aa r X i v : . [ m a t h . C A ] F e b A NEW FORMULA FOR THE L p NORM
QINGSONG GU AND PO-LAM YUNG
Abstract.
Recently, Brezis, Van Schaftingen and the second author [4] established a newformula for the ˙ W ,p norm of a function in C ∞ c ( R N ). The formula was obtained by replacingthe L p ( R N ) norm in the Gagliardo semi-norm for ˙ W s,p ( R N ) with a weak- L p ( R N ) quasi-norm and setting s = 1. This provides a characterization of such ˙ W ,p norms, whichcomplements the celebrated Bourgain-Brezis-Mironescu (BBM) formula [1]. In this paper,we obtain an analog for the case s = 0. In particular, we present a new formula for the L p norm of any function in L p ( R N ), which involves only the measures of suitable levelsets, but no integration. This provides a characterization of the norm on L p ( R N ), whichcomplements a formula by Maz ′ ya and Shaposhnikova [12]. As a result, by interpolation, weobtain a new embedding of the Triebel-Lizorkin space F s,p ( R N ) (i.e. the Bessel potentialspace ( I − ∆) − s/ L p ( R N )), as well as its homogeneous counterpart ˙ F s,p ( R N ), for s ∈ (0 , p ∈ (1 , ∞ ). Introduction
The purpose of this paper is to prove a new characterization of the L p norm on R N , bylifting to the product space R N × R N and considering a weak- L p quasi-norm over thereinstead. Indeed, for a measurable function F ( x, y ) on R N × R N and 1 ≤ p < ∞ , we denotethe weak- L p quasi-norm of F by [ F ] L p, ∞ ( R N × R N ) , where[ F ] L p, ∞ ( R N × R N ) := sup λ> (cid:0) λ p L N { ( x, y ) ∈ R N : | F ( x, y ) | ≥ λ } (cid:1) /p (1.1)and L N denotes the Lebesgue measure on R N (see e.g., [5, 9]). Then our first result reads: Theorem 1.1.
For every N ∈ N , there exist constants c = c ( N ) > and c = c ( N ) > ,such that for all ≤ p < ∞ and all u ∈ L p ( R N ) , c /p k u k L p ( R N ) ≤ " u ( x ) − u ( y ) | x − y | Np L p, ∞ ( R N × R N ) ≤ c /p k u k L p ( R N ) . (1.2) . Primary 26D10; Secondary 26A33,35A23,46E30,46E35.
Key words and phrases.
BBM formula; Maz ′ ya-Shaposhnikova formula; Fractional Sobolev space; weak- L p space.Yung is partially supported by a Future Fellowship FT200100399 from the Australian Research Council. oreover, for ≤ p < ∞ and u ∈ L p ( R N ) , if we write E λ := ( ( x, y ) ∈ R N × R N : x = y, | u ( x ) − u ( y ) || x − y | Np ≥ λ ) , (1.3) then lim λ → + λ p L N ( E λ ) = 2 κ N k u k pL p ( R N ) , (1.4) where κ N := π N/ / Γ( N + 1) is the volume of the unit ball in R N . We remark that the power of | x − y | in the denominator of the quantity in the middleof (1.2) is the natural one dictated by dilation invariance. Furthermore, the main thrust of(1.2) is in the first inequality. In fact, the second inequality has already been observed bye.g. Dominguez and Milman in [8]. On the other hand, the first inequality of (1.2) is aneasy consequence of (1.4), with c ( N ) := 2 κ N (because the supremum over λ > λ → + ). In addition, we emphasize that (1.4) is not true unlesswe assume u ∈ L p ( R N ) to begin with; indeed, if 1 ≤ p < ∞ and u is identically 1, then L N ( E λ ) = 0 for every λ >
0, while k u k L p ( R N ) = + ∞ . So the proof of (1.4) is a little delicate,which we give in detail in Section 2.Our point of view of lifting to the product space R N × R N and using the weak- L p quasi-norm there is motivated by recent work of the second author with Ha¨ım Brezis and Jean VanSchaftingen [4], which established an analog of the above theorem for the Sobolev semi-norm k∇ u k L p ( R N ) . The article [4] in turn drew important inspiration from the BBM formula forthe Sobolev space W ,p , which first appeared in a celebrated paper [1] of Bourgain, Brezisand Mironescu. An analogue of the BBM formula for L p in place of W ,p was first obtainedby Maz ′ ya and Shaposhnikova [12]. Our Theorem 1.1 can be thought of as a counterpart ofthe Maz ′ ya-Shaposhnikova formula for the L p norm, in the same way that the main resultin [4] relates to the BBM formula for W ,p .To describe all these developments in more detail, let’s introduce some notations. Let Ωbe a domain (i.e. an open, connected set) in R N . For 1 ≤ p < ∞ and 0 < s <
1, theGagliardo semi-norm of a function u ∈ L p (Ω) is defined as | u | ˙ W s,p (Ω) := (cid:18)Z Ω Z Ω | u ( x ) − u ( y ) | p | x − y | N + sp dxdy (cid:19) /p , (1.5)where | · | in the denominator on the right hand side denotes the Euclidean norm on R N .(The dot above W s,p indicates that this semi-norm is homogeneous with respect to dilations.)This semi-norm is an important tool in the study of many partial differential equations, andhas found numerous important applications (see e.g. [2, 7, 10]).A well-known ‘defect’ of this semi-norm is that | u | ˙ W s,p (Ω) does not converge to the Sobolevsemi-norm k∇ u k L p (Ω) as s → − . Indeed, it is easy to see (c.f. [1]) that if u is any smooth,non-constant function on a domain Ω ⊂ R N , then k u k p ˙ W s,p (Ω) → ∞ as s → − (see also
3, Proposition 4] for an extension to measurable u ’s that are not necessarily smooth). This‘defect’ was addressed by Bourgain, Brezis and Mironescu in [1]: if Ω is a smooth, boundeddomain in R N , then applying their Theorem 2 with ρ s ( x ) = p (1 − s ) | x |≤ D D p (1 − s ) ω N | x | N − p (1 − s ) , s ∈ (0 ,
1) (1.6)where D is the diameter of Ω and ω N is the surface area of S N − , we see that for 1 ≤ p < ∞ and u ∈ W ,p (Ω) := { u ∈ L p (Ω) : |∇ u | ∈ L p (Ω) } , one has what is now known as the BBMformula: lim s → − (1 − s ) | u | p ˙ W s,p (Ω) = 1 p k ( p, N ) k∇ u k pL p (Ω) , (1.7)where k ( p, N ) := Z S N − | e · ω | p dω = 2Γ(( p + 1) / π ( N − / Γ(( N + p ) / . (1.8)Here e ∈ S N − is any fixed vector, e · ω is the inner product of e with ω , and dω is thesurface measure on S N − induced from the Lebesgue measure on R N . See also D´avila [6] foran extension to the space of functions of bounded variation on Ω.On the other hand, for s ∈ (0 ,
1) and 1 ≤ p < ∞ , let W s,p ( R N ) be the completion of C ∞ c ( R N ) under the Gagliardo semi-norm | · | ˙ W s,p ( R N ) . Parallel to the BBM formula (1.7),Maz ′ ya and Shaposhnikova [12] showed that for any u ∈ S
0, is a directconsequence of (1.12) and (1.1). See also Poliakovsky [11, Lemma 3.1] for an extension of thesecond inequality in (1.10) to functions u ∈ W ,p ( R N ) := { u ∈ L p ( R N ) : |∇ u | ∈ L p ( R N ) } .In light of the formula (1.9) of Maz ′ ya and Shaposhnikova mentioned above, which es-tablishes an analog of the BBM formula (1.7) when s → + , a natural question is whetherone has an analog of (1.10) and (1.12) for L p instead of W ,p . Our Theorem 1.1 can bethought of as an affirmative answer to this question. Our proof is technically simpler thanthe corresponding one for (1.10) and (1.12) in [4], in that our proof relies only on Fubini’stheorem, but not on any covering lemma nor any Taylor expansion. On the other hand,it came as a mild surprise that while (1.12) involves a limit as λ → + ∞ , its cousin (1.4)involves instead a limit where λ → + : the former is natural since large values of λ captureswhat happens to | u ( x ) − u ( y ) | when x and y are close to each other, which in turn relatesto the size of |∇ u ( x ) | , but we do not have a good explanation of the latter.We next turn to two results obtained by interpolating the upper bound in (1.2), with theupper bound in (1.10). The first result can be formulated using the Bessel potential spaces( I − ∆) − s/ L p ( R N ): Theorem 1.2.
For every N ∈ N and p ∈ (1 , ∞ ) , there exists a constant C ′ = C ′ ( p, N ) suchthat for all s ∈ (0 , and all u ∈ ( I − ∆) − s/ L p ( R N ) , we have " u ( x ) − u ( y ) | x − y | Np + s L p, ∞ ( R N × R N ) ≤ C ′ k ( I − ∆) s/ u k L p ( R N ) . (1.13)This theorem follows from complex interpolation by considering the following holomorphicfamily of linear operators u ( x ) T z u ( x, y ) := u ( x ) − u ( y ) | x − y | Np + z (1.14)where z ∈ C takes value in the strip { ≤ Re z ≤ } . Indeed, the second inequality in (1.2)shows that when Re z = 0, T z maps L p ( R N ) to L p, ∞ ( R N × R N ). On the other hand, asobserved by Poliakovsky [11, Lemma 3.1], the second inequality in (1.10) continues to holdfor all u ∈ W ,p ( R N ) = ( I − ∆) − / L p ( R N ). Thus when Re z = 1, T z maps ( I − ∆) − / L p ( R N )to L p, ∞ ( R N × R N ). Theorem 1.2 now follows by complex interpolation.One drawback of Theorem 1.2 is that the left-hand side concerns a homogeneous norm,while the right-hand side contains an inhomogeneous norm. But it is only slightly harder toprove a variant of Theorem 1.2, concerning a homogeneous Triebel-Lizorkin space instead.First, let’s recast Theorem 1.2 in terms of (inhomogeneous) fractional Triebel-Lizorkinspaces F s,pq on R N , which we define as follows. Let S ( R N ) be the Fr´echet space of Schwartzfunctions on R N , and S ′ ( R N ) the space of all tempered distributions on R N . Let F − be the nverse Fourier transform on R N given by F − φ ( x ) = Z R N φ ( ξ ) e πix · ξ dξ, (1.15)for φ ∈ S ( R N ). Let ϕ ∈ C ∞ c ( R N ) be a fixed function supported on {| ξ | ≤ } such that ϕ ( ξ ) = 1 whenever | ξ | ≤
1. Write ψ ( ξ ) = ϕ ( ξ ) − ϕ (2 ξ ) (1.16)so that ψ ∈ C ∞ c ( R N ) is supported on { / ≤ | ξ | ≤ } with ϕ ( ξ ) + X j ∈ N ψ (2 − j ξ ) = 1 for all ξ ∈ R N . (1.17)A corresponding family of Littlewood-Paley projections is given by P u ( x ) := u ∗ F − ϕ ( x ) (1.18)and ∆ j u ( x ) := u ∗ F − ψ j ( x ) , (1.19)where ψ j ( ξ ) := ψ (2 − j ξ ). For s ∈ R , p ∈ (1 , ∞ ) and q ∈ (1 , ∞ ), we define the (inhomoge-neous) Triebel-Lizorkin space F s,pq ( R N ) to be the space of all u ∈ S ′ ( R N ) for which k u k F s,pq ( R N ) := (cid:13)(cid:13)(cid:13)(cid:16) | P u | q + X j ∈ N | js ∆ j u | q (cid:17) /q (cid:13)(cid:13)(cid:13) L p ( R N ) < ∞ . (1.20)Standard Littlewood-Paley theory shows that for 1 < p < ∞ , s ∈ R , we have F s,p ( R N ) = ( I − ∆) − s/ L p ( R N ) (1.21)with comparable norms: for all u ∈ S ′ ( R N ), we have k u k F s,p ( R N ) ≃ p,N k ( I − ∆) s/ u k L p ( R N ) . (1.22)Thus we could have replaced the Bessel potential spaces ( I − ∆) − s/ L p ( R N ) in Theorem 1.2by the inhomogeneous F s,p ( R N ).This motivates us to consider a variant of Theorem 1.2 for homogeneous Triebel-Lizorkinspaces instead. To introduce these spaces, we denote by Z ( R N ) the subspace of all u ∈ S ( R N )for which R R N u ( x ) p ( x ) dx = 0 for every polynomial p ( x ) ∈ R [ x ], and denote by Z ′ ( R N ) thespace of all continuous linear functionals on Z ( R N ), which we identify with the quotient S ′ ( R N ) / { polynomials on R N } . If ψ ∈ S ( R N ) is as in (1.16) and ψ j ( ξ ) := ψ (2 − j ξ ) for j ∈ Z ,then X j ∈ Z ψ j ( ξ ) = 1 for all ξ ∈ R N \ { } . (1.23)We denote by { ∆ j } j ∈ Z the family of Littlewood-Paley projections given by∆ j u ( x ) := u ∗ F − ψ j ( x ) , (1.24) hich is well-defined for all u ∈ Z ′ ( R N ) (because R R N F − ψ j ( x ) p ( x ) dx = 0 for all polynomials p ( x ) ∈ R [ x ].) The homogeneous Triebel-Lizorkin space ˙ F s,pq ( R N ) is then defined to be thespace of all u ∈ Z ′ ( R N ) for which k u k ˙ F s,pq ( R N ) := (cid:13)(cid:13)(cid:13)(cid:16) X j ∈ Z | js ∆ j u | q (cid:17) /q (cid:13)(cid:13)(cid:13) L p ( R N ) < ∞ . (1.25)It was known (c.f. proof of Theorem in [15, Chapter 5.1.5]) that F − [ C ∞ c ( R N \ { } )], thespace of (Schwartz) functions on R N given by inverse Fourier transforms of C ∞ , compactlysupported functions on R N \ { } , is a dense subset of ˙ F s,pq ( R N ) for s ∈ R , p ∈ (1 , ∞ ) and q ∈ (1 , ∞ ) (see Appendix below for a sketch of proof). Also, for 1 < p < ∞ , we have k u k ˙ F s,p ( R N ) ≃ p,N k u k L p ( R N ) if s = 0 k∇ u k L p ( R N ) if s = 1 , (1.26)at least if u ∈ F − [ C ∞ c ( R N \ { } )] (indeed this holds as long as u ∈ S ′ ( R N ) for which theright hand side of the above display equation is finite). This allows us to prove the nextresult, concerning the homogeneous space ˙ F s,p ( R N ): Theorem 1.3.
For every N ∈ N and p ∈ (1 , ∞ ) , there exists a constant C ′ = C ′ ( p, N ) suchthat for all s ∈ (0 , and all u ∈ F − [ C ∞ c ( R N \ { } )] , " u ( x ) − u ( y ) | x − y | Np + s L p, ∞ ( R N × R N ) ≤ C ′ k u k ˙ F s,p ( R N ) . (1.27) As a result, for s ∈ (0 , and p ∈ (1 , ∞ ) , one may define the left-hand side of (1.27) for all u ∈ ˙ F s,p ( R N ) by density, and the inequality (1.27) continues to hold. Theorem 1.3 is the most powerful in the case 1 < p <
2, as one can see by comparing(1.27) with the following known inequality for ˙ F s,pp ( R N ) (so q = p as opposed to q = 2 inTheorem 1.3): Proposition 1.4.
For every N ∈ N , there exists a constant C ′′ = C ′′ ( N ) so that for all u ∈ F − [ C ∞ c ( R N \ { } )] , s ∈ (0 , and p ∈ (1 , ∞ ) , one has (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) u ( x ) − u ( y ) | x − y | Np + s (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( R N × R N ) ≤ C ′′ (cid:18) s + 11 − s (cid:19) k u k ˙ F s,pp ( R N ) . (1.28)The left hand side of (1.27) is smaller than the left hand side of (1.28) by Chebyshev’sinequality, but the norm on the right hand side of (1.27) is also smaller than the norm onthe right hand side of (1.28) if 1 < p < k u k ˙ F s,p ≤ k u k ˙ F s,pp if p < C ′ in (1.27) does not blow up if we fix p and let s → + or 1 − .The proof of Theorem 1.3 will be given in Section 3. For the convenience of the reader,we will also give a proof of Proposition 1.4, which we adapt from [13, Chapter V.5]. n interesting related question is whether the inequality in (1.27) can be reversed. Since˙ F s,p ( R N ) = [ L p ( R N ) , ˙ W ,p ( R N )] s , this question could be reformulated as follows: If N ∈ N , p ∈ (1 , ∞ ), s ∈ (0 ,
1) and u ∈ F − [ C ∞ c ( R N \ { } )], is there a holomorphic family of functions { u z ( x ) : z ∈ C , ≤ Re z ≤ } so that u s ( x ) = u ( x ), and so thatmax (cid:26) sup Re z =0 k u z k L p ( R N ) , sup Re z =1 k∇ u z k L p ( R N ) (cid:27) . " u ( x ) − u ( y ) | x − y | Np + s L p, ∞ ( R N × R N ) ? (1.29) Acknowledgements.
The authors thank Ha¨ım Brezis and Jean Van Schaftingen for theirkind encouragement as we pursued this project. They also thank Ka-Sing Lau for his teachingand inspiration over the years. 2.
Proof of Theorem 1.1
Proof.
As remarked above, the second inequality in (1.2) is essentially known. It was statedwithout proof in [8]. But for completeness, and also because we need to use it to derive thefirst inequality in (1.2), we give its simple proof below. Indeed, we show that for 1 ≤ p < ∞ and all measurable functions u on R N , " u ( x ) − u ( y ) | x − y | Np pL p, ∞ ( R N × R N ) ≤ p +1 κ N k u k pL p ( R N ) (2.1)so that the second inequality of (1.2) holds with c ( N ) := 2 κ N , where κ N is the volume ofthe unit ball in R N .To prove (2.1), given 1 ≤ p < ∞ , a measurable u on R N , and λ >
0, let E λ be as in (1.3).Then by the triangle inequality, L N ( E λ ) ≤ L N (cid:18)(cid:26) ( x, y ) ∈ R N × R N : x = y, | u ( x ) | ≥ λ | x − y | N/p (cid:27)(cid:19) + L N (cid:18)(cid:26) ( x, y ) ∈ R N × R N : x = y, | u ( y ) | ≥ λ | x − y | N/p (cid:27)(cid:19) = Z R N Z R N { ( x,y ): | y − x |≤ (2 | u ( x ) | λ − ) p/N } dydx + Z R Z R N { ( x,y ): | y − x |≤ (2 | u ( y ) | λ − ) p/N } dxdy = κ N Z R N (2 λ − ) p | u ( x ) | p dx + κ N Z R N (2 λ − ) p | u ( y ) | p dy = 2 p +1 κ N λ − p k u k pL p ( R N ) . (2.2)(2.1) now follows by multiplying by λ p on both sides and taking supremum over all λ > u ∈ L p ( R N ), 1 ≤ p < ∞ , which would then imply thefirst inequality in (1.2). We first consider the case under the additional assumption that u iscompactly supported on R N . This extra assumption about u will then be removed by usingsuitable truncations of u , together with (2.1) which handles the error that arises. ase 1. u is compactly supported. For λ >
0, let E λ be as in (1.3). Then L N ( E λ ) = 2 L N ( H λ ) (2.3)where H λ := E λ ∩ { ( x, y ) ∈ R N × R N : | y | > | x |} . (2.4)This is because E λ is the union of its three subsets, one where | y | > | x | , one where | y | < | x | ,and one where | y | = | x | . The last set has L N measure zero, and the first two sets have thesame L N measure by symmetry of the set E λ . Hence we only need to estimate L N ( H λ ).Since u is compactly supported, we may assumesupp u ⊆ B R := { x ∈ R N : | x | < R } (2.5)for some R >
0. Now if ( x, y ) ∈ H λ , then we must have x ∈ B R . This is because otherwiseboth x, y are outside B R , which by our assumption about the support of u implies that u ( x ) = u ( y ) = 0, and hence ( x, y ) / ∈ E λ , contradicting that ( x, y ) ∈ H λ . Moreover, for x ∈ B R , let H λ,x := (cid:26) y ∈ R N : | y | > | x | , | u ( y ) − u ( x ) || y − x | N/p ≥ λ (cid:27) (2.6)and H λ,x,R := ( y ∈ R N : | y | ≥ R, | y − x | ≤ (cid:18) | u ( x ) | λ (cid:19) p/N ) . (2.7)Then Fubini’s theorem gives L N ( H λ ) = Z B R L N ( H λ,x ) dx, (2.8)while H λ,x,R = H λ,x \ B R , (2.9)because for | y | ≥ R , we have u ( y ) = 0 and hence | u ( x ) | = | u ( y ) − u ( x ) | . It follows that H λ,x,R ⊆ H λ,x ⊆ H λ,x,R ∪ B R . (2.10)Writing κ N = L N ( B ), from the first inclusion in (2.10), we have L N ( H λ,x ) ≥ L N ( H λ,x,R ) ≥ κ N | u ( x ) | p λ p − κ N R N . (2.11)On the other hand, from the second inclusion in (2.10), we have L N ( H λ,x ) ≤ κ N | u ( x ) | p λ p + κ N R N . (2.12)Integrating (2.11) and (2.12) over x ∈ B R , and using (2.8), we obtain κ N λ p k u k pL p ( R N ) − κ N R N ≤ L N ( H λ ) ≤ κ N λ p k u k pL p ( R N ) + κ N R N . (2.13) ultiplying both sides by λ p and letting λ → + , we havelim λ → + λ p L N ( H λ ) = κ N k u k pL p ( R N ) , (2.14)as desired. Case 2. u ∈ L p ( R N ) , not necessarily compactly supported. Let u R = u · B R be thetruncation of u with | x | ≤ R for some R >
0. Let v R = u − u R . Later we will crucially usethat k v R k L p ( R N ) → R → ∞ , which holds only because u ∈ L p ( R N ) and 1 ≤ p < ∞ .Now since u = u R + v R , for any σ ∈ (0 , E λ = (cid:26) ( x, y ) ∈ R N × R N : | u ( x ) − u ( y ) || x − y | N/p ≥ λ (cid:27) ⊆ A ∪ A (2.15)where A := (cid:26) ( x, y ) ∈ R N × R N : | u R ( x ) − u R ( y ) || x − y | N/p ≥ λ (1 − σ ) (cid:27) (2.16)and A := (cid:26) ( x, y ) ∈ R N × R N : | v R ( x ) − v R ( y ) || x − y | N/p ≥ λσ (cid:27) . (2.17)Hence L N ( E λ ) ≤ L N ( A ) + L N ( A ) . (2.18)Since u R is compactly supported in B R , by (2.12) with λ replaced by λ (1 − σ ), we obtain L N ( A ) ≤ κ N λ p (1 − σ ) p k u R k pL p ( R N ) + 2( κ N R N ) . (2.19)For A , by using (2.1) for v R , we obtain L N ( A ) ≤ p +1 κ N ( λσ ) p k v R k pL p ( R N ) . (2.20)Combining (2.18), (2.19) and (2.20), and multiplying by λ p , we obtain λ p L N ( E λ ) ≤ κ N (1 − σ ) p k u R k pL p ( R N ) + 2 λ p ( κ N R N ) + 2 p +1 κ N σ p k v R k pL p ( R N ) . (2.21)We now first let λ → + , then let R → ∞ and finally let σ → + . Sincelim R →∞ k u R k L p ( R N ) = k u k L p ( R N ) and lim R →∞ k v R k L p ( R N ) = 0 , (2.22)we obtain lim sup λ → + λ p L N ( E λ ) ≤ κ N k u k pL p ( R N ) . (2.23)Similarly, for any σ >
0, we have E λ = (cid:26) ( x, y ) ∈ R N × R N : | u ( x ) − u ( y ) || x − y | N/p ≥ λ (cid:27) ⊇ A \ A (2.24) here A := (cid:26) ( x, y ) ∈ R N × R N : | u R ( x ) − u R ( y ) || x − y | N/p ≥ λ (1 + σ ) (cid:27) (2.25)and A is as in (2.17). Hence L N ( E λ ) ≥ L N ( A ) − L N ( A ) . (2.26)Since u R is compactly supported in B R , by (2.11) with λ replaced by λ (1 + σ ), we have L N ( A ) ≥ κ N λ p (1 + σ ) p k u R k pL p ( R N ) − κ N R N ) . (2.27)Combining (2.26), (2.27) and (2.20), and multiplying by λ p , we obtain λ p L N ( E λ ) ≥ κ N (1 + σ ) p k u R k pL p ( R N ) − λ p ( κ N R N ) − p +1 κ N σ p k v R k pL p ( R N ) . (2.28)We now first let λ → + , then let R → ∞ and finally let σ → + . We obtainlim inf λ → + λ p L N ( E λ ) ≥ κ N k u k pL p ( R N ) . (2.29)(1.4) then follows from (2.23) and (2.29). (cid:3) Embeddings of homogeneous fractional Triebel-Lizorkin spaces
In this section, we first prove Theorem 1.3. Its proof is similar to that of Theorem 1.2, inthat it also proceeds via complex interpolation, for the holomorphic family of linear operators { T z } defined in (1.14). On the other hand, it is not clear whether T maps ˙ F ,p ( R N ) to L p, ∞ ( R N × R N ). Thus we provide a more careful proof below, explaining why interpolationworks.First, we recall the subspace F − [ C ∞ c ( R N \ { } )] which consists of Schwartz functions on R N and is dense in ˙ F s,p ( R N ) when s ∈ (0 , p ∈ (1 , ∞ ). For u ∈ F − [ C ∞ c ( R N \ { } )], say u = F − b u and b u ∈ C ∞ c ( R N \ { } ), we may define complex powers of Laplacian:( − ∆) z u ( x ) := Z R N (2 π | ξ | ) z b u ( ξ ) e πix · ξ dξ, z ∈ C . (3.1)For every fixed x ∈ R N , this defines an entire function of z ∈ C . Furthermore, for everyfixed z ∈ C , this defines a function in F − [ C ∞ c ( R N \ { } )] ⊂ S ( R N ). Lemma 3.1.
For every N ∈ N and < p < ∞ , there exists a constant A = A ( p, N ) suchthat for any u ∈ F − [ C ∞ c ( R N \ { } )] and any s ∈ (0 , , the following estimates hold.(a) For z ∈ C with Re z = 0 , we have k ( − ∆) ( s − z ) / u k L p ( R N ) ≤ A (1 + | Im z | ) N +1 k u k ˙ F s,p ( R N ) . (3.2) (b) For z ∈ C with Re z = 1 , we have k∇ ( − ∆) ( s − z ) / u k L p ( R N ) ≤ A (1 + | Im z | ) N +1 k u k ˙ F s,p ( R N ) . (3.3) roof. The proof is a standard application of the theory of singular integrals. We verify the z -dependence of the constants in (3.2) and (3.3) by providing the necessary details below.We first prove (a). Let z ∈ C with Re z = 0. Write z = it for some t ∈ R . Let { ˜∆ j } j ∈ Z beanother family of Littlewood-Paley projections, given by˜∆ j u ( x ) := u ∗ F − ˜ ψ j ( x ) , (3.4)where ˜ ψ j ( ξ ) := ˜ ψ (2 − j ξ ) for some C ∞ function supported on { / ≤ | ξ | ≤ } , so that˜ ψ ( ξ ) = 1 on the support of ψ ; this gives ˜∆ j ∆ j = ∆ j for all j ∈ Z . As a result, for u ∈ F − [ C ∞ c ( R N \ { } )], j ∈ Z and s ∈ (0 , j ( − ∆) ( s − z ) / u = (2 js ∆ j u ) ∗ K j (3.5)where K j := F − [2 − js (2 π | ξ | ) s − it ˜ ψ j ] satisfiessup j ∈ Z |∇ K j ( x ) | . N (1 + | t | ) N +1 | x | − ( N +1) . (3.6)(This is because for any multiindices α , one has | ∂ αξ [2 − js (2 π | ξ | ) s − it ˜ ψ j ( ξ )] | . α (1 + | t | ) | α | − j | α | χ | ξ |≃ j (3.7)with implicit constant independent of j ∈ Z , s ∈ (0 ,
1) and t ∈ R ; we may apply this with | α | = N and N +1 to bound | x | N +1 |∇ K j ( x ) | in L ∞ ( R N ).) We may now apply a vector-valuedsingular integral theorem to the operator( f j ( x )) j ∈ Z ( f j ∗ K j ( x )) j ∈ Z , (3.8)which is clearly bounded on L ( ℓ ) with norm .
1; by [13, Chapter II, Theorem 5], or[14, Chapter I.6.4], this operator is also bounded on L p ( ℓ ) for all 1 < p < ∞ , with operatornorm . p,N (1+ | t | ) N +1 . Combined with the Littlewood-Paley inequality (which holds because( − ∆) ( s − z ) / u ∈ L p ( R N )), we now have k ( − ∆) ( s − z ) / u k L p ( R N ) ≃ p,N (cid:13)(cid:13)(cid:13)(cid:16) X j ∈ Z | ∆ j ( − ∆) ( s − z ) / u | (cid:17) / (cid:13)(cid:13)(cid:13) L p ( R N ) . p,N (1 + | t | ) N +1 (cid:13)(cid:13)(cid:13)(cid:16) X j ∈ Z | js ∆ j u | (cid:17) / (cid:13)(cid:13)(cid:13) L p ( R N ) = A ( p, N )(1 + | t | ) N +1 k u k ˙ F s,p ( R N ) (3.9)the middle inequality following from (3.5) and the boundedness of the operator in (3.8) on L p ( ℓ ). This completes the proof of (a).To deduce (b), one can either appeal to the boundedness of the Riesz transform ∇ ( − ∆) − / on L p ( R N ), or repeat the argument above. We omit the details. (cid:3) urthermore, we will need to consider, for 1 < p < ∞ , the Lorentz space L p, ( R N × R N ),which is defined to be the set of all measurable functions g ( x, y ) on R N × R N for which[ g ] L p, ( R N × R N ) := p Z ∞ L N ( { ( x, y ) ∈ R N × R N : | g ( x, y ) | ≥ λ } ) /p dλ < ∞ . (3.10)Just like [ · ] L p, ∞ , the quantity [ · ] L p, is not a norm because it does not satisfy the triangleinequality; it is only a quasi-norm. Nevertheless, for 1 < p < ∞ , both L p, and L p, ∞ admita comparable norm, which make them Banach spaces, and L p, ∞ is the dual space of L p ′ , whenever 1 /p + 1 /p ′ = 1: in fact, the easiest way to norm L p ′ , is to define k g k L p ′ , ( R N × R N ) := sup (cid:26)(cid:12)(cid:12)(cid:12)(cid:12)Z R N × R N g ( x, y ) G ( x, y ) dxdy (cid:12)(cid:12)(cid:12)(cid:12) : [ G ] L p, ∞ ( R N × R N ) = 1 (cid:27) . (3.11)If p ∈ (1 , ∞ ), every g ∈ L p ′ , ( R N × R N ) can be approximated in the L p ′ , norm by functionsin L p ′ , ( R N × R N ) that are compactly supported in the open set { ( x, y ) ∈ R N × R N : x = y } (because such approximation is possible in the comparable L p ′ , ( R N × R N ) quasi-norm bythe dominated convergence theorem). We are now ready to prove Theorem 1.3. Proof of Theorem 1.3.
We fix s ∈ (0 , p ∈ (1 , ∞ ), u ∈ F − [ C ∞ c ( R N \ { } )], and g ∈ L p ′ , ( R N × R N ) with compact support in { ( x, y ) ∈ R N × R N : x = y } . Consider the function H ( z ) = Z R N × R N g ( x, y ) ( − ∆) ( s − z ) / u ( x ) − ( − ∆) ( s − z ) / u ( y ) | x − y | Np + z dxdy. (3.12)This is an entire function of z , and we claim that it is a bounded function on the strip { z ∈ C : 0 ≤ Re z ≤ } . Indeed, for u ∈ F − [ C ∞ c ( R N \ { } )], (3.1) gives( − ∆) ( s − z ) / u ( x ) = Z R N (2 π | ξ | ) s − z b u ( ξ ) e πix · ξ dξ for all z ∈ C , (3.13)so | ( − ∆) ( s − z ) / u ( x ) − ( − ∆) ( s − z ) / u ( y ) | ≤ k ( − ∆) ( s − z ) / u k L ∞ ( R N ) ≤ Z ξ ∈ supp b u (2 π | ξ | ) s − Re z | b u ( ξ ) | dξ . exp( a | Re z | )(3.14)if a > { π | ξ | , π | ξ | } ≤ exp( a ) for all ξ ∈ supp b u . Also, on thesupport of g ( x, y ), we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − y | Np + z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . exp( a | Re z | ) (3.15)if a > {| x − y | , | x − y | − : ( x, y ) ∈ supp g } ≤ exp( a ). Finally,since g ∈ L p ′ , ( R N × R N ) has compact support, it is in L ( R N × R N ) as well. So | H ( z ) | . k g k L exp( a | Re z | ) for all z ∈ C (3.16) here a = a + a . In particular, H ( z ) is bounded on the strip { z ∈ C : 0 ≤ Re z ≤ } , asclaimed.Furthermore, for Re z = 0, the upper bound in (1.2), together with (3.2), show that | H ( z ) | ≤ Z R N × R N | g ( x, y ) | | ( − ∆) ( s − z ) / u ( x ) − ( − ∆) ( s − z ) / u ( y ) || x − y | Np dxdy ≤ k g k L p ′ , ( R N ) h ( − ∆) ( s − z ) / u ( x ) − ( − ∆) ( s − z ) / u ( y ) | x − y | Np i L p, ∞ ( R N ) ≤ c /p k g k L p ′ , ( R N ) k ( − ∆) ( s − z ) / u k L p ( R N ) ≤ c /p A (1 + | Im z | ) N +1 k g k L p ′ , ( R N ) k u k ˙ F s,p ( R N ) . (3.17)On the other hand, for Re z = 1, the upper bound in (1.10), together with (3.3), show that | H ( z ) | ≤ Z R N × R N | g ( x, y ) | | ( − ∆) ( s − z ) / u ( x ) − ( − ∆) ( s − z ) / u ( y ) || x − y | Np +1 dxdy ≤ k g k L p ′ , ( R N ) h ( − ∆) ( s − z ) / u ( x ) − ( − ∆) ( s − z ) / u ( y ) | x − y | Np +1 i L p, ∞ ( R N ) ≤ C /p k g k L p ′ , ( R N ) k∇ ( − ∆) ( s − z ) / u k L p ( R N ) ≤ C /p A (1 + | Im z | ) N +1 k g k L p ′ , ( R N ) k u k ˙ F s,p ( R N ) . (3.18)(The upper bound in (1.10) applies because ( − ∆) ( s − z ) / u ∈ S ( R N ) ⊂ W ,p ( R N ) when u ∈F − [ C ∞ c ( R N \ { } )], allowing us to invoke [11, Lemma 3.1].) This allows us to use the threelines lemma from complex analysis to the bounded holomorphic function H ( z ) / ( z + 1) N +1 on the strip { z ∈ C : 0 ≤ Re z ≤ } , and conclude that | H ( s ) | . p,N k g k L p ′ , ( R N ) k u k ˙ F s,p ( R N ) . (3.19)Taking supremum over g , we get h u ( x ) − u ( y ) | x − y | Np + s i L p, ∞ ( R N ) ≤ C ′ k u k ˙ F s,p ( R N ) (3.20)where C ′ = C ′ ( p, N ), and this inequality holds for all u ∈ F − [ C ∞ c ( R N \ { } )]. This showsthat the left-hand side may be defined by density for all u ∈ ˙ F s,p ( R N ), and that the inequalitycontinues to hold after such extension for all u ∈ ˙ F s,p ( R N ). (cid:3) Proof of Proposition 1.4.
We just note that for u ∈ F − [ C ∞ c ( R N \ { } )], we have u ( x ) = X j ∈ Z ∆ j u ( x ) (3.21) the sum on the right vanishes for all but finitely many j ’s), and (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) u ( x ) − u ( y ) | x − y | Np + s (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( R N ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) u ( x + z ) − u ( x ) | z | Np + s (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( R N ) . N X k ∈ Z ksp sup | z |≃ − k k u ( x + z ) − u ( x ) k pL p ( dx ) ! /p ≤ X k ∈ Z ksp sup | z |≃ − k X j ∈ Z k ∆ j u ( x + z ) − ∆ j u ( x ) k L p ( dx ) ! p ! /p . (3.22)Now if | z | ≃ − k , we write∆ j u ( x + z ) − ∆ j u ( x ) = Z ddt ∆ j u ( x + tz ) dt = Z z · ∇ ∆ j u ( x + tz ) dt, (3.23)so its L p norm with respect to x is bounded by | z |k∇ ∆ j u k L p ( R N ) . − k k∇ ∆ j u k L p ( R N ) . N j − k k ∆ j u k L p ( R N ) . (3.24)This shows sup | z |≃ − k k ∆ j u ( x + z ) − ∆ j u ( x ) k L p ( d x ) . N j − k k ∆ j u k L p ( R N ) . (3.25)We also have the trivial boundsup | z |≃ − k k ∆ j u ( x + z ) − ∆ j u ( x ) k L p ( d x ) ≤ k ∆ j u k L p ( R N ) . (3.26)Then combine these two estimate, we writesup | z |≃ − k k ∆ j u ( x + z ) − ∆ j u ( x ) k L p ( d x ) . N k ∆ j u k L p ( R N ) if j > k, j − k k ∆ j u k L p ( R N ) if j ≤ k. (3.27)Then by substituting (3.27) into (3.22), we obtain (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) u ( x ) − u ( y ) | x − y | Np + s (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( R N ) . N (cid:16) X k ∈ Z ksp (cid:16) X j ∈ Z ( χ j>k + 2 j − k χ j ≤ k ) k ∆ j u k L p ( R N ) (cid:17) p (cid:17) /p = (cid:13)(cid:13)(cid:13) X j ∈ Z (2 ( k − j ) s χ j>k + 2 ( j − k )(1 − s ) χ j ≤ k )(2 js k ∆ j u k L p ( R N ) ) (cid:13)(cid:13)(cid:13) ℓ pk (3.28)which by Young’s convolution inequality is . N (cid:0) s + 11 − s (cid:1) k js k ∆ j u k L p ( R N ) k ℓ pj = (cid:0) s + 11 − s (cid:1) k u k ˙ F s,pp ( R N ) (3.29)since the sequence 2 ks χ k< + 2 − k (1 − s ) χ k ≥ has ℓ norm . s + − s as s varies over (0 , (cid:3) . Appendix: Density in Triebel-Lizorkin spaces
In the proof of Theorem 1.3, we appealed to the case s ∈ (0 ,
1) and q = 2 of the followingproposition. Thus we include a sketch of its proof. Proposition 4.1. F − [ C ∞ c ( R N \ { } )] is dense in ˙ F s,pq ( R N ) for s ∈ R , p ∈ (1 , ∞ ) and q ∈ (1 , ∞ ) .Proof. Fix s ∈ R , p ∈ (1 , ∞ ) and q ∈ (1 , ∞ ). First, for u ∈ ˙ F s,pq ( R N ), sincelim J → + ∞ (cid:13)(cid:13)(cid:13)(cid:16) X | j | >J | js ∆ j u | q (cid:17) /q (cid:13)(cid:13)(cid:13) L p ( R N ) = 0 , (4.1)we see that u J := P | j |≤ J ∆ j u converges in ˙ F s,pq ( R N ) as J → + ∞ .Next, let φ ∈ S ( R N ) with φ (0) = 1 whose Fourier transform b φ is compactly supportedon the unit ball. For every fixed J ∈ N , we let u J,δ ( x ) := φ ( δx ) u J ( x ). Then for δ ≪ − J ,we have u J,δ ∈ F − [ C ∞ c ( R N \ { } )]. Thus it remains to show that u J,δ → u J in ˙ F s,pq as δ →
0. To see this, note that u J,δ is C ∞ on R N , so u J,δ converges pointwisely to u J as δ → u J,δ is dominated by a multiple of u J , which is in L p ( R N ), so by the dominateconvergence theorem, lim δ → k u J,δ − u J k L p ( R N ) = 0 . (4.2)As a result, lim δ → k ∆ j ( u J,δ − u J ) k L p ( R N ) = 0 (4.3)for every j ∈ Z , which implies the desired convergence of u J,δ to u J in ˙ F s,pq as δ → (cid:3) References [1] Jean Bourgain, Ha¨ım Brezis, and Petru Mironescu,
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Department of Mathematics, Nanjing University, Nanjing 210093, China
Email address : [email protected] (P.-L. Yung) Mathematical Sciences Institute, Australian National University, CanberraACT 2601, Australia and
Department of Mathematics, The Chinese University of HongKong, Shatin, Hong Kong
Email address : [email protected]
Email address : [email protected]@math.cuhk.edu.hk