A Fractional Hardy-Sobolev-Maz'ya Inequality on the Upper Halfspace

We prove several Sobolev inequalities, which are then used to establish a fractional Hardy-Sobolev- Maz'ya inequality on the upper halfspace.


Introduction
The present work answers a question by Frank and Seiringer [9] concerning fractional Hardy-Sobolev-Maz'ya inequalities for the upper halfspace in the case p = 2. Let R n + = x = (x ′ , x n ) ∈ R n : x ′ ∈ R n−1 , x n > 0 be the upper halfspace, and let Ω be a domain in R n with nonempty boundary. Then, there exists a fractional Hardy inequality on R n + which states that there exists D n,p,α > 0 so that for all f ∈ C c (R n + ), where 1 ≤ p < ∞, 0 < α < p, and α = 1. See, e.g., [7]. Bogdan and Dyda found, in [3], the sharp constant D n,2,α for the p = 2 case. Later, the sharp constant, D n,p,α , for general p was found in [9]. Therein, the authors posed the question whether there existed a lower bound to the remainder for the inequality in (1) that is a positive multiple of the L p * -norm of f , where p * = np/(n − α) is the critical Sobolev exponent.
Such an inequality would be a fractional analogue to the Hardy-Sobolev-Maz'ya inequality on the halfspace.
Maz'ya [13] was the first to show the general result More recently the existence of minimizers for dimensions greater than or equal to 4 [14] and the sharp constant for dimension 3 [4] has been established. Further improvements in the general case have been shown in [10].
The following Theorem for the sharp fractional hardy inequality with remainder was proven by Frank and Seiringer in [9]. THEOREM 1.1. Let n ≥ 1, 2 ≤ p ≤ ∞, and 0 < α < p with α = 1. Then for all f ∈ C ∞ c (R n + ), where 0 < c p ≤ 1 is given by If p = 2 then this is an equality with c p = 1.
For notational convenience, we write Since we are primarily concerned with the case p = 2, we further denote I Ω α = I Ω α,2 and J Ω α = J Ω α,2 . Thus, for p = 2, we can rewrite (2) as The main result of this paper is the following fractional Hardy-Sobolev-Maz'ya inequality for p = 2.
There exists a n,α > 0 so that Alternatively, we write (4) as J R n + α (f ) ≥ a n,α f 2 2 * , where · p refers to the L p -norm with usual Lebesgue measure.

Sobolev Inequalities
Herein, we establish two Sobolev-type inequalities that we'll need in the proof of Theorem 1.2. We prove each for the more general p-case. The first inequality we prove is for I Ω α,p with respect to convex sets Ω.
The next inequality is a weighted inequality for the term J R n + α,p (f ). We leave the proof to the appendix.
In the next section, we'll show we can minimize over a certain class of functions that are decreasing, albeit not symmetrically. The crux of the proof of Theorem 1.2 in the following section is to decompose this function by truncation and use these two Sobolev inequalities to appropriately bound the L 2 * -norms of the resulting "upper" and "lower" functions.

Class of Minimizing Functions
In this section, we determine the properties of those functions that minimize our Rayleigh quotient To minimize this functional, we'd like to do a rearrangement, but we have the restriction that the function must have support in the upper halfspace. Still, we can rearrange the function along hyperplanes parallel to the boundary of R n + . In addition, we consider the conformal transformation T : Note that T is an involution, and its Jacobian is η(ω) n . See, e.g., Appendix, [6]. We define We use these results throughout the remainder of this paper. These provide a new "ball" picture in which to consider our inequality and minimization problem. Among other things, we can also perform a rotation of f on the ball. It turns out that repeated application of this rotation, along with the rerrangement mentioned above, result in a limiting function that is radial in the ball picture and whose Rayleigh quotient is always smaller than that of the original.
This first result follows from Carlen and Loss [6].
2. F is nonnegative, symmetric decreasing in hyperplanes parallel to the boundary of R n + , 3. F is rotationally symmetric, and we can assume f is nonnegative. Then, the first three items of this theorem are a direct result of [6], Theorem 2.4. Indeed, let U f be the transformation of f obtained by a certain fixed rotation of f , as described in [6]. In particular, using the Note how the last transformation mimics the map f → f . Further, let V be the symmetric decreasing rearrangement in hyperplanes parallel to the boundary of R n + . Define F k := (V U ) k f , then, by Theorem 2.4 in [6], there exists F ∈ L 2 * (R n + ) that is nonnegative and symmetric decreasing in hyperplanes parallel to the boundary of R n + and such that f 2 * = F 2 * , F is radial on the unit ball, and lim k→∞ F k = F in L 2 * (R n + ). By passing to a subsequence, we can assume, without loss of generality, F k → F almost everywhere.
As was calculated in [3], there exists a constant c > 0 so that we can write the remainder term as We By a modification of Theorem 7.17 of [11], Next, as the rearrangement under V is only along hyperplanes parallel to the boundary it is invariant under U as well. Finally, the L 2 * -norm is clearly invariant under U and V .
Therefore, applying Fatou's lemma, As a result, we can explicitly write the limiting function in Theorem 3.1 as the product of two radial functions in the ball picture, a specific, known symmetrically increasing function and a symmetrically decreasing function.

Proof of Main Result
and I R n α (f ), f 2 2 * < ∞. However, we note that f is no longer necessarily in C ∞ c (R n + ). We decompose h = h 1 + h 0 by truncation, by fixing R ∈ (0, 1) so h 0 (r) = min{h(r), h(R)}. Then, f = f 1 + f 0 , with the definitions f 1 , f 0 following from the above. We claim there exists c, d > 0, each dependent on R, n and α, such that and Then, using the triangle and arithmetic-geometric mean inequalities, for all 0 < λ < 1, we obtain Clearly, by fixing λ, R not equal to zero or one, the constant is greater than zero. So, by taking the supremum over λ and R, the result follows.
First, we prove (7). Note that supp h 1 ⊆ [0, R], so supp f 1 ⊆ B R , where B R is as in (5) above. Thus, for all x, y ∈ B R , It is easy to see that for any 0 < R < 1, where A 1 is dependent on R, n and α. Thus, . We claim we can apply Theorem 2.1 to x Using Theorem 2.2, where A 2 is also dependent on R, n and α. Therefore, which proves (7).
In establishing (8), we use the inequality Then, we calculate As 2 * > q, the claim follows.
Finally, we show that we can approximate x almost everywhere. Then, by monotone convergence, as c → 0. Now, supp g c B R and g c ∈ L 2 (B R ), so I R n α (g c ) < ∞ from (6). Denote · W α/2,2 (R n ) = · 2 2 + I R n α (·), and let W α/2,2 0 (R n ) be the completion of C ∞ c (R n ) with respect to · W α/2,2 (R n ) . Then it is known that see, e.g. [1], [2]. Since supp g c B R , there exists a sequence {g j as j → ∞.

Conclusion
Consider the general Hardy-Sobolev-Maz'ya inequality where p ≥ 2, 1 < α < min{n, p}. It is still unknown whether a n,p,α > 0 for p > 2. Still, other than Theorem 3.1, the elements of this paper have either already been proven for general p or extend quite easily. Indeed, Theorem 3.1 cannot be extended because Lemma 6.1 is not true for p = 2. However, for functions that are symmetrically decreasing about a point in the upper halfspace or that can be rerranged about a point in the upper halfspace, while still maintaining support there, and where the Hardy term increases due to the rearrangement, then a fixed a n,p,α > 0 can be found.
Acknowledgement. I am very thankful to Michael Loss for countless valuable discussions and especially for collaboration on Theorem 2.2. This work was partially supported by NSF Grant DMS 0901304.

Appendix
Proof of Theorem 2.2: This proof uses the idea from Theorem 4.3, [11] to write the integral in terms of its layer cake representation . We can assume that f ≥ 0, since We need a few preliminary results. Let 1 Ω be the indicator function on the set Ω, then, for any s ∈ R, These next results follow from the Appendix in [6]. Let t ≥ 0, so t p = p(p − 1) ∞ 0 (t − a) + a p−2 da. Then, letting a ≥ 0, . Then, where d n,p,α is a generic constant, and using the results above, From [11], Theorem 1.13, g q (x) = ∞ 0 qa q−1 1 {g(x)>a} da. Thus, we denote Using the substitution c = λ(b) D 2 2n+α−1 t, and the identity 1 − p q = α+1 2n+α−1 , then Since α−1 2 − q α−1 p = −n + nq p * , we are done.
LEMMA 6.1. Let n ≥ 2, 0 < α < 2, and f ∈ C c (R n ). Then, Proof. First, let us define the set Using the transformation T , as discussed in the remarks prior to Theorem 3.1, and results from [6], We show that the limit on the last line is zero for all ǫ > 0; thus, establishing our result. We write x = (x ′ , x n ), x ′ ∈ R n−1 , x n ∈ R. Then, where the complement in the last integral is with respect to the half line [0, ∞). This product of integrals is zero as the left integral is finite, while the right integral is zero, for all ǫ > 0. Indeed, for the right integral, note that there is no singularity so long as t = 1. If we split the integral into t above and below 1, then the latter must be finite, so the integral is finite if the sum is. We compute In fact, the sum is zero, as (1 + t 2 − 2st) (n+α)/2 .
Lastly, we consider the left integral. From [8], there exists c > 0 so that Hence, if we assume that I R n α (f ) < ∞, then as desired. If I R n α (f ) is not finite, then we need to ask whether R n f 2 (x ′ , x n − 1)|x| −α dx < ∞. If so, then the result still holds. If this is not true, then since R n f 2 (x ′ , x n − 1)|x| −α dx ≤ cI R n α ( f ), then I R n α ( f ) is not finite as well.