Chapter 5 Integral Hardy Inequalities on Homogeneous Groups

In this chapter we discuss the integral form of Hardy inequalities where instead of estimating a function by its gradient, we estimate the integral by the function itself.

where p > 1, b > 0, and f ≥ 0 is a non-negative function. We analyse the weighted versions of such inequalities in the setting of general homogeneous groups. Most of the results of this chapter have been obtained in [RY18a] and here we follow the presentation of this paper.

Two-weight integral Hardy inequalities
Here we discuss the weighted Hardy inequalities in the integral form extending that in (5.1). It turns out that one can actually derive the necessary and sufficient conditions on weights for these inequalities to hold.
Theorem 5.1.1 (Integral Hardy inequalities for p ≤ q). Let G be a homogeneous group of homogeneous dimension Q and let 1 < p ≤ q < ∞. Let φ 1 > 0, φ 2 > 0 be positive functions on G. Then we have the following properties: (1) The inequality G B(0,|x|) f (z)dz q φ 1 (x)dx holds for all f ≥ 0 a.e. on G if and only if

e. on G if and only if
are the smallest constants for which (5.2) and (5.4) hold, then Before we prove this theorem let us give a few comments.
1. In the Abelian case G = (R n , +) and Q = n, if we take p = q > 1, and φ 1 (x) = |B(0, |x|)| −p and ψ 1 (x) = 1 in (5.2), then we have A 1 = (p − 1) −1/p and R n 1 |B(0, |x|)| B(0,|x|) f (z)dz where |B(0, |x|)| is the volume of the ball B(0, |x|). The inequality (5.7) was obtained in [CG95]. 2. Theorem 5.1.1 was obtained in [RY18a] and here we follow the proof from that paper. However, due to the fact that the formulations do not make use of the differential structure, the statement can be actually extended to general metric measure spaces with polar decomposition. More specifically, consider a metric space X with a Borel measure dx allowing for the following polar decomposition at a ∈ X: we assume that there is a locally integrable function λ ∈ L 1 loc such that for all f ∈ L 1 (X) we have for some set Σ ⊂ X with a measure on it denoted by dω, and (r, ω) → a as r → 0. In the case of homogeneous groups such a polar decomposition is given in Proposition 1.2.10.
Let us denote by B(a, r) the ball in X with centre a and radius r, i.e., B(a, r) := {x ∈ X : d(x, a) < r}, where d is the metric on X. Once and for all we will fix some point a ∈ X, and we will write |x| a := d (a, x).
Then the following result was obtained in [RV18] which we record here without proof.
Theorem 5.1.3 (Integral Hardy inequality in metric measure spaces). Let 1 < p ≤ q < ∞ and let s > 0. Let X be a metric measure space with a polar decomposition (5.8) at a. Let u, v > 0 be measurable functions positive a.e. in X such that u ∈ L 1 (X\{a}) and v 1−p ∈ L 1 loc (X). Denote holds for all measurable functions f : X → C if and only if any of the following equivalent conditions hold: Moreover, the constant C for which (5.9) holds and quantities D 1 -D 5 are related by D 1 ≤ C ≤ D 1 (p ) 1 p p 1 q , (5.10) and 3. As such, Theorem 5.1.3 is an extension of (5.1) to the setting of metric measures spaces X with the polar decomposition (5.8): in particular, for p = q and real-valued non-negative measurable f ≥ 0, inequality (5.9) becomes as an extension of (5.1). Indeed, in this case we can take u(x) = 1 where σ is the area of the unit sphere in G with respect to the quasi-norm | · |.
Let us show how Theorem 5.1.3 implies Corollary 5.1.4. If we take a = 0, and the power weights then the inequality (5.9) holds for 1 < p ≤ q < ∞ if and only if where σ is the area of the unit sphere in G with respect to the quasi-norm | · |. For this supremum to be well defined we need to have α + Q < 0 and β(1 − p ) + Q > 0. Consequently, we can calculate which is finite if and only if the power of r is zero. Consequently, Corollary 5.1.4 follows from Theorem 5.1.3.
Proof of Theorem 5.1.1. We will prove Part (1) of the theorem since the proof of Part (2) is similar. The obtained estimates will also show the corresponding part of the statement in Part (3). Thus, let us first show that (5.3) implies (5.2). Using the polar decomposition in Proposition 1.2.10 and denoting r = |x|, we write and using Hölder's inequality, we can estimate (5.14) Let us introduce the following notations: for s, r > 0. Plugging (5.14) into (5.12) we obtain We now recall the following continuous version of the Minkowski inequality (see, e.g., [DHK97, Formula 2.1]): Using this inequality with θ = q/p ≥ 1 in the right-hand side of (5.18), we can estimate (5.20) Let us introduce one more temporary notation T (s) := ℘ s Q−1 (ψ 1 (sy)) 1−p dσ(y).
The next case is the version of Theorem 5.1.1 for the indices p > q: Theorem 5.1.5 (Integral Hardy inequalities for p > q). Let G be a homogeneous group of homogeneous dimension Q and let 1 < q < p < ∞ and 1/δ = 1/q − 1/p. Let φ 3 and φ 4 be positive functions on G. Then we have the following properties: (1) The inequality (5.26) (2) The inequality (5.28) Proof of Theorem 5.1.5. We will prove Part (1) of the theorem since Part (2) is similar. We will denote First we prove that if A 3 < ∞, then we have inequality (5.25). Denote for h ≥ 0 on G to be chosen later. Using the polar decomposition in Proposition 1.2.10 we have the following equalities: Here, using Hölder's inequality with three factors with indices 1 Leaving K 1 as it is, we will estimate K 2 and K 3 . We rewrite K 2 as We want to apply (5.2) to K 2 with indices p = q, and with functions f (x) = (ψ 3 (x)) 1−p h(x) and For that, we will check the condition that holds uniformly for all R > 0. Assuming (5.35) uniformly in R > 0 for a moment, the inequality (5.2) would imply that which is something we will use later. So, let us check (5.35). For this, we denote

Using integration by parts we have
so that (5.36) is confirmed. Next, for K 3 , taking into account and using (5.26), we have Now, plugging (5.32), (5.36) and (5.37) into (5.31), we obtain Let us now show the converse, namely, that (5.25) implies (5.26). For this, we consider a sequence of functions Inserting these functions in the place of f (x) in (5.25), we obtain (5.26), if we take 0 < α k < β k with α k 0 and β k ∞ for k → ∞.

Convolution Hardy inequalities
In this section we discuss integral Hardy inequalities in the convolution form. Such inequalities are particularly useful if we make particular choices of the convolution kernels. For example, by taking the Riesz kernels of hypoelliptic differential operators on graded groups, such inequalities can be used to derive a number of hypoelliptic versions of Hardy inequalities. While this topic falls outside the scope of this book, we refer to [RY18a] for such applications. The inequalities that we will present here have been established in [RY18a] and we follow the proofs there in our exposition.
holds for all x = 0. Then there exists a positive constant holds for all f ∈ L p (G).
The critical case b = Q of Theorem 5.2.1 will be shown in Theorem 5.2.5.
Remark 5.2.2 (Riesz kernels). Let us briefly describe a typical situation when condition (5.38) is satisfied. Without going much into detail, let us assume that R is a positive homogeneous left invariant hypoelliptic differential operator on G of homogeneous degree ν. (5.45) The fractional powers R −a/ν for {a ∈ R, 0 < a < Q} and (I + R) −a/ν for a ∈ R + are called Riesz and Bessel potentials, respectively, and they are well defined, see [FR16,Chapter 4.2]. Let us denote their respective kernels by I a and B a . Then we have the relations for a > 0, where Γ denotes the Gamma function. Consequently, it can be shown (see [FR16,Section 4.3.4]) that for any 0 < a < Q there exists a positive constant holds for all x = 0. Therefore, the Riesz kernel I a gives a typical example of an operator satisfying condition (5.38).
Proof of Theorem 5.2.1. We split the integral in the left-hand side of (5.39) into three parts: First, let us estimate M 1 . We can assume without loss of generality that |·| is a norm (such a norm always exists, see Proposition 1.2.4, Part (2)) since replacing the seminorm by an equivalent one only changes the appearing constants.
Observe that by the reverse triangle inequality and the assumption 2|y| < |x| we have which is |x| < 2|y −1 x|. Taking into account this and that T (1) a (x) is bounded by a radial function which is non-increasing with respect to |x|, we can estimate (5.51) We will now apply Theorem 5.1.1, Part (1), to estimate M 1 . For this we need to check condition (5.3), that is, that To check this, we consider two cases: R ≥ 1 and 0 < R < 1. For R ≥ 1, we can estimate which is uniformly bounded since a Q = 1 p − 1 q + b qQ and (a−Q)q −b+Q = − Qq p = 0. Now let us check the condition (5.52) for 0 < R < 1. Here, taking into account Since |x| 2|y| 4|x| and 2 k |x| < 2 k+1 , we have 2 k−1 |y| < 2 k+2 . As in (5.50), assuming | · | is the norm and using the triangle inequality, we have then by the assumption we have Taking into account these observations and applying Young's inequality in Proposition 1.2.13 with 1 + 1 since (a−Q)pq pq+p−q + Q = bp pq+p−q > 0 and q ≥ p. Now let us estimate M 3 . Without loss of generality, we may assume again that | · | is a norm. Then, similarly to (5.50) we note that 2|x| < |y| implies |y| < 2|y −1 x|. Consequently we can estimate M 3 as We will apply Theorem 5.1.1, Part (2), to estimate M 3 . For this, we need to check that To verify this, we consider two cases: R ≥ 1 and 0 < R < 1. First, for R ≥ 1, using the assumption |T (1) qQ . Now let us check the condition (5.59) for the range 0 < R < 1. In this case, noting that ap − Q < 0 we have For p = q, the condition a Q = 1 p − 1 q + b qQ in Theorem 5.2.1 implies that b = ap, so we are interested in the L p -boundedness of the operator The adjoint, defined by (f, S * a g) = (S a f, g), is given by S * a g := (|x| −a g) * |x| a−Q . We now recall again the Schur test: Assume that the integral operator S has a positive integral kernel, and that there exist a positive function h and constants A p and B p such that Let This condition is true if 0 < a < Q/p. Thus, it follows from Schur's test that Taking into account this and |T (1) Let us now show the critical case b = Q of Theorem 5.2.1.

Theorem 5.2.5 (Critical convolution Hardy inequality). Let G be a homogeneous Lie group of homogeneous dimension Q with a homogeneous quasi-norm | · |.
Let 1 < p < r < ∞ and p < q < (r − 1)p , where 1/p + 1/p = 1. Assume that for a = Q/p we have for some positive C 2 = C 2 (a, Q). Then there exists a positive constant C 1 = C 1 (p, q, r, Q) > 0 such that holds for all f ∈ L p (G).
Remark 5.2.6. We note that compared to the condition (5.38), the decay assumption in (5.65) for large x is stronger. Continuing with the notation of Remark 5.2.2, we observe that the Bessel kernel (5.47) of the operator (I + R) −a/ν for 0 < a < Q satisfies (5.65): there exists a positive constant C = C(Q, a) > 0 such that we have, in particular, We refer to [RY18a] for further details, as well as to the original proof of Theorem 5.2.5 that we follow here.
Proof of Theorem 5.2.5. Let us split the integral in the left-hand side of (5.66) into three parts, We begin by estimating N 1 . Similar to the argument in (5.50), in the region 2|y| < |x| we have which is non-increasing with respect to |x|, then using (5.69) we get We will now apply Theorem 5.1.1, Part (1), to estimate N 1 . For this we have to check the condition (5.3), that is, that holds uniformly for all R > 0. To verify this uniform boundedness, we consider two cases: R ≥ 1 and 0 < R < 1. First, for R ≥ 1, using the second equality in (5.70), we can estimate Next, let us check (5.71) for 0 < R < 1. We split the integral into two terms, We note that the second integral in the right-hand side of (5.73) is finite by the second equality in (5.70). For the first integral, using the first equality in (5.70), we can estimate Combining this with (5.73), we obtain uniformly for all 0 < R < 1. Thus, we have verified (5.71), so that applying Theorem 5.1.1, Part (1), to N 1 we obtain (5.74) Now let us estimate N 2 . We decompose it into the sum Since the function log 1 |x| r |x| Q is non-decreasing with respect to |x| near the origin, there exists an integer k 0 ∈ Z with k 0 −3 such that this function is non-decreasing in |x| ∈ (0, 2 k0+1 ). Fixing this k 0 , we decompose N 2 further as (5.75) where Let us first estimate N 22 . Since |x| 2|y| 4|x| and 2 k |x| < 2 k+1 , we must also have 2 k−1 |y| < 2 k+2 . Before starting to estimate N 22 , using (5.65) and q > p, let us show that wherer ∈ [1, ∞] is such that 1 + 1 q = 1 r + 1 p . Then, (5.76) and Young's inequality in Proposition 1.2.13 with 1 + 1 q = 1 r + 1 p andr ∈ [1, ∞] imply that Next, let us estimate N 21 . As in (5.69), assuming | · | is the norm and using the triangle inequality and |y| 2|x|, we can estimate 3|x| = |x| + 2|x| ≥ |x| + |y| ≥ |y −1 x|. (5.78) Since log 1 |x| r |x| Q is non-decreasing in |x| ∈ (0, 2 k0+1 ) and 3|x| |y −1 x|, we have log 1 |x| Consequently, this and (5.65) yield Since |x| ≤ 2|y| ≤ 4|x| and 2 k ≤ |x| < 2 k+1 with k ≤ k 0 , we must also have 2 k−1 ≤ |y| < 2 k+2 and |y −1 x| ≤ 3|x| < 3 · 2 k0+1 ≤ 3/4, using (5.78) and k 0 ≤ −3. Taking into account these and setting Since p < q < (r − 1)p , we use Young's inequality in Proposition 1.2.13 with 1 + 1 q = 1 r + 1 p andr ∈ [1, ∞), to get (5.79) provided that g ∈ Lr(G). Since Q q + Q p r = Q, rr q = rp p +q and q < (r − 1)p , then changing variables, we obtain Let us estimate N 3 now. Without loss of generality, we may assume again that | · | is the norm. Similarly to (5.69) we obtain |y| < 2|y −1 x| from 2|x| < |y|. Then, we have for N 3 that We will apply Theorem 5.1.1, Part (2), for the required estimate of N 3 . For this we have to check the following condition: ⎛ (5.80) To check this, let us consider the cases: R ≥ 1 and 0 < R < 1. Then, for R ≥ 1 by the second equality in (5.70), we get Now let us check the condition (5.80) for 0 < R < 1. We split the integral into two terms: (5.83) We note that the second integral in the right-hand side of above is finite by the second equality in (5.70). Then, using the first equality in (5.70) we get for the first integral that , and (5.84), and taking into account r > 1 and q < (r − 1)p we obtain that ⎛ (5.85) Thus, we have checked (5.80). Consequently, applying Theorem 5.1.1, Part (2), for the term N 3 , we obtain Finally, a combination of (5.74), (5.86), (5.75), (5.77), (5.79) and (5.68) completes the proof of Theorem 5.2.5.

Hardy-Littlewood-Sobolev inequalities on homogeneous groups
In this section we discuss the Hardy-Littlewood-Sobolev inequality on homogeneous groups. We show that it can be obtained as a simple consequence of the convolution Hardy inequality in Theorem 5.2.1. In fact, the argument implies a little more.
Remark 5.3.2 (Stein-Weiss inequality).  (5.88) and proved that if 1 < p < q < ∞ and 1 q = 1 p + λ − 1, then there is C > 0 such that holds for all f ∈ L p (0, ∞). The N -dimensional analogue of (5.88) can be written by the formula (5.89) Consequently, if was shown by Sobolev in [Sob38] that if 1 < p < q < ∞ and holds for all f ∈ L p (R N ). In [SW58], Stein and Weiss obtained the following two-weight extension of the Hardy-Littlewood-Sobolev inequality, which is nowadays called the Stein-Weiss inequality. More specifically, they have shown that if 0 < λ < N, 1 < p < ∞, α < N (p−1) (5.92) 5. (Differential Stein-Weiss inequality on graded groups). Continuing with the notation of Remark 5.2.2, letL p a (G) be the homogeneous Sobolev space over L p of order a associated to a Rockland operator. Such spaces are well defined on graded groups and do not depend on a particular choice of a Rockland operator, we refer to [FR17] or to [FR16, Section 4.4] for the extensive analysis and exposition of their properties. The following differential version of the Stein-Weiss inequality was obtained in [RY18a]: Theorem 5.3.3 (Differential Stein-Weiss inequality). Let G be a graded group of homogeneous dimension Q and let |·| be a quasi-norm on G. Let 1 < p, q < ∞, 0 ≤ a < Q/p and 0 ≤ b < Q/q. Let 0 < λ < Q, 0 ≤ α < a + Q/p and 0 ≤ β ≤ b be such that (Q − ap)/(pQ) + (Q − q(b − β))/(qQ) + (α + λ)/Q = 2 and α + λ ≤ Q, where 1/p + 1/p = 1. Then there exists a positive constant C = C (Q, λ, p, α, β, a, b) such that holds for all f ∈L p a (G) and g ∈L q b (G). While the setting of graded groups falls outside the scope of this book, we follow [RY18a] in the proof of Theorem 5.3.1 below. as ε → 0 + , where we have used that Note that in (5.96) we can take also the limit as A → +∞ since it is valid for all A > 0. Then, when θ > 1, that is, for p < Q/(Q + λ), taking A → +∞ in (5.96) we see that C Q,λ,p = 0. In the case θ = 1, that is, for p = Q/(Q + λ), taking the limit as A → +∞ in (5.96) we get Finally, we show that the right-hand side of (5.97) goes to zero as R → ∞ if we insert the function for any R > 1. Indeed, in this case p = Q/(Q + λ), and from (5.97) we obtain that as R → ∞, where |℘| is a Q − 1-dimensional surface measure of the unit quasisphere in G.
Theorem 5.4.1 (Maximal integral weighted Hardy inequality). Let G be a homogeneous group of homogeneous dimension Q with a homogeneous quasi-norm | · |. Let φ and ψ be positive functions defined on G. Then there exists a constant C > 0 such that holds for all positive f ≥ 0 if and only if since χ is the cut-off function. Thus, from this, by plugging (5.105) into the lefthand side of (5.104) we calculate which implies (5.101), where we have used |℘| |B(0,1)| = Q, 2R Q s Q − 2R Q Qs Q > 0, and (5.102) in the last two lines.
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