An optimal Hardy-Littlewood-Sobolev inequality on R n−k × R n and its consequences
aa r X i v : . [ m a t h . F A ] S e p AN OPTIMAL
HARDY–LITTLEWOOD–SOBOLEV INEQUALITY ON R n − k × R n AND ITS CONSEQUENCES
QU ´ˆOC ANH NG ˆO, QUOC-HUNG NGUYEN, AND VAN HOANG NGUYEN
Dedicated to Professor Bidaut-V´eron and Professor V´eron on the occasion of their 70th birthday A BSTRACT . For n > k ≥ , λ > , and p, r > , we establish the following optimal Hardy–Littlewood–Sobolev inequality (cid:12)(cid:12)(cid:12) ZZ R n × R n − k f ( x ) g ( y ) | x − y | λ | y ′′ | β dxdy (cid:12)(cid:12)(cid:12) . k f k L p ( R n − k ) k g k L r ( R n ) with y = ( y ′ , y ′′ ) ∈ R n − k × R k under the two conditions β < ( k − k/r if < λ ≤ n − k,n − λ − k/r if n − k < λ, and n − kn p + 1 r + β + λn = 2 − kn . Remarkably, there is no upper bound for λ , which is quite different from the case with theweight | y | − β , commonly known as Stein–Weiss inequalities. We also show that the abovecondition for β is sharp . Apparently, the above inequality includes the classical Hardy–Littlewood–Sobolev inequality when k = 0 and the HLS inequality on the upper halfspace R n + when k = 1 . In the unweighted case, namely β = 0 , our finding immediatelyleads to the sharp HLS inequality on R n − k × R n with the optimal range < λ < n − k/r, which has not been observed before, even for the case k = 1 . Improvement to the Stein–Weiss inequality in the context of R n − k × R n is also considered. The existence of anoptimal pair for this new inequality is also studied.
1. I
NTRODUCTION
In the existing literature, the classical Hardy–Littlewood–Sobolev inequality on R n ,named after Hardy and Littlewood [ HL28, HL30 ] and Sobolev [
Sob38 ], states that forany n ≥ , p, r > , and λ ∈ (0 , n ) satisfying the balance condition /p + 1 /r + λ/n = 2 , (1.1)there exists a sharp constant N > depending on n , λ , and p such that (cid:12)(cid:12)(cid:12) Z Z R n × R n f ( x ) g ( y ) | x − y | λ dxdy (cid:12)(cid:12)(cid:12) ≤ N k f k L p ( R n ) k g k L r ( R n ) (1.2)for any f ∈ L p ( R n ) and g ∈ L r ( R n ) . The inequality (1.2) is also referred to as the weakform of the classical Young inequality (cid:12)(cid:12)(cid:12) Z R n f ( x )( h ∗ g )( x ) dx (cid:12)(cid:12)(cid:12) . k f k L p ( R n ) k h k L q ( R n ) k g k L r ( R n ) Date : nd Sept, 2020 at 00:49.
Mathematics Subject Classification.
Key words and phrases.
Hardy-Littlewood-Sobolev inequality; Stein-Weiss inequality; sharp constant; opti-mal function. with p, q, r ≥ and /p + 1 /q + 1 /r = 2 , since | x | − λ belongs to the weak space L n/λw ( R n ) ; see [ LL01 , Chapter 4]. Here and inwhat follows, by . and & we mean inequalities up to universal constants such as n , λ , p , r , etc.Although the rough form of (1.2) was proved rather earlier, it took quite a long timeto find the their sharp form until a seminal work of Lieb in 1983; see [ Lie83 ]. Amongother things, Lieb proved the existence of the optimal functions to the inequality (1.2) andcompute the sharp constant N in several special cases.In the last two decades, the sharp HLS inequality (1.2) has captured the attention ofmany mathematicians and many remarkable results have already been drawn. For ex-ample, there are new methods to prove the inequality (1.2) and new arguments to provethe existence of the optimal functions; see [ Lio84, CL92, LL01, FL10, CCL10, FL11,FL12b, DQZ17 ]. In addition, one has the sharp HLS inequalities on the upper half space R n + = R n − × (0 , + ∞ ) in [ DZ15h, Dou16, Glu20 ], on bounded domains in [
GZ19 ],on the Heisenberg group in [
FL12a, HLZ12 ], and on compact Riemannian manifolds in[
HZ15 ]. The interaction between the HLS inequality and other important inequalities hasalso been exploited; see [
Bec93, DJ14, JN14 ].In this work, we look for a possible weighted HLS inequality on R n − k × R n of thefollowing form (cid:12)(cid:12)(cid:12) Z Z R n × R n − k f ( x ) g ( y ) | x − y | λ | y ′′ | β dxdy (cid:12)(cid:12)(cid:12) . k f k L p ( R n − k ) k g k L r ( R n ) , (1.3)where k is a non-negative integer less than n , x ∈ R n − k , y = ( y ′ , y ′′ ) ∈ R n − k × R k , andthe “distance” | x − y | is being understood as follows | x − y | = p | x − y ′ | + | y ′′ | . There is a number of reasons supporting us to work on the weighted HLS inequality (1.3).For clarity, let us just mention a few connection between (1.3) and some known results,while a detailed discussion and interesting consequences will be exploited in subsection 4below. Clearly, the inequality (1.3) with β = 0 , if true, becomes (1.2) if k = 0 . In the case k = 1 , if we let g be such that g ≡ on the lower half space R n − = R n − × ( −∞ , ,then (1.2) becomes (cid:12)(cid:12)(cid:12) Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ dxdy (cid:12)(cid:12)(cid:12) . k f k L p ( R n − ) k g k L r ( R n + ) . (1.4)Inequality (1.4) is known that the HLS inequality on the upper half space R n + first provedby Dou and Zhu in [ DZ15 ] under the balance condition ( n − / ( np ) + 1 /r + λ/n = 2 − /n. (1.5)Relaxing the condition β = 0 gives (cid:12)(cid:12)(cid:12) Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ y βn dxdy (cid:12)(cid:12)(cid:12) . k f k L p ( R n − ) k g k L r ( R n + ) , (1.6)which was proved by Gluck in [ Glu20 ] and Liu in [
Liu20 ] under the technical condition β ≤ and the balance condition ( n − / ( np ) + 1 /r + ( λ + β ) /n = 2 − /n. (1.7)On one hand, the restriction β ≤ in (1.6) seems to be not natural from the validity of theinequality. This indicates that (1.6) could be true for certain β > . On the other hand,since the balance conditions (1.1), (4.3), and (1.7) have a similar form, it is natural to ask N OPTIMAL
HLS INEQUALITY ON R n − k × R n whether or not there is a unification of (1.2), (1.4), and (1.6). In this paper, we aim toaddress these points and we are successful leading to the inequality (1.3) above.That said, in this work, we aim to study (1.3). Toward a complete picture of (1.3),our first step concerns to the validity of (1.3). We summarize this step as the followingtheorem. Theorem 1.1 (weighted HLS inequality on R n − k × R n ) . Let n ≥ , ≤ k < n , λ > , p, r > , and β < k − kr if < λ ≤ n − k,n − λ − kr if n − k < λ, (1.8) satisfying the balance condition n − kn p + 1 r + β + λn = 2 − kn . (1.9) Then there exists a sharp constant N k,βn,λ,p > such that (cid:12)(cid:12)(cid:12) Z Z R n × R n − k f ( x ) g ( y ) | x − y | λ | y ′′ | β dxdy (cid:12)(cid:12)(cid:12) ≤ N k,βn,λ,p k f k L p ( R n − k ) k g k L r ( R n ) (1.10) k,β for any functions f ∈ L p ( R n − k ) and g ∈ L r ( R n ) . Moreover, the two conditions < λ < n and (1.8) are sharp. Before moving on, Theorem 1.1 deserves some comments. First, it is important to notethat there is no upper bound for λ , namely (1.10) k,β holds for all λ > so long as β issuitably small. Next we list a few further comments. • When k = 0 , the weight | y ′′ | − β does not appear, hence the inequality (1.10) ,β becomes the classical HLS inequality (1.2) on R n . • When k = 1 , the inequality (1.10) ,β is essentially the same as that on R n + exceptthe fact that the domain of the double integration is no longer R n + but the wholespace R n . As a matter of fact, (1.10) ,β deals with a larger class of functions. Butthe sharp constant N , n,λ,p and that of (1.4) are related; see subsection 4.2 below. • In all existing works on R n + in the literature, the condition < λ < n − is alwaysassumed. But our optimal inequality (1.10) k,β shows that this is not necessary. Ifwe let β = 0 and k = 1 , then (1.10) , holds for < λ < n − /r but does not if n − /r ≤ λ < n ; see subsection 4.1 below. • An interesting consequence of (1.10) k,β is that it holds for any λ > , not just < λ < n , so long as β < n − λ − k/r ; see subsection 4.3 below. • Although it is not explicitly stated in Theorem 1.1, there is a lower bound for β because it can be easily seen from (1.9). To be more precise, we must have β > − λ , thanks to p, r > . • Our inequality (1.10) ,β remains valid if we replace | y ′′ | − β by | y | − β . But in thisscenario, there are some minor changes including the condition for λ ; see subsec-tion 4.4 below. We leave this for future research.More interesting applications of (1.10) k,β will be discussed in section 4. From now on,we call (1.10) k,β the optimal HLS inequality to highlight the fact that all parameters for(1.10) k,β are in the optimal range.As routine, the proof Theorem 1.1 is carried through two steps; see section 2 below.In the first step, we prove a rough form of (1.10) k,β , namely without the sharp constant N k,βn,λ,p . Then the existence of the sharp constant N k,βn,λ,p is guaranteed through the following Q.A. NG ˆO, Q.H. NGUYEN, AND V.H. NGUYEN variational problem N k,βn,λ,p := sup f ≥ ,g ≥ (cid:8) F βλ,k ( f, g ) : k f k L p ( R n − k ) = 1 , k g k L p ( R n ) = 1 (cid:9) , (1.11)where F βλ,k ( f, g ) = Z Z R n × R n − k f ( x ) g ( y ) | x − y | λ | y ′′ | β dxdy. As we shall soon see, in the present work, we present two different proofs for the roughinequity. These new proofs do not make use the layer cake representation technique northe Marcinkiewicz interpolation inequality. Instead, we borrow some ideas from harmonicanalysis and the theory of maximal functions.Once we establish Theorem 1.1, it is natural to ask whether an optimal pair ( f ♯ , g ♯ ) forthe weighted HLS inequality (1.10) k,β , which consists of non-negative, non-trivial func-tions, actually exists, namely Z Z R n × R n − k f ♯ ( x ) g ♯ ( y ) | x − y | λ | y ′′ | β dxdy = N k,βn,λ,p k f ♯ k L p ( R n − k ) k g ♯ k L r ( R n ) . To this purpose, let us first formally introduce an “extension” operator E βλ,k , which turns afunction f on R n − k to a function on R n via the following rule E βλ,k [ f ]( y ) = Z R n − k f ( x ) dx | x − y | λ | y ′′ | β a.e. y ∈ R n . Using this operator, we may rewrite (1.10) k,β as (cid:12)(cid:12)(cid:12) Z R n ( E βλ,k [ f ])( y ) g ( y ) dy (cid:12)(cid:12)(cid:12) ≤ N k,βn,λ,p k f k L p ( R n − k ) k g k L r ( R n ) . Then, by duality, the HLS inequality (1.10) k,β is equivalent to the following inequality (cid:13)(cid:13)(cid:13) Z R n − k f ( x ) dx | x − ·| λ | · ′′ | β (cid:13)(cid:13)(cid:13) L q ( R n ) ≤ N k,βn,λ,p k f k L p ( R n − k ) (1.12)for any function f ∈ L p ( R n − k ) with q being the number q = 1 − r = n − kn (cid:16) p − n − k − λ − βn − k (cid:17) . (1.13)It is important to note that q > p , see (3.1) below.Similarly, one can consider the “restriction” operator R βλ,k , which maps a function g on R n to a function on R n − k via the following rule R βλ,k [ g ]( x ) = Z R n g ( y ) dy | x − y | λ | y ′′ | β a.e. x ∈ R n − k . Clearly, the two operators E βλ,k and R βλ,k are dual in the sense that for any functions f on R n − k and g on R n , the following identity Z R n ( E βλ,k [ f ])( y ) g ( y ) dy = Z R n − k f ( x )( R βλ,k [ g ])( x ) dx holds, thanks to Tonelli’s theorem. Once we introduce R βλ,k , we can easily see that theweighted HLS inequality (1.10) k,β is also equivalent to the following inequality (cid:13)(cid:13)(cid:13) Z R n g ( y ) dy | · − y | λ | y ′′ | β (cid:13)(cid:13)(cid:13) L q ( R n − k ) ≤ N k,βn,λ,p k g k L r ( R n ) for any function g ∈ L r ( R n ) with q > satisfies q = 1 − p = nn − k (cid:16) r − n − λ − βn (cid:17) . N OPTIMAL
HLS INEQUALITY ON R n − k × R n Now we turn our attention to the existence of optimal pairs ( f ♯ , g ♯ ) for the variationalproblem (1.11). In view of (1.12), to study the existence of optimal pairs for (1.11), westudy the following maximizing problem N k,βn,λ,p := sup f ≥ (cid:8)(cid:13)(cid:13) E βλ,k [ f ] (cid:13)(cid:13) L q ( R n ) : k f k L p ( R n − k ) = 1 (cid:9) . (1.14)Clearly, the two maximizing problems (1.11) and (1.14) are actually equivalent; see section2.1 below. In the next result, we prove that the maximizing problem (1.14) always admitsa solution f ♯ ∈ L p ( R n − k ) , thus giving a solution ( f ♯ , ( E βλ,k [ f ♯ ]) q − ) to the maximizingproblem (1.11). Theorem 1.2 (existence of optimal functions for (1.14)) . Suppose that all conditionsin Theorem 1.1 hold. Let q be given by (1.13) . Then, there exists a function f ♯ ∈ L p ( R n − k ) such that f ♯ ≥ , k f ♯ k L p ( R n − k ) = 1 , and (cid:13)(cid:13) E βλ,k [ f ♯ ] (cid:13)(cid:13) L q ( R n ) = N k,βn,λ,p . Moreover, the function f ♯ is strictly decreasing and radially symmetric with respectto some point in R n − k . We prove Theorem 1.2 in section 3 below. This is done by following Talenti’s proof ofthe sharp Sobolev inequality by considering (1.14) within the set of symmetric decreasingrearrangements. In view of the constraint in the maximizing problem (1.14), if we denoteby f ⋆ the symmetric decreasing rearrangement with respect to R n − k of a function f ∈ L p ( R n − k ) , then on one hand, it is well-known that k f ⋆ k L p ( R n − k ) = k f k L p ( R n − k ) while on the other hand, there holds (cid:13)(cid:13) E βλ,k [ f ] (cid:13)(cid:13) L q ( R n ) ≤ (cid:13)(cid:13) E βλ,k [ f ⋆ ] (cid:13)(cid:13) L q ( R n ) ; see (3.2) below. Hence, it suffices to look for an optimal function within the set of sym-metric decreasing rearrangements.A quick consequence of Theorem 1.2 is the following. Proposition 1.3 (existence of optimal pairs for (1.11)) . Assume all conditions in Theorem1.1. Then, the sharp constant N k,βn,λ,p for the inequality (1.10) k,β is achieved by someoptimal pair ( f ♯ , g ♯ ) ∈ L p ( R n − k ) × L r ( R n ) . The functions f ♯ and g ♯ (cid:12)(cid:12) R n − k are radiallysymmetric with respect to some point in R n − k and monotone decreasing. In a future work, we shall study a reverse HLS inequality on R n − k × R n . The paper isorganized as follows: C ONTENTS
1. Introduction 12. The weighted HLS inequality on R n − k × R n ( f ♯ , g ♯ ) for the HLS inequality 124. Discussions 174.1. The HLS inequality on R n − k × R n with optimal range < λ < n − k/r Q.A. NG ˆO, Q.H. NGUYEN, AND V.H. NGUYEN R n + with extended kernel (1.6) inthe regime < λ < n − and < β < − /r R n − k × R n to R n − k × R n − k +1+ R n − k × R n ( f ♯ , g ♯ ) ℓ , we denote by B ℓR ( x ) the open ball in R ℓ centered at x and radius R , namely B ℓR ( x ) = { ξ ∈ R ℓ : | ξ − x | < R } . For simplicity, we often write B ℓR (0) as B ℓR .2. T HE WEIGHTED
HLS
INEQUALITY ON R n − k × R n In this section, we give a proof of Theorem 1.1, namely to prove (1.10) k,β without thesharp constant N k,βn,λ,p .As mentioned in Introduction, this is equivalent to showing that the supremum in (1.11)is finite. Toward this purpose, we first prove in subsection 2.1 below that the two maxi-mizing problems (1.11) and (1.14) are equivalent. Therefore, to prove the rough inequality(1.10) k,β , it suffices to prove the rough inequality (1.12), which will be done in subsection2.2. Finally, we spend subsection 2.3 to verify the necessity of the two conditions for λ and β .2.1. The equivalence between (1.11) and (1.14).
We now prove that the two maximizingproblems (1.11) and (1.14) are equivalent. Such a result seems to be foreseeable and stan-dard. We provide a short proof for completeness. Denote by N the supremun in (1.14),namely N := sup f ≥ (cid:8)(cid:13)(cid:13) E βλ,k [ f ] (cid:13)(cid:13) L q ( R n ) : k f k L p ( R n − k ) = 1 (cid:9) with q = (1 − /r ) − . (Conventionally, we also write (1 − /r ) − = r ′ .) In the first stepof the proof, we show that N = N k,βn,λ,p . Indeed, let ( f ♯ , g ♯ ) be an optimal pair for (1.11).By definition, there holds k f ♯ k L p ( R n − k ) = k g ♯ k L r ( R n ) = 1 . As /q + 1 /r = 1 , we use H¨older’s inequality to get N k,βn,λ,p = Z R n ( E βλ,k [ f ♯ ])( y ) g ♯ ( y ) dy ≤ (cid:13)(cid:13) E βλ,k [ f ♯ ] (cid:13)(cid:13) L q ( R n ) k g ♯ k L r ( R n ) ≤ N. Hence, we necessarily have N k,βn,λ,p ≤ N . Now let h ♯ be an optimal function for (1.14).Obviously, we must have k h ♯ k L p ( R n − k ) = 1 and k ( E βλ,k [ h ♯ ]) q − k L r ( R n ) = (cid:16) Z R n ( E βλ,k [ h ♯ ]) q ( y ) dy (cid:17) /r = N q − , N OPTIMAL
HLS INEQUALITY ON R n − k × R n thanks to ( q − r = q . Then using (1.10) k,β applied to ( h ♯ , ( E βλ,k [ h ♯ ]) q − ) we obtain N q = Z R n ( E βλ,k [ h ♯ ])( y )( E βλ,k [ h ♯ ]) q − ( y ) dy ≤ N k,βn,λ,p (cid:13)(cid:13) h ♯ (cid:13)(cid:13) L p ( R n ) k ( E βλ,k [ h ♯ ]) q − k L r ( R n ) = N k,βn,λ,p N q − . Hence, we now get N ≤ N k,βn,λ,p . Thus, we have just shown that N k,βn,λ,p = N as claimed.Now, we show that each optimal pair ( f ♯ , g ♯ ) for (1.11) gives rise to an optimal function h ♯ for (1.14) and vice versa. By seeing the above calculation, this fact is quite clear.Obviously, if ( f ♯ , g ♯ ) is an optimal pair for (1.11), then the function f ♯ is also an optimalfunction for (1.14). Conversely, if if h ♯ is an optimal function for (1.14), then the pair ( f ♯ , ( E βλ,k [ f ♯ ]) q − ) is an optimal pair for (1.11).2.2. Proof of the weighted HLS inequality (1.12).
As mentioned in Introduction, toprove (1.10) k,β , it suffices to prove (1.12). As we shall soon see, in the present work,we present two different proofs for the rough inequity (1.12). While the idea of the secondproof stems from harmonic analysis and the theory of maximal functions, see Remark 3.3below, the idea of the first proof, which is presented in this section, demonstrates an in-triguing connection between the weighted and unweighted versions of the HLS inequality;see Lemma 2.2 below.To begin, recall that < k < n , < p, r < + ∞ , λ > , and β satisfies β < k − kr if < λ ≤ n − k,n − λ − kr if n − k < λ. We do not consider the case k = 0 since (1.10) ,β becomes the classical HLS inequality.The balance identity (1.9) can be rewritten as follows p + 1 r + λ − k/q + βn − k = 2 , (2.1)where q is given by (1.13). Now we denote γ = λ − k/q + β. (2.2)As p, r > we deduce from (2.1) that γ > . Now we estimate γ from the above. As β < k/q if λ ≤ n − k , we easily obtain γ < λ ≤ n − k in this range. Now for λ > n − k ,it follows from β < n − λ − k/r that γ < n − k/r − k/q = n − k . Hence, we obtain thefollowing important estimate < γ < n − k. for all λ > . For clarity, we split the proof into several steps. First we start with thefollowing simple observation. Lemma 2.1.
Let h be a non-negative, non-decreasing function. Then we have Z + ∞ (cid:16) ρ γ h ( ρ ) (cid:17) τ dρρ ≤ (2 γ +1) τ (cid:16) Z + ∞ ρ γ h ( ρ ) dρρ (cid:17) τ for any τ ≥ .Proof. To see the inequality, we first decompose the left hand side as follows Z + ∞ (cid:16) ρ γ h ( ρ ) (cid:17) τ dρρ = X j ∈ Z Z j +1 j (cid:16) ρ γ h ( ρ ) (cid:17) τ dρρ Q.A. NG ˆO, Q.H. NGUYEN, AND V.H. NGUYEN ≤ X j ∈ Z Z j +1 j (cid:16) jγ h (2 j +1 ) (cid:17) τ dρ j = X j ∈ Z (cid:16) jγ h (2 j +1 ) (cid:17) τ , thanks to the monotonicity of h . Hence, by changing the index of the sum, we arrive at Z + ∞ (cid:16) ρ γ h ( ρ ) (cid:17) τ dρρ ≤ γτ X j ∈ Z (cid:16) jγ h (2 j ) (cid:17) τ ≤ γτ (cid:16) X j ∈ Z jγ h (2 j ) (cid:17) τ , thanks to τ ≥ . Again by the monotonicity of h , we see that jγ h (2 j ) = 1log 2 12 jγ Z j +1 j h (2 j ) dρρ ≤ γ log 2 Z j +1 j ρ γ h ( ρ ) dρρ . From this we obtain Z + ∞ (cid:16) ρ γ h ( ρ ) (cid:17) τ dρρ ≤ (cid:16) γ log 2 (cid:17) τ (cid:16) Z + ∞ ρ γ h ( ρ ) dρρ (cid:17) τ , giving the first inequality. The proof is complete. (cid:3) Our next step is the key step to prove (1.12). The idea is to transform a weightedinequality to a suitable an unweighted inequality.
Lemma 2.2.
Let n > k ≥ , p, r > , and λ > . Suppose that β satisfies (1.8) . Then forany non-negative function h , we have Z R n (cid:16) Z R n − k h ( x ) | x − y | λ | y ′′ | β dx (cid:17) q dy ≤ (cid:16) λ γ +1 γ λ − γ (cid:17) q | S k − | Z R n − k (cid:16) Z R n − k h ( x ) | z − x | γ dx (cid:17) q dz, (2.3) where γ is given in (2.2) and q is given in (1.13) .Proof. To see this, first we notice that | x − y | λ = λ Z + ∞| x − y | ρ λ dρρ (2.4)with λ > , which, together with Fubini’s theorem applied for the 90-degree cone (cid:8) ( x, t ) : t > | x − y ′ | + | y ′′ | (cid:9) ⊂ R n − k +1+ , implies Z R n − k h ( x ) | x − y | λ | y ′′ | β dx = λ | y ′′ | β Z R n − k (cid:16) Z + ∞| x − y | ρ λ dρρ (cid:17) h ( x ) dx ≤ λ | y ′′ | β Z + ∞| y ′′ | ρ λ (cid:16) Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) dρρ , (2.5)as the above cone is contained in the ungula (cid:8) ( x, t ) : t < | y ′′ | , | x − y ′ | < t (cid:9) ⊂ R n − k +1+ . We still need some work on (2.5). For some ǫ > to be determined later, we applyH¨older’s inequality to get Z + ∞| y ′′ | ρ λ (cid:16) Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) dρρ ≤ h Z + ∞| y ′′ | (cid:16) ρ λ − ǫ Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) q dρρ i /q h Z + ∞| y ′′ | ρ − ǫq ′ dρρ i /q ′ = h ǫq ′ i /q ′ h Z + ∞| y ′′ | (cid:16) ρ λ − ǫ Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) q dρρ i /q h | y ′′ | ǫq ′ i /q ′ (2.6) N OPTIMAL
HLS INEQUALITY ON R n − k × R n with /q + 1 /q ′ = 1 . Hence, combing (2.5) and (2.6) and making use of Fubini’s theoremapplied for the 90-degree cone (cid:8) ( z, t ) : t ≥ | z | (cid:9) ⊂ R k +1+ , the left hand side of (2.3) can be estimated as follows Z R n (cid:16) Z R n − k h ( x ) | x − y | λ | y ′′ | β dx (cid:17) q dy ≤ λ q Z R n − k Z R k (cid:16) Z + ∞| y ′′ | ρ λ (cid:16) Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) dρρ (cid:17) q dy ′′ | y ′′ | βq dy ′ ≤ λ q h ǫq ′ i q/q ′ Z R n − k Z R k h Z + ∞| y ′′ | (cid:16) ρ λ − ǫ Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) q dρρ i dy ′′ | y ′′ | βq + ǫq dy ′ = λ q h ǫq ′ i q/q ′ Z R n − k Z + ∞ Z B kρ (0) (cid:16) ρ λ − ǫ Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) q dy ′′ | y ′′ | βq + ǫq dρρ dy ′ = λ q h ǫq ′ i q/q ′ Z R n − k Z + ∞ (cid:16) ρ λ − ǫ Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) q h Z B kρ (0) dy ′′ | y ′′ | βq + ǫq i dρρ dy ′ . Notice that the condition β < k ( r − /r always holds, which yields βq = β (1 − /r ) −
Having all the preparations above, to conclude (1.12), we simply apply Lemma 2.2 andthe classical HLS inequality (1.2) on R n − k × R n − k , namely (cid:13)(cid:13) E βλ,k [ f ] (cid:13)(cid:13) L q ( R n ) . h Z R n − k (cid:16) Z R n − k f ( x ) | z − x | γ dx (cid:17) q dz i /q (1.2) . (cid:13)(cid:13) f (cid:13)(cid:13) L p ( R n − k ) , thanks to (2.1) and the key estimate < γ < n − k for all λ > .Before closing this part, we prove a reverse version of (2.3), see (2.9) below, which hasits own interest. We do not directly use this result in the proof of (1.10) k,β , but we shalluse it in section 4.1 to consider the HLS inequality on R n − k × R n with the optimal rangefor λ , which is quite remarkable.The strategy of proving (2.9) is similar to that of (2.3), however, instead of using Lemma2.1, we use a technical result from [ PV08 ], see also [
BVNV14 ], which concerns the seriesof equivalent norms concerning Radon measures.
Lemma 2.3.
Let n > k ≥ , p, r > , and λ > . Suppose that β satisfies (1.8) . Then forany non-negative function h we have Z R n − k (cid:16) Z R n − k h ( x ) | z − x | γ dx (cid:17) q dz . Z R n (cid:16) Z R n − k h ( x ) | x − y | λ | y ′′ | β dx (cid:17) q dy, (2.9) where γ is given in (2.2) and q is given in (1.13) .Proof. Our starting point is the equality in (2.5) together with Fubini’s theorem applied forthe 90-degree cone (cid:8) ( x, t ) : t ≥ | x − y ′ | + | y ′′ | (cid:9) ⊂ R n − k +1+ , which helps us to write Z R n − k h ( x ) | x − y | λ | y ′′ | β dx = λ | y ′′ | β Z + ∞| y ′′ | ρ λ (cid:16) Z B n − kρ −| y ′′| ( y ′ ) h ( x ) dx (cid:17) dρρ ≥ λ | y ′′ | β Z + ∞ | y ′′ | ρ λ (cid:16) Z B n − kρ −| y ′′| ( y ′ ) h ( x ) dx (cid:17) dρρ N OPTIMAL
HLS INEQUALITY ON R n − k × R n ≥ λ | y ′′ | β (cid:16) Z B n − k | y ′′| ( y ′ ) h ( x ) dx (cid:17)(cid:16) Z + ∞ | y ′′ | ρ λ dρρ (cid:17) & λ | y ′′ | β + λ (cid:16) Z B n − k | y ′′| ( y ′ ) h ( x ) dx (cid:17) . In the above estimate, the non-decreasing property of the function ρ Z B n − kρ ( y ′ ) h ( x ) dx have used once. Hence, we arrive at Z R n (cid:16) Z R n − k h ( x ) | x − y | λ | y ′′ | β dx (cid:17) q dy ≥ λ q Z R n − k Z R k (cid:16) Z B n − k | y ′′| ( y ′ ) h ( x ) dx (cid:17) q dy ′′ | y ′′ | ( β + λ ) q dy ′ . (2.10)Keep in mind that < λ − k/q + β < n − k . Arguing as in (2.8) and making use of [ PV08 ,Proposition 5.1], we easily get Z R n − k Z R k (cid:16) Z B n − k | y ′′| ( y ′ ) h ( x ) dx (cid:17) q dy ′′ | y ′′ | ( β + λ ) q dy ′ = | S k − | Z R n − k Z + ∞ (cid:16) ρ λ − k/q + β Z B n − kρ ( y ′ ) h ( x ) dx (cid:17) q dρρ dy ′ & Z R n − k (cid:16) Z + ∞ ρ λ − k/q + β h Z B n − kρ ( y ′ ) h ( x ) dx i dρρ (cid:17) q dy ′ = (cid:16) λ − k/q + β (cid:17) q Z R n − k (cid:16) Z R n − k h ( x ) | x − y ′ | λ − k/q + β dx (cid:17) q dy ′ . Combining the previous two estimates gives (2.9) as claimed. (cid:3)
An immediate consequence of Lemmas 2.2 and 2.3 is the following, which is quitesimilar to [
BVHNV15 , estimate (2.15)].
Corollary 2.4.
Let n > k ≥ , p, r > , and < λ < n . Suppose that β satisfies (1.8) .Then for any non-negative function h we have Z R n − k (cid:16) Z R n − k h ( x ) | z − x | γ dx (cid:17) q dz ∼ Z R n (cid:16) Z R n − k h ( x ) | x − y | λ | y ′′ | β dx (cid:17) q dy, where γ is given in (2.2) and q is given in (1.13) . Necessity of the condition (1.8).
We spend this part to discuss the necessity of thecondition β < k − kr if < λ ≤ n − k,n − λ − kr if n − k < λ. The argument performed in this part essentially follows from [
Ngo20 ].First we establish the necessity of the condition β < k ( r − /r regardless of the sizeof λ . In this case, we may take f ≡ χ B n − k the characteristic function of B n − k . Then (cid:13)(cid:13) E βλ,k [ χ B n − k ] (cid:13)(cid:13) qL q ( R n ) ≥ Z B k | y ′′ | βq h Z Z ( B n − k ) dxdy ′ ( | x − y ′ | + | y ′′ | ) λ/ i q dy ′′ & Z B k dy ′′ | y ′′ | βq = + ∞ , as βq ≥ k . Here we also use | x − y ′ | + | y ′′ | ≤ | x | + | y ′ | ) + | y ′′ | ≤ to bound the double integral of | x − y | − λ from below.Notice that the above argument does not cover the range λ > n − k since n − λ − k/r
HLS
INEQUALITY
In this section, we prove the existence of an optimal pair ( f ♯ , g ♯ ) for the optimal HLSinequality (1.10) k,β in the full regime of the parameters. Again, we do not treat the case k = 0 . Recall that n > k ≥ , p, r > , λ > , and β satisfies (1.8). In particular, thereholds < λ − k/q + β < n − k for all λ > . This together with (2.1) helps us to deduce that /p + 1 /r > N OPTIMAL
HLS INEQUALITY ON R n − k × R n for all λ > . Hence, we obtain q := (cid:16) − r (cid:17) − > p, (3.1)which is very important in the proof.In this section, we prove Theorem 1.2 and Proposition 1.3, namely there exists an op-timal function f ♯ to maximizing problem (1.14); see subsection 2.1. This is equivalent toproving that there exists a radially symmetric, strictly decreasing function f ♯ such that (cid:13)(cid:13) E βλ,k [ f ♯ ] (cid:13)(cid:13) L q ( R n ) = N k,βn,λ,p , k f ♯ k L p ( R n − k ) = 1 . This is done within the first three steps of the proof. Finally, to conclude Proposition 1.3and as any optimal function f ♯ for (1.14) gives rise to an optimal pair ( f ♯ , ( E βλ,k [ f ♯ ]) q − ) for (1.11), we shall show that the function E βλ,k [ f ♯ ] is radially symmetric and strictly de-creasing, which is the last step in the proof.Throughout this section, for a function h , we denote by h ⋆ the symmetric decreasingrearrangement of h with respect to the first n − k coordinates; see [ LL01 ] or [
Bur09 ] forthe definition. Now we prove the existence of a non-trivial maximizer f ♯ for the problem(1.14). For the sake of clarity, we divide our proof into several steps. Step 1 . Selecting a suitable minimizing sequence for (1.14) . We start our proof by letting ( f j ) j be a maximizing sequence in L p ( R n − k ) for theproblem (1.14) such that f j is non-negative. Keep in mind that k f j k L p ( R n − k ) = k ( f j ) ⋆ k L p ( R n − k ) Now by using Riesz’s rearrangement inequality, see [
LL01 , chapter 3], H¨older’s inequality,and /q + 1 /r = 1 , we know that (cid:13)(cid:13) E βλ,k [ f j ] (cid:13)(cid:13) L q ( R n ) = sup k h k Lr ( R n ) =1 Z R k | y ′′ | β h Z Z ( R n − k ) f j ( x ) h ( y ′ , y ′′ ) dxdy ′ (cid:0) | x − y ′ | + | y ′′ | (cid:1) λ/ i dy ′′ ≤ sup k h k Lr ( R n ) =1 Z R k | y ′′ | β h Z Z ( R n − k ) ( f j ) ⋆ ( x ) h ⋆ ( y ′ , y ′′ ) dxdy ′ (cid:0) | x − y ′ | + | y ′′ | (cid:1) λ/ i dy ′′ ≤ sup k h k Lr ( R n ) =1 (cid:13)(cid:13) E βλ,k [( f j ) ⋆ ] (cid:13)(cid:13) L q ( R n ) k h ⋆ k L r ( R n ) = (cid:13)(cid:13) E βλ,k [( f j ) ⋆ ] (cid:13)(cid:13) L q ( R n ) . (3.2)Notice that Z R n | h | r dy = Z R k (cid:16) Z R n − k | h | r ( y ′ , y ′′ ) dy ′ (cid:17) dy ′′ = Z R k (cid:16) Z R n − k | h ⋆ | r ( y ′ , y ′′ ) dy ′ (cid:17) dy ′′ = Z R n | h ⋆ | r dy. Hence, as k h k L r ( R n ) = 1 we deduce that k h ⋆ k L r ( R n ) = 1 . Thus (cid:13)(cid:13) E βλ,k [ f j ] (cid:13)(cid:13) L q ( R n ) ≤ (cid:13)(cid:13) E βλ,k [( f j ) ⋆ ] (cid:13)(cid:13) L q ( R n ) . Putting the above two estimates between f j and ( f j ) ⋆ together, we may further assumethat f j is radially symmetric with respect to the origin and non-increasing. By abusing no-tations, we shall write f j ( x ) by f j ( | x | ) or even by f j ( r ) where r = | x | . We can normalize f j in such a way that k f j k L p ( R n − k ) = 1 . From this and the monotonicity of f j , we have | S n − k − | Z ∞ f j ( r ) p r n − k − dr ≥ | S n − k − | n − k f j ( R ) p R n − k for any R > . From this, we obtain the following estimate ≤ f j ( r ) ≤ (cid:16) n − k | S n − k − | (cid:17) /p r − ( n − k ) /p (3.3)for any r > . Step 2 . Existence of a potential maximizer f ♯ for the problem (1.14) . For each non-negative function h on R n − k , we denote k h k ∗ = sup x ∈ R n − k ,ρ> h ρ − n − kp ′ Z B n − kρ ( x ) h ( z ) dz i with p ′ = p/ ( p − . Suppose that h ∈ L p ( R n − k ) . By H¨older’s inequality we have ρ − n − kp ′ Z B n − kρ ( x ) h ( z ) dz . k h k L p ( R n − k ) . for arbitrary x ∈ R n − k and for any ρ > . Hence, by definition we get k h k ∗ . k h k L p ( R n − k ) . To go further, we need an auxiliary result, an analogue of [
Lie83 , Lemma 2.4] concerningthe behavior of (cid:13)(cid:13) E βλ,k [ f ] (cid:13)(cid:13) L q ( R n ) , whose proof is located in Appendix A Lemma 3.1.
Suppose that f ∈ L p ( R n − k ) is non-negative. Then there exists a constant C > , independent of f such that Z R n ( E βλ,k [ f ]) q dy ≤ C k f k q − p ∗ Z R n − k f p dx where q = r/ ( r − > p > . Going back to the maximizing sequence ( f j ) j in L p ( R n − k ) for the problem (1.14), foreach j we set a j = sup r> (cid:2) r n − kp f j ( r ) (cid:3) . In view of (3.3) we know that < a j ≤ (cid:16) n − k | S n − k − | (cid:17) /p for all j . Using the monotonicity of f j , we deduce that Z B n − kρ ( x ) f j ( z ) dz ≤ a j Z B n − kρ ( x ) | z | − n − kp dz ≤ a j Z B n − kρ (0) | z | − n − kp dz = a j ( n − k )(1 − /p ) ρ ( n − k )(1 − p ) for arbitrary x ∈ R n − k and for any ρ > . Consequently, there holds k f j k ∗ ≤ a j ( n − k )(1 − /p ) for all j . Recall from the choice of f j the following k f j k L p ( R n − k ) = 1 , (cid:13)(cid:13) E βλ,k [ f j ] (cid:13)(cid:13) L q ( R n ) → N k,βn,λ,p . Making use of Lemma 3.1 above, we obtain the following estimate ( N k,βn,λ,p ) q ≤ Z R n ( E βλ,k [ f j ]) q dy + o (1) j ր + ∞ N OPTIMAL
HLS INEQUALITY ON R n − k × R n ≤ C k f j k q − p ∗ Z R n − k f pj dx + o (1) j ր + ∞ = C k f j k q − p ∗ + o (1) j ր + ∞ . Keep in mind that q > p . Hence, k f j k ∗ is bounded from below away from zero. Thistogether with k f j k ∗ . a j allows us to assume that a j ≥ c for some c > and for all j .Consequently, for each j , we can choose λ j > in such a way that λ n − kp j f j ( λ j ) > c . Then we set g j ( x ) = λ n − kp j f j ( λ j x ) . From this, it is routine to check that ( g j ) j is also a minimizing sequence for problem(1.14), however, g j (1) > c for any j by our choice for λ j . Consequently, by replacingthe sequence ( f j ) j by the new sequence ( g j ) j , if necessary, we can further assume that oursequence ( f j ) j obeys f j (1) > c for any j. Similar to Lieb’s argument in [
Lie83 ], which is based on Helly’s theorem, by passingto a subsequence, we have f j → f ♯ a.e. in R n − k . It is now evident that f ♯ is non-negative, radially symmetric, non-increasing, and is in L p ( R n − k ) . Of course, there holds f ♯ . The rest of our arguments is to show that f ♯ isindeed the desired minimizer for (1.14). Step 3 . The function f ♯ is a maximizer for (1.14) . Recall that ( f j ) j is a minimizing sequence for the problem (1.14) and f j → f ♯ a.e. in R n − k . The limit function f ♯ satisfies k f ♯ k L p ( R n − k ) > because f j ( x ) > c for any j and all | x | ≤ . To go further, we need the following auxiliary result. Lemma 3.2.
Suppose that ( f j ) j is a sequence of non-negative functions satisfying f j ( x ) ≤ C | x | − n − kp for all x ∈ R n − k and for some C > . Then, if f j → f ♯ a.e. in R n − k , then we have E βλ,k [ f j ]( y ) → E βλ,k [ f ♯ ]( y ) for almost every y ∈ R n . Lemma 3.2 above simply follows from the dominated convergence theorem. It is worthnoting that in order to apply the dominated convergence theorem, we make use of theinequality λ + ( n − k ) /p > n − k, which always holds true under our assumption (1.9). Hence, we omit the details and itsproof is left for interested readers.Using Lemma 3.2 above, we further know that E βλ,k [ f j ] → E βλ,k [ f ♯ ] for a.e. in R n .The rest of the proof is more or less standard; see [ Lie83 , Lemma 2.7]. Applying theBrezis–Lieb lemma to get Z R n − k (cid:12)(cid:12) | f j | p − | f ♯ | p − | f j − f ♯ | p (cid:12)(cid:12) dx → as j ր + ∞ . So, one one hand we have k f j − f ♯ k pL p ( R n − k ) = 1 − k f ♯ k pL p ( R n − k ) + o (1) j ր + ∞ . (3.4) However, on the other hand, we can estimate ( N k,βn,λ,p ) q + o (1) j ր + ∞ = (cid:13)(cid:13) E βλ,k [ f j ] (cid:13)(cid:13) qL q ( R n ) = (cid:13)(cid:13) E βλ,k [ f ♯ ] (cid:13)(cid:13) qL q ( R n ) + (cid:13)(cid:13) E βλ,k [ f j − f ♯ ] (cid:13)(cid:13) qL q ( R n ) + o (1) j ր + ∞ ≤ ( N k,βn,λ,p ) q (cid:2) k f ♯ k qL p ( R n ) + k f j − f ♯ k qL p ( R n ) (cid:3) + o (1) j ր + ∞ . Thus, dividing both sides of the preceding computation by ( N k,βn,λ,p ) q gives ≤ k f ♯ k qL p ( R n − k ) + k f j − f ♯ k qL p ( R n − k ) + o (1) j ր + ∞ . (3.5)Combining (3.4) and (3.5) and sending j ր + ∞ , we arrive at ≤ k f ♯ k qL p ( R n − k ) + (cid:0) − k f ♯ k pL p ( R n − k ) (cid:1) q/p . From the fact that q > p seeing (3.1) and that k f ♯ k L p ( R n − k ) > , we deduce that k f ♯ k L p ( R n − k ) = 1 . This shows that f ♯ is a minimizer for (1.14); hence finishing the proof of Step 3. Step 4 . The function E βλ,k [ f ♯ ] has two symmetries in y ′ and y ′′ and is strictly decreasing in y ′ This step is for the proof of Proposition 1.3. We show that E βλ,k [ f ♯ ] of variable y hastwo symmetries in y ′ and y ′′ . While the symmetry with respect to y ′′ is obvious fromthe definition of E βλ,k [ f ♯ , the symmetry with respect to y ′ is also clear since E βλ,k [ f ♯ ] isessentially the convolution of two radially symmetric functions f ♯ and ( | · | + | y ′′ | ) − λ/ ;see [ Lie83 , Lemma 2.2(i)]. Since the argument is simple and short, we provide a proof forcompleteness. Indeed, let A ∈ O ( n − k ) be arbitrary. Then ( E βλ,k [ f ♯ ])( Ay ′ , y ′′ ) = 1 | y ′′ | β Z R n − k f ♯ ( x )( | x − Ay ′ | + | y ′′ | ) λ/ dx = 1 | y ′′ | β Z R n − k f ♯ ( x )( | A ( A t x − y ′ ) | + | y ′′ | ) λ/ dx = 1 | y ′′ | β Z R n − k f ♯ ( A t x )( | A t x − y ′ | + | y ′′ | ) λ/ dx = 1 | y ′′ | β Z R n − k f ♯ ( x )( | x − y ′ | + | y ′′ | ) λ/ | det A | dx = ( E βλ,k [ f ♯ ])( y ′ , y ′′ ) , where A t is the transpose of A . Finally, the monotonicity of E βλ,k [ f ♯ ] in y ′ follows from[ Lie83 , Lemma 2.2(ii)].Notice that the monotonicity of E βλ,k [ f ♯ ] in y ′ can also be derived from a general resultof Anderson applied to the function h = ( | · | + | y ′′ | ) − λ/ f ♯ ( · + y ′ ) for y fixed; see[ And55 , Theorem 1]. This is because ( E βλ,k [ f ♯ ])( τ y ′ , y ′′ ) ≥ Z R n − k f ♯ ( x + (1 − τ ) y ′ )( | x − τ y ′ | + | y ′′ | ) λ/ dx = Z R n − k h ( x − τ y ′ ) dx ≥ Z R n − k h ( x − y ′ ) dx = ( E βλ,k [ f ♯ ])( y ′ , y ′′ ) N OPTIMAL
HLS INEQUALITY ON R n − k × R n for any ≤ τ ≤ . Here the radial symmetry and monotonicity of f ♯ are crucial to get f ♯ ( x ) ≥ f ♯ ( x + (1 − τ ) y ′ ) . If β ≥ , then the monotonicity of E βλ,k [ f ♯ ] in y ′′ is clear. But it is not clear if this stillholds when β < .Before closing this section, we have the following remark. Remark . As k f ∗ k . k f k L p ( R n − k ) , Lemma 3.1 gives us another proof of the roughHLS inequality (1.10) k,β . 4. D ISCUSSIONS
This section is devoted to a number of discussion and application from simple to com-plex around our main inequality (1.10) k,β .4.1.
The HLS inequality on R n − k × R n with optimal range < λ < n − k/r . Westart this section with a quite surprise application of Theorem 1.1. To be more precise, with β = 0 , which is possible because λ < n − k/r , we obtain from Theorem 1.1 the followingoptimal result. Theorem 4.1 (optimal HLS inequality on R n − k × R n ) . Let n ≥ , ≤ k < n , p, r > , and λ ∈ (0 , n − k/r ) satisfying the balance condition n − kn p + 1 r + λn = 2 − kn . Then there exists a sharp constant N k, n,λ,p > such that (cid:12)(cid:12)(cid:12) Z Z R n × R n − k f ( x ) g ( y ) | x − y | λ dxdy (cid:12)(cid:12)(cid:12) ≤ N k, n,λ,p k f k L p ( R n − k ) k g k L r ( R n ) (4.1) for any functions f ∈ L p ( R n − k ) and g ∈ L r ( R n ) . For arbitrary r > , Theorem 4.1 is optimal in the sense that it does not hold if λ ≥ n − k/r by seeing (1.8). However, if we fix < λ < n , then resolving the inequality λ < n − k/r gives r > max n , kn − λ o . This condition tells us that the closer to n the parameter λ is, the bigger r is. In the specialcase k = 1 , Theorem 4.1 helps us to revisit the HLS inequality (1.4) on the upper halfspace R n + with the optimal range < λ < n − /r. This improves the result of Dou and Zhu in [
DZ15 ], which is stated for < λ < n − .4.2. An improvement of the HLS inequality on R n + with extended kernel (1.6) in theregime < λ < n − and < β < − /r . Our motivation of working on this problemalso comes from the fact that the HLS inequality (1.6) on R n + with extended kernel is“weaker” than the HLS inequality (1.4) on R n + . Here by “weaker” we mean we can use(1.4) to derive (1.6). Indeed, as β ≤ , we clearly have y − βn ≤ | x − y | − β , giving Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ y βn dxdy ≤ Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ + β dxdy, where, for simplicity, all the functions f and g are being non-negative. Notice that thebalance condition (1.7) allows us to apply (1.4) with λ replaced by λ + β . To be moreprecise, fixing any f ∈ L p ( ∂ R n + ) and any g ∈ L r ( R n + ) , we have Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ y βn dxdy ≤ Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ + β dxdy (1.4) . k f k L p ( ∂ R n + ) k g k L r ( R n + ) , provided (1.7) and β ≤ hold. This explains why (1.6) is weaker than (1.4).From the above discussion, it is natural to ask if (1.6) still holds for suitable β > .If there is such an inequality, then it implies that we will have a “stronger” version of theHLS inequality (1.4) on R n + in the following sense Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ dxdy ≤ Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ − β y βn dxdy ? . k f k L p ( ∂ R n + ) k g k L r ( R n + ) , so long as (1.4) holds.To be able to compare, we limit ourselves to the case < λ < n − , hence β < − /r .Clearly, under the above setting, our inequality (1.10) ,β becomes (cid:12)(cid:12)(cid:12) Z Z R n × R n − f ( x ) g ( y ) | x − y | λ | y n | β dxdy (cid:12)(cid:12)(cid:12) ≤ N ,βn,λ,p k f k L p ( R n − ) k g k L r ( R n ) with < λ < n − and β < − /r ; hence providing us an improvement of (1.4) for β ,if we let g | R n − ≡ .In subsection 4.3 below, we show that there is another way, which is quite intriguing,to obtain (1.6) directly from (1.10) k,β without assuming k = 1 . Moreover, it is quiteinteresting to note that from the argument leading to (1.10) k,β , we can relate the sharpconstant N k,βn,λ,p and that of (1.4).4.3. From R n − k × R n to R n − k × R n − k +1+ . An other idea to improve (1.6) for possible β > is to transform (1.10) k,β into (1.6). To fix the idea and for simplicity, we still limitourselves to the case < λ < n − k , hence < β < k (1 − /r ) .To transform (1.10) k,β into (1.6), we simply make use of (1.10) k,β for function g beingradially symmetric in the last k coordinates, namely g ( y ′ , y ′′ ) = g ( y ′ , | y ′′ | ) . Observe that Z R n (cid:0) g ( y ) (cid:1) r dy = | S k − | Z R n − k Z + ∞ (cid:0) g ( y ′ , ρ ) (cid:1) r ρ k − dρdy ′ . Hence, by setting G ( y ′ , | y ′′ | ) = | S k − | /r g ( y ′ , y ′′ ) | y ′′ | ( k − /r , on one hand, it is easy to verify that Z R n g r ( y ) dy = Z R n − k +1+ G r ( z, ρ ) dzdρ. On the other hand, by using R R k = | S k − | R + ∞ , we get F ,βλ,k ( f, g ) = | S k − | Z R n − k Z R n − k Z + ∞ f ( x ) g ( y ′ , ρ ) (cid:2) | x − y ′ | + ρ (cid:3) λ/ ρ β ρ k − dρdy ′ dx N OPTIMAL
HLS INEQUALITY ON R n − k × R n = | S k − | − /r Z R n − k Z R n − k +1+ f ( x ) G ( y ′ , ρ ) (cid:2) | x − y ′ | + ρ (cid:3) λ/ ρ β − ( k − − /r ) dρdy ′ dx. Hence, the inequality (1.10) k,β for ( f, g ) becomes the following inequality for ( f, G ) Z R n − k Z R n − k +1+ f ( x ) G ( y ′ , ρ ) (cid:2) | x − y ′ | + ρ (cid:3) λ/ ρ b β dρdy ′ dx ≤ | S k − | /r − N k,βn,λ,p k f k L p ( R n − k ) k G k L r ( R n − k +1+ ) (4.2)with b β = β − ( k − (cid:0) − /r (cid:1) < − /r. Obviously, if ( f ♯ , g ♯ ) is an optimal pair for (1.10) k,β , then ( f ♯ , G ♯ ) is also an optimalpair for (4.2). However, due to the above transformation, the regularity for ( f ♯ , g ♯ ) and ( f ♯ , G ♯ ) are quite different. This could shed some light on the problem of classification ofall optimal pairs for the inequality on the upper half space.Nevertheless, from the above derivation and the existence of an optimal pair ( f ♯ , g ♯ ) for(1.10), we obtain the following improvement of (1.6), which is optimal. Theorem 4.2.
Let n ≥ , λ ∈ (0 , n − , β < ( r − /r , and p, r > satisfying thebalance condition n − n p + 1 r + β + λn = 2 − n . (4.3) Then there exists a sharp constant P βn,λ,p > such that (cid:12)(cid:12)(cid:12) Z Z R n + × R n − f ( x ) g ( y ) | x − y | λ y βn dxdy (cid:12)(cid:12)(cid:12) ≤ P βn,λ,p k f k L p ( R n − ) k g k L r ( R n + ) (4.4) β for any functions f ∈ L p ( R n − ) and g ∈ L r ( R n + ) . In the last part of this subsection, we show that the transformed inequality (4.2) revealsan connection between the two sharp constants P βn,λ,p and N k,βn,λ,p . We turn this observationinto a proposition as follows. Proposition 4.3.
There holds P βn − k +1 ,λ,p = | S k − | /r − N k,β +( k − − /r ) n,λ,p . In particular, with k = 1 we have P βn,λ,p = 2 /r − N ,βn,λ,p , relating the sharp constant of the HLS inequalities on R n − × R n + and on R n − × R n .Proof. We apply (4.2) for ( f ♯ , G ♯ ) being an optimal pair for (4.2) to get P b βn − k +1 ,λ,p k f ♯ k L p ( R n − k ) k G ♯ k L r ( R n − k +1+ ) ≥ Z R n − k Z R n − k +1+ f ♯ ( x ) G ♯ ( y ′ , ρ ) (cid:2) | x − y ′ | + ρ (cid:3) λ/ ρ b β dρdy ′ dx = | S k − | /r − N k,βn,λ,p k f ♯ k L p ( R n − k ) k G ♯ k L r ( R n − k +1+ ) . From this we obtain P b βn − k +1 ,λ,p ≥ | S k − | /r − N k,βn,λ,p . Now we use (4.4) b β with ( f ♯ , g ♯ ) being an optimal pair for (4.4) b β to get P b βn − k +1 ,λ,p k f ♯ k L p ( R n − k ) k g ♯ k L r ( R n − k +1+ ) = Z R n − k Z R n − k +1+ f ♯ ( x ) g ♯ ( y ′ , ρ ) (cid:2) | x − y ′ | + ρ (cid:3) λ/ ρ b β dρdy ′ dx ≤ | S k − | /r − N k,βn,λ,p k f ♯ k L p ( R n − k ) k g ♯ k L r ( R n − k +1+ ) . Hence we get P b βn − k +1 ,λ,p ≤ | S k − | /r − N k,βn,λ,p . The proof follows by putting the above estimates together. (cid:3)
An Stein–Weiss type inequality on R n − k × R n . In the literature, another weightedversion of the inequality (1.2), or the doubly weighted HSL inequality, also known asthe SW inequality, named after Stein and Weiss, was also obtained in [
SW58 ]. Roughlyspeaking, for < λ < n and for suitable α, β satisfying α < n ( p − /p, β < n ( r − /r, α + β ≥ , the following rough inequality holds (cid:12)(cid:12)(cid:12) Z Z R n × R n f ( x ) g ( y ) | x | α | x − y | λ | y | β dxdy (cid:12)(cid:12)(cid:12) . k f k L p ( R n ) k g k L r ( R n ) (4.5)for any f ∈ L p ( R n ) and any g ∈ L r ( R n ) together with /p + 1 /r ≥ and the new balance condition /p + 1 /r + ( α + β + λ ) /n = 2 . Inequality (4.5) was extended by Dou to the case of R n + ; see [ Dou16 ]. To be more precise,under the condition < λ < n − the following sharp inequality was proved (cid:12)(cid:12)(cid:12) Z Z R n + × R n − f ( x ) g ( y ) | x | α | x − y | λ | y | β dxdy (cid:12)(cid:12)(cid:12) . k f k L p ( R n − ) k g k L r ( R n + ) (4.6)for any f ∈ L p ( R n − ) and any g ∈ L r ( R n + ) together with α < ( n − p − /p, β < n ( r − /r, α + β ≥ , and /p + 1 /r ≥ and the new balance condition ( n − / ( np ) + 1 /r + ( α + β + λ ) /n = 2 − /n. In the context of R n − k × R n , it is expected that the following inequality holds (cid:12)(cid:12)(cid:12) Z Z R n × R n − k f ( x ) g ( y ) | x | α | x − y | λ | y | β dxdy (cid:12)(cid:12)(cid:12) ≤ M k,α,βn,λ,p k f k L p ( R n − k ) k g k L r ( R n ) (4.7)for any functions f ∈ L p ( R n − k ) and g ∈ L r ( R n ) under the following balance condition n − kn p + 1 r + α + β + λn = 2 − kn . (4.8)We do not prove (4.7) and leave it for future research. Instead, our aim is to replace theweight | y | − β by the weight | y ′′ | − β . In light of the discussion in subsection 4.2, this givesan improvement of (4.7) in the regime β ≥ . N OPTIMAL
HLS INEQUALITY ON R n − k × R n Interestingly, our method of proving Theorem 1.1 is flexible enough to obtain a similarresult. To be more precise, we easily obtain the following SW type inequality on R n − k × R n with < λ < n − k . Theorem 4.4 (SW type inequality on R n − k × R n ) . Let n ≥ , ≤ k < n , p, r > , < λ < n − k , ≤ α < ( n − k )( p − /p , and β < k ( r − /r satisfying /p + 1 /r ≥ and the balance condition (4.8) . Then there exists a sharp constant M k,α,βn,λ,p > suchthat (cid:12)(cid:12)(cid:12) Z Z R n × R n − k f ( x ) g ( y ) | x | α | x − y | λ | y ′′ | β dxdy (cid:12)(cid:12)(cid:12) ≤ M k,α,βn,λ,p k f k L p ( R n − k ) k g k L r ( R n ) (4.9) k,α,β for any functions f ∈ L p ( R n − k ) and g ∈ L r ( R n ) .Proof. Notice that the balance condition (4.8) can be rewritten as follows p + 1 r + λ − k/q + β + αn − k = 2 . Now we can use Lemma 2.2 with the function h ( x ) = | x | − α f ( x ) and the classical SWinequality (4.5) with β = 0 to get Z R n (cid:16) Z R n − k f ( x ) | x | α | x − y | λ | y ′′ | β dx (cid:17) q dy (2.3) . Z R n − k (cid:16) Z R n − k f ( x ) | x | α | x − y | λ − k/q + β dx (cid:17) q dy (4.5) . k f k L p ( R n − k ) with q = r/ ( r − . Notice that in the regime < λ < n − k , to be able to apply (2.3), werequire β < k ( r − /r, < λ − k/q + β < n − k. The latter inequality holds thanks to λ > , β < k/q , and α ≥ . Similarly, to be able toapply (4.5) we need ≤ α ≤ ( n − k )( p − /p, /p + 1 /r ≥ . By duality, we obtain (4.9) k,α,β as expected. (cid:3)
The requirement /p + 1 /r ≥ is necessary for Theorem 4.4. In the case α = 0 , it isautomatically satisfied thanks to λ − k/q + β ≤ n − k . Apparently, Theorem 4.4 can beimproved for any λ > , but we do not consider this case here and also leave it for futureresearch.Using the computation in subsection 4.3 above, it is not hard to see that our SW typeinequality (4.9) k,α,β also leads us to Z Z R n − k × R n − k +1+ f ( x ) G ( y ′ , ρ ) dρdy ′ dx | x | α (cid:2) | x − y ′ | + ρ (cid:3) λ/ ρ b β . k f k L p ( R n − k ) k G k L r ( R n − k +1+ ) , (4.10)which is an analogue of (4.6). Clearly, (4.10) is also an improvement of (4.6) if b β ≥ .4.5. Characterization of any optimal pair ( f ♯ , g ♯ ) . We now turn our attention to anoptimal pair ( f ♯ , g ♯ ) for the maximizing problem (1.11) found by Proposition 1.3 above.To gain further properties on ( f ♯ , g ♯ ) , it is often to study the Euler–Lagrange equation for N k,βn,λ,p . Let us compute the Euler–Lagrange equation for F λ,k together with the constraint k f k L p ( R n − k ) = k g k L r ( R n ) = 1 . Clearly, with respect to the function f , the first variation of the functional F λ,k is D f ( F λ,k )( f, g )( h ) = Z R n − k (cid:16) Z R n g ( y ) | x − y | λ | y ′′ | β dy (cid:17) h ( x ) dx while the first variation of the constraint R R n − k | f ( x ) | p dx = 1 is p Z R n − k | f ( x ) | p − f ( x ) h ( x ) dx. Therefore, by the Lagrange multiplier theorem, there exists some constant α such that Z R n − k (cid:16) Z R n g ( y ) | x − y | λ | y ′′ | β dy (cid:17) h ( x ) dx = α Z R n − k | f ( x ) | p − f ( x ) h ( x ) dx holds for all function h defined in R n − k . From this we know that f and g must satisfy thefollowing equation α | f ( x ) | p − f ( x ) = Z R n g ( y ) | x − y | λ | y ′′ | β dy. Interchanging the role of f and g , we also know that f and g must fulfill the following β | g ( y ) | r − g ( y ) = Z R n − k f ( x ) | x − y | λ | y ′′ | β dx for some new constant β . Using a suitable test function, we know that α = β = F λ,k ( f, g ) .Hence, up to a constant multiple and simply using the following changes u = f p − and v = g r − , the two relations above lead us to the following system of integral equations u ( x ) = Z R n v κ ( y ) | x − y | λ | y ′′ | β dy,v ( y ) = Z R n − k u θ ( x ) | x − y | λ | y ′′ | β dx, . (4.11) k,β with κ = 1 r − > , θ = 1 p − > . Using the balance condition (1.9), it is not hard to see that κ and θ fulfill n − kn θ + 1 + 1 κ + 1 = λ + βn . (4.12)In this sense, the condition (4.12) usually refers to the critical condition for (4.11) k,β . Fromthe above derivation, any optimal pair ( f ♯ , g ♯ ) for the weighted HLS inequality (1.10) k,β must solve the system (4.11) k,β . Hence, we can naively ask the following: • a regularity result for solutions to (4.11) k,β with λ > and • a classification for solutions to (4.11) k,β with λ > .As far as we know, there is no such a result for the above questions, even in the case k = 1 .A CKNOWLEDGMENTS
This project initiated when the authors were visiting ICTP in 2017, whose support isgreatly acknowledged. QAN is supported by the Tosio Kato Fellowship awarded in 2018.QHN is supported by the ShanghaiTech University startup fund. VHN would like to thankVIASM for support during the finalizing of the paper. N OPTIMAL
HLS INEQUALITY ON R n − k × R n A PPENDIX
A. P
ROOF OF L EMMA Z R n ( E βλ,k [ f ]) q ( y ) dy . k f k q − p ∗ Z R n − k f p ( x ) dx for any f ∈ L p ( R n − k ) , which can be assumed to be non-negative. As mentioned earlier,Lemma 3.1 also provides us another way to prove (1.10) k,β . To prove Lemma 3.1, wemimic the proof of Adams for Riesz potentials; see [ Ada75 ]. Recall from the definition of E βλ,k [ f ] the following E βλ,k [ f ]( y ) = Z R n − k f ( x ) dx | x − y | λ | y ′′ | β . Hence, our starting point is the following (cid:13)(cid:13) E βλ,k [ f ] (cid:13)(cid:13) qL q ( R n ) . Z R n − k Z + ∞ " R B n − kρ ( y ′ ) f ( x ) dxρ λ − k/q + β q dρρ dy ′ , (A.1)thanks to (2.7). In the sequel, we prove that Z R n − k Z + ∞ " R B n − kρ ( y ′ ) f ( x ) dxρ λ − k/p + β q dρρ dy ′ . k f k q − p ∗ k f k pL p ( R n − k ) . To this purpose, recall the following estimate Z B n − kρ ( y ′ ) f ( x ) dx . ρ n − k ( M f )( y ′ ) , where M f is the Hardy–Littlewood maximal function of f ≥ in R n − k , defined by M f ( z ) = sup r> | B n − k ( z, r ) | Z | x − z |≤ r f ( x ) dx. In addition, from the definition, we also have Z B n − kρ ( y ′ ) f ( x ) dx . ρ n − kp ′ k f k ∗ for any ρ > and any y ′ ∈ R n − k . Thus, for some δ > to be determined later, we canestimate Z + ∞ h R B n − kρ ( y ′ ) f ( x ) dxρ λ − k/q + β i q dρρ = (cid:16) Z δ + Z + ∞ δ (cid:17)h R B n − kρ ( y ′ ) f ( x ) dxρ λ − k/q + β i q dρρ . Z δ h ρ n − k ( M f )( y ′ ) ρ λ − k/q + β i q dρρ + Z + ∞ δ h ρ ( n − k ) /p ′ k f k ∗ ρ λ − k/q + β i q dρρ . δ ( n − λ − β ) q − ( q − k f ∗ ( y ′ ) q + δ ( n − λ − β ) q − ( q − k − q ( n − k ) p k f k q ∗ . To obtain the last line in the above estimate, we also note that λ − kq + β − n − kp ′ > , thanks to (1.9); otherwise, the integral R + ∞ δ diverges. The trick is first to select δ such that δ ( n − λ − β ) q − ( q − k ( M f )( y ′ ) q = k f k q − p ∗ ( M f )( y ′ ) p (A.2) and then to select q such that δ ( n − λ − β ) q − ( q − k − q ( n − k ) p k f k q ∗ = k f k q − p ∗ ( M f )( y ′ ) p . (A.3)Indeed, to fulfill (A.2), we simply choose δ = h k f k ∗ M f ( y ′ ) i q − p ( n − λ − β ) q − ( q − k . From this choice of δ we deduce δ ( n − λ − β ) q − ( q − k = h k f k ∗ M f ( y ′ ) i q − p , which immediately implies (A.2). Notice that ( n − λ − β ) q − ( q − k − q ( n − k ) p = qq − p (cid:2) ( n − λ − β ) q − ( q − k (cid:3) − q ( n − k ) p − pq − p (cid:2) ( n − λ − β ) q − ( q − k (cid:3) = − pq − p (cid:2) ( n − λ − β ) q − ( q − k (cid:3) , thanks to (1.9). Hence, we obtain δ ( n − λ − β ) q − ( q − k − q ( n − k ) p = h k f k ∗ M f ( y ′ ) i − p , which yields δ ( n − λ ) q − ( q − k − q ( N − k ) p k f k q ∗ = k f k q − p ∗ ( M f )( y ′ ) p , which is nothing but (A.3). Thus, we arrive at Z + ∞ " R B n − kρ ( y ′ ) | f ( x ) | dxρ λ − k/q + β q dρρ . k f k q − p ∗ ( M f )( y ′ ) p . Since k f ∗ k L p ( R n − k ) . k f k L p ( R n − k ) , we deduce that Z R n − k Z + ∞ " R B n − kρ ( y ′ ) | f ( x ) | dxρ λ − k/q + β q dρρ dy ′ . k f k q − p ∗ Z R n − k ( M f )( y ′ ) p dy ′ . (A.4)Combining (A.1) and (A.4) gives the desired estimate.R EFERENCES[Ada75] D. A
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