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Hardy type inequalities for the fractional relativistic operator

  • We prove Hardy type inequalities for the fractional relativistic operator by using two different techniques. The first approach goes through trace Hardy inequalities. In order to get the latter, we study the solutions of the associated extension problem. The second develops a non-local version of the ground state representation in the spirit of Frank, Lieb, and Seiringer.

    Citation: Luz Roncal. Hardy type inequalities for the fractional relativistic operator[J]. Mathematics in Engineering, 2022, 4(3): 1-16. doi: 10.3934/mine.2022018

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  • We prove Hardy type inequalities for the fractional relativistic operator by using two different techniques. The first approach goes through trace Hardy inequalities. In order to get the latter, we study the solutions of the associated extension problem. The second develops a non-local version of the ground state representation in the spirit of Frank, Lieb, and Seiringer.



    Let 0<s<2. The goal of this note is to prove Hardy type inequalities for the fractional relativistic operator, which is defined as

    H:=(Δ+m2)s/2ms,m0

    where Δ is the Euclidean Laplacian. More precisely, if we denote

    L:=Δ+m2,

    so that we can write Hs:=Ls/2ms, we will study such inequalities for Ls/2. The operator Hs has interest from physical, probabilistic and mathematical analysis point of view. It is involved in the description of the kinetic energy of the relativistic particle with mass m, see e.g., [17,21] and in the dynamics of relativistic boson stars [13,14]. It plays an important role in the theory of interpolation spaces of Bessel potentials and applications in harmonic analysis [18,27], and it has been studied in the context of potential theory of s-stable relativistic processes [15,26].

    On the other hand, Hardy inequalities for fractional powers of the Laplacian Δ on Rn (i.e., the case m=0) have been investigated by many authors and there is a vast literature on the topic, see for instance [4,5,11,12,29]. We remark that in [17], a Hardy inequality for the operator H1 was already implicitly shown. Hardy inequality for the fractional Laplacian reads as

    ((Δ)s/2f,f)2sΓ(n+s4)2Γ(ns4)2Rn|f(x)|2|x|sdx, (1.1)

    for 0<s<2 and suitable functions. The constant is known to be sharp, but never achieved in the space of functions for which both sides of the inequality are finite.

    In operator terms, the sharp Hardy inequality in (1.1) can be interpreted, formally, as

    0(Δ)s/22sΓ(n+s4)2Γ(ns4)21|x|s,

    and therefore (Δ)s/2ν|x|s is not bounded from below, for all ν>2sΓ(n+s4)2Γ(ns4)2. One important consequence of this is the determination of existence and nonexistence of positive solutions to fractional elliptic and parabolic problems involving singular weights, see for instance [1,2,6]. In the local case, i.e., s=2, we refer to the work by Baras–Goldstein [3] (a first reference might be back to the seminal paper by J. Leray [20]). Another interesting example is presented in [22], where the authors study a solid combustion model in which the Hardy inequality arises naturally.

    Apart from the potential applications to partial differential equations, Hardy inequality is an interesting object of investigation and its study goes beyond the Euclidean setting or the Laplacian operator. This inequality plays an important role in many areas such as the spectral theory, geometric estimates and analyticity of functions. We refer to the famous book by Hardy–Littlewood–Pólya [16] as a primitive reference, but it is impossible to make just a fair glimpse of the huge literature on the topic.

    Returning to the case of the fractional Laplacian in the Euclidean case, we can also consider another version of Hardy inequality, where the homogeneous weight function |x|s is replaced by a non-homogeneous one:

    ((Δ)s/2f,f)2sΓ(n+s2)Γ(ns2)δsRn|f(x)|2(δ2+|x|2)sdx,δ>0. (1.2)

    Here again the constant is sharp and equality is achieved for the functions (δ2+|x|2)(ns)/2 and their translates (see [7]). Observe that these functions can be regarded as a generalised Poisson kernel solving a generalised harmonic extension (and therefore related to the solutions of the so called extension problem [8]).

    In this note we will prove Hardy inequalities for the operator Ls/2 associated to the fractional relativistic operator. We will assume that m>0 (since the case m=0 reduces to the usual Laplacian). Such inequalities will be obtained in two different ways: first, through a trace Hardy inequality, and second, through a ground state representation. In the first case, the framework will be rather general and we will apply the results to two particular cases which will lead to Hardy inequalities that may be understood like the natural analogues to (1.1) and (1.2).

    Our results may be seen as a revisit and generalisation of the Hardy inequality shown in [12,Subsection 2.2], and an attempt to write together several facts that seem to be around in the literature. Moreover, the two particular examples that we provide will produce sharp Hardy inequalities (see the statement of Corollary 2.7 and Remark 3.3). As explained above, sharp Hardy inequalities for the fractional relativistic operator may imply consequences on existence and nonexistence of solutions to problems involving Hs and different potentials. These will not be discussed here.

    We split the note into two parts: in the first one, we will obtain trace Hardy and Hardy type inequalities via the corresponding extension problem. We will obtain general results, see Theorem 2.1 and Corollary 2.2, and from there we will deduce Hardy inequalities for two particular instances of the functions involved. These are contained in Corollary 2.7 and Corollary 2.9. In the second part, we will show a Hardy inequality from ground state representations, following the ideas by Frank, Lieb, and Seiringer in the Euclidean setting in [11], see Corollary 3.2. The latter will coincide with the Hardy inequality in Corollary 2.7 in the first part. Actually, the Hardy inequality obtained in the second part is an improvement in the sense that the error in the inequality is explicitly computed, allowing the discussion on the sharpness.

    The contents of this part can be stated in the setting of fractional powers of general operators given by sums of squares of vector fields, see [7]. Therefore, most of the results shown here can be formulated in a more general form, but we will just stick ourselves to the case of the relativistic operator (which can be seen as the fractional power of an operator given as sum of squares of vector fields perturbed by a mass). Actually, the proofs are easy modifications of the proofs in [7].

    In order to prove a trace Hardy inequality for L we need to find solutions of the extension problem

    (L+2ρ+1sρρ)v(x,ρ)=0,xRn,ρ>0;v(x,0)=f(x),xRn. (2.1)

    The extension problem (2.1) falls into the general theory developed in [28], we also highlight the works in [9,10]. Let us introduce the gradient

    =(1,,n,ρ)

    on Rn×[0,). We define |u(x,ρ)|2=|1u(x,ρ)|2++|nu(x,ρ)|2+|ρu(x,ρ)|2. For 0<s<2, let Ws0(Rn×[0,)) be the completion of C0(Rn×[0,)) with respect to the norm (see [9,10])

    u2(s)=0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρ.

    The following theorem is our first main result: a trace Hardy inequality related to the relativistic operator.

    Theorem 2.1 (General trace Hardy inequality). Let 0<s<2 and let φ be a real valued function in the domain of Ls/2. Assume also that φ1Ls/2φ is locally integrable. Then for any real valued function uWs0(Rn×[0,)), we have the inequality

    0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρ21sΓ(1s/2)Γ(s/2)Rnu2(x,0)Ls/2φ(x)φ(x)dx.

    It is enough to prove the inequality in Theorem 2.1 for functions that belong to C0(Rn×[0,)). Then standard density arguments guarantee the validity for uWs0(Rn×[0,)). Equality is attained when u is a solution of the extension problem, see Proposition 2.5 below. From Theorem 2.1, we can prove the following Hardy type inequality for Ls/2.

    Corollary 2.2 (General Hardy inequality). Let 0<s<2. Let fL2(Rn) be such that Ls/2fL2(Rn). Then

    (Ls/2f,f)Rnf2(x)Ls/2φ(x)φ(x)dx,

    for any real valued function φ in the domain of Ls/2 such that the right hand side is finite.

    Theorem 2.1 and Corollary 2.2 are general results. In Subsection 2.2 we will provide two significative examples of functions φ for which we can deduce Hardy inequalities with specific weights (one will be with a non-homogeneous weight and another with a homogeneous weight).

    In this subsection we prove the results related to general trace Hardy and Hardy inequalities in Theorem 2.1 and Corollary 2.2.

    The proof of the trace Hardy inequality in Theorem 2.1 depends on Lemma 2.3, which is a sort of Picone identity [23]. We include it here for the convenience of the readers. It is initially stated for C0 functions, but it remains valid for functions u coming from the Sobolev space Ws0.

    Lemma 2.3. Let 0<s<2. Let u(x,ρ) be a real valued function in C0(Rn×[0,)) and let v(x,ρ) be another real valued function for which limρ0ρ1sρv(x,ρ) exists and limρ0ρ1sρv(x,ρ)(v(x,0))1L1loc(Rn). We have

    0Rn|u(x,ρ)u(x,ρ)v(x,ρ)v(x,ρ)|2ρ1sdxdρm20Rnu2(x,ρ)ρ1sdxdρ=0Rn|u(x,ρ)|2ρ1sdxdρ+0Rnu(x,ρ)2v(x,ρ)(Lsv(x,ρ))ρ1sdxdρ+Rnu(x,0)2v(x,0)limρ0ρ1sρv(x,ρ)dx,

    where Ls is the operator

    Ls:=L+2ρ+1sρρ.

    Proof. Consider the following integral:

    Rn(juuvjv)2dx=Rn((ju)22uvjujv+(uvjv)2)dx.

    Integrating by parts, we get

    Rnuvjujvdx=Rnuj(uvjv)dx=RnuvjujvdxRnu2j(1vjv)dx.

    Since Rnu2j(1vjv)dx=Rnu2v2(jv)2dx+Rnu2v2jvdx, the above gives

    Rn(u2v2(jv)22uvjujv)dx=Rnu2v2jvdx.

    On the other hand, a similar calculation with the ρ-derivative gives

    0(u2v2(ρv)22uvρuρv)ρ1sdρ=0u2vρ(ρ1sρv)dρ+u(x,0)2v(x,0)limρ0(ρ1sρv)(x,ρ).

    Adding and then taking all integrations into account we get our result.

    In Lemma 2.3, if v satifies the extension problem (2.1), i.e., the equation Lsv=0 on Rn×[0,) (with a given initial condition v(x,0)=φ(x)), then we get the inequality

    0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρRnu2(x,0)v(x,0)limρ0ρ1sρv(x,ρ)dx.

    In view of the above, in order to prove Theorem 2.1 we need to solve the extension problem for L with a given initial conditon φ. We also need to compute limρ0ρ1sρv(x,ρ) in terms of L and φ.

    Before proceeding with the proofs, we first introduce some well known facts about modified Bessel functions and Macdonald's functions that will be needed in a moment. Let Iν(z) be the modified Bessel function of first kind given by the formula (see [19,Chapter 5,Section 5.7])

    Iν(z)=k=0(z/2)ν+2kΓ(k+1)Γ(k+ν+1),|z|<,|argz|<π (2.2)

    and let Kν be the Macdonald's function of order ν defined by (see also [19,Chapter 5,Section 5.7])

    Kν(z)=π2Iν(z)Iν(z)sinνπ,|argz|<π,ν0,±1,±2, (2.3)

    and Kn(z)=limνnKν(z), n=0,±1,±2,. Even more, from (2.2), (2.3) and [19,Chapter 5,Section 5.11] the following asymptotics for the modified Bessel functions Kν, for ν>0, hold

    Kν(r)Γ(ν)2(r2)νasr0andKν(r)π2r1/2erasr. (2.4)

    We recall the following integral representation for the Macdonald's functions, see for instance [19,Chapter 5,(5.10.25)]

    Kν(z)=2ν1zν0e(t+14tz2)tν1dt,|argz|<π/4. (2.5)

    It is clear from (2.5) that Kν(z) is positive for real z>0.

    As explained above, we will make use of the solutions to the extension problem (2.1) to prove the results. Our operator falls into the scope of the general framework developed in [28] and from there one can write a formula for the solution to such an extension problem. Let

    qt(x)=etm2(4πt)n/2e14t|x|2

    be the heat kernel associated to the operator L. Observe that qt(x)=etm2˜qt(x), with Rn˜qt(x)dx=1. For 2<s<2, s0, we define

    us,ρ(x)=ρs2s|Γ(s/2)|0eρ24tqt(x)ts/21dt. (2.6)

    The identity is a Poisson type formula and it was introduced in a general setting in [28]. The integral in (2.6) defines an L1 function and

    Rnus,ρ(x)dx=ρs2s|Γ(s/2)|0eρ24tts/21etm2(Rn(4πt)n/2e14t|x|2dx)dt=ρs2s|Γ(s/2)|0e(ρ24t+tm2)ts/21dt=ρs2s|Γ(s/2)|ms0e(u+14u(ρm)2)us/21du=ρs2s|Γ(s/2)|ms2s/2+1(ρm)s/22s/21(ρm)s/20e(u+14u(ρm)2)us/21du=(ρm)s/22s/21|Γ(s/2)|Ks/2(ρm),

    where we used the identity (2.5) in the last equality. Even more, since qt2Cetm2tγ,γ>1, the integral defining us,ρ defines an L2 function. Indeed,

    us,ρ2Csρs0eρ24tqt2ts/21dtCsρ2γ. (2.7)

    Actually, a better decay in ρ could be provided, in view of (2.5) and (2.4), but this is enough for our purposes.

    As mentioned before, the function us,ρ may be regarded as a generalised Poisson kernel and we can give a result relating this function and the solution of the extension problem, the latter seen as a generalised harmonic extension. Indeed, Theorem 2.4 below was proved in a more abstract setting, but we state it here in the particular case of the relativistic operator and the solution of the corresponding extension problem.

    Theorem 2.4 ([28] Theorem 1.1). For fLp(Rn),1p, the function u(x,ρ)=fus,ρ(x) solves the extension problem (2.1). Moreover, for 0<s<2,

    limρ0ρ1sρ(fus,ρ)=21sΓ(1s/2)Γ(s/2)Ls/2f (2.8)

    where the convergence is understood in the Lp sense, under the extra assumption that Ls/2fLp(Rn), 1p<.

    Lemma 2.3 and Theorem 2.4 are the main ingredients to prove the first main result.

    Proof of Theorem 2.1. As already remarked, it is enough to prove the result when uC0(Rn×[0,)). We take v=φus,ρ and observe that, by Theorem 2.4, v solves the equation (L+2ρ+1sρρ)v=0, with v(x,0)=φ(x). Then, by taking this v in Lemma 2.3 and taking into account (2.8) in Theorem 2.4, we obtain the inequality

    0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρ21sΓ(1s/2)Γ(s/2)Rnu2(x,0)Ls/2φ(x)φ(x)dx,

    as desired.

    Moreover, we claim the equality in Theorem 2.1 when u is the solution of the extension problem with initial condition φ.

    Proposition 2.5. Let 0<s<2 and let φ be a real valued function such that Ls/2φL2(Rn). If u is the solution of the extension problem (2.1) with initial condition φ, then

    0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρ=21sΓ(1s/2)Γ(s/2)Rnφ(x)Ls/2φ(x)dx.

    Proof. Note that if f and g belong both to L2(Rn), then their convolution is uniformly continuous and vanishes at infinity. This can be proved by approximating f and g by a sequence of compactly supported smooth functions. Since φ and us,ρ belong to L2(Rn), due to (2.7), it follows that the solution u of the extension problem vanishes at infinity as a function of x for any fixed ρ. Moreover, jus,ρL2(Rn), and the same is true for ρ[ρsus,ρ]. Integrating by parts and using the fact that u vanishes at infinity, we have

    Rn|ju(x,ρ)|2dx=Rnu(x,ρ)2ju(x,ρ)dx.

    Furthermore, by (2.7), |u(x,ρ)|Cφ2us,ρ2Csρ2γφ2 which goes to 0 as ρ tends to infinity. The same is true for ρu(x,ρ). A similar computation with the ρ-derivative yields

    0(ρu(x,ρ))2ρ1sdρ=0u(x,ρ)ρ(ρ1sρu(x,ρ))dρu(x,0)limρ0(ρ1su)(x,ρ).

    Now, we sum up and use the fact that u solves the extension problem with initial condition φ. The result follows.

    For s>0, let Hs(Rn) be the Sobolev space defined in the following way: fHs(Rn) if and only if fL2(Rn) and Ls/2fL2(Rn). We observe that Proposition 2.5 says that the "energy norm" of the solution u is a constant multiple of the Hs(Rn) norm of the initial condition.

    We now give the proof of Corollary 2.2.

    Proof of Corollary 2.2. Let u(x,ρ)=fus,ρ(x,ρ). By Theorem 2.4, u solves the equation (L+2ρ+1sρρ)u=0, with u(x,0)=f(x). By Lemma 2.3 with v(x,ρ)=u(x,ρ), and taking into account that u=fus,ρ solves the differential equation, we have that

    0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρ=Rnu(x,0)limρ0(ρ1su)(x,ρ)dx.

    Then, by Theorem 2.4, the right hand side of the above identity reduces to

    21sΓ(1s/2)Γ(s/2)Rnf(x)Ls/2f(x)dx.

    On the other hand, by Theorem 2.1, we have that

    0Rn(|u(x,ρ)|2+m2u2(x,ρ))ρ1sdxdρ21sΓ(1s/2)Γ(s/2)Rnu2(x,0)Ls/2φ(x)φ(x)dx.

    Combining all these facts, we conclude the result.

    In this subsection we provide some examples of functions φ which lead to Hardy inequalities with concrete weights. Observe that, in view of Corollary 2.2, the task boils down to finding functions for which the action of Ls/2 can be performed and the quotient Ls/2φφ is simplified. In general, one can obtain inequalities with a weight function w if there exists a function φ such that Ls/2φwφ. Moreover, optimality of the constants is susceptible to be studied if we indeed have Ls/2φ=wφ. Unfortunately, only few examples are known in which these computations can be accomplished.

    We will give two examples that will produce two different Hardy inequalities, one with a "non-homogeneous" weight and another with a "homogeneous" weight. They will be the weights analogous to the corresponding Poisson-type kernel and singular potential in the Euclidean case, and the Hardy inequalities will be the counterpart of (1.2) and (1.1), respectively.

    The common starting point will be the following result relating the functions us,ρ and us,ρ defined in (2.6), via Ls/2.

    Lemma 2.6. For 2<s<2, s0, we have

    ρsLs/2us,ρ=2s|Γ(s/2)||Γ(s/2)|us,ρ.

    Proof. For a function f in the Schwartz class, we define the Fourier transform as

    ˆf(ξ)=1(2π)n/2Rnf(x)eixξdx,ξRn.

    The operator Ls/2 is then defined as the pseudo-differential operator

    ^Ls/2f(ξ)=(|ξ|2+m2)s/2ˆf(ξ),

    so we have to prove that ρs(|ξ|2+m2)s/2^us,ρ(ξ)=Cs^us,ρ(ξ), where Cs is the constant in the statement of the lemma.

    From the definition of us,ρ in (2.6) and since

    ^qt(ξ)=(2π)n/2et(|ξ|2+m2), (2.9)

    it follows that

    ^us,ρ(ξ)=ρs(2π)n/22s|Γ(s/2)|0eρ24tet(|ξ|2+m2)ts/21dt.

    The change of variables t(|ξ|2+m2|)ρ24u turns the above integral into

    1(2π)n/2|Γ(s/2)|(|ξ|2+m2|)s/20eρ24ueu(|ξ|2+m2)us/21du=ρs(|ξ|2+m2|)s/22s|Γ(s/2)||Γ(s/2)|^us,ρ(ξ).

    The proof is complete.

    Lemma 2.6 was proved in [28] for more general operators L, by considering the inner product (Ls/2fus,ρ,g) and using the spectral definition of Ls/2. In fact, the proof of Lemma 2.6 depends essentially on the numerical identity

    ρs2s0eρ24tetλts/21dt=λs/20eρ24tetλts/21dt

    valid for λ>0 (which is true by the change of variables in the proof of Lemma 2.6).

    The first example of Hardy inequality will be an application of Lemma 2.6 with the choice φ=us,ρ, where us,ρ is the function given in (2.6). Thus, it is easy to obtain a "non-homogeneous Hardy type inequality" from the general Hardy inequality in Corollary 2.2.

    Corollary 2.7. Let 0<s<2. Let f be a real valued function on Rn such that f and Ls/2f are in L2(Rn). Then we have

    (Ls/2f,f)msρsRnf2(x)(ρ2+|x|2)sKn+s2(mρ2+|x|2)Kns2(mρ2+|x|2)dx,ρ>0, (2.10)

    where equality is achieved for the function us,ρ.

    Proof. In the case of the operator L, recall that the heat kernel is explicitly given by

    qt(x)=(4πt)n/2e14t|x|2etm2

    and hence the expression in (2.6) can be explicitly computed. Indeed,

    us,ρ(x)=ρs2s|Γ(s/2)|0eρ24t(4πt)n/2e14t|x|2etm2ts/21dt=ρs2s|Γ(s/2)|(4π)n/20eρ2+|x|24tetm2t(s+n)/21dt.

    From (2.5) we can deduce that

    us,ρ(x)=22n+s2πn/2|Γ(s/2)|mn+s2ρsKn+s2(mρ2+|x|2)(ρ2+|x|2)(n+s)/2 (2.11)

    for 2<s<2. Then observe that inequality (2.10) follows from Corollary 2.2 (indeed, if fL2 and Ls/2fL2, the solution of the extension problem given by u(x,ρ)=fus,ρ(x) belongs to Ws0(Rn×[0,)), see [25,Proposition 3.13]), after choosing φ=us,ρ as above, and taking into account Lemma 2.6.

    It is verified that equality holds when we take f(x)=ρsus,ρ by Lemma 2.6 and perform a direct computation.

    Remark 2.8. From the computation in the proof of Corollary 2.7 we see that the solution of the extension problem u(x,ρ)=fus,ρ(x) has an expression as an integral with the explicit Poisson-type kernel (2.11).

    The second example will lead us to a "homogeneous" Hardy inequality. Observe that in the Hardy inequality in Corollary 2.7 we cannot just take limit as ρ goes to zero to get something non-trivial. For any δ>0 and α>0, let Rα,δ to be the function defined by

    Rα,δ(x)=1Γ(α/2)0etδ2qt(x)tα/21dt. (2.12)

    In view of (2.9) and by using the definition of the Gamma function Γ(λ)=0evvλ1dv, we have that

    ^Rα,δ(ξ)=(2π)n/2Γ(α/2)0et(δ2+|ξ|2+m2)tα/21dt=(2π)n/2(δ2+|ξ|2+m2)α/2.

    For 0<α<n, we can consider also the functions Rα:=Rα,0

    Rα(x)=1Γ(α/2)0qt(x)tα/21dt

    and analogously,

    ^Rα(ξ)=(2π)n/2(|ξ|2+m2)α/2. (2.13)

    The functions Rα satisfy RαRβ=Rα+β. This semigroup property can be easily deduced from (2.13), see also [27,page 135]. We will denote the functions Rα as "Riesz potentials for the operator L", but actually they are essentially nothing but the kernels of the classical Bessel potentials for the Euclidean Laplacian (cf. [27,Chapter V,Section 3]). Moreover, they can be explicitly calculated by using the expression for the heat kernel and the integral representation for Kν. Indeed,

    Rα(x)=1Γ(α/2)0qt(x)tα/21dt=1Γ(α/2)0(4πt)n/2e14t|x|2etm2tα/21dt=22α+n2Γ(α/2)πn/2mnα2Knα2(m|x|)|x|nα2. (2.14)

    The second example of Hardy inequality will be with the choice φ=Rα,δus,ρ, where Rα,δ is the one in (2.12) and us,ρ is the function given in (2.6). We will need the properties of the kernels Rα of the Riesz potentials Lα/2, and the resulting Hardy inequality will be with an "homogeneous" weight (homogeneity to be understood near the origin, see Remark 2.10).

    Corollary 2.9. Let 0<s<2. Let fL2(Rn) be such that Ls/2fL2(Rn). Then,

    (Ls/2f,f)2s/2ms/2Γ((n+s)/4)Γ((ns)/4)Rnf2(x)|x|s/2Kn+s4(m|x|)Kns4(m|x|)dx.

    Proof. We consider the inequality in Corollary 2.2 with φ(x)=Rα,δus,ρ. In view of the relation in Lemma 2.6 we obtain that

    (Ls/2f,f)ρs2sΓ(s/2)|Γ(s/2)|Rnf2(x)Rα,δus,ρ(x)Rα,δus,ρ(x)dx.

    Recalling the definition of us,ρ in (2.6) we see that ρsus,ρ(x) converges pointwise to 2s|Γ(s/2)||Γ(s/2)|Rs(x) as ρ tends to zero. We also have ρsRα,δus,ρ(x)2s|Γ(s/2)||Γ(s/2)|Rα,δRs(x) and consequently

    (Ls/2f,f)Rnf2(x)Rα,δus,ρ(x)Rα,δRs(x)dx.

    As us,ρ is an approximate identity, see for instance [7,Proof of Theorem 2.4], by passing first to the limit as ρ goes to zero, then letting δ0, and finally noting that Rα has the semigroup property, we obtain

    (Ls/2f,f)Rnf2(x)Rα(x)Rα+s(x)dx.

    Recalling the explicit expression for Rα in (2.14) and simplifying we get

    (Ls/2f,f)2s/2ms/2Γ((α+s)/2)Γ(α/2)Rnf2(x)|x|s/2Knα2(m|x|)Knαs2(m|x|)dx.

    The choice α=ns2 leads to the desired inequality

    (Ls/2f,f)2s/2ms/2Γ((n+s)/4)Γ((ns)/4)Rnf2(x)|x|s/2Kn+s4(m|x|)Kns4(m|x|)dx.

    This completes the proof.

    Remark 2.10. We could get a lower bound with a more specific weight in the right hand side of the inequality in Corollary 2.9, by using estimates for the ratio of Macdonald's functions. Observe that, in particular, the ratio of Macdonald's functions behaves, near the origin, like

    Kn+s4(m|x|)Kns4(m|x|)(m|x|)s/2,

    and the weight in the right hand side in Corollary 2.9 resembles its counterpart in the Hardy inequality for the Euclidean Laplacian.

    Recall that the fractional relativistic operator is the operator Ls/2ms. We have an integral representation, namely

    Ls/2f(x)msf(x)=cn,smn+s2p.v.Rnf(x)f(y)|xy|n+s2Kn+s2(m|xy|)dy,xRn,

    where the positive constant is given by

    cn,s=2sn2+1πn/21Γ(s/2),

    see [10,24]. Recall also that the Riesz kernel Rα(x) is given by

    Rα(x)=22α+n2Γ(α/2)πn/2mnα2Knα2(m|x|)|x|nα2

    and we also have

    ^Rα(ξ)=(|ξ|2+m2)α/2.

    Finally, let the corresponding ground state representation for the operator Ls/2 be given by

    Hs[f]=(Ls/2f,f)En,sms/2Rnf(x)2|x|s/2Kn+s4(m|x|)Kns4(m|x|)dx

    where

    En,s=2s/2Γ((n+s)/4)Γ((ns)/4).

    The following theorem contains a formula for the ground representation showing that Hs[f] is positive. It is the same as in [11,Proposition 4.1] with the obvious modification, and we sketch the proof.

    Theorem 3.1. Let 0<s<min{1,n}. If uC0(Rn{0}) and v(x)=u(x)(Rn+s2(x))1, then

    Hs[u]=cn,smn+s2RnRn|v(x)v(y)|2Rn+s2(x)Rn+s2(y)Kn+s2(m|xy|)|xy|n+s2dxdy. (3.1)

    Proof. The integral representation for Ls/2f(x)msf(x) gives

    ((Ls/2ms)f,g)=cn,smn+s2RnRn(f(x)f(y))¯g(x)|xy|n+s2Kn+s2(m|xy|)dxdy.

    In view of the symmetry of the kernel, we can also write

    ((Ls/2ms)f,g)=cn,smn+s2RnRn(f(x)f(y))¯g(y)|xy|n+s2Kn+s2(m|xy|)dxdy.

    Adding both identities, we obtain that

    ((Ls/2ms)f,g)=cn,smn+s212RnRn(f(x)f(y))(¯g(x)g(y))|xy|n+s2Kn+s2(m|xy|)dxdy. (3.2)

    For α>s to be chosen later, take g(x)=Rα(x) and f(x)=|u(x)|2Rα(x). By Plancherel identity, the left hand side of (3.2) reads as

    Rn((|ξ|2+m2)s/2ms)ˆf(ξ)ˆg(ξ)dξ=Rn(|ξ|2+m2)s/2α/2ˆf(ξ)dξmsRn(|ξ|2+m2)α/2ˆf(ξ)dξ=Rn|u(x)|2Rαs(x)Rα(x)dxmsRn|u(x)|2dx=2s/2Γ(α/2)Γ((αs)/2)ms/2Rn|u(x)|2|x|s/2Knα+s2(m|x|)Knα2(m|x|)dxmsRn|u(x)|2dx,

    where we used the explicit expression for Rα. With the choice α=n+s2, we arrive at

    2s/2Γ(α/2)Γ((αs)/2)ms/2Rn|u(x)|2|x|s/2Knα+s2(m|x|)Knα2(m|x|)dx=En,sms/2Rnf(x)2|x|s/2Kn+s4(m|x|)Kns4(m|x|)dx.

    The right hand side of (3.2), after simplification, reduces to

    cn,s12RnRn(|u(x)u(y)|2|u(x)Rα(x)u(y)Rα(y)|2Rα(x)Rα(y))Kn+s2(m|xy|)|xy|n+s2dxdy.

    By taking (3.2) into account with f=g, the proof is completed.

    As an immediate corollary of Theorem 3.1, we recover the same fractional Hardy inequality as in Corollary 2.9.

    Corollary 3.2. Let 0<s<2. Let fL2(Rn) be such that Ls/2fL2(Rn). Then

    (Ls/2f,f)2s/2ms/2Γ((n+s)/4)Γ((ns)/4)Rnf(x)2|x|s/2Kn+s4(m|x|)Kns4(m|x|)dx.

    A final remark concerning sharpness is in order.

    Remark 3.3. The constant in Corollary 3.2 is not achieved in the class of functions for which both sides of the inequality are finite. This can be deduced from the ground state representation in Theorem 3.1, which represents the error obtained in the Hardy inequality. It can be also seen that the constant Cn,s,m=2s/2ms/2Γ((n+s)/4)Γ((ns)/4) is sharp, just by the same reasoning as in [11,Remark 4.2]. Indeed, consider a sequence of functions uj, supported in the unit ball, approximating Rn+s2(x) close to the origin. The right hand side in (3.1) remains finite as j, but Rnuj(x)2|x|s/2Kn+s4(m|x|)Kns4(m|x|)dx diverges.

    The author is grateful to Krzysztof Bogdan for posing the question and for clarifications on the topic, and to Sundaram Thangavelu for substantial remarks. She also acknowledges discussions with Agnid Banerjee.

    Many thanks to the referee for the thorough reading of the manuscript and for valuable suggestions and corrections that helped to improve the presentation in an essential way.

    This research was supported by the Basque Government through BERC 2018–2021 program, by Spanish Ministry of Economy and Competitiviness through BCAM Severo Ochoa accreditation SEV-2017-2018 and the project MTM2017-82160-C2-1-P funded by AEI/FEDER, UE. She also acknowledges the project RYC2018-025477-I and IKERBASQUE.

    The author declares no conflict of interest.



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