An extension problem and Hardy's inequality for the fractional Laplace-Beltrami operator on Riemannian symmetric spaces of noncompact type
aa r X i v : . [ m a t h . F A ] J a n AN EXTENSION PROBLEM AND HARDY’S INEQUALITY FOR THEFRACTIONAL LAPLACE-BELTRAMI OPERATOR ON RIEMANNIANSYMMETRIC SPACES OF NONCOMPACT TYPE
MITHUN BHOWMIK AND SANJOY PUSTI
Abstract.
In this paper we study an extension problem for the Laplace-Beltrami operatoron Riemannian symmetric spaces of noncompact type and use the solution to prove Hardy-type inequalities for fractional powers of the Laplace-Beltrami operator. Next, we study themapping properties of the extension operator. In the last part we prove Poincar´e-Sobolevinequalities on these spaces. Introduction
In recent years there has been intensive research on various kinds of inequalities for fractionalorder operators because of their applications to many areas of analysis (see for instance [8, 19,39] and the references therein). The classical definitions of the fractional operator in termsof the Fourier analysis involve functional analysis and singular integrals. They are nonlocalobjects. This fact does not allow to apply local PDE techniques to treat nonlinear problemsfor the fractional operators. To overcome this difficulty, in the Euclidean case, Caffarelliand Silvestre [11] studied the extension problem associated to the Laplacian and realised thefractional power as the map taking Dirichlet data to the Neumann data. On a certain class ofnoncompact manifolds, this definition of the fractional Laplacian through an extension problemhas been studied by Banika et al. [6].In the first part of this article we will concern with the Hardy-type inequalities for thefractional operators. Let ∆ R n = P nj =1 ∂ ∂x j denote the Euclidean Laplacian on R n . For 0 < s Primary 43A85; Secondary 26A33, 22E30. Key words and phrases. Hardy’s inequality, fractional Laplacian, extension problem, Riemannian symmetricspaces. There is another version of Hardy’s inequality where the homogeneous weight function | x | − s is replaced by non-homogeneous one:(1.3) Z R n | f ( x ) | ( δ + | x | ) s dx ≤ − s Γ (cid:0) n − s (cid:1) Γ (cid:0) n +2 s (cid:1) δ − s h ( − ∆ R n ) s f, f i , δ > . Here also the constant is sharp and equality is achieved for the functions ( δ + | x | ) − ( n − s ) / and their translates [10].Generalization of the classical Hardy’s inequality (1.2) to Riemannian manifolds was inten-sively pursued after the seminal work of Carron [12], see for instance [9, 18, 28, 29, 30, 42].In [12], the following weighted Hardy’s inequality was obtained on a complete noncompactRiemannian manifold M : Z M η α |∇ g φ | dv g ≥ (cid:18) C + α − (cid:19) Z M η α φ η dv g , where φ ∈ C ∞ c ( M − η − { } ) , α ∈ R , C > , C + α − > η satisfies |∇ M η | = 1 and | ∆ M η | ≥ C/η in the sense of distribution. Here ∇ g , dv g denote respectivelythe Riemannian gradient and Riemannian measure on M . In the case of Cartan-Hadamardmanifold M of dimension N (namely, a manifold which is complete, simply-connected, and haseverywhere non-positive sectional curvature), the geodesic distance function d ( x, x ), where x ∈ M , satisfies all the assumptions of the weight η and the above inequality holds with thebest constant ( N − / 4, see [30]. Analogues of Hardy-type inequalities for fractional powersof the sublaplacian are also known, for instance, the work by P. Ciatti, M. Cowling and F. Riccifor stratified Lie groups [14]. There the authors have not paid attention to the sharpness ofthe constants. Recently, in [37], Roncal and Thangavelu have proved analogues of Hardy-typeinequalities with sharp constants for fractional powers of the sublaplacian on the Heisenberggroup. For recent results on the Hardy-type inequalities for the fractional operators we refer[10, 36, 38].Our first aim in this article is to prove analogues of Hardy’s inequalities (1.1) and (1.3)for fractional powers of the Laplace-Beltrami operator ∆ on Riemannian symmetric space X of noncompact type. We have the following analogue of Hardy’s inequality in the non-homogeneous case. Theorem 1.1. Let < σ < and y > . Then there exists a constant C σ > such that for F ∈ H σ ( X ) h ( − ∆) σ F, F i ≥ C σ y σ Z { x : | x | + y < } | F ( x ) | ( y + | x | ) σ dx + Z { x : | x | + y ≥ } | F ( x ) | ( y + | x | ) σ dx ! . Remark 1.2. In contrast with the inequality (1.3) for the Euclidean space, we get an improve-ment in the theorem above. This comes as a consequence of the geometry of the symmetricspace. In the following theorem also we get similar improvement.For the homogeneous weight function, we prove the following analogue of Hardy’s inequalityon X . Theorem 1.3. Let < σ < . Then there exists a constant C ′ σ > such that for F ∈ C ∞ c ( X ) h ( − ∆) σ F, F i ≥ C ′ σ Z { x : | x | < } | F ( x ) | | x | σ dx + Z { x : | x |≥ } | F ( x ) | | x | σ dx ! . XTENSION PROBLEM AND HARDY’S INEQUALITY 3 Given σ ∈ (0 , − ∆ R n ) σ on R n is defined as a pseudo-differentialoperator by F (( − ∆ R n ) σ f ) ( ξ ) = | ξ | σ F f ( ξ ) , ξ ∈ R n , where F f is the Fourier transform of f given by F f ( ξ ) = (2 π ) − n/ Z R n f ( x ) e − ix · ξ dx, ξ ∈ R n . It can also be written as the singular integral( − ∆ R n ) σ f ( x ) = c n,σ P.V. Z R n f ( x ) − f ( y ) | x − y | n +2 σ dy, where c n,σ is a positive constant. Caffarelli and Silvestre have developed in [11] an equivalentdefinition of the fractional Laplacian ( − ∆ R n ) σ , σ ∈ (0 , R n +1+ . For a function f : R n → R , consider the solution u : R n × [0 , + ∞ ) → R of the following differential equation∆ R n u + (1 − σ ) y ∂u∂y + ∂ u∂y = 0 , y > u ( x, 0) = f ( x ) , x ∈ R n . Then the fractional Laplacian of f can be computed as( − ∆ R n ) σ f = − σ − Γ( σ )Γ(1 − σ ) lim y → + y − σ ∂u∂y . The Poisson kernel for the fractional Laplacian ( − ∆ R n ) σ in R n is K σ ( x, y ) = c n,σ y σ ( | x | + y ) σ + n , and then u ( x, y ) = f ∗ R n K σ . Therefore( − ∆ R n ) σ f = − σ − Γ( σ )Γ(1 − σ ) lim y → + y − σ ∂∂y ( f ∗ R n K σ )( x ) . Later, Stinga and Torrea [39] showed that one can define the fractional Laplacian on a domainΩ ⊂ R n through the extension (1.4) using the heat-diffusion semigroup generated by theLaplacian ∆ Ω provided that the heat kernel associated with ∆ Ω exists and it satisfies somedecay properties. Since the heat kernel on general noncompact manifolds has been extensivelystudied depending on the underlying geometry, Banica et al. in [6] take this approach to definethe fractional Laplace-Beltrami operator on some noncompact manifolds which in particular,include the Riemannian symmetric spaces of noncompact type. Let d be a Riemannian metricon a Riemannian symmetric space X and ∆ be the corresponding Laplace-Beltrami operatoron X . Also, let g be the product metric on X × R + given by g = d + dy . For σ > 0, let H σ ( X )denote the Sobolev space on X (defined in Section 2). In [6, Theorem 1.1], the following resultis proved for the Riemannian symmetric space X of noncompact type of arbitrary rank. Theorem 1.4. ( Banica; Gon´zalez; S´aez ) Let σ ∈ (0 , . Then for any given f ∈ H σ ( X ) , thereexists a unique solution of the extension problem ∆ u + (1 − σ ) y ∂u∂y + ∂ u∂y = 0 , y > u ( x, 0) = f ( x ) , x ∈ X. MITHUN BHOWMIK AND SANJOY PUSTI Moreover, the fractional Laplace-Beltrami operator on X can be recovered through (1.6) ( − ∆) σ f ( x ) = − σ − Γ( σ )Γ(1 − σ ) lim y → + y − σ ∂u∂y ( x, y ) . The following theorem gives an alternative expression of a solution of the extension problem(1.5), which will be useful for us. The proof is similar to [39, Theorem 1.1]. See also [6,Theorem 3.1] . For the sake of completeness we give a proof in section 3. Theorem 1.5. Let f ∈ Dom ( − ∆) σ . A solution of (1.5) is given by (1.7) u ( x, y ) = 1Γ( σ ) Z ∞ e t ∆ ( − ∆) σ f ( x ) e − y / t dtt − σ , and u is related to ( − ∆) σ f by the equation (1.6). Moreover, the following Poisson formula for u holds: (1.8) u ( x, y ) = Z X f ( ζ ) P σy ( ζ − x ) dζ = ( f ∗ P σy )( x ) , where (1.9) P σy ( x ) = y σ σ Γ( σ ) Z ∞ h t ( x ) e − y / t dtt σ . All these identities in theorem above are to be understood in the L sense. The mappingproperties of the Poisson operator P σ on R n which maps boundary value f to the solution u of the extension problem (1.4) were studied by M¨ollers et al. [34]. In the same paper, theauthors have also obtained a similar result for Heisenberg groups. On the Euclidean spaces,they proved the following Theorem 1.6 (M¨ollers; Ørsted; Zhang) . Let < σ < n . Then (1) P σ : H σ ( R n ) → H σ +1 / ( R n × R + ) is isometric up to a constant. (2) P σ extends to a bounded operator from L p ( R n ) to L q ( R n × R + ) , for < p ≤ ∞ and q = n +1 n p (Figure 1, (a)). In [13], Chen proved that for particular values p = nn − σ and q = n +2 n − σ , there exists a sharpconstant C such that k P σ f k L q ( R n ) ≤ C k f k L p ( R n ) , for f ∈ L p ( R n ) , and the optimizer of this inequality are translations, dilations and multiples of the function f ( x ) = (cid:0) | x | (cid:1) − n + σ . Our second main aim in this article is to study the mapping properties of the “Poissonoperator” T σ given by(1.10) T σ f ( x, y ) = f ∗ P σy , x ∈ X, y > , which maps f to the solution u of the extension problem (1.5) related to the Laplace-Beltramioperator on Riemannian symmetric spaces of noncompact type. The following analogue ofTheorem 1.6 is our main result in this direction. Theorem 1.7. Let dim X = n and < σ < . Then (1) T σ : H σ ( X ) → H σ +1 / ( X × R + ) is isometric up to a constant. (2) T σ extends to a bounded operator from L p ( X ) to L q ( X × R + ) , for < p < ∞ and p < q ≤ n +1 n p ; and from L ( X ) to L q ( X ) , for < q < n +1 n (Figure 1, (b)). XTENSION PROBLEM AND HARDY’S INEQUALITY 5 q = ( + / n ) p ( a ) + / nq q = ( + / n ) p q = p ( b ) + / nq Figure 1. (a) Euclidean (b) Symmetric spaces Remark 1.8. In contrast with Theorem 1.6 on Euclidean space, the exponents p, q in The-orem 1.7 on X can vary over a much larger region (see in the figure 1 above). This strikingphenomenon comes as a consequence of the Kunze-Stein phenomenon. The Kunze-Stein phe-nomenon, proved by Cowling [16] on connected semi-simple Lie groups G with finite center,says that the convolution inequality L ( G ) ∗ L p ( G ) ⊂ L ( G ) , holds for p ∈ [1 , p = 1. We use the following generalize version [17, Theorem 2.2, (ii)]: let k ∈ L q ( X ), for1 < q ≤ ≤ p < q . Then the map f f ∗ k is bounded from L p ( X ) to L q ( X ).An explicit expression of the heat kernel is known for certain symmetric spaces. Using thisin section 5, we write the precise expression of the kernel P σy in the case of complex and rankone symmetric spaces.The final topic we shall deal with here is analogues of the Poincar´e-Sobolev inequalitiesfor the fractional Laplace-Beltrami operator on X . In [33], Mancini and Sandeep proved thefollowing optimal Poincar´e-Sobolev inequalities for the Laplace-Beltrami operator ∆ H n on thereal hyperbolic space H n of dimension n ≥ Theorem 1.9. ( Mancini; Sandeep ) Let n ≥ . Then for < p ≤ nn − , there exists S = S n,p > such that for all u ∈ C ∞ c ( H n ) , k (cid:0) − ∆ H n − ( n − / (cid:1) / u k L ( H n ) ≥ S k u k L p ( H n ) . In case of real hyperbolic space H of dimension three, Benguria, Frank and Loss [8] provedthat the best constant S in the theorem above is the same as the best sharp Sobolev constantfor the first order Sobolev inequality on H . Recently, using Green kernel estimates Li, Lu,Yang [31, Theorem 6.2] proved the following Poincar´e-Sobolev inequalities for the fractionalLaplace-Beltrami operator ∆ H n on H n . Theorem 1.10. ( Li; Lu; Yang ) Let n ≥ and ≤ σ < . Then there exists a constant C = C n,σ,p > such that k (cid:0) − ∆ H n − ( n − / (cid:1) σ u k L ( H n ) ≥ C k u k L nn − σ ( H n ) , for u ∈ H σ ( H n ) . For related results and their sharpness, we refer the reader to [32, 40]. Our aim in the finalsection is to prove an analogue of the Poincar´e-Sobolev inequality for the fractional Laplace-Beltrami operator ∆ on X which generalizes the above mentioned theorems. The idea of theproof is to use the estimate of the Bassel-Green-Riesz kernel due to Anker-Ji [4]. Since weare working on general Riemannian symmetric spaces of noncompact type, it is difficult to getthe explicit values of the constants involve and we do not make attempt to get the optimal MITHUN BHOWMIK AND SANJOY PUSTI constant. Here is our final result. We refer the reader to the next section for the unexplainednotation used in the theorem below. Theorem 1.11. Let dim X = n ≥ and < σ < min { l + 2 | Σ +0 | , n } . Then for < p ≤ nn − σ there exists S = S n,σ,p > such that for all u ∈ H σ ( X ) , k ( − ∆ − | ρ | ) σ/ u k L ( X ) ≥ S k u k L p ( X ) . Preliminaries In this section, we describe the necessary preliminaries regarding semisimple Lie groups andharmonic analysis on Riemannian symmetric spaces. These are standard and can be found,for example, in [20, 24, 25, 26]. To make the article self-contained, we shall gather only thoseresults which will be used throughout this paper.2.1. Notations. Let G be a connected, noncompact, real semisimple Lie group with finitecentre and g its Lie algebra. We fix a Cartan involution θ of g and write g = k ⊕ p where k and p are +1 and − θ respectively. Then k is a maximal compact subalgebraof g and p is a linear subspace of g . The Cartan involution θ induces an automorphism Θ ofthe group G and K = { g ∈ G | Θ( g ) = g } is a maximal compact subgroup of G . Let B denotethe Cartan Killing form of g . It is known that B | p × p is positive definite and hence inducesan inner product and a norm k · k B on p . The homogeneous space X = G/K is a smoothmanifold. The tangent space of X at the point o = eK can be naturally identified to p andthe restriction of B on p then induces a G -invariant Riemannian metric d on X . For x ∈ X and r > 0, we denote B ( x, r ) to be the ball of radius r centered at x in this metric.Let a be a maximal subalgebra in p ; then a is abelian. We assume that dim a = l , calledthe real rank of G . We can identify a endowed with the inner product induced from p with R d and let a ∗ be the real dual of a . The set of restricted roots of the pair ( g , a ) is denoted by Σ.It consists of all α ∈ a ∗ such that g α = { X ∈ g | [ Y, X ] = α ( Y ) X, for all Y ∈ a } is nonzero with m α = dim( g α ). We choose a system of positive roots Σ + and with respect toΣ + , the positive Weyl chamber a + = { X ∈ a | α ( X ) > , for all α ∈ Σ + } . We also let Σ +0 bethe set of positive indivisible roots. We denote by n = ⊕ α ∈ Σ + g α . Then n is a nilpotent subalgebra of g and we obtain the Iwasawa decomposition g = k ⊕ a ⊕ n .If N = exp n and A = exp a then N is a Nilpotent Lie group and A normalizes N . For thegroup G , we now have the Iwasawa decomposition G = KAN , that is, every g ∈ G can beuniquely written as g = κ ( g ) exp H ( g ) η ( g ) , κ ( g ) ∈ K, H ( g ) ∈ a , η ( g ) ∈ N, and the map ( k, a, n ) kan is a global diffeomorphism of K × A × N onto G . Let n be the dimension of X then n = l + X α ∈ Σ + m α . XTENSION PROBLEM AND HARDY’S INEQUALITY 7 We always assume that n ≥ 2. Let ρ denote the half sum of all positive roots counted withtheir multiplicities: ρ = 12 X α ∈ Σ + m α α. It is known that the L -spectrum of the Laplace-Beltrami operator ∆ on X is the half-line( −∞ , −| ρ | ]. Let M ′ and M be the normalizer and centralizer of a in K respectively. Then M is a normal subgroup of M ′ and normalizes N . The quotient group W = M ′ /M is afinite group, called the Weyl group of the pair ( g , k ). The Weyl group W acts on a by theadjoint action. It is known that W acts as a group of orthogonal transformations (preservingthe Cartan-Killing form) on a . Each w ∈ W permutes the Weyl chambers and the action of W on the Weyl chambers is simply transitive. Let A + = exp a + . Since exp : a → A is anisomorphism we can identify A with R d . If A + denotes the closure of A + in G , then one hasthe polar decomposition G = KAK , that is, each g ∈ G can be written as g = k (exp Y ) k , k , k ∈ K, Y ∈ a . In the above decomposition, the A component of g is uniquely determined modulo W . Inparticular, it is well defined in A + . The map ( k , a, k ) k ak of K × A × K into G inducesa diffeomorphism of K/M × A + × K onto an open dense subset of G . We extend the innerproduct on a induced by B to a ∗ by duality, that is, we set h λ, µ i = B ( Y λ , Y µ ) , λ, µ ∈ a ∗ , Y λ , Y µ ∈ a , where Y λ is the unique element in a such that λ ( Y ) = B ( Y λ , Y ) , for all Y ∈ a . This inner product induces a norm, again denoted by | · | , on a ∗ , | λ | = h λ, λ i , λ ∈ a ∗ . The elements of the Weyl group W acts on a ∗ by the formula sY λ = Y sλ , s ∈ W, λ ∈ a ∗ . Let a ∗ C denote the complexification of a ∗ , that is, the set of all complex-valued real linearfunctionals on a . The usual extension of B to a ∗ C , using conjugate linearity is also denoted by B . Through the identification of A with R d , we use the Lebesgue measure on R d as the Haarmeasure da on A . As usual on the compact group K , we fix the normalized Haar measure dk and dn denotes a Haar measure on N . The following integral formulae describe the Haarmeasure of G corresponding to the Iwasawa and polar decomposition respectively. For any f ∈ C c ( G ), Z G f ( g ) dg = Z K Z a Z N f ( k exp Y n ) e ρ ( Y ) dn dY dk = Z K Z A + Z K f ( k ak ) J ( a ) dk da dk , where dY is the Lebesgue measure on R d and for H ∈ a + (2.1) J (exp H ) = c Y α ∈ Σ + (sinh α ( H )) m α ≍ Y α ∈ Σ + (cid:18) α ( H )1 + α ( H ) (cid:19) m α e ρ ( H ) , MITHUN BHOWMIK AND SANJOY PUSTI where c (in the equality above) is a normalizing constant. If f is a function on X = G/K then f can be thought of as a function on G which is right invariant under the action of K . Itfollows that on X we have a G invariant measure dx such that(2.2) Z X f ( x ) dx = Z K/M Z a + f ( k exp Y ) J (exp Y ) dY dk M , where dk M is the K -invariant measure on K/M .2.2. Fourier analysis on X . For a sufficiently nice function f on X , its Fourier transform e f is a function defined on a ∗ C × K given by e f ( λ, k ) = Z G f ( g ) e ( iλ − ρ ) H ( g − k ) dg, λ ∈ a ∗ C , k ∈ K, whenever the integral exists [25, P. 199]. As M normalizes N the function k e f ( λ, k ) is right M -invariant. It is known that if f ∈ L ( X ) then e f ( λ, k ) is a continuous function of λ ∈ a ∗ ,for almost every k ∈ K (in fact, holomorphic in λ on a domain containing a ∗ ). If in addition, e f ∈ L ( a ∗ × K, | c ( λ ) | − dλ dk ) then the following Fourier inversion holds, f ( gK ) = | W | − Z a ∗ × K e f ( λ, k ) e − ( iλ + ρ ) H ( g − k ) | c ( λ ) | − dλ dk, for almost every gK ∈ X [25, Chapter III, Theorem 1.8, Theorem 1.9]. Here c ( λ ) denotesHarish Chandra’s c -function. Moreover, f e f extends to an isometry of L ( X ) onto L ( a ∗ + × K, | c ( λ ) | − dλ dk ) [25, Chapter III, Theorem 1.5]: Z X | f ( x ) | dx = | W | − Z a ∗ × K | e f ( λ, k ) | | c ( λ ) | − dλ dk. It is known [26, Ch. IV, prop 7.2] that there exists a positive number C and d ∈ N such thatfor all λ ∈ a ∗ + | c ( λ ) | − ≤ C (1 + | λ | ) n − l , for | λ | ≥ ≤ C (1 + | λ | ) d , for | λ | < . We now specialize in the case of K -biinvariant function f on G . Using the polar decompositionof G we may view a K -biinvariant integrable function f on G as a function on A + , or by usingthe inverse exponential map we may also view f as a function on a solely determined byits values on a + . Henceforth, we shall denote the set of K -biinvariant functions in L ( G ) by L ( K \ G/K ). If f ∈ L ( K \ G/K ) then the Fourier transform e f reduces to the spherical Fouriertransform b f ( λ ) which is given by the integral(2.4) e f ( λ, k ) = b f ( λ ) := Z G f ( g ) φ − λ ( g ) dg, for all k ∈ K where(2.5) φ λ ( g ) = Z K e − ( iλ + ρ ) (cid:0) H ( g − k ) (cid:1) dk, λ ∈ a ∗ C , is Harish Chandra’s elementary spherical function. We now list down some well-known prop-erties of the elementary spherical functions which are important for us ([4, Prop. 2.2.12], [20,Prop. 3.1.4]; [25, Lemma 1.18, P. 221]). Theorem 2.1. (1) φ λ ( g ) is K -biinvariant in g ∈ G and W -invariant in λ ∈ a ∗ C . XTENSION PROBLEM AND HARDY’S INEQUALITY 9 (2) φ λ ( g ) is C ∞ in g ∈ G and holomorphic in λ ∈ a ∗ C . (3) The elementary spherical function φ satisfies the following global estimate: (2.6) φ (exp H ) ≍ Y α ∈ Σ +0 (1 + α ( H )) e − ρ ( H ) , for all H ∈ a + . (4) For all λ ∈ a ∗ + we have (2.7) | φ λ ( g ) | ≤ φ ( g ) ≤ . Function spaces on X . For 1 ≤ p < ∞ we define L p ( X × R ) = (cid:26) u | k u k pL p ( X × R ) := Z X × R | u ( x, y ) | p dx dy < ∞ (cid:27) , and L p ( X × R + ) to be the subspace of L p ( X × R ) consisting of all functions u ( x, y ) which areeven in the y -variable. We also define L ∞ ( X × R + ) analogously. For σ > 0, the Sobolev spaceof order σ on X is defined by H σ ( X ) = (cid:8) f ∈ L ( X ) | k f k H σ ( X ) := Z a ∗ × K | ˜ f ( λ, k ) | ( | λ | + | ρ | ) σ | c ( λ ) | − dλ dk < ∞ (cid:9) . Similarly, for σ > H σ ( X × R ) as the space of all functions u ∈ L ( X × R ) suchthat k u k H σ ( X × R ) := Z R Z a ∗ × K |F (˜ u ( λ, k, · )( ξ )) | (cid:0) | λ | + | ρ | + ξ (cid:1) σ | c ( λ ) | − dλ dk dξ < ∞ , where F ˜ u ( λ, k, · )( ξ ) denotes the Euclidean Fourier transform of the function y ˜ u ( λ, k, y )at the point ξ ∈ R , for almost every ( λ, k ) ∈ a ∗ × K . Let H σ ( X × R + ) be the subspace of H σ ( X × R ) consisting of all elements u ( x, y ) which are even in the y -variable.2.4. Heat kernel on X . For the details of the heat kernel h t on X = G/K we refer [3, 4]. Itis a family { h t : t > } of smooth functions with the following properties:(a) h t ∈ L p ( K \ G/K ) , p ∈ [1 , ∞ ], for each t > t > h t is positive with(2.8) Z G h t ( g ) dg = 1 . (c) h t + s = h t ∗ h s , t, s > . (d) For each f ∈ L p ( G/K ) , p ∈ [1 , ∞ ) the function u ( x, t ) = f ∗ h t ( x ), for x ∈ X solvesthe heat equation ∆ x u ( x, t ) = ∂∂t u ( x, t ) u ( · , t ) → f in L p ( X ) , as t → . (e) The spherical Fourier transform of h t is given by(2.9) b h t ( λ ) = e − t ( | λ | + | ρ | ) , λ ∈ a ∗ . We need the following both side estimates of the heat kernel [4, Theorem 3.7]. Theorem 2.2. Let κ be an arbitrary positive number. Then there exists positive constants C , C (depending on κ ) such that C ≤ h t (exp H ) t − n (1 + t ) n − l −| Σ +0 | nQ α ∈ Σ +0 (1 + α ( H ) o e −| ρ | t − ρ ( H ) − | H | t ≤ C , for all t > , and H ∈ a + , with | H | ≤ κ (1 + t ) . For H ∈ a + with t ≪ H , we will use the following global upper bound [3, Theorem 3.1](2.10) | h t (exp H ) | ≤ t − d (1 + | H | ) d e −| ρ | t − ρ ( H ) −| H | / (4 t ) , where d and d are positive constants depending on the position of H ∈ a + with respect tothe walls and on the relative size of t > | H | .3. Extension problem and kernel estimates Since we are dealing with fractional operators, it is natural to relate the fractional Laplace-Beltrami operator acting on f to the solution u in (1.5). We proceed by proving Theorem 1.5which will provide us an expression for the Poisson kernel of the extension operator. This willcrucially be used throughout this paper. Proof of Theorem 1.5. Using the heat-diffusion semigroup generated by the Laplace-Beltramioperator, the first part of the theorem follows exactly as in [39, Theorem 1.1]. We will provethe second part. Let f ∈ Dom ( − ∆) σ , and u be the solution of the extension problem (1.5)given by equation (1.7). It now follows that h u ( · , y ) , g i = 1Γ( σ ) Z ∞| ρ | Z ∞ e − tλ λ σ e − y / t dtt − σ dE f,g ( λ ) , for all g ∈ L ( X ), where dE f,g ( λ ) is the regular Borel complex measure of bounded variationconcentrated on the spectrum [ | ρ | , ∞ ) of − ∆ with d | E f,g | ( | ρ | , ∞ ) ≤ k f k L ( X ) k g k L ( X ) . Bythe Fubini’s theorem, putting rλ = y / t the above equation yields h u ( · , y ) , g i = y σ σ Γ( σ ) Z ∞ Z ∞| ρ | e − y / r e − rλ dE f,g ( λ ) drr σ = y σ σ Γ( σ ) Z ∞ (cid:10) e r ∆ f, g (cid:11) L ( X ) e − y / r drr σ = (cid:28) y σ σ Γ( σ ) Z ∞ e r ∆ f e − y / r drr σ , g (cid:29) . This proves that u ( x, y ) = y σ σ Γ( σ ) Z ∞ e r ∆ f ( x ) e − y / r drr σ . Now, using e t ∆ f = f ∗ h t and Fubini’s theorem, we get from the above equation that u ( x, y ) = y σ σ Γ( σ ) Z X Z ∞ f ( xz − ) h t ( z ) e − y / t dtt σ dz = f ∗ P σy ( x ) , where the kernel P σy is given by the equation (1.9). (cid:3) As in [39, Theorem 2.1], we have the following consequences of the theorem above. XTENSION PROBLEM AND HARDY’S INEQUALITY 11 Corollary 3.1. Let u ( x, y ) = ( f ∗ P σy )( x ) , for x ∈ X, y > be the solution of the extensionproblem (1.5) given in Theorem 1.5. Then (a) sup y ≥ | u ( x, y ) | ≤ sup t ≥ | f ∗ h t | in X . (b) k u ( · , y ) k L p ( X ) ≤ k f k L p ( X ) , for all y ≥ and p ∈ [1 , ∞ ] . (c) lim y → + u ( · , y ) = f in L p ( X ) , for p ∈ [1 , ∞ ) .Proof. Part (a) follows from the expression of the Poisson kernel given in equation (1.9). For(b), we observe that e t ∆ has the contraction property in L p ( X ). Hence, by equation (1.9) andMinkowski’s integral inequality it follows that k u ( · , y ) k L p ( X ) ≤ y σ σ Γ( σ ) Z ∞ k f ∗ h t k L p ( X ) e − y / t dtt σ ≤ k f k L p ( X ) . Similarly, for part (c) we observe that k u ( · , y ) − f k L p ( X ) ≤ y σ σ Γ( σ ) Z ∞ k f ∗ h t − f k L p ( X ) e − y / t dtt σ . Since k f ∗ h t − f k L p ( X ) ≤ k f k L p ( X ) , using dominated convergence theorem the result followsfrom the fact that lim t → + f ∗ h t = f in L p ( X ), for p ∈ [1 , ∞ ). (cid:3) For 0 < σ < y > 0, let us define the function P − σy given by the equation (1.9), that is P − σy ( x ) = y − σ − σ Γ( − σ ) Z ∞ h t ( x ) e − y / t dtt − σ , for x ∈ X. By the estimate of the heat kernel (Theorem 2.2), it follows that P − σy is well defined. For0 < σ < 1, we observe that Γ( − σ ) := Γ(1 − σ ) − σ < P − σy ≤ 0. Since the heat kernel h t is K -biinvariant so is the function P − σy . By (2.4) the spherical Fourier transform is given by(3.1) d P − σy ( λ ) = Z X P − σy ( x ) φ − λ ( x ) dx = y − σ − σ Γ( − σ ) Z ∞ b h t ( λ ) e − y / t dtt − σ , for λ ∈ a ∗ . Interchange of the integration is possible by the Fubini’s theorem. Indeed, by (2.7) and (2.9) Z ∞ Z X h t ( x ) | φ − λ ( x ) | dx e − y / t dtt − σ ≤ Z ∞ Z X h t ( x ) φ ( x ) dx e − y / t dtt − σ = Z ∞ e − t | ρ | e − y / t dtt − σ < ∞ . Moreover, P − σy is contained in the Sobolev space H σ ( X ). Indeed, by using (3.1), (2.9) andMinkowski’s integral inequality we get that k P − σy k H σ ( X ) = (cid:18)Z a ∗ | d P − σy ( λ ) | (cid:0) | λ | + | ρ | (cid:1) σ | c ( λ ) | − dλ (cid:19) ≤ y − σ − σ | Γ( − σ ) | Z ∞ (cid:18)Z a ∗ | b h t ( λ ) | ( | λ | + | ρ | ) σ | c ( λ ) | − dλ (cid:19) e − y / t dtt − σ ≤ y − σ − σ | Γ( − σ ) | Z ∞ (cid:18)Z a ∗ e − t ( | λ | + | ρ | ) ( | λ | + | ρ | ) σ | c ( λ ) | − dλ (cid:19) e − | ρ | t e − y / t dtt − σ = I + I , where(3.2) I = y − σ − σ | Γ( − σ ) | Z (cid:18)Z a ∗ e − t ( | λ | + | ρ | ) ( | λ | + | ρ | ) σ | c ( λ ) | − dλ (cid:19) e − | ρ | t e − y / t dtt − σ , and I is defiend as above with the integration in the t -variable over the interval [1 , ∞ ). It isenough to show that both I and I are finite. We consider I first. Using the property (2.3)of | c ( λ ) | − , we estimate the inner integral in the equation above as follows Z { λ ∈ a ∗ : | λ | < } e − t | λ | ( | λ | + | ρ | ) σ + d dλ + Z { λ ∈ a ∗ : | λ |≥ } e − t | λ | ( | λ | + | ρ | ) σ + n − l dλ ≤ C + C Z ∞ e − tr r σ + n − l ) r l − dr ≤ C + C t − ( σ + n − l/ . It now follows from (3.2) that I ≤ C Z t − ( σ + n − l/ e − | ρ | t e − y / t dtt − σ < ∞ . On the other hand I ≤ C Z ∞ (cid:18)Z a ∗ e − | λ | + | ρ | ) ( | λ | + | ρ | ) σ | c ( λ ) | − dλ (cid:19) e − | ρ | t e − y / t dtt − σ ≤ C k h / k H σ ( X ) . This completes the proof that P − σy ∈ H σ ( X ).The proofs of Hardy’s inequalities is crucially depend on the following lemma. Lemma 3.2. For < σ < and y > we have, ( − ∆) σ P − σy ( x ) = σ Γ( σ ) y σ Γ( − σ ) P σy ( x ) .Proof. Let f ∈ H σ ( X ) and u ( x, y ) = f ∗ P σy ( x ) be the solution of the extension problem (1.5).For any g ∈ L ( X ) we have by equation (1.7) that h u ( · , y ) , g i = 1Γ( σ ) Z ∞ (cid:10) e t ∆ ( − ∆) σ f, g (cid:11) e − y / t dtt − σ (3.3) = 1Γ( σ ) Z ∞ Z ∞| ρ | e − tλ λ σ dE f,g ( λ ) e − y / t dtt − σ = 1Γ( σ ) Z ∞| ρ | (cid:18) λ σ Z ∞ e − tλ t σ − e − y / t dt (cid:19) dE f,g ( λ ) . By using change of variable t → y / (4 λr ) we get the following formula ([10], p. 2582, equa-tion(2.5)) λ σ Z ∞ e − tλ t σ − e − y / t dt = y σ σ Z ∞ e − tλ t − σ − e − y / t dt. Using this in the equation above it follows that h u ( · , y ) , g i = y σ σ Γ( σ ) Z ∞ t − σ − e − y / t Z ∞| ρ | e − tλ dE f,g ( λ ) ! dt = y σ σ Γ( σ ) Z ∞ (cid:10) e t ∆ f, g (cid:11) t − σ − e − y / t dt. (3.4) XTENSION PROBLEM AND HARDY’S INEQUALITY 13 Therefore, from equations (3.3) and (3.4) we have Z ∞ e t ∆ ( − ∆) σ f ( x ) e − y / t t σ − dt = y σ σ Z ∞ e t ∆ f ( x ) e − y / t t − σ − dt. If we take the function f to be the heat kernel h t , t > 0, then the equation above reduces to Z ∞ ( − ∆) σ ( h t + t )( x ) e − y / t t σ − dt = y σ σ Z ∞ h t + t ( x ) e − y / t t − σ − dt. Taking t → 0, we get from dominated convergent theorem that(3.5) Z ∞ ( − ∆) σ h t ( x ) e − y / t t σ − dt = y σ σ Z ∞ h t ( x ) e − y / t t − σ − dt. Using (3.5) and (1.9) we get( − ∆) σ P − σy ( x ) = y − σ − σ Γ( − σ ) Z ∞ ( − ∆) σ h t ( x ) e − y / t t σ − dt = 1Γ( − σ ) Z ∞ h t ( x ) e − y / t t − σ − dt = 4 σ Γ( σ ) y σ Γ( − σ ) P σy ( x ) . This completes the proof. (cid:3) We will now compute the asymptotic behaviour of the Poisson kernel P σy for arbitrary rankRiemannian symmetric spaces of noncompact type. We use this estimate crucially for theremaining part of this article. Theorem 3.3. For − < σ < , σ = 0 and y > we have Γ( σ ) P σy ( x ) ≍ y σ σ p | x | + y − l/ − / − σ −| Σ +0 | φ ( x ) e −| ρ | √ | x | + y , for | x | + y ≥ , ≍ y σ (cid:0) | x | + y (cid:1) − n/ − σ , for | x | + y < . Proof. We first assume that | x | + y < 1. In this case, we will use the following local expansionof the heat kernel h t ( x )(3.6) h t ( x ) = e −| x | / t t − n/ v ( x ) + e − c | x | /t O (cid:16) t − n/ (cid:17) , where v ( x ) = (4 π ) − n/ + O ( | x | ) and c < / σ ) P σy ( x ) = y σ σ Z (cid:16) e −| x | / t t − n/ v ( x ) + e − c | x | /t O (cid:16) t − n/ (cid:17)(cid:17) e − y / t dtt σ + y σ σ Z ∞ h t ( x ) e − y / t dtt σ = y σ σ v ( x ) Z e − ( | x | + y ) / t t − n/ − − σ dt + y σ σ Z e − ( c | x | + y / /t O ( t − n/ ) t − − σ dt + y σ σ Z ∞ h t ( x ) e − y / t dtt σ . We write the right-hand side of the equation above as I + I + I , where I , I and I are the first,second and third term respectively. Then applying change of variable s = (cid:0) | x | + y (cid:1) / (4 t ), wehave I = 4 n y σ (cid:0) | x | + y (cid:1) − n/ − σ v ( x ) Z ∞ ( | x | + y ) / e − s s n/ σ − ds. As | x | + y < Z ∞ e − s s n/ σ − ds ≤ Z ∞ ( | x | + y ) / e − s s n/ σ − ds ≤ Z ∞ e − s s n/ σ − ds. This implies that for | x | + y < I ≍ y σ (cid:0) | x | + y (cid:1) − n/ − σ , as v ( x ) = (4 π ) − n/ + O ( | x | ). For I , using c < / I ≤ C y σ σ Z e − c ( | x | + y ) /t O ( t − n/ ) t − − σ dt ≤ C y σ (cid:0) | x | + y (cid:1) − n/ − σ +1 Z ∞ c ( | x | + y ) e − s s n/ σ − ds ≤ C y σ (cid:0) | x | + y (cid:1) − n/ − σ Z ∞ e − s s n/ σ − ds ≤ C y σ (cid:0) | x | + y (cid:1) − n/ − σ . For the integral I , using Theorem 2.2 we get that for | x | + y < I = y σ σ Z ∞ h t ( x ) e − y / t dtt σ ≤ Cy σ ≤ Cy σ (cid:0) | x | + y (cid:1) − n/ − σ . This proves that for | x | + y < σ ) P σy ( x ) ≍ y σ ( | x | + y ) − n/ − σ . We will now assume that | x | + y ≥ 1. Let us fix a positive number κ > 4. We proceed as inthe proof of [4, Theorem 4.3.1].Γ( σ ) P σy ( x ) = y σ σ Z ∞ h t ( x ) e − y / t dtt σ = y σ σ { I + I + I } , where the quantities I , I and I are defined by the integration of the above integrand h t ( x ) e − y / t t − − σ over the intervals (cid:2) , κ − b (cid:1) , (cid:2) κ − b, κb (cid:1) and (cid:2) κb, ∞ (cid:1) with b = p | x | + y / (2 | ρ | )respectively. For the integral I , using Theorem 2.2 and the asymptotic of φ in Theorem 2.1 XTENSION PROBLEM AND HARDY’S INEQUALITY 15 (3) we get the following: I ≍ Z κ √ | x | y | ρ | κ − √ | x | y | ρ | t − n/ (1 + t ) ( n − l ) / −| Σ +0 | Y α ∈ Σ +0 (1 + α ( x )) e −| ρ | t − ρ (log x ) −| x | / t e − y / t dtt σ ≍ Z κ √ | x | y | ρ | κ − √ | x | y | ρ | t − l/ −| Σ +0 | φ ( x ) e −| ρ | t e − ( | x | + y ) / t dtt σ = Z κκ − ( s p | x | + y / | ρ | ) − l/ − σ − −| Σ +0 | φ ( x ) e − s | ρ | √ | x | + y / e −| ρ | √ | x | + y / (2 s ) p | x | + y | ρ | ! ds ≍ p | x | + y | ρ | ! − l/ − σ −| Σ +0 | φ ( x ) Z κκ − e − √ | x | + y | ρ | ( s +1 /s ) / ds. The last both side estimate follows because κ − ( l/ σ +1+ | Σ +0 | ) ≤ s − ( l/ σ +1+ | Σ +0 | ) ≤ κ ( l/ σ +1+ | Σ +0 | ) . Now, using the fact that Z κκ − e −| ρ | √ | x | + y ( s +1 /s ) / ds ≍ | ρ | − / ( | x | + y ) − / e −| ρ | √ | x | + y , (this follows by the Laplace method [15, Ch 5]) we get from the above equation that I ≍ (cid:16)p | x | + y (cid:17) − l/ − / − σ −| Σ +0 | φ ( x ) e −| ρ | √ | x | + y . For the third integral I , we will use the fact that κ > 4. Using Theorem 2.2, we get I ≤ φ ( x ) Z ∞ κ √ | x | + y / (2 | ρ | ) t − l/ −| Σ +0 |− − σ e −| ρ | t e − ( | x | + y ) / t dt ≤ φ ( x ) (cid:16) κ p | x | + y / (2 | ρ | ) (cid:17) − l/ −| Σ +0 |− Z ∞ κ √ | x | + y / (2 | ρ | ) t − σ e −| ρ | t e − ( | x | + y ) / (4 t ) dt ≤ Cφ ( x ) (cid:16)p | x | + y (cid:17) − l/ −| Σ +0 |− e −| ρ | k √ | x | + y / (4 | ρ | ) Z ∞ κ √ | x | + y / (2 | ρ | ) t − σ e −| ρ | t/ e − ( | x | + y ) / (4 t ) dt ≤ C (cid:16)p | x | + y (cid:17) − l/ −| Σ +0 |− / φ ( x ) e − ( | ρ | + η ) √ | x | + y , where η = | ρ | κ/ − | ρ | > 0. For the first integral I , we use heat kernel Gaussian estimate(2.10) and the estimate of φ in Theorem 2.1 to obtain the following I ≤ Z κ − √ | x | + y / (2 | ρ | )0 t − d (1 + | x | ) d e −| ρ | t − ρ (log x ) e − ( | x | + y ) / (4 t ) dtt σ ≤ (1 + | x | ) d −| Σ +0 | φ ( x ) Z κ − √ | x | + y / (2 | ρ | )0 e − ( | x | + y ) / (4 t ) t − − σ − d dt = (1 + | x | ) d −| Σ +0 | φ ( x ) Z κ − √ | x | + y / (2 | ρ | )0 e − ( | x | + y ) / (8 t ) e − ( | x | + y ) / (8 t ) t − − σ − d dt ≤ C (1 + | x | ) d −| Σ +0 | φ ( x ) e −| ρ | κ √ x y Z κ − √ | x | + y / (2 | ρ | )0 e − ( | x | + y ) / (8 t ) t − − σ − d dt ≤ C (1 + | x | ) d −| Σ +0 | φ ( x ) e − ( | ρ | + ǫ ) √ | x | + y ( | x | + y ) − σ − d , for some ǫ > 0, as κ > (cid:3) To prove Hardy’s inequalities we use an integral representation for the operator ( − ∆) σ . Thefollowing function(3.7) P σ ( x ) = Z ∞ h t ( x ) dtt σ , serves as the kernel of the integral representation. We state both sides estimate of P σ , whoseproof is exactly the same as of Theorem 3.3. Theorem 3.4. For any α > − n/ the following asymptotic estimates holds: P α ( x ) ≍ | x | − l/ − / − α −| Σ +0 | φ ( x ) e −| ρ || x | , for | x | ≥ , ≍ | x | − n − α , for | x | < . Corollary 3.5. Let χ be the characteristic function of the unit ball in X and α > . Then thefunction (1 − χ ) P α is in L p ( X ) for ≤ p ≤ ∞ .Proof. For 1 < p ≤ ∞ , the result follows trivially from the asymptotic formula in Theorem3.4. We prove the case p = 1. We recall from (2.2) that Z { x ∈ X : | x | > } P α ( x ) dx ≤ C Z { H ∈ a + : | H | > } P α (exp H ) e ρ ( H ) dH. Let Γ be a small circular cone in a + around the ρ -axis. By introducing polar coordinates in Γand using (2.6) we get Z { H ∈ Γ: | H | > } P α (exp H ) e ρ ( H ) dH ≤ C Z { H ∈ Γ: | H | > } | H | − l/ − / − α e ρ ( H ) −| ρ || H | dH ≤ C Z ∞ r − l/ − / − α r l − Z ν (sin ξ ) l − e − r (1 − cos ξ ) dξ dr. XTENSION PROBLEM AND HARDY’S INEQUALITY 17 Since sin ξ ∼ ξ and 1 − cos ξ ∼ ξ , the inner integral behaves like r / − l/ . Consequently, theintegral above is finite. On the other hand, e ρ ( H ) −| ρ || H | decreases exponentially outside Γ, andtherefore Z { H ∈ a + \ Γ: | H | > } P α (exp H ) e ρ ( H ) dH = Z { H ∈ a + \ Γ: | H | > } H − l/ − / − α e ρ ( H ) −| ρ || H | dH < ∞ . This completes the proof. (cid:3) Fractional Hardy inequalities This section aims to prove two versions of the Hardy’s inequalities for fractional powers ofthe Laplace-Beltrami operator on X , namely Theorem 1.1 and Theorem 1.3 with homogeneousand non-homogeneous weight functions respectively. In order to prove these inequalities, wewill follow similar ideas used by Frank et al. [19] in the case of Euclidean Laplacian. Therefore,we need to establish ground state representations for the operators ( − ∆) σ . We start with thefollowing integral representations of ( − ∆) σ on X . For the cases of real hyperbolic spaces,analogues integral representations were proved in [6, Theorem 2.5]. Lemma 4.1. Let < σ < / . Then for all f ∈ C ∞ c ( X ) we have ( − ∆) σ f ( x ) = 1 | Γ( − σ ) | Z X ( f ( x ) − f ( z )) P σ ( z − x ) dz, where P σ is defined in (3.7).Proof. Let f ∈ C ∞ c ( X ). Using the numerical identity λ σ = 1 | Γ( − σ ) | Z ∞ (cid:16) − e − tλ (cid:17) dtt σ , λ > , and the spectral theorem we have( − ∆) σ f ( x ) = 1 | Γ( − σ ) | Z ∞ (cid:0) f ( x ) − e t ∆ f ( x ) (cid:1) dtt σ . By (2.8) it follows that(4.1) f ( x ) − e t ∆ f ( x ) = f ( x ) − f ∗ h t ( x ) = Z X ( f ( x ) − f ( xz − )) h t ( z ) dz. Thus, we have the following representation( − ∆) σ f ( x ) = 1 | Γ( − σ ) | Z ∞ Z X (cid:0) f ( x ) − f ( xz − ) (cid:1) h t ( z ) dz dtt σ . We now show that the right-hand side is absolutely integrable and hence, interchange of theorder of integral is possible. Then the result follows by the change of variable z z − x . Toshow absolute integrability let us define I = 1 | Γ( − σ ) | Z ∞ Z { z ∈ X : | z | < } (cid:12)(cid:12) f ( x ) − f ( xz − ) (cid:12)(cid:12) h t ( z ) dz dtt σ ,I = 1 | Γ( − σ ) | Z ∞ Z { z ∈ X : | z |≥ } (cid:12)(cid:12) f ( x ) − f ( xz − ) (cid:12)(cid:12) h t ( z ) dz dtt σ . For the integral I , we use the fact that P σ ∈ L ( X ) away from the origin (Corollary 3.5).Indeed, we have that Z { z ∈ X : | z |≥ } Z ∞ | f ( x ) − f ( xz − ) | h t ( z ) dtt σ dz ≤ k f k L ∞ ( X ) Z { z ∈ X : | z |≥ } P σ ( z ) dz < ∞ . Therefore, by Fubini’s theorem I is also finite. For I we first observe by the fundamentaltheorem of calculus (see the proof of equation (34) in [2]) that(4.2) | f ( x ) − f ( x (exp H )) | ≤ | H | Z |∇ f ( x exp( sH )) | ds ≤ | H | k∇ f k L ∞ ( X ) , for x ∈ X, H ∈ a . Using the above estimate and the fact that P σ ( x ) ≍ | x | − n − σ around theorigin (Theorem 3.4) it follows that Z | z | < Z ∞ | f ( x ) − f ( xz − ) | h t ( z ) dtt σ dz ≤ C k∇ f k L ∞ ( X ) Z { H ∈ a + : | H | < } | H | | H | − n − σ J (exp H ) dH ≤ C k∇ f k L ∞ ( X ) Z r − n − σ r n − dr, and the right-hand side is finite if 0 < σ < / 2. This completes the proof. (cid:3) Remark 4.2. If rank ( X ) = 1, then for 1 / ≤ σ < a = span { H } with | H | = 1. Clearly, for σ > 0, theintegral I is absolutely convergent and we can interchange the order of the integral. On theother hand the formula (2.2) yields I = 1 | Γ( − σ ) | Z ∞ Z − (cid:0) f ( x ) − f ( x exp( − sH )) (cid:1) h t (exp( sH )) J (exp( sH )) ds dtt σ . We now define F ( s ) := f ( x exp( sH )), for s ∈ R . Since f ∈ C ∞ c ( X ), it follows that for each x ∈ X , the function F ∈ C ∞ c ( R ). By using the Taylor development of F , we get that I = 1 | Γ( − σ ) | Z ∞ Z − (cid:18) sF ′ ( x ) + s F ′′ ( x )2! + O ( s ) (cid:19) h t (exp( sH )) J (exp( sH )) ds dtt σ . Since the heat kernel h t and the Jacobian J is even, the first order term vanishes. Hence, usingthe fact that P σ ( x ) ∼ | x | − n − σ , around the origin (Theorem 3.4), it follows that I ≤ C f Z ∞ Z s h t (exp( sH )) s n − ds dtt σ = C f Z s n +1 s − n − σ ds, which is finite if 0 < σ < 1. Hence, the required integral formula exists as a principal valuesense. For the case of higher rank symmetric spaces, neither the heat kernel h t (exp( · )) nor theJacobian J (exp( · )) is, in general, radial function on a . They are only Weyl group invariant.This is the main difficulty that we could not prove the integral formula in the lemma abovefor 1 / ≤ σ < rank ( X ) > Lemma 4.3. Let < σ < . Then, for all f ∈ H σ ( X ) h ( − ∆) σ f, f i = 12 | Γ( − σ ) | Z X Z X | f ( z ) − f ( x ) | P σ ( z − x ) dz dx. XTENSION PROBLEM AND HARDY’S INEQUALITY 19 Proof. We first prove that for 0 < σ < f ∈ C ∞ c ( X ) the quantity(4.3) 12 | Γ( − σ ) | Z X Z X | f ( z ) − f ( x ) | P σ ( z − x ) dz dx < ∞ . To show this let us assume supp f ⊂ B ( o, m ) for some m > I = 12 | Γ( − σ ) | Z B ( o, m ) Z X | f ( z ) − f ( x ) | P σ ( z − x ) dz dx,I = 12 | Γ( − σ ) | Z X \ B ( o, m ) Z X | f ( z ) − f ( x ) | P σ ( z − x ) dz dx. Since supp f ⊂ B ( o, m ) it follows that I = 12 | Γ( − σ ) | Z X \ B ( o, m ) Z B ( o,m ) | f ( z ) | P σ ( z − x ) dz dx ≤ k f k L ∞ ( X ) | Γ( − σ ) | Z B ( o,m ) Z X \ B ( o, m ) P σ ( z − x ) dx dz ≤ k f k L ∞ ( X ) | Γ( − σ ) | | B ( o, m ) | Z X \ B ( o,m ) P σ ( x ) dx < ∞ . The last term is finite because of the fact that P σ is integrable away from the origin (Corollary3.5). To show that I is finite we write it as follows I = Z B ( o, m ) Z B (0 , m ) | f ( z ) − f ( x ) | P σ ( z − x ) dz dx + Z B ( o, m ) Z X \ B (0 , m ) | f ( z ) − f ( x ) | P σ ( z − x ) dz dx. Using change of variable z xz − in the first integral, the estimate (4.2) and the asymptoticestimates of P σ in Theorem 3.4 it follows that I ≤ Z B ( o, m ) Z B ( o, m ) (cid:12)(cid:12) f ( xz − ) − f ( x ) (cid:12)(cid:12) P σ ( z ) dz dx + Z B ( o, m ) Z X \ B ( o, m ) | f ( z ) − f ( x ) | P σ ( z − x ) dz dx ≤ C k∇ f k L ∞ ( X ) Z B ( o, m ) dx Z { H ∈ a + : | H | < m } | H | H − n − σ J (exp H ) dH + C Z B ( o, m ) k f k L ∞ ( X ) Z X \ B ( o, m ) P σ ( z − x ) dz dx ≤ C k∇ f k L ∞ ( X ) Z m r − n − σ r n − dr + C k f k L ∞ ( X ) Z X \ B ( o,m ) P σ ( z ) dz. The first term of the above quantity is finite provided σ < < σ < / f ∈ C ∞ c ( X ). By the integral representation in Lemma 4.1 it followsthat h ( − ∆) σ f, f i = 1 | Γ( − σ ) | Z X Z X ( f ( x ) − f ( z )) P σ ( z − x ) f ( x ) dz dx. As the kernel P σ is symmetric, that is P σ ( x ) = P σ ( x − ), the above quantity is also equals to1 | Γ( − σ ) | Z X Z X ( f ( z ) − f ( x )) P σ ( z − x ) f ( z ) dx dz. By adding them up we get that h ( − ∆) σ f, f i = 12 | Γ( − σ ) | Z X Z X | f ( z ) − f ( x ) | P σ ( z − x ) dz dx. The justification of the change of order of integration follows from (4.3). By the analyticcontinuation, we extend the range of σ to 0 < σ < f ∈ C ∞ c ( X ). Indeed, thefunctions σ 7→ − Γ( − σ ) and σ 7→ h ( − ∆) σ f, f i are holomorphic on S = { w ∈ C : 0 < ℜ w < } .Hence their product F ( σ ) = − Γ( − σ ) h ( − ∆) σ f, f i is also holomorphic on S . On the otherhand, since right-hand side of (4.3) is finite for 0 < σ < 1, by the Morera’s theorem it followsthat the function G defined by G ( σ ) = 12 Z X Z X | f ( z ) − f ( x ) | P σ ( z − x ) dz dx. is holomorphic on S . Since F ( σ ) = G ( σ ) for 0 < σ < / F ( σ ) = G ( σ ) for all σ ∈ S , in particular, for 0 < σ < f ∈ H σ ( X ) by a sequence of functions f k ∈ C ∞ c ( X ), wecomplete the proof. This uses the fact that P σ ( x ) ≍ | x | − n − σ around the origin and the restfollows as in the proof of Lemma 5.1 in [37]. (cid:3) We now establish ground state representation for the operator ( − ∆) σ as a consequence ofthe integral representation proved in Lemma 4.3. As in the Euclidean case, we define thefollowing error term. For 0 < σ < y > H σy [ F ] = h ( − ∆) σ F, F i − σ Γ( σ ) y σ Γ( − σ ) Z X | F ( x ) | (cid:18) P σy ( x ) P − σy ( x ) (cid:19) dx. Theorem 4.4. Let < σ < and y > . If F ∈ C ∞ c ( X ) and G ( x ) = F ( x ) (cid:0) P − σy ( x ) (cid:1) − then H σy [ F ] = 12 | Γ( − σ ) | Z X Z X | G ( x ) − G ( z ) | P − σy ( x ) P − σy ( z ) P − σ ( z − x ) dx dz. Proof. Let f, g ∈ H σ ( X ). From Lemma 4.3 we get that(4.4) h ( − ∆) σ f, g i = 12 | Γ( − σ ) | Z X Z X ( f ( z ) − f ( x )) ( g ( z ) − g ( x )) P σ ( z − x ) dz dx. Let us assume g = P − σy , and f ( x ) = | F ( x ) | g ( x ) − . Then the right-hand side of (4.4) reducesto 12 | Γ( − σ ) | Z X Z X (cid:18) | F ( z ) | g ( z ) − | F ( x ) | g ( x ) (cid:19) ( g ( z ) − g ( x )) P σ ( z − x ) dz dx (4.5) = 12 | Γ( − σ ) | Z X Z X | F ( x ) − F ( z ) | − (cid:12)(cid:12)(cid:12)(cid:12) F ( x ) g ( x ) − F ( z ) g ( z ) (cid:12)(cid:12)(cid:12)(cid:12) g ( x ) g ( z ) ! P σ ( z − x ) dz dx. XTENSION PROBLEM AND HARDY’S INEQUALITY 21 Also, using Lemma 3.2 the left-hand side of (4.4) reduces to h ( − ∆) σ f, g i = (cid:10) ( − ∆) σ ( | F ( x ) | /g ( x )) , g ( x ) (cid:11) = (cid:10)(cid:0) | F ( x ) | /g ( x ) (cid:1) , ( − ∆) σ P − σy (cid:11) = 4 σ Γ( σ ) y σ Γ( − σ ) (cid:10) ( | F ( x ) | /g ( x )) , P σy (cid:11) = 4 σ Γ( σ ) y σ Γ( − σ ) Z X | F ( x ) | P σy ( x ) P − σy ( x ) dx. Therefore, equating the left-hand and right-hand sides of the equation (4.4) we have4 σ Γ( σ ) y σ Γ( − σ ) Z X | F ( x ) | P σy ( x ) P − σy ( x ) dx = 12 | Γ( − σ ) | Z X Z X | F ( x ) − F ( z ) | P σ ( z − x ) dx dz − | Γ( − σ ) | Z X Z X (cid:12)(cid:12)(cid:12)(cid:12) F ( x ) g ( x ) − F ( z ) g ( z ) (cid:12)(cid:12)(cid:12)(cid:12) g ( x ) g ( z ) P σ ( z − x ) dz dx. By Lemma 4.3 the first term in the right-hand side of the above equation is equals to h ( − ∆) σ F, F i .Hence, it follows that h ( − ∆) σ F, F i − σ Γ( σ ) y σ Γ( − σ ) Z X | F ( x ) | P σy ( x ) P − σy ( x ) dx = 12 | Γ( − σ ) | Z X Z X | G ( x ) − G ( z ) | P − σy ( x ) P − σy ( x ) P − σ ( z − x ) dx dz, where G ( x ) = F ( x ) P − σy ( x ) − . This completes the proof. (cid:3) We have already observed that for 0 < σ < 1, Γ( − σ ) < P − σy ≤ 0. Therefore, asa corollary of Theorem 4.4 we get the following result. Corollary 4.5. For a fixed y > and < σ < we have h ( − ∆) σ F, F i ≥ σ y σ Z X | F ( x ) | (cid:18) Γ( σ )Γ( − σ ) P σy ( x ) P − σy ( x ) (cid:19) dx, for F ∈ H σ ( X ) . Remark 4.6. By Lemma 3.2 it follows that the equality in the expression above is achieved forthe function F = P − σy . Therefore, the constant 4 σ Γ( σ ) /y σ | Γ( − σ ) | appeared in the corollaryabove is sharp.Now, using the estimate of P σy (Theorem 3.3) in Corollary 4.5 we get Theorem 1.1. Proof of Theorem 1.1. From Theorem 3.3 we haveΓ( σ )Γ( − σ ) P σy ( x ) P − σy ( x ) ≍ ( y σ ( | x | + y ) σ if | x | + y ≥ y σ ( | x | + y ) σ if | x | + y < . Therefore, from Corollary 4.5 we have h ( − ∆) σ F, F i ≥ C σ y σ Z { x : | x | + y < } | F ( x ) | ( y + | x | ) σ dx + Z { x : | x | + y ≥ } | F ( x ) | ( y + | x | ) σ dx ! . (cid:3) We now prove Hardy’s inequality corresponding to the homogeneous weight function (The-orem 1.3). To prove this theorem we need the following expression of the error term. Theorem 4.7. Let < σ < and α > (2 σ + n ) / . Then for F ∈ C ∞ c ( X ) and G ( x ) = F ( x ) (cid:0) P − α ( x ) (cid:1) − we have h ( − ∆) σ F, F i − Γ( α )Γ( α − σ ) Z X | F ( x ) | (cid:18) P σ − α ( x ) P − α ( x ) (cid:19) dx = 12 | Γ( − σ ) | Z X Z X | G ( x ) − G ( z ) | P − α ( x ) P − α ( z ) P σ ( z − x ) dx dz, (4.6) where the function P − α is defined by (3.7).Proof. Since α > n/ 4, we observe from Theorem 3.4 that P − α ∈ L ( X ). As before by Fubinitheorem the spherical Fourier transform of P − α is given by d P − α ( λ ) = Z ∞ e − t ( | λ | + | ρ | ) dtt − α = Γ( α ) (cid:0) | λ | + | ρ | (cid:1) − α , λ ∈ a ∗ . Since α > (2 σ + n ) / 4, it follows that P − α ∈ H σ ( X ). Indeed, using (2.3) we get that Z a ∗ | d P − α ( λ ) | ( | λ | + | ρ | ) σ | c ( λ ) | − dλ ≤ C + C ′ Z { a ∗ : | λ |≥ } ( | λ | + | ρ | ) − α + σ (1 + | λ | ) n − l dλ, which is finite. We recall from (4.4) that for f, g ∈ H σ ( X )(4.7) h ( − ∆) σ f, g i = 12 | Γ( − σ ) | Z X Z X ( f ( z ) − f ( x )) ( g ( z ) − g ( x )) P σ ( z − x ) dz dx. If we put g ( x ) = P − α ( x ) and f ( x ) = | F ( x ) | ( P − α ( x )) − in the equation above, then theleft-hand side reduces to h ( − ∆) σ f, g i = Z a ∗ (cid:0) | λ | + | ρ | (cid:1) σ b f ( λ ) b g ( λ ) | c ( λ ) | − dλ = Γ( α ) Z a ∗ (cid:0) | λ | + | ρ | (cid:1) σ − α b f ( λ ) | c ( λ ) | − dλ = Γ( α )Γ( α − σ ) Z a ∗ \ P σ − α ( λ ) b f ( λ ) | c ( λ ) | − dλ = Γ( α )Γ( α − σ ) Z X | F ( x ) | P σ − α ( x ) P − α ( x ) dx. The right-hand side of the equation (4.7) becomes (see (4.5))(4.8) 12 | Γ( − σ ) | Z X Z X | F ( x ) − F ( z ) | − (cid:12)(cid:12)(cid:12)(cid:12) F ( x ) g ( x ) − F ( z ) g ( z ) (cid:12)(cid:12)(cid:12)(cid:12) g ( x ) g ( z ) ! P σ ( z − x ) dz dx. Hence, equating both sides of the equation (4.7) we haveΓ( α )Γ( α − σ ) Z X | F ( x ) | P σ − α ( x ) P − α ( x ) dx = 12 | Γ( − σ ) | Z X Z X | F ( x ) − F ( z ) | P σ ( z − x ) dz dx − | Γ( − σ ) | Z X Z X (cid:12)(cid:12)(cid:12)(cid:12) F ( x ) g ( x ) − F ( z ) g ( z ) (cid:12)(cid:12)(cid:12)(cid:12) g ( x ) g ( z ) P σ ( z − x ) dz dx. By Lemma 4.3 the first term in the right-hand side of the above equation is equals to h ( − ∆) σ F, F i and hence the required identity follows. (cid:3) XTENSION PROBLEM AND HARDY’S INEQUALITY 23 Proof of Theorem 1.3. Since σ < n ≥ 2, we can choose a positive α such that 2 σ + n/ <α < n/ 2. From Theorem 3.4 above it follows that P σ − α ( x ) P − α ( x ) ≍ | x | − σ , for | x | < ≍ | x | − σ , for | x | ≥ . Therefore, it follows from Theorem 4.7 that h ( − ∆) σ F, F i ≥ C σ Z | x | < | F ( x ) | | x | σ dx + Z | x |≥ | F ( x ) | | x | σ dx ! . (cid:3) Mapping properties of Poisson Operator In this section we prove Theorem 1.7. We start with the following lemma. Lemma 5.1. For < σ < and < q < n +1 n , the function ( x, y ) P σy ( x ) ∈ L q ( X × R + ) .Proof. We first observe from (2.1) that for H ∈ a with | H | < 1, the Jacobian J (exp H )corresponding to the polar decomposition is of order | H | n − l . From Theorem 3.3 it follows that Z | x | + y < | P σy ( x ) | q dx dy ≤ C Z | x | + y < y σq ( | x | + y ) − nq/ − σq dx dy ≤ C Z y =0 Z { H ∈ a + : | H | < } y σq ( | H | + y ) − nq/ − σq | H | n − l dH dy = Z Z y σq ( r + y ) − nq/ − σq r n − l r l − dr dy ≤ Z (cid:18)Z ∞ (1 + s ) − nq/ − σq s n − ds (cid:19) y n − nq dy. We now use the following fact from [22, 3.251, (2); p.324](5.1) Z ∞ x µ − (1 + x ) ν − dx = 12 B ( µ/ , (1 − ν − µ/ , if ℜ µ > , and ℜ ( ν + µ/ < . In our case, µ = n and ν = − nq/ − σq + 1. Hence, ν + µ/ < q > n/ ( n + 2 σ ).Therefore, if q > n/ ( n + 2 σ ) the above integral reduces to12 B ( n/ , ( nq/ σq − n/ Z y n − nq dy. This is finite only if q < (1 + n ) /n . Hence, for n/ ( n + 2 σ ) < q < n , Z | x | + y ≤ | P σy ( x ) | q dx dy < ∞ . On the other hand for q > 1, using the estimate of Jacobian in (2.1) and the asymptoticbehaviour of φ given in (2.6), it follows from Theorem 3.3 that Z | x | + y ≥ | P σy ( x ) | q dx dy ≤ Z | x | + y ≥ y σq (4 σ Γ( σ )) q (cid:16)p | x | + y (cid:17) − ( l/ | Σ +0 | + σ + 1 / q e −| ρ | q √ | x | + y | φ ( x ) | q dx dy ≤ C Z | x | + y ≥ y σq e − | ρ | ( q +1)2 √ | x | + y e − | ρ | ( q − √ | x | + y | φ ( x ) | q dx dy ≤ C Z { ( H,y ) ∈ a + × (0 , ∞ ): | H | + y ≥ } y σq e − | ρ | ( q − | y | e − | ρ | ( q +1) | H | | H | | Σ +0 | q e − q ρ ( H ) e ρ ( H ) dH dy ≤ (cid:18)Z ∞ y σq e −| ρ | ( q − | y | / dy (cid:19) (cid:18)Z a + | H | | Σ +0 | q e − ( q − ρ ( H ) dH (cid:19) < ∞ . This completes the proof. (cid:3) We are now in a position to prove Theorem 1.7. We follow similar ideas which are used tothe proof of [34, Theorem B]. Proof of Theorem 1.7. We first prove (1). Let u be the solution of (1.5) with boundary value f ∈ H σ ( X ), and let U ( λ, k, η ) = F ( e u ( λ, k )) ( η ) , for λ ∈ a ∗ , k ∈ K, η ∈ R + be the composition of the Helgason and the Euclidean Fourier transform on X × R . Multiplying y on both sides of the equation (1.5) and taking the composition of Helgason and EuclideanFourier transform on X × R it follows that ∂ ∂η (cid:0) ( | λ | + | ρ | + η ) U ( λ, k, η ) (cid:1) − (1 − σ ) ∂∂η ( η U ( λ, k, η )) = 0which is equivalent to(5.2) (cid:26) ( | λ | + | ρ | + η ) ∂ ∂ η + (3 + 2 σ ) η ∂∂η + (1 + 2 σ ) (cid:27) U ( λ, k, η ) = 0 . Let t = η √ | λ | + | ρ | and we define v ( λ, k, t ) = U ( λ, k, η ) . Then equation (5.2) reduces to D σ,t v ( λ, k, t ) := (cid:26) (1 + t ) d dt + (2 σ + 3) t ddt + (2 σ + 1) (cid:27) v ( λ, k, t ) = 0 . Since f ( x ) = u ( x, 0) for x ∈ X , by the Euclidean Fourier inversion formula we have e f ( λ, k ) = u ( · , e ( λ, k ) = 1 √ π Z R U ( λ, k, η ) dη = p | λ | + | ρ | √ π Z R v ( λ, k, t ) dt. Therefore, the function v satisfies D σ,t v ( λ, k, t ) = 0 , and Z R v ( λ, k, t ) dt = √ π p | λ | + | ρ | e f ( λ, k ) , XTENSION PROBLEM AND HARDY’S INEQUALITY 25 for almost every ( λ, k ) ∈ a ∗ × K . Hence, the function v is given by(5.3) v ( λ, k, t ) = √ π p | λ | + | ρ | e f ( λ, k ) ψ ( t ) , where ψ satisfies(5.4) D σ,t ψ = 0 , and Z R ψ ( t ) dt = 1 . The equation D σ,t ψ = 0 has a fundamental system of solutions spanned by ψ ( t ) = F (cid:18) , σ + 12 ; 12 ; − t (cid:19) = (1 + t ) − σ − / ,ψ ( t ) = t F (cid:18) , σ + 1; 32 ; − t (cid:19) . Using (5.3) it is now easy to check that Z a ∗ × K × R |U ( λ, k, η ) | ( | λ | + | ρ | + η ) σ + | c ( λ ) | − dλ dk dη = Z a ∗ × K × R | v ( λ, k, t ) | ( | λ | + | ρ | ) σ +1 (cid:0) t (cid:1) σ + | c ( λ ) | − dλ dk dt = 2 π Z a ∗ × K | b f ( λ, k ) | ( | λ | + | ρ | ) σ | c ( λ ) | − dλ dk Z R | ψ ( t ) | (1 + t ) σ + dt. (5.5)Since f ∈ H σ , it follows that u ∈ H σ + if and only if ψ ∈ L ( R , (1 + t ) σ + dt ). It is easy tocheck from the asymptotic properties of hypergeometric function that ψ / ∈ L ( R , (1+ t ) σ + dt )(see [1, Theorem 2.3.2]). Hence, we choose ψ ( t ) to be a constant multiple of ψ ( t ) = (1 + t ) − σ − . From (5.1) we get that k ψ k L ( R ) = √ π Γ( σ ) / Γ( σ + ). Hence, using (5.4) it followsthat ψ ( t ) = Γ( σ + 1 / √ π Γ( σ ) ψ ( t ) . We now observe that Z R | ψ ( t ) | (1 + t ) σ + dz = Γ( σ + 1 / √ π Γ( σ ) , and hence from (5.5), k u k H σ + 12 ( X × R + ) = 2 √ π Γ( σ + )Γ( σ ) k f k H σ ( X ) . This completes the proof of part (1). We now prove part (2). We first observe that k T σ f k qL q ( X × R + ) = Z ∞ k f ∗ P σy k qL q ( X ) dy. Also, from Theorem 3.3 it follows that for each y > P σy ∈ L q ( X ), for all q > ≤ p < q ≤ k f ∗ P σy k L q ( X ) ≤ C k f k L p ( X ) k P σy k L q ( X ) . Therefore, by Lemma 5.1 it follows that(5.6) T σ : L p ( X ) → L q ( X × R + ) , is a bounded map, for 1 ≤ p < q < ( n + 1) /n . We also observe that(5.7) T σ : L ∞ ( X ) → L ∞ ( X × R + ) , is a bounded map, as the integral R X P σy ( x ) dx = 1 for all y > 0. By Riesz Thorin interpolationtheorem it now follows from (5.6) and (5.7) that(5.8) T σ : L p ( X ) → L q ( X × R + ) , is bounded for 1 ≤ p < ∞ and p < q < ( n +1 n ) p . We now prove that k T σ f k L q ( X × R + ) ≤ C k f k L p ( X ) , for p > q = ( n +1 n ) p . By (5.7) and Marcinkiewicz interpolation theorem it is enough toshow that T σ : L ( X ) → L (1+ n ) /n, ∞ ( X × R + ) . Using Theorem 3.3 and the boundedness of the function φ we get that | T σ f ( x, y ) | ≤ Z X | f ( z ) | P σy ( z − x ) | f ( z ) | dz ≤ Cy − n k f k L ( X ) + Cy σ Z | z − x | + y ≥ p ( | z − x | + y ) ( − l/ − / − σ −| Σ +0 | ) e −| ρ | √ ( | z − x | + y ) | | f ( z ) | dz ≤ Cy − n k f k L ( X ) + Cy − n k f k L ( X ) sup y ∈ R + (cid:16) y σ + n e −| ρ | y (cid:17) ≤ Cy − n k f k L ( X ) . Hence, | T a f ( x, y ) | > λ implies that y ≤ (cid:18) C k f k L X ) λ (cid:19) n = b (say). Then Chebyshev’s inequalityyields m ( { ( x, y ) ∈ X × R + : | T a f ( x, y ) | > λ } )= m ( { ( x, y ) ∈ X × R + : y < b, | T a f ( x, y ) | > λ } ) ≤ λ Z { ( x,y ) ∈ X × R + : y 0. Thiscompletes the proof. (cid:3) Expression of the kernel P σy In the case of R n and of the Heisenberg groups the function P σy is the classical Poissonkernel. In the case of symmetric spaces, we only have the integral expression as in Theorem1.5 and the both-sides estimates (Theorem 3.3) for P σy . In this section we write the preciseexpression of P σy for complex and rank one symmetric spaces using the expression of the heatkernel. XTENSION PROBLEM AND HARDY’S INEQUALITY 27 G is complex. In this case we have the following formula for the heat kernel [5] h t (exp H ) = (4 πt ) − n/ e −| ρ | t Y α ∈ Σ + α ( H )sinh α ( H ) e − H / t , t > , H ∈ a . It now follows from the definition (1.9) of P σy P σy (exp H ) = y σ σ Γ( σ ) (4 πt ) − n/ Y α ∈ Σ + α ( H )sinh α ( H ) Z ∞ t − n/ − σ − e −| ρ | t e − ( | H | + y ) / t dtt σ = y σ Γ( σ ) 2 − n/ − σ π − n/ Y α ∈ Σ + α ( H )sinh α ( H ) p | H | + y | ρ | ! − ( n +2 σ ) / K − n/ − σ ( p | H | + y | ρ | ) . Here the last equality follows from the formula [22, 3.471(9), p. 368], and K − n/ − σ is themodified Bessel function (defined in [22, 8.407 (1), p. 911]).6.2. X is of rank one. Let F = R , C , H , or O be the real numbers, the complex numbers,the quaternions or the Cayley octonions respectively. The rank one symmetric spaces can berealized as the hyperbolic space H n ( F ). Here the subscript n denotes the dimension over thebase field F . Using the expression of the heat kernel [5, 21] we have the following results.(1) X = H n ( R ), and n ≥ P σy ( x ) = c Z ∞ t − / e − ρ t e − y / t (cid:18) − x ∂∂x (cid:19) ( n − / e −| x | / t dtt σ = c (cid:18) − x ∂∂x (cid:19) ( n − / Z ∞ t − / − σ e − ρ t e − ( | x | + y ) / t dt = c (cid:18) − x ∂∂x (cid:19) ( n − / p | x | + y ρ ! − σ − / K − σ − / ( ρ p | x | + y ) . (2) X = H n ( R ), and n ≥ P σy ( x )= c Z ∞ t − / e − ρ t e − y / t Z ∞ x sinh z p cosh z − cosh x (cid:18) − z ∂∂z (cid:19) n/ e −| z | / t dz dtt σ = c Z ∞ x sinh z p cosh z − cosh x (cid:18) − z ∂∂z (cid:19) n/ Z ∞ t − / − σ e − ρ t e − ( | z | + y ) / t dt dz = c Z ∞ x sinh z p cosh z − cosh x (cid:18) − z ∂∂z (cid:19) n/ p | z | + y ρ ! − σ − / K − σ − / ( ρ p | z | + y ) dz. (3) X = H n ( F ) where F = C , H or O . Then there exist constants c , c , · · · , c n/ suchthat P σy ( x ) = Z ∞ t − / e − ρ t n/ X j =1 c j Z ∞ x sinh z p cosh z − sinh x (cosh z ) j +1 − d (cid:18) − π sinh z ∂∂z (cid:19) j + m α / e −| z | / t dz dtt σ = c σ d/ X j =1 c j Z ∞ x sinh z p cosh z − sinh x (cosh z ) j +1 − d ρ σ (cid:18) − π sinh z ∂∂z (cid:19) j + m α / p | z | + y ρ ! σ +1 / K − σ − / ( ρ p | z | + y ) dz, where the constant c σ depends only on σ .7. Poincar´e-Sobolev inequality In this section we prove Theorem 1.11. For the convenience of the reader we restate thetheorem here. Theorem 7.1. Let dim X = n ≥ and < σ < min { l + 2 | Σ +0 | , n } . Then for < p ≤ nn − σ there exists S = S n,σ,p > such that for all f ∈ H σ ( X )(7.1) k ( − ∆ − | ρ | ) σ/ f k L ( X ) ≥ S k f k L p ( X ) . Proof. We first observe that it is enough to prove the result for f ∈ C ∞ c ( X ). It also suffices toshow that(7.2) Z X f ( x ) ( − ∆ − | ρ | ) − σ/ f ( x ) dx ≤ C k f k L p ′ ( X ) . Indeed, if (7.2) holds, then by H¨older’s inequality |h f, g i| = (cid:12)(cid:12)(cid:12)D(cid:0) − ∆ − | ρ | (cid:1) σ/ f, (cid:0) − ∆ − | ρ | (cid:1) − σ/ g E(cid:12)(cid:12)(cid:12) ≤ (cid:13)(cid:13)(cid:13)(cid:0) − ∆ − | ρ | (cid:1) σ/ f (cid:13)(cid:13)(cid:13) L ( X ) (cid:13)(cid:13)(cid:13)(cid:0) − ∆ − | ρ | (cid:1) − σ/ g (cid:13)(cid:13)(cid:13) L ( X ) = D(cid:0) − ∆ − | ρ | (cid:1) σ/ f, f E / D(cid:0) − ∆ − | ρ | (cid:1) − σ/ g, g E / ≤ C D(cid:0) − ∆ − | ρ | (cid:1) σ/ f, f E k g k L p ′ ( X ) , and hence k f k L p ( X ) ≤ C n D(cid:0) − ∆ − | ρ | (cid:1) σ/ f, f E . We now prove (7.2). Let k σ be the Schwartz kernel for the operator ( − ∆ − | ρ | ) − σ/ . We havethe following well-known estimates due to Anker and Ji [4, Theorem 4.2.2], for 0 < σ < l +2 | Σ +0 | k σ ( x ) ≍ | x | σ − l − | Σ +0 | φ ( x ) , | x | ≥ , (7.3) ≍ | x | σ − n , | x | < . XTENSION PROBLEM AND HARDY’S INEQUALITY 29 To prove (7.2), it is enough to show that k f ∗ k σ k L p ( X ) ≤ C k f k L p ′ ( X ) . Let χ be the characteristic function of the unit ball B ( o, 1) and k σ ( x ) = χ ( x ) k σ ( x ) and k ∞ σ = k − k . Now, by Young’s inequality we have that k f ∗ k σ k L p ( X ) ≤ C k f k L p ′ ( X ) k k σ k L p/ ( X ) , and k k σ k p L p/ ( X ) ≍ Z | t | ( σ − n ) p/ | t | | Σ + | t l − dt. The right-hand side is finite if p < nn − σ . Using the fact that for r < 1, the volume of the ball B ( o, r ) in X is of order r n , it is easy to check that k σ ∈ L nn − σ , ∞ ( X ). By Young’s inequalityfor weak type spaces [23, Theorem 1.4.24. page 63] it follows that k f ∗ k σ k L nn − σ ( X ) ≤ C k f k L nn + σ ( X ) . Therefore, we have for all p ≤ nn − σ ,(7.4) k f ∗ k σ k L p ( X ) ≤ C k f k L p ′ ( X ) . Next, we shall show that for p > k f ∗ k ∞ σ k L p ( X ) ≤ C p k f k L p ′ ( X ) . To prove this we shall use complex interpolation theorem and the idea of [35, Theorem 4.1].For ℜ z ≥ − , we define an analytic family of linear operators T z from ( X, dx ) to itself asfollows: T z f = f ∗ ( k ∞ σ ) z . For z = − + iy , we have k T z f k L ∞ ( X ) = k f ∗ ( k ∞ σ ) + iy k L ∞ ( X ) ≤ C sup { x ∈ X : | x |≥ } ϕ ( x ) | x | ( σ − l ) / −| Σ +0 | k f k L ( X ) ≤ C k f k L ( X ) . For z = ǫ + iy, ǫ > 0, we have k T z f k L ( X ) = Z R Z K (cid:12)(cid:12)(cid:12) e f ( λ, k ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) \ ( k ∞ σ ) ǫ + iy ( λ ) (cid:12)(cid:12)(cid:12) | c ( λ ) | − dλ dk ≤ sup (cid:12)(cid:12)(cid:12) \ ( k ∞ σ ) ǫ + iy ( λ ) (cid:12)(cid:12)(cid:12) k f k L ( X ) . Now, by Theorem 2.1 it follows that for λ ∈ a ∗ and ǫ > | \ ( k ∞ σ ) ǫ + iy ( λ ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z { x ∈ X : | x |≥ } | x | ( σ − l − | Σ +0 | )(1+ ǫ + iy ) ( φ ( x )) ǫ + iy φ − λ ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z { x ∈ X : | x |≥ } φ ( x ) ǫ dx < ∞ , and hence k T z f k L ∞ ( X ) ≤ k f k L ( X ) . Hence, by analytic interpolation for p > k f ∗ k ∞ σ k L p ( X ) = k T f k L p ( X ) ≤ C k f k L p ′ ( X ) . Therefore, from (7.4) and from (7.5), it follows that for all 2 < p ≤ nn − σ , k f ∗ k σ k L p ( X ) ≤ C k f k L p ′ ( X ) . This completes the proof. (cid:3) As a corollary of the theorem above we have the following Corollary 7.2. Let < p ≤ nn − and dim X = n ≥ . Then there exists S n,p > such thatfor all u ∈ H ( X ) , k∇ u k L ( X ) − | ρ | k u k L ( X ) ≥ S n,p k u k L p ( X ) . Acknowledgement: The first author is supported by the Post Doctoral fellowship from IITBombay. 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