Sharp pointwise gradient estimates for Riesz potentials with a bounded density

We establish sharp inequalities for the Riesz potential and its gradient in Rn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^{n}$$\end{document} and indicate their usefulness for potential analysis, moment theory and other applications.


Introduction
Given a measurable function f (x) on R n , its Riesz potential of order 0 ≤ α < n is defined 1 by where d ω x denotes the n-dimensional Lebesgue measure on R n normalized by ω n := π n/2 /Γ( n 2 + 1) being the n-dimensional volume of the unit ball in R n .Let ρ(x), 0 ≤ ρ ≤ 1, be a measurable function with compact support in R n and let be the exponential transform of ρ [9], [10].Then E ρ (y) is the exponential of a Riesz potential of first nonintegrable index (α = 0).If ρ = χ D is a characteristic function of a bounded domain D ⊂ R n , Putinar and Gustafsson [9] established that E χD (x) is superharmonic in the complement of D and it tends to zero at smooth points of the boundary ∂D.It has also been conjectured in [9] that in fact a stronger result holds: for any density ρ(x), the function /n , if n ≥ 3, is subharmonic everywhere outside supp ρ.The conjecture has been settled in the affirmative by the author in [23].A key ingredient in the proof of the subharmonicity was the following sharp inequality.Theorem 1.1 (Corollary 2.2. in [23]).For any density function 0 ≤ ρ(x) ≤ 1, 0 ∈ supp ρ, the inequality holds where M n (t) is the unique solution of the initial problem The inequality (2) is sharp and the equality holds when ρ(x) is the characteristic function of a ball B with a center on the x 1 -axes and 0 ∈ B.
Inequality (2) is remarkable in many respects.First notice that it implies a sharp gradient estimate for the Newtonian potential ( 4) for the Newtonian potential with a bounded density ρ.Since M n (t) < 1, (5) yields a 'truncated version' x ∈ supp ρ.
It is well-known, see for example Proposition 3.1.7 in [3], that for where is the Hardy-Littlewood maximal function of f .As Adams remarks in [2], while the maximal function is an important tool for estimates involving L p measures f , it is not a sharp tool for analyzing their pointwise behaviour.Some generalizations involving the Hardy-Littlewood maximal function for a complex order α ∈ C can be found in [19].This estimate also appears for the Poisson equation and quasilinear equations, see [20], [21], see also [7].In fact, a straightforward application of Cauchy's inequality yields just (7) which is optimal in the class of arbitrary (not uniformly bounded) nonnegative measurable densities ρ and the equality attains asymptotically for a suitable approximation of a δ-function distribution with a single-point support on the x 1 -axes.
In this respect, (2) is a considerable refinement of (7) for uniformly bounded Observe that the sharp inequality (2) has no longer symmetry of Cauchy's inequality, see (7).This symmetry breaking can appropriately be explained in the moment problem context.Namely, (2) can be thought as a natural extension of the Markov inequalities in the L-problem [4], [15] for the critical negative powers.
Recall that given L > 0, the L-problem of moments concerns the existence of a density function 0 ≤ ρ ≤ L with a given sequence of the power moments (9) s k (ρ where I ⊂ R is an arbitrary fixed (finite or infinite) interval.By a celebrated result of A.A. Markov, the solvability of the L-problem is equivalent to the solvability of the corresponding classical moment problem for where dµ(x) is a positive measure, and the correspondence is given by the (onedimensional) exponential transform see [4, p. 72] or [15, p. 243] for more details.[8].Setting L = 1 and I = [0, ∞) in ( 9), the solvability of the corersponding L-moment problem is equivalent to the solvability of the Stieltjes problem for the sequence {a k } k≥0 defined by (10) which is, in its turn, is equivalent to the nonnegativity of the Hankel determinants sequence ∆ m := det(a i+j ) m i,j=0 ≥ 0 and ∆ ′ m := det(a i+j+1 ) m i,j=0 ≥ 0, m ≥ 0. For example, ∆ ′ 1 ≥ 0 readily yields (11) s 2 0 ≤ 12(s 0 s 2 − s 2 1 ).Furthermore, the inversion of the variable x → 1/x in (9) implies a correspondence between the negative power moments s m for m = −2, −3, . . .and the classical ones: This implies the classical Markov inequalities for all power moments s m except for the critical exponent m = −1.
In this respect, in the one-dimensional case (2) provides a novel inequality for power L-moments involving the critical exponent s −1 .Indeed, in the notation of Theorem 1.1 one has M 1 (t) = tanh t and d ω x = 1 2 dx, hence (2) yields for any density function 0 ≤ ρ ≤ 1.In the moment notation this yields a sharp inequality ( 14) Remark that in contrast to the algebraic character of the classical Markov inequalities [5], (14) has a different, transcentental nature.The analogous two-dimensional L-problem is much less explored, recent works point out some direct applications of this problem to tomography, geophysics, the problem in particular has to do with the distribution of pairs of random variables or the logarithmic potential of a planar domain, see [11], [10], [16], [12].

Main results
In this paper we extend (2) on the Riesz potentials of a general index.Then we have for its gradient (15) 1 When y is fixed it is natural to assume that y = 0, hence after a suitable orthogonal transformation of R n the above integrals amount respectively to with a new density function ρ.We are interested in the gradient estimates, i.e. those involving both F α ρ and H α ρ.In this paper, we consider the following variational problem.
where the supremum is taken over all measurable functions 0 ≤ ρ ≤ 1 with support outside of the origin.We refer to such a ρ(x) as an admissible density function.A pair (u, v) ⊂ R 2 ≥0 is said to be admissible for the variational problem ( 16) if there exists an admissible density function ρ such that It is easy to see that N α (u, v) is well-defined and finite for any α and any admissible pair (u, v).Indeed, it follows from Cauchy's inequality that ( 17) We point out, however, that the estimate ( 17) provide a correct approximation only when u and v are infinitesimally small.By virtue of (15), N α (u, v) implies the following pointwise gradient estimate on the Riesz potential I α ρ by means of I α ρ itself and the contiguous potential I α−2 ρ.Corollary 2.2.In the above notation, the following pointwise estimate holds: and the inequality is sharp.
Our main result provides an explicit form of the goal function N α (u, v).
Theorem 2.3.Let n ≥ 1 and 0 < α ≤ 2. Then the set of admissible pairs coincides with the nonnegative quadrant R 2 ≥0 and for any u, v > 0 , We are in particular interested in the shape structute of the goal function N α (u, v), i.e. how it depends on the variables u and v. Combining ( 20) and ( 19) yields the following alternative representation.

Corollary 2.4. Under assumptions of Theorem 2.3,
where ψ(s) well-defined by the parametric representation Some further remarks are in order.The condition α ≤ 2 in Theorem 2.3 is of a technical character.In the complementary case 2 < α < n, the set of admissible pairs is a proper subset of the quadrant R 2 ≥0 .The corresponding analysis requires some more care, and will be done elsewhere.
The borderline case α = 2 corresponds to inequality (2) established in [23] and in the present notation the goal function here becomes We derive also it as an application of Theorem 2.3 in Corollary 6.1 below.Remarkably, that both Cauchy's inequality estimate (17) and its sharp version (22) separate into one-variable factors.This separable form, however, no longer holds for a general α, when the shape of N α has a more complicated structure, see (21).Another interesting case is α = 1.Under this condition, the Riesz potentials in the right hand side of (18) have in fact the same exponent because for α = 1 the contiguous potential amounts to I −1 ρ = I 1 ρ, where ρ is the inversion of ρ.A further remarkable feature of this case is that N 1 (u, v) becomes a symmetric function of u and v. Indeed, a straightforward analysis of (21) implies that for α = 1 the goal function N 1 (u, v) depends only on the product uv.More precisely, we have the following Theorem 2.5.For any measurable function 0 ≤ ρ(x) ≤ 1, 0 ∈ supp ρ, the sharp inequality holds where Φ n (t) is the unique solution of the initial problem subject to the asymptotic condition .
Remark 2.6.Concerning the definition of the shape function Φ n , we note that the initial problem (24) itself does not determine a unique solution because the initial condition Φ ′ n (0) = 1 is singular.One can prove that if φ(x) solves ( 24) then any solution of ( 24) is obtained by the homothetic scaling 1 c φ(cx), c > 0. Note also that all thus obtained solutions have the logarithmic growth at infinity, so a further normalization like (25) is needed.
The proof of Theorem 2.3 relies on a refinement of the technique initiated in [23] and uses the Bathtub principle.We obtain some preliminary results in section 3 and finish the proof in section 5. Then we prove Theorem 2.5 in section 7.
Remark also that the obtained gradient estimates are sharp for Euclidean balls with constant density.The latter symmetry phenomenon is natural for the Riesz and Newton potentials [6], [18], and studied recently in connection with Riesz potential integral equations on exterior domains [22], [14], [13], [24].
We finally mention that our results can also be thought of as an analogue of the polynomial moment inequalities for the singular Riesz potential dµ α (x) = |x| α−n in R n .Then the above functionals are recognized as the lower order polynomial moments: It is natural to speculate what is the natural extension of the Hankel determinant inequalities for dµ α .We pursue this theme elsewhere.

Auxiliary identities for spherical integrals
Let us decompose R n = R×R n−1 such that x = (x 1 , y), where y = (x 2 , . . ., and let D(t) := B(t, 1).We refer to B(τ, σ) as an x 1 -ball.It is easy to see and will be used later that the inversion x → x * = x/|x| 2 acts on x 1 -balls as follows: Let us fix some notation: where f α (t) := F α (t, 1), h α (t) := H α (t, 1).All the introduced functions depend also on the ambient dimension n.
First notice that f α (t) and h α (α) are real analytic functions of t > 0, and for any real α we get the following identity: Further, applying the inversion readily yields by virtue of (26) that Lemma 3.1.For any α ∈ R, the following identities hold: Proof.Let us consider an axillary function λ(x) = (|x| 2 + 1)/2x 1 .Then Notice that the s-level set {x ∈ R n : λ(x) = s} is exactly the boundary sphere ∂D(s), therefore λ foliates the punctured ball D(t) \ {(1, 0)} into the family of spheres {∂D(s) : 1 < s ≤ t}.Applying the co-area formula one obtains from ( 27) and (36) where dA is the (n − 1)-dimensional surface measure on ∂D(s) and d ω A = 1 ωn dA.In particular, (37)

Similarly one finds
Differentiating the obtained identity and applying (37) yields (32).Next, λ| ∂D(s) = s, therefore the outward normal vector ν along the boundary ∂D(s) is found from (35) by and applying Stoke's formula we obtain by virtue of (37) Then (34) follows from f α (1) = 0 and the previous identity by virtue of l'Hospital's rule.
Lemma 3.2.If 0 < γ < 1 then for any p, q there holds that Proof.A straightforward corollary of ( 27) and the Hölder inequality.

The reduced functions via hypergeometric functions
Differentiating the first identity in (31) followed by elimination of and h ′ α by virtue of ( 32) and (33) readily yields the following identity: (39) the equation ( 39) is transformed to the hypergeometric differential equation ( 40) By (34), φ α (1) = 1.Since φ α (z) is regular at z = 1, it follows from a Kummer transformation (see formula 15.5.5 in [1, p. 563]) that The latter representation is useful for the asymptotic behavior of f α at ∞. From (42) we obtain ).In particular, using the Gauss type identity (44) lim we have from (43) the following asymptotic growth: .
Also, applying a linear transformation (formula 15.3.5 in [1, p. 559]) to (43) yields ).A similar argument also works for h α : eliminating f α−2 from (32) by virtue of (31) 1 yields (47) The a substitution h α (t) = t(t 2 − 1) n/2 ψ α (t 2 ) transforms the latter equation into a hypergeometric one: For the same reasons as above, we obtain ).Let us consider some particular cases when f α can be explicitly specified.When n = 1 and α is arbitrary, one easily finds from (27) that and h α (t) = f α−1 (t), where This yields When α = 2, the spherical mean property for harmonic functions |x| 2−n and x 1 |x| −n was used in [23, Sec.2.2] to obtain explicit expressions Another interesting particular case is α = 1, we have by (31) 1 that f −1 = f 1 , and the reduced functions f 1 and h 1 can be determined explicitly at least when n is an odd integer.
Since the latter integral is an increasing function of τ and since τ 0 satisfies (60), we conclude that ρ 0 = χ B(τ,σ0) is a maximizer in (59).This yields Combining the obtained inequalities and using (58) implies (61) Step 2. We claim that N + α (u, v) defined implicitly by ( 61) is an increasing function of each argument separately.It suffices to verify this for an auxiliary function G(u, v) = (N + α (u, v)) α/2 .On eliminating σ 0 by virtue of (56) we obtain (62) . Using (31) and (32) yields for the logarithmic derivatives which implies that g 1 and g 2 are increasing functions of t for any 0 < α ≤ 2. Now suppose that u 1 ≥ u and v 1 ≥ v. Let t 1 be the unique solution of g(t 1 ) = u α−2 1 v α 1 .Since g(t) in ( 57) is increasing we conclude that t 1 ≥ t.First let us consider 1 ≤ α ≤ 2. Then using the first equality in (62) yields Similarly, if 0 < α ≤ 1 then using the second equality in (62) yields The claim follows.
Step 3. Now suppose that ρ is an arbitrary density with a compact support not containing the origin.We claim that N + α (u, v) ≤ N α (u, v).Let U ± and U 0 denote the classes of densities satisfying respectively the conditions (a) supp ρ ∩ R n ∓x1>0 has measure zero; (b) both supp ρ ∩ R n x1>0 and supp ρ ∩ R n x1<0 have nonzero measures.Let N * α (u, v) denote the supremum in ( 16) taken over the corresponding class of densities.Then where ρ 1 is one of ρ + and ρ − , and ).This proves our claim and, thus, finishes the proof of the theorem.
where M n (t) is the unique solution of the initial problem (3).