I1(τ, ǫτ ) ≤C

τ||Tf ||qq (3.32)

where C = C(n, α,K).

16

Assuming Lemma 4, let τ0 = mint0, t1. To prove (3.7) it is enough to show that

∫ τ0

0

exp

[1

Ag

((Tf)∗(t)

)q′]

1 +((Tf)∗(t)

)q′ dt ≤ C||Tf ||qq (3.33)

and then show that if t1 < t0,

∫ t0

t1

exp

[1

Ag

((Tf)∗(t)

)q′]

1 +((Tf)∗(t)

)q′ dt ≤ C||Tf ||qq. (3.34)

To prove (3.33), we take τ = τ0 in (3.32) and (3.31) to get

I1(τ0, ǫτ0) ≤C

τ0||Tf ||qq and

∫ τ0

0

I2(τ0, t, ǫτ0)dt ≤ Cτ0. (3.35)

Therefore, using (3.30) it is immediate that

∫ τ0

0

exp

[1

Ag

((Tf)∗(t)

)q′]

1 +((Tf)∗(t)

)q′ dt ≤

∫ τ0

0

CI1(τ0, ǫτ0)I2(τ0, t, ǫτ0)dt ≤ C||Tf ||qq (3.36)

and (3.33) follows.

Next, to show (3.34), we take τ = t for t1 ≤ t ≤ t0, and λ = 18in the definition of

I2 in Lemma 3. Then by definition of O’Neil’s operator and the fact that the supportft has measure less than or equal t,

Uft(t) = C0t− 1

q′

∫ t

0

f ∗t (u)du ≤ C||ft||q ≤ C.

So we have

I2

(t, t,

1

8

)≤ C. (3.37)

Since t1 ≤ t ≤ t0 and ǫt1 =14, by definition ǫt =

14. Take θ = (6/7)

αn−α < 1. Hence

I1

(t,1

8

)=

exp

[(7/8)1−q

′

Ag

(Wt +Mt

)q′]

1 + (Wt +Mt)q′

=

exp

[(7/6)1−q

′ (3/4)1−q′

Ag

(Wt +Mt

)q′]

1 + (Wt +Mt)q′

=

(exp

[(3/4)1−q

′

Ag

(Wt +Mt

)q′])θ

1 + (Wt +Mt)q′

≤

(exp

[(3/4)1−q

′

Ag

(Wt +Mt

)q′]

1 + (Wt +Mt)q′

)θ

= Iθ1

(t,1

4

)= Iθ1 (t, ǫt) ≤

C

tθ||Tf ||θqq .

(3.38)

17

Using (3.37), (3.38) we get

∫ t0

t1

CI1

(t,1

8

)I2

(t, t,

1

8

)dt ≤ C

∫ t0

t1

1

tθ||Tf ||θqq dt ≤ Ct1−θ0 ||Tf ||θqq

≤ C||Tf ||(1−θ)qq ||Tf ||θqq = C||Tf ||qq,

(3.39)

where the last inequality is by the fact that

||Tf ||qq ≥ t0 (3.40)

since (Tf)∗(t) ≥ 1 for all t < t0, by the definition of t0. So (3.34) follows from (3.30).In order to complete the proof of Theorem 1, we are left to prove Lemma 4.

4. Proof of Lemma 4

It is enough to show that

exp

[(1− ǫτ )

− αn−α

Ag(W (τ, x2) +M(τ, x1))

nn−α

]

1 + (W (τ, x2) +M(τ, x1))n

n−α

≤ C||Tf ||qqτ

(4.1)

for all x1, x2 ∈ Eτ . Now let us state the following key lemma in [MS1]-[MS3], [LT].

Lemma 5. Given any sequence a = akk≥0. Let q > 1 and

||a||1 =∞∑

k=0

|ak|, ||a||q =( ∞∑

k=0

|ak|q)1/q

(4.2)

define

µd(h) = inf

∞∑

k=0

|ak|qeqk : ||a||1 = h, ||a||q ≤ 1.

Then for h > 1, we have

C1(q)exp

[qhq

′]

hq′≤ µd(h) ≤ C2(q)

exp[qhq

′]

hq′. (4.3)

As a consequence of the above optimal growth lemma, we deduce that for anyq > 1 and any µ > 0, h > 1 there is C = C(q) such that for any sequence aksatisfying

∞∑

k=0

|ak| = h

∞∑

k=0

|ak|q ≤ µ (4.4)

we haveexp

[qµ1−q′hq

′]

hq′≤ Cµ−q′

∞∑

k=0

|ak|qeqk. (4.5)

18

The next task is to find a number h1, depending on f and x1, and a sequence a = ak,also depending on f and x1, such that

q− 1

q′A− 1

q′

g (W (τ, x2) +M(τ, x1)) ≤ h1 (4.6)

∞∑

k=0

|ak| = h1,∞∑

k=0

|ak|q ≤ 1− ǫτ ,

∞∑

k=0

|ak|qeqk ≤

C

τ||Tf ||qq. (4.7)

Clearly (4.1) follows from (4.4)-(4.7), with µ = 1− ǫτ ≥34and h = h1.

From now on, throughout the proof of Lemma 4, we fix 0 < τ ≤ t0, and x1, x2 ∈ Eτas defined in (3.11). First let us introduce some notation. Recall that r(τ) is thenumber such that |B(0, r(τ))| = τ . Define for each j = 0, 1, 2...

rj = r(τ)eqnj , Dj = B(x1, rj)

αj = ||f ′τχDj+1\Dj

||q, α−1 = ||f ′τχD0 ||q

αj = max α−1, α0, ..., αj, βj = ||f ′τχDc

j||q.

Notice that for any jαj ≤ βj ≤ 1.

Clearly βj is decreasing, and it vanishes for all j large enough, since f has compactsupport. In particular, there is an integer N so that

supp f ⊆ DN = B(x1, rN).

Now we are ready to state the main estimates on M(τ, x2) and W (τ, x1):

Proposition 1. There exist constants C2, C3 independent of f and an integer J suchthat

q− 1

q′A− 1

q′

g (W (τ, x2) +M(τ, x1)) ≤

J∑

j=0

αj + C2αJ + C2βJ (4.8)

andJ∑

j=0

αqjeqj + βqJe

qJ ≤C3

τ||Tf ||qq. (4.9)

The proof of Proposition 1 will be given in section 5. Assuming the proposition,we now show how to derive (3.32), and hence finish the proof of Lemma 4, using(4.4)-(4.7) together with Proposition 1.

Our goal is to find a number h1 and a sequence a = ak that satisfies (4.6) and(4.7). Let

h1 =J∑

j=0

αj + C2 αJ + C2βJ . (4.10)

Clearly, by Proposition 1, we have

q− 1

q′A− 1

q′

g (W (τ, x2) +M(τ, x1)) ≤ h1. (4.11)

19

Let J∗ be the smallest integer such that αJ∗ = αJ . It is clear that J∗ ≤ J . Toconstruct the sequence a that satisfies (4.7), let us first define Ni, i = 1, ..., 4 asfollows:

N1 = J∗

N2 = N1 + ⌈(1 + C2)q′⌉

N3 = N2 + J − 1− J∗

N4 = N3 + ⌈(1 + C2)q′⌉.

(4.12)

Let a = ak be the following:

ak =

αk−1 if J∗ 6= −1; 0 if J∗ = −1 for k = 0, ..., N1

(1 + C2)αJN2 −N1

if J∗ 6= −1;C2αJ

N2 −N1if J∗ = −1 for k = N1 + 1, ..., N2

αk−N2+J∗ for k = N2 + 1, ..., N3

αJ + C2βJN4 −N3

for k = N3 + 1, ..., N4.

(4.13)With this definition of ak we have

||a||1 =

N4∑

k=0

|ak| =

N4∑

k=0

ak = h1. (4.14)

If J∗ 6= −1,

N4∑

k=0

|ak|q =

N1∑

k=0

αqk−1 +

N2∑

k=N1+1

((1 + C2)αJN2 −N1

)q+

N3∑

k=N2+1

αqk−N2+J∗

+N4∑

k=N3+1

(αJ + C2βJN4 −N3

)q

≤J∗−1∑

k=0

αqk−1 + αqJ +J−1∑

k=J∗+1

αqk + βqJ

= ||f ′τ ||

qq ≤ (1− ǫτ )||f ||

qq ≤ 1− ǫτ .

(4.15)

Likewise for J∗ = −1,

N4∑

k=0

|ak|q =

N2∑

k=N1+1

(C2αJ

N2 −N1

)q+

N3∑

k=N2+1

αqk−N2+J∗

+

N4∑

k=N3+1


)q

≤ αq−1 +J−1∑

k=0

αqk + βqJ

= ||f ′τ ||

qq ≤ (1− ǫτ )||f ||

qq ≤ 1− ǫτ .

(4.16)

20

And using (4.9) in Proposition 1, we also have, if J∗ 6= −1

N4∑

k=0

|ak|qeqk =

N1∑

k=0

αqk−1eqk +

N2∑

k=N1+1

((1 + C2)αJN2 −N1

)qeqk +

N3∑

k=N2+1

αqk−N2+J∗eqk

+

N4∑

k=N3+1


)qeqk

≤J∗−1∑

k=0

αqk−1eqk + CαqJe

q(J∗+C4) +J−1∑

k=J∗+1

αqkeq(k+C4) + βqJe

q(J+2C4)

≤ Ce2C4

(αq−1 +

J∑

k=0

αqkeqk + βqJe

qJ

)≤ Ce2C4

(C +

J∑

k=0

αqkeqk + βqJe

qJ

)

= Ce2C4

(Cτ

τ+

J∑

k=0

αqkeqk + βqJe

qJ

)≤C

τ||Tf ||qq

(4.17)where C4 in the above inequality is C4 = ⌈(1 + C2)

q′⌉, and in the last inequality weused the fact that τ ≤ ||Tf ||qq since (Tf)∗(t) ≥ 1 for 0 < t ≤ τ < t0. Similarly forJ∗ = −1,we also have

N4∑

k=0

|ak|qeqk ≤ Ce2C4

(αq−1 +

J∑

k=0

αqkeqk + βqJe

qJ

)≤C

τ||Tf ||qq. (4.18)

Finally, (4.15)-(4.18) shows that the sequence a satisfies (4.6) and (4.7). Hence(3.32) follows and the proof is concluded.

5. Proof of Proposition 1

In the following proof we will set for any measurable function φ : Rm → R

Sjφ = φχDcj= φχ|y−x1|≥rj

.

With this notation we then have

(Sj − Sj+1)f′τ = f ′

τχDj+1\Dj= f ′

τχrj≤|y−x1|<rj+1

andαj = ‖(Sj − Sj+1)f

′τ‖q, βj = ‖Sjf

′τ‖q.

Also note that

f ′τ = f ′

τχB(x1,r(τ))+ f ′

τχBc(x1,r(τ))= f ′

τχD0+ S0f

′τ . (5.1)

For the rest of the proof we assume that

TS0f′τ (x1) ≥ 0. (5.2)

21

If, on the other hand,TS0f

′τ (x1) < 0, (5.3)

we replace T by −T , and the proof is exactly the same.We first give some preliminary estimates on W (τ, x2) and M(τ, x1). We have that

W (τ, x2) =

∫ 2τ

τ

k∗1(u)(f′τχB(x,r(τ)))

∗(u− τ)du

≤ C

∫ 2τ

τ

u− 1

q′ (f ′τχB(x2,r(τ))

)∗(u− τ)du ≤ C||(f ′τχB(x2,r(τ))

)∗||q

= C||f ′τχB(x2,r(τ))

||q.

(5.4)

Since f is supported in DN , we also have that

supp f ′τχB(x2,r(τ))

⊆ DN =

N−1⋃

j=0

(Dj+1 \Dj) ∪D0

by the definition ofDj it is clear that B(x2, r(τ)) can only have nonempty intersectionwith at most two elements in the set

D0, Dj+1 \Dj , for j = 0, 1, ..., N − 1

,

therefore we have||f ′

τχB(x2,r(τ))||q ≤ αj1 + αj2 (5.5)

for some j1, j2 ∈ −1, 0, 1, ..., N. Then by the definitions of α, α, β, we have for anyJ ∈ 0, 1, ..., N

αj ≤ αJ if J ≥ j

αj ≤ βJ if J ≤ jj = j1, j2 (5.6)

so that by combining (5.4),(5.5) and (5.6), we have

W (τ, x2) ≤ CαJ + CβJ (5.7)

where C = C(n, α,K). Next, recall that

M(τ, x) = |T (f ′τχBc(x,r(τ)))(x)|. (5.8)

By (5.1) and (5.2), we can write, for any J ∈ 0, 1, ..., N

M(τ, x1) = |TS0f′τ (x1)| = TS0f

′τ (x1)

=

J∑

j=0

(TSjf

′τ (x1)− TSj+1f

′τ (x1)

)+ TSJ+1f

′τ (x1)

=

J∑

j=0

T(Sjf

′τ − Sj+1f

′τ

)(x1) + TSJ+1f

′τ (x1).

22

Next, for any integer j, we have the estimate

T(Sjf

′τ − Sj+1f

′τ

)(x1) ≤ |T (Sjf

′τ − Sj+1f

′τ )(x1)|

≤

( ∫

rj≤|y|<rj+1

|K(y)|q′

dy

)1/q′

‖Sjf′τ − Sj+1f

′τ‖q.

(5.9)

Using (A1), (A4) and the inequality (a+ b)β ≤ aβ + β2β−1(aβ−1b+ bβ) for β > 1 (seeAdams [A1, inequality (17)] or use mean value theorem) we get

|K(y)|q′

≤ |g(y∗)|q′

|y|−n + Cmin|y|−n+δ1, |y|−n−δ2 (5.10)

for some C > 0, C = C(n, α,H1, H2, B, δ1, δ2). Since rj+1 = eqn rj ,

T(Sjf

′τ − Sj+1f

′τ

)(x1) ≤

(qAg + Cminrδ1j , r

−δ2j

)1/q′αj

≤

(q

1q′A

1q′

g + Cminrδ1/q′

j , r−δ2/q′

j

)αj.

(5.11)

Using (5.11), we then get that

M(τ, x1) ≤J∑

j=0

(q

1q′A

1q′

g + Cminrδ1/q′

j , r−δ2/q′

j

)αj + TSJ+1f

′τ (x1)

= q1q′A

1q′

g

J∑

j=0

αj + CαJ

∞∑

j=0

minrδ1/q′

j , r−δ2/q′

j + TSJ+1f′τ (x1)

≤ q1q′A

1q′

g

J∑

j=0

αj + CαJ

∞∑

j=0

(e− q

n

δ1q′j+ e

− qn

δ2q′j) + TSJ+1f

′τ (x1)

= q1q′A

1q′

g

J∑

j=0

αj + CαJ + TSJ+1f′τ (x1).

(5.12)

Note that (5.6) and (5.12) are true for any J ∈ 0, 1, ..., N. The main task nowis prove that there exists J ∈ 0, 1, ..., N such that

TSJ+1f′τ (x1) ≤ CαJ + CβJ . (5.13)

This will be effected by a double stopping time argument, which will simultaneouslyyield (4.9) in Proposition 1.

Recall that N is an integer such that suppf ⊆ DN . Let J1 ∈ 1, ..., N be suchthat

βqj+1 ≤

(–

∫

Dj+1\Dj

|Tf(x)|dx

)q

for j = 0, ..., J1 − 1 (5.14)

βqJ1+1 >

(–

∫

DJ1+1\DJ1

|Tf(x)|dx

)q

. (5.15)

23

If condition (5.14) is never satisfied we let J1 = 0, and if (5.15) is never satisfied letJ1 = N + 1. Next, let J2 ∈ 1, ..., Nbe such that

TSj+1f′τ (x1) ≥

(eq−1 + 1

2eq−1

)TSjf

′τ (x1) for j = 0, ..., J2 − 1 (5.16)

TSJ2+1f′τ (x1) <

(eq−1 + 1

2eq−1

)TSJ2f

′τ (x1). (5.17)

As in the definition of J1, we let J2 = 0 if condition (5.16) is never satisfied , and letJ2 = N + 1 if (5.17) is never satisfied.

We will first prove (5.13), and hence (4.8), in three cases depending on J1, J2, thenwe will show that (4.9) holds with the chosen J in each case.

Case 1: J2 ≤ J1 ≤ N + 1 and J2 6= N + 1.

Case 2: J2 ≥ J1 + 1.

Case 3: J1 = J2 = N + 1.

Proof of (5.13) in the case J2 ≤ J1 ≤ N + 1 and J2 6= N + 1:

In this case, by (5.17) we have

TSJ2+1f′τ (x1) <

(eq−1 + 1

2eq−1

)TSJ2f

′τ (x1)

=

(eq−1 + 1

2eq−1

)(TSJ2+1f

′τ (x1) + T (SJ2 − SJ2+1)f

′τ (x1)

)

≤

(eq−1 + 1

2eq−1

)(TSJ2+1f

′τ (x1) +

∣∣T (SJ2 − SJ2+1)f′τ (x1)

∣∣)

≤

(eq−1 + 1

2eq−1

)(TSJ2+1f

′τ (x1) + CαJ2

)

(5.18)

where the last inequality is by (5.11). So we have

TSJ2+1f′τ (x1) < C

(eq−1 + 1

2eq−1

)αJ2 = CαJ2. (5.19)

Hence by taking J = J2 in (5.12), we obtain

q− 1

q′A− 1

q′

g (W (τ, x2) +M(τ, x1)) ≤

J2∑

j=0

αj + CαJ2 + CβJ2 + TSJ2+1f′τ (x1)

≤

J2∑

j=0

αj + CαJ2 + CβJ2 + CαJ2

≤

J2∑

j=0

αj + CαJ2 + CβJ2.

(5.20)

24

Therefore we get (4.8) with J = J2 and C2 = C in the above inequality.

Proof of (5.13) in the case J2 ≥ J1 + 1:

We will need the following lemma to handle this case. Let us state it here, andits proof will be postponed to the Appendix.

Lemma 6. There is a constant C1 = C1(n, α,K) such that for any J ≤ N − 1

–

∫

DJ+1\DJ

|Tfτ (x)|dx ≤ C1

(1

eq−1

)J, (5.21)

∣∣∣∣ –∫

DJ+1\DJ

TSJ+2f′τ (x)dx− TSJ+1f

′τ (x1)

∣∣∣∣ ≤ C1βJ+1, (5.22)

∣∣∣∣ –∫

DJ+1\DJ

T (S0 − SJ+2)f′τ (x)dx

∣∣∣∣ ≤ C1αJ+1, (5.23)

and ∣∣∣∣ –∫

DJ+1\DJ

T (f ′τχD0)(x)dx

∣∣∣∣ ≤ C1α−1. (5.24)

Assuming Lemma 6, let us first make a reduction. Recall that 0 < τ ≤ t0. We willassume that

M(τ, x1) ≥ max4C1, 1. (5.25)

where C1 is the constant which is defined in Lemma 6. If the above is not true, thenwe have that M(τ, x1) ≤ C, and on the other hand, by (5.4)

W (τ, x2) ≤ C||f ′τχB(x2,r(τ))||q ≤ C. (5.26)

Therefore, W (τ, x2) +M(τ, x1) ≤ C, and hence

exp

[(1− ǫτ )

− αn−α

Ag(W (τ, x2) +M(τ, x1))

nn−α

]

1 + (W (τ, x2) +M(τ, x1))n

n−α

≤ C = Cτ

τ≤ C

||Tf ||qqτ

(5.27)

which is (4.1), and the last inequality is by (3.40).By (5.22) in Lemma 6 and recalling that

f = fτ + f ′τ = fτ + f ′

τχD0+ f ′

τχDJ1+2\D0+ f ′

τχDcJ1+2

,

25

we have

TSJ1+1f′τ (x1) ≤ –

∫

DJ1+1\DJ1

TSJ1+2f′τ (x)dx+ CβJ1+1

≤

∣∣∣∣ –∫

DJ1+1\DJ1

TSJ1+2f′τ (x)dx

∣∣∣∣ + CβJ1+1

=

∣∣∣∣ –∫

DJ1+1\DJ1

(Tf − Tfτ − T (f ′

τχD0)− T (S0 − SJ1+2)f

′τ

)(x)dx

∣∣∣∣ + CβJ1+1

≤

∣∣∣∣ –∫

DJ1+1\DJ1

Tf(x)dx

∣∣∣∣ +∣∣∣∣ –∫

DJ1+1\DJ1

Tfτ (x)dx

∣∣∣∣ +∣∣∣∣ –∫

DJ1+1\DJ1

T (f ′τχD0

)(x)dx

∣∣∣∣

+

∣∣∣∣ –∫

DJ1+1\DJ1

T (S0 − SJ1+2)f′τ (x)dx

∣∣∣∣ + CβJ1+1

≤ –

∫

DJ1+1\DJ1

|Tf(x)|dx+ –

∫

DJ1+1\DJ1

|Tfτ (x)|dx+ Cα−1 + CαJ1+1 + CβJ1+1

(5.28)where the second last inequality is by Lemma 6 (5.23),(5.24). To estimate the secondintegral, note first that by reduction (5.25) we have

M(τ, x1) = TS0f′τ (x1) ≥ 4C1. (5.29)

Using (5.21) in Lemma 6, and condition (5.16), we get

–

∫

DJ1+1\DJ1

|Tfτ(x)|dx ≤ C1

(1

eq−1

)J1≤

1

4

(1

eq−1

)J1TS0f

′τ (x1)

≤1

4

(1

eq−1

)J1 ( 2eq−1

eq−1 + 1

)J1+1

TSJ1+1f′τ (x1)

=1

4

(2

eq−1 + 1

)J1 ( 2eq−1

eq−1 + 1

)TSJ1+1f

′τ (x1) ≤

1

2TSJ1+1f

′τ (x1).

(5.30)Hence we have

TSJ1+1f′τ (x1) ≤ –

∫

DJ1+1\DJ1

|Tf(x)|dx+1

2TSJ1+1f

′τ (x1) + CαJ1+1 + CβJ1+1.

So the above inequality along with the condition (5.15) give us

TSJ1+1f′τ (x1) ≤ 2 –

∫

DJ1+1\DJ1

|Tf(x)|dx+ 2CαJ1+1 + 2CβJ1+1

≤ 2CαJ1+1 + (2C + 2)βJ1+1.

(5.31)

26

By taking J = J1 in (5.12), we get

q− 1

q′A− 1

q′

g (W (τ, x2) +M(τ, x1)) ≤

J1∑

j=0

αj + CαJ1 + CβJ1 + TSJ1+1f′τ (x1)

≤

J1∑

j=0

αj + CαJ1+1 + CβJ1+1

≤

J1∑

j=0

αj + CαJ1 + CβJ1

(5.32)

where the last inequality is by the fact that αJ+1 ≤ αJ+αJ+1 ≤ αJ+βJ+1 ≤ αJ+βJ .Therefore we get (4.8) with J = J1.

Proof of (5.13) in the case J1 = J2 = N + 1:

In this case, we will simply write the entire series, that is, we will take J = N .Since TSN+1f

′τ (x1) = 0 we have

q− 1

q′A− 1

q′

g (W (τ, x2) +M(τ, x1)) ≤

N∑

j=0

αj + CαN + CβN + TSN+1f′τ (x1)

≤N∑

j=0

αj + CαN .

(5.33)

To check (4.9), note that we take J = J2 in case 1, J = J1 in case 2 and J = N incase 3. Assume first that J1 6= 0. Then in all the cases we have that (5.14) is true forall j ≤ J1 − 1, so

J∑

j=0

αqjeqj + βqJe

qJ ≤ 3e2qJ1−1∑

j=0

βqj eqj ≤ C

J1−1∑

j=0

(–

∫

Dj+1\Dj

|Tf(x)|dx

)q

eqj

≤ C

J1−1∑

j=0

(1

rnj

∫

Dj+1\Dj

|Tf(x)|dx

)q

eqj

≤ CJ1−1∑

j=0

r−qnj

(∫

Dj+1\Dj

|Tf(x)|qdx

)(∫

Dj+1\Dj

dx

)q/q′

eqj

= C

J1−1∑

j=0

r−qnj r(n−α)qj

(∫

Dj+1\Dj

|Tf(x)|qdx

)eqj

≤ C

J1−1∑

j=0

r−nj eqj∫

Dj+1\Dj

|Tf(x)|qdx ≤C

τ||Tf ||qq

(5.34)

27

where in the last inequality we used the fact that rnj = rn0 eqj and τ = |B(x1, r0)|, so

τ = Crn0 .If J1 = 0, then we just need to check (4.9) for J = 0:

αq0 + βq0 ≤ 2βq0 ≤ C = Cτ

τ≤C

τ||Tf ||qq, (5.35)

where the last inequality is by (3.40). Proposition 1 is proved.

6. Proofs of the sharpness statements in Theo-

rem 1.

We will make use of the extremal family of functions constructed in [FM2, Section6], that the authors used to prove the sharpness of the exponential constants in(1.27) and (1.15). Specifically, under the hypothesis that K is n-regular, the authorsproduced a family of compactly supported functions ψǫ,r ∈ Lq(B(0, r)) such that

max||ψǫ,r||qq , ||Tψǫ,r||

qq ≤ 1

|Tψǫ,r(x)|q′ ≥ Ag log

1

(ǫr)n+ br

(1−

C

log 1ǫn

)− C, |x| ≤ ǫr/2, (6.1)

||Tψǫ,r||qq ≤ Crn(log

1

ǫn)−1, (6.2)

where

br :=

∫

1≤|y|≤r

|K(y)|q′

dy

and

1 ≤ rn ≤Ag2C4

(log

1

ǫn

). (6.3)

Note that by the assumptions (A1), (A4), we have

br ≤ Ag log rn + C. (6.4)

Note also that for ǫ small (6.1) and (6.3) imply

|Tψǫ,r(x)| ≥ 1, ∀x ∈ Bǫr/2. (6.5)

To prove that the exponential constant sharp, i.e. it cannot be replaced by alarger constant, pick

rn =Ag2C4

(log

1

ǫn

)

and for any fixed θ > 1 estimate

28

Note that we have

2q′

|B1| log rn − C ≤ br ≤ 2q

′

|B1| log rn + C.

Choose

rn =Ag2C4

(log

1

ǫn

), (6.9)

which satisfies (6.3). Hence we have

∫

Bǫr/2

exp

[1

|B1||Tψǫ,r(x)|

nn−α

]

1 + |Tψǫ,r(x)|n

n−α

dx ≥ |Bǫr/2|

exp

[log

1

(ǫr)n+

br|B1|

(1−

C

log 1ǫn

)− C

]

1 + C log1

(ǫr)n+ br − C

≥ Cr2

q′n

1 + Crn→ ∞

(6.10)as ǫ→ 0+.

On the other hand, since K is a Riesz-like kernel, the inequality (1.27) [FM2,Theorem 5] holds under the Ruf condition.

7. Proof of Corollary 1

Assume that||f ||n/α ≤ 1. (7.1)

Let q = n/α. It is enough to show that∫

|Tf |≥1

exp

[θ

Ag|Tf(x)|

nn−α

]dx ≤

C

1− θ||Tf ||qq (7.2)

since (1.15) is then a direct consequence of the exponential regularization Lemma A.To show (7.2), write

exp

[θ

Ag|Tf(x)|

nn−α

]

=exp

[1Ag

|Tf(x)|n

n−α

]

1 + |Tf |n

n−α

1 + |Tf |n

n−α

exp[1−θAg

|Tf(x)|n

n−α

] .(7.3)

Observe that1 + y

e(1−θ)y/Ag≤

C

1− θ, for y ≥ 0

So by Theorem 1,

∫

|Tf |≥1

exp

[θ

Ag|Tf(x)|

nn−α

]dx ≤

C

1− θ

∫

|Tf |≥1

exp

[1Ag|Tf(x)|

nn−α

]

1 + |Tf |n

n−α

dx

≤C

1− θ||Tf ||qq.

(7.4)

30

Obviously (7.4) also follows under the more restrictive condition

||f ||pn/αn/α + ||Tf ||

pn/αn/α ≤ 1, p <∞.

The proof of sharpness is the same as in [FM2]. We use the family of functions ψǫ,rin section 6, and choose

rn =Ag2C4

(log

1

ǫn

).

8. Proof that Theorem 1 implies (1.27)

It is enough to show that

∫

|Tf |≥1

exp

[1

Ag|Tf(x)|

nn−α

]dx ≤ C (8.1)

under the Ruf condition||f ||

n/αn/α + ||Tf ||

n/αn/α ≤ 1.

Let τ = ||Tf ||qq. Clearly we can assume that τ ∈ (0, 1). We consider two cases:

Case 1: τ ≥ 1− (2/3)q−1.

Case 2: τ < 1− (2/3)q−1.

Proof of (8.1) in case 1: In this case,

||f ||qq ≤ 1− τ ≤(23

)q−1, (8.2)

so letting f = f/(23)q−1q =

(32

) q−1q f gives ||f ||qq ≤ 1. We can write

∫

|Tf |≥1

exp

[1

Ag|Tf(x)|

nn−α

]dx =

∫

|Tf |≥1

exp

[2

3Ag|T f(x)|

nn−α

]dx. (8.3)

So by taking θ =2

3in Adachi-Tanaka result, we have

∫

|Tf |≥1

exp

[2

3Ag|T f(x)|

nn−α

]dx ≤

C

1− 2/3||T f ||qq = 3(

3

2)q−1C||Tf ||qq ≤ C. (8.4)

Combining (8.3) and (8.4) finishes the proof in case 1.

Proof of (8.1) in case 2: In this case,

||f ||qq ≤ 1− τ ∈

((23

)q−1, 1

). (8.5)

31

Let p > 1 be such thatp(1− τ)

αn−α = 1.

Rewrite (8.1) and apply Holder’s inequality,

∫

|Tf |≥1

exp

[1

Ag|Tf(x)|

nn−α

]dx =

∫

|Tf |≥1

exp

[1

Ag|Tf(x)|

nn−α

]

1 + |Tf |n

(n−α)p

(1 + |Tf |

n(n−α)p

)dx

≤

∫

|Tf |≥1

exp

[1

Ag|Tf(x)|

nn−α

]

1 + |Tf |n

(n−α)p

p

dx

1p (∫

|Tf |≥1

(1 + |Tf |

n(n−α)p

) pp−1

dx

) p−1p

= I ′ · I ′′.(8.6)

Let

f =f

(1− τ)1/q

so that ||f ||qq ≤ 1. Applying Theorem 1 gives

I ′ ≤C

(1− τ)1/q

∫

|Tf |≥1

exp

[(1− τ)

αn−α

Ag|T f(x)|

nn−α

]

1 + |T f |n

(n−α)p

p

dx

1p

≤ C

∫

|Tf |≥1

exp

[1

Ag|T f(x)|

nn−α

]

1 + |T f |n

(n−α)

dx

1p

≤ C||T f ||qpq ≤

C

1− τ||Tf ||

qpq ≤ C||Tf ||

qpq .

(8.7)

Next, to estimate I ′′ we start with the following Adachi-Tanaka inequality:

∫

|Tf |≥1

exp

[1

2Ag|Tf(x)|

nn−α

]dx ≤ C||Tf ||qq. (8.8)

Let

P =1

p− 1, P0 =

⌈1

p− 1

⌉− 1, P1 =

⌈1

p− 1

⌉

and define

F (x) :=

Tf(x) if |Tf(x)| ≥ 1

0 otherwise.(8.9)

32

By the power series expansion of the exponential function, we have that the inequality(8.8) implies that for any integer N ≥ 1,

∫

|Tf |≥1

|Tf(x)|nNn−α = ||F ||q

′Nq′N ≤ (2Ag)

NN !

∫

|Tf |≥1

exp

[1

2Ag|Tf(x)|

nn−α

]dx

≤ C(2Ag)NN !||Tf ||qq ≤ C(2Ag)

NNN ||Tf ||qq.(8.10)

By (8.5), we have P0, P1 ≥ 1, hence

||F ||q′

q′P0≤ 2AgC

1/P0P0||Tf ||q/P0q ≤ CP0||Tf ||

q/P0q

||F ||q′

q′P1≤ 2AgC

1/P1P1||Tf ||q/P1q ≤ CP1||Tf ||

q/P1q .

Let a ∈ [0, 1] be the number such that

1

P=

a

P0+

1− a

P1.

By interpolation [Fol, Proposition 6.10] we have

||F ||q′

q′P ≤ ||F ||q′aq′P0

||F ||q′(1−a)q′P1

≤ CP a0 P

1−a1 ||Tf ||q(a/P0)

q ||Tf ||q(1−a)/P1q

= CP a0 P

1−a1 ||Tf ||q/Pq ≤ CP ||Tf ||q/Pq .

(8.11)

Hence, since p > 1

I ′′ =

(∫

|Tf |≥1

|Tf(x)|n

n−α1

p−1dx

) p−1p

= ||F ||q′/pq′P ≤ C

(1

p− 1

)1/p

||Tf ||q(p−1

p)

q

≤C

p− 1||Tf ||

q(p−1p

)q .

(8.12)So combining (8.7) and (8.12), we get

I ′ · I ′′ ≤ C1

p− 1||Tf ||qq = C

τ

p− 1≤ C

τ

1− (1− τ)α

n−α

≤ C (8.13)

where the second inequality is by (8.5).

9. Proof of Theorem 2 and Corollary 2

In Theorem 1 we assume that the functions f are compactly supported, with bothf and Tf in the space Lq(Rn). We denote this space of functions by D0:

D0 := f ∈ Lq(Rn) : suppf is compact and Tf ∈ Lq(Rn).

In the following Theorem [FM2, Theorem 7] we see that T has a smallest closedextension, which enables us to extend Theorem 1 to all functions in the domain ofthe extension. In particular, Theorem 2 is a consequence of the following Theorem:

33

Theorem B ([FM2, Theorem 7]). If K is a Riesz-like kernel, then the operatorT : D0(T ) → Lq(Rn) is closable, and its smallest closed extension (still denoted T )has domain

D(T ) = f ∈ Lq(Rn) : ∃fk ⊆ D0(T ), ∃h ∈ Lq(Rn) with fkLq

−→ f, TfkLq

−→ h(9.1)

andTf = h.

In the case of Riesz potential we have

W α,q(Rn) = Iαf, f ∈ D(Iα) (9.2)

and the operator (−∆)α2 is a bijection between W α,q(Rn) and D(Iα), with inverse

cαIα.

By using the above Theorem B and Fatou’s lemma, we easily deduce that Theorem1 is still valid for all functions f in D(T ). Also (9.2) tells us that the Riesz potentialfor all functions f in the extended domain D(Iα) is the space W α,q(Rn). Thereforeby the fact that the inverse of (−∆)

α2 is cαIα, we have (1.20).

In the case of elliptic operator, by the formula (1.18), the kernel of the integraloperator is homogeneous of order α− n, therefore we have (1.20).

It is enough to assume u ∈ C∞c (Rn) since α is an integer for the remaining cases.

For P = ∇(−∆)α−12 and α is an odd integer, since u = cα+1Iα+1(−∆)

α+12 u, we can

write

u(x) =

∫

Rn

cα+1(n− α− 1)|x− y|α−n−1(x− y) · f(y), f = ∇(−∆)α−12 u. (9.3)

Clearly the kernel in the above formula satisfies our assumptions (A1)-(A4), so (1.20)follows.

For the proof of Corollary 2, it is clear that the inequality (1.23) is a directconsequence of (1.20) since Ω ⊆ Rn.

Proof of sharpness: To prove the sharpness, let ψǫ,r be the function as in theproof of sharpness (section 6). If P = (−∇)

α2 , consider the functions

uǫ,r = cαIαψǫ,r.

Similarly, for P an elliptic operator, let uǫ,r = gP ∗ ψǫ,r.Lastly, we construct the extremal family of functions that proves sharpness for the

case P = ∇(−∆)α−12 in Theorem 2, as well as sharpness for Corollary 2. Note that

in all these cases α is an integer. We use the same extremal functions as in Adams([A1], see also [FM1], [FM2], [MS2]). Let ϕ ∈ C∞([0, 1]) such that ϕ(k)(0) = 0 for0 ≤ k ≤ α − 1, and ϕ(1) = ϕ′(1) = 1, ϕ(k)(1) = 0 for 2 ≤ k ≤ m− 1. Let ǫ be small

34

enough, define

vǫ(y) =

0 for |y| ≥ 34

ϕ(log 1|y|) for 1

2≤ |y| ≤ 3

4

log 1|y|

for 2ǫ ≤ |y| ≤ 12

log 1ǫ− ϕ(log |y|

ǫ) for ǫ ≤ |y| ≤ 2ǫ

log 1ǫ

for |y| ≤ ǫ.

Then we have that

||vǫ||βq ≤ C, ||∇αvǫ||q′

q =γ(P )

n(log

1

ǫ)q

′−1 +O(1).

Letuǫ =

vǫ||∇αvǫ||q

,

it is clear that

||∇uǫ||q ≤ 1, ||uǫ||qq ≤ C(log

1

ǫ)−1, (9.4)

and

|uǫ|q′ ≥ γ(P )−1 log

1

ǫn, |y| ≤ ǫ.

For the sharpness of the exponential constant, we take θ > 1 and estimate

∫

Rn

exp⌈nα−2⌉

[θγ(P )|uǫ|

nn−α

]

1 + |uǫ|n

n−α

dy ≥

∫

|y|≤ǫ

exp[θγ(P )|uǫ|

nn−α

]

1 + |uǫ|n

n−α

dy ≥ Cǫnexp

[θ log

1

ǫn+ C

]

1 + C log1

ǫ

= Cǫ(1−θ)n

1 + C log1

ǫ

→ ∞

as ǫ→ 0+.For the sharpness of the power of the denominator, we take θ < 1 and get

∫

Rn

exp⌈nα−2⌉

[γ(P )|uǫ|

nn−α

]

1 + |uǫ|θn

n−α

dy ≥

∫

|y|≤ǫ

exp[γ(P )|uǫ|

nn−α

]

1 + |uǫ|θn

n−α

dy ≥ Cǫnexp

[log

1

ǫn+ C

]

1 + C(log1

ǫ)θ

≥ C(log1

ǫ)−θ.

Hence by (9.4) we have that the quotient of the above integral over the norm ||uǫ||qq

goes to infinity as ǫ→ 0+, so the sharpness follows.

35

Appendix: Proof of Lemma 6

Proof of (5.21): Using (A3) we get

–

∫

DJ+1\DJ

|Tfτ(x)|dx ≤C

|DJ+1 \DJ |

∫

Rn

|fτ (y)|

∫

DJ+1\DJ

|x− y|α−ndxdy

≤C

rnJ

∫

Rn

|fτ (y)|rαJdy.

(9.5)

Here the second inequality above is by the straightforward computation:∫

DJ+1\DJ

|x− y|α−ndx

=

∫

|x−y|≤rJ∩(DJ+1\DJ )

|x− y|α−ndx+

∫

|x−y|>rJ∩(DJ+1\DJ )

|x− y|α−ndx

≤

∫

|x|≤rJ

|x|α−ndx+

∫

DJ+1\DJ

rα−nJ dx ≤ CrαJ + Crα−nJ rnJ = CrαJ .

(9.6)

Recall that |Fτ | = τ = |D0|, we have

C

rnJ

∫

Rn

|fτ (y)|rαJdy = Crα−nJ

∫

Rn

|fτ (y)|dy = Crα−nJ

∫

Rn

|f |χFτdy

≤ Crα−nJ |Fτ |1/q′||f ||q ≤ Crα−nJ rn−α0 = C1

(r0rJ

)n−α= C1

(1

eq−1

)J.

(9.7)

Proof of (5.22): First write

TSJ+2f′τ (x)− TSJ+1f

′τ (x1) = TSJ+2f

′τ (x)− TSJ+2f

′τ (x1)− T

(SJ+1 − SJ+2)f

′τ (x1)

so|TSJ+2f

′τ (x)− TSJ+1f

′τ (x1)|

≤ |TSJ+2f′τ (x)− TSJ+2f

′τ (x1)|+ |T

(SJ+1 − SJ+2)f

′τ (x1)|.

(9.8)

Use similar argument as in (5.9), we get

|T(SJ+1 − SJ+2)f

′τ (x1)| =

∣∣∣∣∫

DJ+2\DJ+1

K(x1 − y)f ′τ(y)dy

∣∣∣∣

≤ C

∣∣∣∣∫

DJ+2\DJ+1

|x1 − y|α−nf ′τ (y)dy

∣∣∣∣

≤ Crα−nJ+1

∫

DJ+2\DJ+1

|f ′τ (y)|dy ≤ Crα−nJ+1 r

n−αJ+1αJ+1 ≤ CβJ+1.

(9.9)Next, by the regularity assumption (A3), since x1 ∈ D0, we have for x ∈ DJ+1 \DJ

and y ∈ DcJ+2

|K(x− y)−K(x1 − y)| ≤ C|x− x1|(eq/n)n+1−α|x1 − y|α−n−1.

36

Hence

|TSJ+2f′τ (x)− TSJ+2f

′τ (x1)| =

∣∣∣∣∫

DcJ+2

(K(x− y)−K(x1 − y))f ′τ(y)dy

∣∣∣∣

≤ C|x− x1|

∫

DcJ+2

|f ′τ (y)||x1 − y|α−n−1dy

≤ C|x− x1|βJ+2

(∫

DcJ+2

|x1 − y|−n−n

n−αdy

)n−αn

≤ CrJ+1βJ+2C

rJ+2≤ CβJ+2 ≤ CβJ+1.

(9.10)So we have (5.22) by (9.8)-(9.10).

Proof of (5.23) and (5.24): Let j ∈ 0, 1, ..., J, then

∣∣∣∣ –∫

DJ+1\DJ

T (Sj − Sj+1)f′τ (x)dx

∣∣∣∣ =∣∣∣∣ –∫

DJ+1\DJ

∫

Dj+1\Dj

K(x− y)f ′τ (y)dydx

∣∣∣∣

≤ C –

∫

DJ+1\DJ

∫

Dj+1\Dj

|x− y|α−n|f ′τ (y)|dydx

= C

∫

Dj+1\Dj

|f ′τ (y)| –

∫

DJ+1\DJ

|x− y|α−ndxdy

≤ Crα−nJ

∫

Dj+1\Dj

|f ′τ (y)|dy ≤ Crα−nJ rn−αj αj

= Ceqn(j−J)αj

(9.11)where the second inequality is by (9.6). Therefore,

∣∣∣∣ –∫

DJ+1\DJ

T (S0 − SJ+2)f′τ (x)dx

∣∣∣∣ =∣∣∣∣ –∫

DJ+1\DJ

(J+1∑

j=0

T (Sj − Sj+1)f′τ (x)

)dx

∣∣∣∣

≤

J+1∑

j=0

∣∣∣∣ –∫

DJ+1\DJ

T (Sj − Sj+1)f′τ (x)dx

∣∣∣∣

≤ C

J+1∑

j=0

eqn(j−J)αj ≤ CαJ+1.

(9.12)

37

So we have (5.23). For (5.24), by similar calculations as in (9.11), we have∣∣∣∣ –∫

DJ+1\DJ

T (f ′τχD0)(x)dx

∣∣∣∣ =∣∣∣∣ –∫

DJ+1\DJ

∫

D0

K(x− y)f ′τ(y)dydx

∣∣∣∣

≤ C –

∫

DJ+1\DJ

∫

D0

|x− y|α−n|f ′τ (y)|dydx

= C

∫

D0

|f ′τ(y)| –

∫

Dj+1\Dj

|x− y|α−ndxdy

≤ Crα−nJ

∫

D0

|f ′τ (y)|dy ≤ Crα−nJ rn−α0 α−1

≤ Cα−1.

(9.13)

Acknowledgments. The results presented in this paper are part of the author’sPh.D. dissertation at University of Missouri, Columbia. The author is grateful to heradvisor Carlo Morpurgo for his advice and useful suggestions.

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