Hypercontractivity of the semigroup of the fractional laplacian on the n-sphere

For $1<p\leq q$ we show that the Poisson semigroup $e^{-t\sqrt{-\Delta}}$ on the $n$-sphere is hypercontractive from $L^{p}$ to $L^{q}$ in dimensions $n \leq 3$ if and only if $e^{-t\sqrt{n}} \leq \sqrt{\frac{p-1}{q-1}}$. We also show that the equivalence fails in large dimensions.


Poisson semigroup on the sphere. Let
be the unit sphere in R n+1 , where x = x 2 1 + . . . + x 2 n+1 for x = (x 1 , . . . , x n+1 ) ∈ R n+1 . Let ∆ be the Laplace-Beltrami operator on S n . We will be working with spherical polynomials f : S n → C, i.e., finite sums The heat semigroup e t∆ is defined by e t∆ f = d≥0 e −d(d+n−1)t H d . The hypercontractivity result for the heat semigroup on S n states that for any 1 ≤ p ≤ q < ∞, any integer n ≥ 1, and any t ≥ 0 we have where f p p = f p L p (S n ,dσ n ) = S n |f | p dσ n , and dσ n is the normalized surface area measure of S n . The case n = 1 was solved independently in [9] and [10], and the general case n ≥ 2 was settled in [7]. We remark that the condition e −tn ≤ p−1 q−1 in (1) is different from the classical hypercontractivity condition e −t ≤ p−1 q−1 in Gauss space due to Nelson [8], and on the hypercube due to Bonami [2]. The appearance of the extra factor n in (1) can be explained from the fact that the spectral gap (the smallest nonzero eigenvalue) of −∆ equals n.
In [7] the authors ask what the corresponding hypercontractivity estimates are for the Poisson semigroup on S n . As pointed out in [7], there are two natural Poisson semigroups on S n one can consider: 1) e −t √ −∆ f , and 2) P r f = r d H d , r ∈ [0, 1]. Notice that when n = 1 both of these semigroups coincide (with r = e −t ). It was conjectured by E. Stein that holds on S n for all n ≥ 1. Besides the case n = 1 mentioned above, the case n = 2 was confirmed in [4], and the general case n ≥ 2 in [1]. The question of hypercontractivity for the semigroup e −t √ −∆ on S n for n ≥ 2, however, has remained open. Since the spectral gap of √ −∆ equals √ n, it is easy to see that a necessary condition for the estimate q−1 ; see Section 2.1. One might conjecture that this necessary condition is also sufficient. Surprisingly, it turns out the answer is positive in small dimensions and negative in large dimensions.
It remains an open problem to find a necessary and sufficient condition on t > 0 in dimensions n ≥ 4 for which the semigroup e −t √ −∆ is hypercontractive from L p (S n ) to L q (S n ).

The sufficiency part
The case n = 1 was confirmed in [10]. In what follows we assume n ∈ {2, 3}. First we need the fact that the heat semigroup e t∆ has a nonnegative kernel. Indeed, for each t > 0 there exists where ξ · η = n+1 j=1 ξ j η j for ξ = (ξ 1 , . . . , ξ n+1 ) and η = (η 1 , . . . , η n+1 ), see, for example, Proposition 4.1 in [7]. Next, we recall the subordination formula By the functional calculus, we deduce that the Poisson semigroup e −t √ −∆ has a positive kernel with total mass 1. The latter fact together with the convexity of the map Next we claim that it suffices to verify (5) only for the powers p, q such that 2 ≤ p ≤ q. Indeed, assume (5) holds for 2 ≤ p ≤ q. By duality and the symmetry of the semigroup (5) to all p, q such that 1 < p ≤ q ≤ 2. It remains to extend (5) for those powers p, q when p ≤ 2 ≤ q. To do so, let p ≤ 2 ≤ q, and let t ≥ 0 be such e −2t √ n = p−1 q−1 . Choose t 1 , t 2 ≥ 0 so that t = t 1 + t 2 and e −2t 1 √ n = p − 1 and e −2t 2 √ n = 1 q−1 . Then we have In what follows we assume 2 ≤ p ≤ q. We will use a standard argument to deduce the validity of the hypercontractivity estimate from a log Sobolev inequality. Nonnegativity of the kernel for the Poisson semigroup combined with the triangle inequality implies |e −t √ −∆ f | ≤ e −t √ −∆ |f | for any f . Thus by continuity and standard density arguments we can assume that f ≥ 0, f is not identically zero, and f is smooth in (5).
Let g = k≥0 H d be the decomposition of g into its spherical harmonics. Then the estimate (7) for q = 2 takes the form S n g 2 ln g 2 dσ n − S n g 2 dσ n ln S n It follows from Beckner's conformal log Sobolev inequality [1] (which is a consequence of Lieb's sharp Hardy-Littlewood-Sobolev inequality [6]) that for any smooth nonnegative g = k≥0 H k we have Proof. We first check the inequality for k ≤ 3 by direct computation. Indeed, the case k = 1 is an equality. The case k = 2 can be checked as follows, which is true because n(2 + 2n) ≤ (2 + n) 2 holds for n = 2, 3. The case k = 3 can be checked similarly: 2 + 2n 2 + n + n 4 + n ≤ 6 + 3n n holds for n = 2, 3 (notice that this inequality fails for n = 4). Next, we assume k ≥ 4. We have

HYPERCONTRACTIVITY OF THE SEMIGROUP OF THE FRACTIONAL LAPLACIAN ON THE n-SPHERE 5
Thus it suffices to show Notice that the left hand side, call it h(k), is decreasing in k. Indeed, we have On the other hand, we have for n = 2, 3, where ⌊ d 2 ⌋ denotes the largest integer m such that m ≤ d 2 , and Γ (x) is the Gamma function. Notice that if we let Y d (ξ) = C (λ) d (ξ · e 1 ), where e 1 = (1, 0, . . . , 0) ∈ R n+1 , then Y d (ξ) is a spherical harmonic of degree d on S n . In particular, for t ≥ 0 such that e −2t Next, we need where dγ(y) = e −y 2 /2 √ 2π dy is the standard Gaussian measure on the real line, and h d (x) is the probabilistic Hermite polynomial Proof. Indeed, notice that where c λ = Γ(λ+1) Γ( 1 2 )Γ(λ+ 1 2 ) . In particular, after the change of variables t = s √ 2λ in (12), and multiplying both sides in (12) by (d!/(2λ) d/2 ) p we obtain Notice that by Stirling's formula for any j ≥ 0, and any d ≥ 0 we have Therefore, (11) and (8) together with (13) imply that for all s ∈ R we have Invoking Stirling's formula again we have Finally, to apply Lebesgue's dominated convergence theorem it suffices to verify that for all s ∈ R and all λ ≥ λ 0 we have the following pointwise estimates where λ 0 , C, c 1 (d), c 2 (d) are some positive constants independent of λ and s. ≤ C for all λ ≥ λ 0 , where λ 0 is a sufficiently large number.
To verify b) it suffices to show that for all λ ≥ λ 0 > 0 and all integers j such that d ≥ j ≥ 0 one has where C(d − j) depends only on d − j. The latter inequality follows from (13) provided that λ ≥ λ 0 where λ 0 is a sufficiently large number. Thus, it follows from the Lebesgue's dominated convergence theorem that lim n→∞ d! (n − 1) d/2 Y d L p (S n ,dσ n ) = h d L p (R,dγ) .
The lemma is proved. Now we fix q > max{p, 2} and, in order to prove the failure of (ii) ⇒ (i) for all sufficiently large n, we argue by contradiction and assume that there is a sequence of dimensions {n j } j≥1 going to infinity such that (ii) ⇒ (i) in Theorem 1.1 does hold. Then, by combining (9) and (10) we have On the other hand, a consequence of the main result in [5] and the assumption q > max{p, 2} is that