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Sbornik: Mathematics, 2023, Volume 214, Issue 3, Pages 396–410
DOI: https://doi.org/10.4213/sm9786e
(Mi sm9786)
 

On a class of interpolation inequalities on the 2D sphere

S. V. Zelikab, A. A. Ilyinc

a Department of Mathematics, University of Surrey, Guildford, UK
b School of Mathematics and Statistics, Lanzhou University, Lanzhou, P. R. China
c Keldysh Institute of Applied Mathematics of Russian Academy of Sciences, Moscow, Russia
References:
Abstract: We prove estimates for the Lp-norms of systems of functions and divergence-free vector functions that are orthonormal in the Sobolev space H1 on the 2D sphere. As a corollary, order sharp constants for the embedding H1Lq, q<, are obtained in the Gagliardo-Nirenberg interpolation inequalities.
Bibliography: 25 titles.
Keywords: Gagliardo-Nirenberg inequalities, sphere, orthonormal systems.
Received: 25.04.2022
Bibliographic databases:
Document Type: Article
MSC: 26D10, 46E35
Language: English
Original paper language: Russian

§ 1. Introduction

The following interpolation inequality holds on the sphere Sd (see [1] and also [2]):

q2dSd|φ|2dμ+Sd|φ|2dμ(Sd|φ|qdμ)2/q.
Here dμ is the normalized Lebesgue measure on Sd:
dμ=dσσd=dσ2π(d+1)/2Γ((d+1)/2),
so that μ(Sd)=1 (the gradient is calculated with respect to the natural metric). Next, q[2,) for d=1,2 and q[2,2d/(d2)] for d3. The remarkable fact about (1.1) is that the constant (q2)/d is sharp for all admissible q. The inequality clearly degenerates and turns to equality on constants. The fact that the constant (q2)/d is sharp is verified by means of the minimizing sequence φε(s)=1+εv(s), where ε0 and v(s) is an eigenfunction of the Laplacian on Sd, corresponding to the first positive eigenvalue d; see [3] and the references there.

However, in applications (for instance, to the Navier-Stokes equations on the 2D sphere) the functions φ usually play the role of stream functions of divergence-free vector functions u, u=φ, and therefore without loss of generality φ can be chosen to be orthogonal to constants.

In this work we consider the two-dimensional sphere S2 only and are interested in writing the Sobolev embedding H1(S2)Lq(S2) as a multiplicative inequality of Gagliardo-Nirenberg type involving the L2-norms of φ and φ on the right-hand side: φL2(S2)=:φ and φL2(S2)=:φ.

It is also well known that in the case of Rd interpolation inequalities in the additive and multiplicative form are equivalent, and the passage from the former to the latter is realized by introducing a parameter m into the inequality (by scaling xmx) and then minimizing with respect to m. To go other way round one can use Young’s inequality (with parameter) for products to obtain the interpolation inequality in the additive form.

This scheme obviously does not work on a manifold due to the lack of scaling. One possible way to introduce a parameter into Sobolev’s inequality is to consider the Sobolev space H1 with norm and inner product

φ2H1:=m2φ2+φ2and(φ1,φ2)H1:=m2(φ1,φ2)+(φ1,φ2),
which depend on a parameter m>0, and then to trace down the explicit dependence of the embedding constant on m. In this work this is done in a much more general framework of the inequalities for H1-orthonormal families proved in [4].

We can now state and discuss our main result.

Theorem 1. Let a family of zero-mean functions {φj}nj=1˙H1(S2) be orthonormal with respect to the inner product

m2(φi,φj)+(φi,φj)=δij.
Then for 1p< the function
ρ(x):=nj=1|φj(x)|2
satisfies the inequality
ρLpBpm2/pn1/p,
where
Bp(p14π)(p1)/p.

These inequalities were proved in the case of Rd in [4] for p= and d=1, for 1p< and d=2 and for the critical p=d/(d2) and d3. No expressions for constants were given, the dependence on m was again uniquely defined by scaling, and the main interest was in the dependence of the right-hand side on n.

For p=2 inequality (1.3) has played an essential role in finding explicit optimal bounds for the attractor dimension in the damped regularized Euler-Bardina-Voight system for various boundary conditions, both in the two- and three-dimensional cases: see [5]–[7]. More precisely, it was shown in [5] and [7] that B2(4π)1/2 for T2, S2 and R2; this was based on the following two inequalities for a lattice sum over Z20=Z2{0,0} and a series with respect to the spectrum of the Laplacian on S2, which were proved there in the special case when p=2:

Jp(m):=(p1)m2(p1)πnZ201(m2+|n|2)p<1,
Ip(m):=m2(p1)(p1)n=12n+1(m2+n(n+1))p<1.

The case p=2 is not at all specific in the general scheme of the proof of Theorem 1, and the general case in this theorem immediately follows for both T2 and S2 once we have at our disposal inequalities (1.5) and (1.6) for all 1<p<.

Inequality (1.5) and Theorem 1 for the torus T2 were recently proved in [8], and the main result of this work is the proof of (1.6) and Theorem 1 for the sphere.

We point out that in the case of R2, instead of (1.5) and (1.6), we have the equality

(p1)m2(p1)πR2dx(m2+|x|2)p=1.

For one function (that is, for n=1) Theorem 1 turns to Sobolev’s inequality with parameter for the embedding H1Lq, q=2p[2,), which can equivalently be written as the Gagliardo-Nirenberg inequality

fLq(14π)(q2)/(2q)(q2)1/2f2/qf12/q,
which holds for R2, T2 and S2, see the corollary.

For the torus T2 inequality (1.8) can be proved in a direct way (see [8]), by using the Hausdorff-Young inequality for Fourier series and estimate (1.5) again. In the case of R2 this approach is well known, and with the additional use of the Babenko-Beckner inequality (see [9] and [10]) for the Fourier transform (and equality (1.7)), it gives us (1.8) for R2 with the best to date estimate for the constant (see [11]):

φLq(R2)(14π)(q2)/(2q)q(q2)/q(q1)(q1)/q(q2)1/2φ2/qφ12/q,q2;
also see [17], Theorem 8.5, where an equivalent result was obtained for the inequality in the additive form.

Of course, both inequality (1.9) for R2 and inequality (2.4) for T2 are special cases of the Gagliardo-Nirenberg inequality.

For R2 the best constant is known for every q2 and is expressed in terms of the norm of the ground state solution of the corresponding nonlinear Euler-Lagrange equation (see [13]). However, not in an explicit form. As mentioned above, inequality (1.9) was known before, while inequality (2.4) (more precisely, an estimate for the constant in it) for the torus T2 was recently obtained in [8].

As far as the case of the sphere S2 is concerned, we do not know how to prove (1.8) in a way other than the one-function corollary of the general result of Theorem 1. The main difference from T2 is that the orthonormal spherical functions are not uniformly bounded in L.

Our approach makes it possible to prove similar inequalities in the vector case. Namely, we show that for uH10(Ω){divu=0}

uLq(S2)(14π)(q2)/(2q)(q2)1/2u2/qrotu12/q.
Here ΩS2 is an arbitrary domain on S2. This inequality looks very similar to (1.8), the important difference being that, unlike the scalar case, the vector Laplacian on S2 is positive definite, and we can freely use the extension by zero for Ω.

Finally, it is natural to compare inequalities (1.1) for d=2 and (1.8) for functions with mean value zero. To do this we go over to the natural measure \mathbb{S}^2 in (1.1) and then use Poincaré’s inequality \|\varphi\|^2\leqslant2^{-1}\|\nabla\varphi\|^2. Then we obtain

\begin{equation*} \begin{aligned} \, \|\varphi\|_{L^q(\mathbb{S}^2)} &\leqslant\biggl(\frac{1}{4\pi}\biggr)^{(q-2)/(2q)} \biggl(\frac{q-2}2\|\nabla\varphi\|^2+\|\varphi\|^2\biggr)^{1/2} \\ &\leqslant \biggl(\frac{1}{4\pi}\biggr)^{(q-2)/(2q)} \biggl(\frac{q-1}2\biggr)^{1/2}\|\nabla\varphi\|. \end{aligned} \end{equation*} \notag
We obtain from (1.8) that
\begin{equation*} \|\varphi\|_{L_q(\mathbb{S}^2)}\leqslant\biggl(\frac{1}{4\pi}\biggr)^{(q-2)/(2q)} \biggl(\frac q2\biggr)^{1/2}\frac1{2^{1/q}}\|\nabla \varphi\|. \end{equation*} \notag
The constant in the second inequality is smaller, since
\begin{equation*} 2^{-2/q}\leqslant1-\frac 1q, \qquad q\geqslant2. \end{equation*} \notag

Since (1.1) turns to equality on constants, this inequality may not be sharp on the subspace of zero-mean functions on \mathbb{S}^2, and the constant in (1.8) is not sharp. However, looking at (1.8) and (1.9) for \mathbb{T}^2 and \mathbb{S}^2 and for \mathbb{R}^2, respectively, one can suggest that that the sharp constant \mathrm{c}_q satisfies

\begin{equation*} \mathrm{c}_q\sim\biggl(\frac1{8\pi}\biggr)^{1/2}q^{1/2} \quad\text{as } q\to\infty. \end{equation*} \notag

The expression on the right-hand side here curiously coincides with the sharp constant in Sobolev’s inequality for the limiting exponent

\begin{equation*} \|\varphi\|_{L^q({\mathbb{R}^d})}\leqslant \frac{\sqrt{q}}{d\sqrt{2\pi}}\biggl[\frac{\Gamma(d)}{\Gamma(d/2)}\biggr]^{1/d} \|\nabla\varphi\|_{L^2({\mathbb{R}^d})}, \qquad\frac1q=\frac12-\frac1d \end{equation*} \notag
(see [14] and [15]), if we set formally d=2. Of course, this inequality does hold in \mathbb{R}^2, since d\geqslant3 there.

Theorem 1 and a similar result in the vector case are proved in § 2, while the key estimate for the series (1.6) is proved in § 3.

§ 2. Proof of the main result

Proof of Theorem 1. First we recall the basic facts concerning the spectrum of the scalar Laplace operator \Delta=\operatorname{div}\nabla on the sphere \mathbb{S}^{2} (see, for instance, [16]):
\begin{equation} -\Delta Y_n^k=n(n+1) Y_n^k, \qquad k=1,\dots,2n+1, \quad n=0,1,2,\dotsc\,. \end{equation} \tag{2.1}
Here the Y_n^k are the orthonormal real-valued spherical harmonics and each eigenvalue \Lambda_n:=n(n+1) has multiplicity 2n+1.

The following identity is essential in what follows: for any s\in\mathbb{S}^{2}

\begin{equation} \sum_{k=1}^{2n+1}Y_n^k(s)^2=\frac{2n+1}{4\pi}. \end{equation} \tag{2.2}
Since inequality (1.3) with bound (1.4) clearly holds for p=1, we assume below that 1<p<\infty. Let us define two operators,
\begin{equation} \mathbb{H}= V^{1/2}(m^2-{\Delta})^{-1/2}\Pi\quad\text{and} \quad \mathbb{H}^*=\Pi(m^2-{\Delta})^{-1/2}V^{1/2}, \end{equation} \tag{2.3}
where V\in L^p is a nonnegative scalar function and \Pi is the projection onto the space of functions with mean value zero:
\begin{equation*} \Pi\varphi=\varphi-\frac1{4\pi}\int_{\mathbb{S}^2}\varphi(s)\,d\sigma. \end{equation*} \notag
Then {\bf K}= \mathbb{H}^*\mathbb{H} is a compact self-adjoint operator in {L}^2({\mathbb{S}}^2), and for r=p'=p/(p-1)\in(1,\infty)
\begin{equation*} \begin{aligned} \, \operatorname{Tr} \mathbf{K}^r &=\operatorname{Tr}\bigl(\Pi(m^2-{\Delta})^{-1/2}V(m^2-{\Delta})^{-1/2}\Pi\bigr)^r \\ &\leqslant\operatorname{Tr}\bigl(\Pi(m^2-{\Delta})^{-r/2}V^r(m^2-{\Delta})^{-r/2}\Pi\bigr) =\operatorname{Tr}\bigl(V^r(m^2-{\Delta})^{-r}\Pi\bigr), \end{aligned} \end{equation*} \notag
where we have used the Araki-Lieb-Thirring inequality for traces (see [17]–[19])
\begin{equation*} \operatorname{Tr}(BA^2B)^p\leqslant\operatorname{Tr}(B^pA^{2p}B^p), \qquad p\geqslant1, \end{equation*} \notag
and the cyclicity property of the trace, together with the facts that \Pi commutes with the Laplacian and that \Pi is a projection: \Pi^2=\Pi. Using the basis of orthonormal eigenfunctions (2.2), in view of the key estimate (3.1) we find that
\begin{equation*} \begin{aligned} \, \operatorname{Tr} \mathbf{K}^r &\leqslant \operatorname{Tr}\bigl(V^r(m^2-{\Delta})^{-r}\Pi\bigr) \\ &=\int_{\mathbb{S}^2}V(s)^r\sum_{n=1}^\infty\sum_{k=1}^{2n+1} \frac1{(m^2+n(n+1))^r}Y_n^k(s)^2\,d\sigma \\ &=\frac1{4\pi}\sum_{n=1}^\infty\frac{2n+1}{(m^2+n(n+1))^r} \int_{\mathbb{S}^2}V(s)^r\,d\sigma\leqslant \frac1{4\pi}\,\frac{m^{-2(r-1)}}{{r-1}}\|V\|^r_{L^r}. \end{aligned} \end{equation*} \notag
Now we can complete the proof as in [4]. Note that
\begin{equation*} \int_{\mathbb{S}^2}\rho(s)V(s)\,d\sigma=\sum_{i=1}^n\|\mathbb{H}\psi_i\|^2_{L^2}, \end{equation*} \notag
where
\begin{equation*} \psi_j=(m^2-{\Delta})^{1/2}\varphi_j, \qquad j=1,\dots,n. \end{equation*} \notag
It follows from (1.2) that the \psi_j are orthonormal in L^2:
\begin{equation*} \begin{aligned} \, (\psi_i,\psi_j) &=\bigl((m^2-\Delta)^{1/2}\varphi_i,(m^2-\Delta)^{1/2}\varphi_j\bigr) =(\varphi_i,(m^2-\Delta)\varphi_j) \\ &=m^2(\varphi_i,\varphi_j)+(\nabla\varphi_i,\nabla\varphi_j)=\delta_{ij}, \end{aligned} \end{equation*} \notag
and in view of the variational principle,
\begin{equation*} \sum_{i=1}^n\|\mathbb{H}\psi_i\|^2_{L^2}=\sum_{i=1}^n(\mathbf{K}\psi_i,\psi_i) \leqslant\sum_{i=1}^n\lambda_i, \end{equation*} \notag
where \lambda_i>0 are the eigenvalues of the operator \mathbf{K}. Therefore,
\begin{equation*} \begin{aligned} \, \int_{\mathbb{S}^2}\rho(s)V(s)\,d\sigma &\leqslant\sum_{i=1}^n\lambda_i\leqslant n^{1/p}(\operatorname{Tr} K^r)^{1/r} \\ &\leqslant n^{1/p}\biggl(\frac{p-1}{4\pi m^{2/(p-1)}}\biggr)^{(p-1)/p}\|V\|_{L^{p/(p-1)}} \\ &=n^{1/p}m^{-2/p}\biggl(\frac{p-1}{4\pi}\biggr)^{(p-1)/p}\|V\|_{L^{p/(p-1)}}. \end{aligned} \end{equation*} \notag
Finally, setting V(x)=\rho(x)^{p-1} we obtain (1.3) and (1.4).

Theorem 1 is proved.

Corollary. The following interpolation inequality holds for \varphi\in \dot{H}^1(\mathbb {S}^2):

\begin{equation} \|\varphi\|_{L^q(\mathbb{S}^2)}\leqslant\biggl(\frac1{4\pi}\biggr)^{(q-2)/(2q)} \biggl(\frac q2\biggr)^{1/2} \|\varphi\|^{2/q}\|\nabla\varphi\|^{1-2/q}, \qquad q\geqslant2. \end{equation} \tag{2.4}

Proof. For n=1 inequality (1.3) goes over to
\begin{equation*} \|\varphi\|_{L^{2p}}^2\leqslant\mathrm{B}_p \bigl(m^{2-2/p}\|\varphi\|^2+m^{-2/p}\|\nabla\varphi\|^2\bigr). \end{equation*} \notag
Minimizing with respect m we obtain
\begin{equation*} \|\varphi\|_{L^{2p}}^2 \leqslant\mathrm{B}_p\frac p{(p-1)^{(p-1)/p}}\|\varphi\|^{2/p}\|\nabla\varphi\|^{2-2/p} =\biggl(\frac1{4\pi}\biggr)^{(p-1)/p}\!\!\!p\|\varphi\|^{2/p}\|\nabla\varphi\|^{2-2/p}, \end{equation*} \notag
which is just (2.4).

The corollary is proved.

The inequality for H^1-orthonormal divergence-free vector functions on \mathbb{S}^2 and the corresponding one-function interpolation inequality are similar to the scalar case.

Theorem 2. Let \Omega\subseteq\mathbb{S}^2 be an arbitrary domain on the sphere and let the family of vector functions \{u_j\}_{j=1}^n\in\mathbf{H}^1_0(\Omega), \Omega\subseteq\mathbb{S}^2, \operatorname{div} u_j=0, be orthonormal in \mathbf{H}^1:

\begin{equation*} m^2(u_i,u_j)+(\operatorname{rot} u_i,\operatorname{rot} u_j)=\delta_{ij}. \end{equation*} \notag
Then for 1\leqslant p<\infty the function
\begin{equation*} \rho(x):=\sum_{j=1}^n|u_j(x)|^2 \end{equation*} \notag
satisfies the inequality
\begin{equation*} \|\rho\|_{L^p}\leqslant\mathrm{B}_pm^{-2/p}n^{1/p}, \end{equation*} \notag
where
\begin{equation*} \mathrm{B}_p\leqslant\biggl(\frac{p-1}{4\pi}\biggr)^{(p-1)/p}. \end{equation*} \notag

Proof. As we now have (3.1) for all 1<p<\infty, the proof of the theorem is completely analogous. To make the paper self contained we provide some details.

In the vector case identity (2.2) is replaced by its vector analogue (see [22])

\begin{equation} \sum_{k=1}^{2n+1}|\nabla Y_n^k(s)|^2=n(n+1)\frac{2n+1}{4\pi}. \end{equation} \tag{2.5}
In fact, substituting \varphi(s)=Y_n^k(s) into the identity
\begin{equation*} \Delta\varphi^2=2\varphi\Delta\varphi+2|\nabla\varphi|^2 \end{equation*} \notag
we sum the results over k=1,\dots,2n+1. In view of (2.2) the left-hand side vanishes and we obtain (2.5) since the Y_n^k(s) are the eigenfunctions corresponding to n(n+1).

Next, by the vector Laplace operator acting on (tangent) vector fields on \mathbb{S}^2, identifying 1-forms and vectors, we mean the Laplace-de Rham operator -d\delta-\delta d. Then for a two-dimensional manifold we have (see [23])

\begin{equation*} \mathbf{\Delta} u=\nabla\operatorname{div} u-\operatorname{rot}\operatorname{rot} u, \end{equation*} \notag
where the operators \nabla=\operatorname{grad} and \operatorname{div} have the conventional meaning. The operator \operatorname{rot} u at a vector u is a scalar and for a scalar \psi, \operatorname{rot}\psi is a vector:
\begin{equation*} \operatorname{rot} u:=\operatorname{div}(u^\perp)\quad\text{and} \quad \operatorname{rot}\psi:=\nabla^\perp\psi, \end{equation*} \notag
where in the local frame u^\perp=(u_2,-u_1), that is, it is the \pi/2 clockwise rotation of u in the local tangent plane. Integrating by parts we obtain
\begin{equation*} (-\mathbf{\Delta} u,u)=\|{\operatorname{rot} u}\|^2_{L^2}+\|{\operatorname{div} u}\|^2_{L^2}. \end{equation*} \notag

Corresponding to the eigenvalue \Lambda_n=n(n+1), where n=1,2,\dots, there is a family of 2n+1 orthonormal vector-valued eigenfunctions w_n^k(s) of the vector Laplacian on the invariant space of divergence-free vector functions (that is, the Stokes operator on \mathbb{S}^2):

\begin{equation*} w_n^k(s)=(n(n+1))^{-1/2}\nabla^\perp Y_n^k(s), \qquad -\mathbf{\Delta}w_n^k=n(n+1)w_n^k, \quad \operatorname{div} w_n^k=0; \end{equation*} \notag
where k=1,\dots,2n+1, and (2.5) implies the following identity:
\begin{equation} \sum_{k=1}^{2n+1}|w_n^k(s)|^2=\frac{2n+1}{4\pi}. \end{equation} \tag{2.6}
Finally, we observe that -\mathbf{\Delta} is strictly positive -\mathbf{\Delta}\geqslant \Lambda_1I=2I.

Turning to the proof we consider first the whole of the sphere \Omega=\mathbb{S}^2 and, as in (2.3), we set

\begin{equation*} \mathbb{H}= V^{1/2}(m^2-\mathbf{\Delta})^{-1/2}\mathbf{\Pi}\quad\text{and} \quad \mathbb{H}^*=\mathbf{\Pi}(m^2-\mathbf{\Delta})^{-1/2}V^{1/2}, \end{equation*} \notag
where \mathbf{\Pi} is the orthogonal Helmholtz-Leray projection onto the subspace of divergence-free vectors \{u\in \mathbf{L}^2(\mathbb{S}^2),\operatorname{div} u=0\}. From this point, using (2.6), we can complete the proof as in the scalar case.

Finally, if \Omega\varsubsetneq\mathbb{S}^2 is a proper domain on \mathbb{S}^2, then we extend the u_j by zero outside \Omega and denote the results by \widetilde{u}_j, so that \widetilde{u}_j\in {\bf H}^1(\mathbb{S}^2) and \operatorname{div}\widetilde{u}_j=0. We also set \widetilde\rho(x):=\sum_{j=1}^n|\widetilde{u}_j(x)|^2. Setting \widetilde{\psi}_i:=(m^2-\mathbf{\Delta})^{1/2}\widetilde{u}_i we obtain that the system \{\widetilde\psi_j\}_{j=1}^n is orthonormal in \mathbf L^2(\mathbb{S}^2) and \operatorname{div}\widetilde\psi_j=0. Since \|\widetilde\rho\|_{ L^2(\mathbb{S}^2)}=\|\rho\|_{ L^2(\Omega)}, the proof reduces to the case of the whole sphere.

Theorem 2 is proved.

Remark 1. For q=4 inequality (2.4) is Ladyzhenskaya’s inequality on the 2D sphere \mathbb{S}^2

\begin{equation*} \|\varphi\|_{L^4}^4\leqslant \mathrm{c}_\mathrm{Lad}\|\varphi\|^2\|\nabla\varphi\|^2 \end{equation*} \notag
with estimate \mathrm{c}_\mathrm{Lad}\leqslant1/\pi for the constant. A recent estimate for this constant in [20] in terms of the Lieb-Thirring inequality is slightly better: \mathrm{c}_\mathrm{Lad}\leqslant3\pi/32. On the other hand (2.4) works for all q\geqslant2 and provides a simple expression for the constant.

Remark 2. The rate of growth as q\to\infty of the constant in both (2.4) and (1.9), namely, q^{1/2}, is optimal on the power scale. If we had not imposed the zero-mean condition for the sphere, it would have immediately followed from (1.1) with d=2.

In the general case, if the growth exponent in (2.4) and (1.9) were less than 1/2, then the Sobolev space H^1 in two dimensions would be embedded into the Orlicz space with Orlicz function e^{t^{2+\varepsilon}}-1, \varepsilon>0, which is impossible (see [21]).

Furthermore, while for every fixed q<\infty the constant in (2.4) and (1.9) is not sharp, we think, as mentioned before, that the sharp constant \mathrm{c}_q behaves like

\begin{equation*} \frac{\mathrm{c}_q}{\sqrt{q}}\to \frac1{\sqrt{8\pi}} \quad \text{as } q\to\infty. \end{equation*} \notag

§ 3. Proof of estimate (1.6)

Proposition. The following inequality holds for p>1 and m\geqslant0

\begin{equation} I_p(m):=m^{2(p-1)}(p-1)\sum_{n=1}^\infty\frac{2n+1}{(m^2+n^2+n)^p}<1. \end{equation} \tag{3.1}

Proof. We point out first that inequality (1.6) holds for all sufficiently large m as follows from the asymptotic expansion for this type of series. In fact, we observe that we can write I_p(m) in the form
\begin{equation*} I_p(m)=\frac{p-1}{m^2}\sum_{n=1}^\infty (2n+1)g\biggl(\frac{n(n+1)}{m^2}\biggr), \qquad g(t)=\frac1{(1+t)^p}. \end{equation*} \notag
The following asymptotic expansion as m\to \infty holds for this type of series (see [25], Lemma 3.5):
\begin{equation*} \begin{aligned} \, I_p(m) &=(p-1)\biggl[\int_0^\infty g(t)\,dt-\frac1{m^2}\,\frac23 g(0)+O(m^{-4})\biggr] \\ &=1-\frac1{m^2}\,\frac{2(p-1)}3+O(m^{-4}). \end{aligned} \end{equation*} \notag
Therefore, for fixed p\,{>}\,1 there exists a sufficiently large m\,{=}\,m_p such that inequality (3.1) holds for all m\geqslant m_p.

The proof that it holds for all p>1 and m\geqslant0 requires some specific work. We shall use the Euler-Maclaurin summation formula (see, for example, [24]). Namely, we use the forth-order formula

\begin{equation} \sum_{n=1}^\infty f(n)=\int_0^\infty f(x)\,dx-\frac12 f(0)-\frac1{12}f'(0)+\frac1{720}f'''(0)+R_4 \end{equation} \tag{3.2}
with remainder term
\begin{equation*} R_4=-\frac1{4!}\int_0^\infty f''''(x)B_4(x)\,dx, \end{equation*} \notag
where B_4(x) is the periodic Bernoulli polynomial. The remainder term R_4 in this formula can be estimated as
\begin{equation} |R_4|\leqslant\frac{2\zeta(4)}{(2\pi)^4}\int_0^\infty|f''''(x)|\,dx= \frac1{720}\int_0^\infty|f''''(x)|\,dx, \end{equation} \tag{3.3}
where \zeta(4)=\frac{\pi^4}{90} and \zeta(s) is the Riemann zeta function.

We use this formula for relatively large m and

\begin{equation*} f_m(x)=\frac{m^{2(p-1)}(p-1)(2x+1)}{(m^2+x^2+x)^p}. \end{equation*} \notag
A straightforward calculation gives
\begin{equation*} \begin{gathered} \, \int_0^\infty f_m(x)\,dx=1, \\ f_m(0)=\frac{p-1}{m^2}, \qquad f_m'(0)=\frac{(p-1)(2m^2-p)}{m^4}, \\ f_m'''(0)=-\frac{(p - 1)p(12m^4 - 12m^2p - 12m^2 + p^2 + 3p + 2)}{m^8} \end{gathered} \end{equation*} \notag
and
\begin{equation*} \begin{aligned} \, f_m''''(x) &=32p(p^2 - 1)m^{2p - 2} \biggl(\frac{(x + 1/2)^5(p + 2)(p + 3)}{(m^2 + x^2 + x)^{p + 4}} \\ &\qquad-\frac{5(x + 1/2)^3(p + 2)}{(m^2 + x^2 + x)^{p +3}} + \frac{15(x+1/2)}{4(m^2 + x^2 + x)^{p+2}}\biggr). \end{aligned} \end{equation*} \notag
We change the sign of the second term in the above expression and set
\begin{equation*} \begin{aligned} \, g(x) &=32p(p^2 - 1)m^{2p - 2} \biggl(\frac{(x + 1/2)^5(p + 2)(p + 3)}{(m^2 + x^2 + x)^{p + 4}} \\ &\qquad+\frac{5(x + 1/2)^3(p + 2)}{(m^2 + x^2 + x)^{p +3}} + \frac{15(x+1/2)}{4(m^2 + x^2 + x)^{p+2}}\biggr). \end{aligned} \end{equation*} \notag
Then, obviously, |f_m''''(x)|\leqslant g(x) for all x. On the other hand, the integral of g(x) can be computed explicitly (since g(x) contains odd powers of (x-1/2) in the numerators, hence the corresponding antiderivatives are expressed in elementary functions):
\begin{equation*} \int_0^\infty g(x)\,dx=\frac{p(p-1)\bigl(172m^4+28(p+1)m^2+p^2+3p+2\bigr)}{m^8}. \end{equation*} \notag
Thus, the Euler-Maclaurin formula (3.2) gives us the estimate
\begin{equation} \begin{aligned} \, \notag I_p(m) &<1-\frac23(p-1)m^{-2}+\frac{11}{36}p(p-1)m^{-4}+\frac1{18}p(p^2-1)m^{-6} \\ &=1-\frac1{36}(p-1)m^{-6}\bigl(24m^4-11pm^2-2p(p+1)\bigr). \end{aligned} \end{equation} \tag{3.4}
Therefore,
\begin{equation*} I_p(m)<1 \end{equation*} \notag
if 24m^4-11pm^2-2p(p+1)>0, that is, if
\begin{equation} m>\frac{\sqrt{3\sqrt{313p^2 + 192p} + 33p}}{12}=:m_0(p). \end{equation} \tag{3.5}

Now consider two cases, p\in(1,2] and p>2. Let p\in(1,2]. The maximum value of m_0(p) on p\in(1,2] is attained at p=2, so we have proved the desired inequality (3.1) for all p\in (1,2] and

\begin{equation*} m>m_0:=\frac{\sqrt{66 + 6\sqrt{409}}}{12}\approx1.1406. \end{equation*} \notag
Thus, we only need to verify the desired inequality for m<m_0. We single out the first term of the series and drop the dependence on m in the remaining terms. Then we obtain
\begin{equation} \begin{aligned} \, \notag I_p(m) &=m^{2(p-1)}(p-1)\biggl(\frac3{(m^2+2)^p}+ \sum_{n=2}^\infty\frac{2n+1}{(m^2+n^2+n)^p}\biggr) \\ \notag &<m^{2(p-1)}(p-1)\biggl(\frac3{(m^2+2)^p}+\sum_{n=2}^\infty\frac{2n+1}{(n^2+n)^p}\biggr) \\ &=m^{2(p-1)}(p-1)\biggl(\frac3{(m^2+2)^p}+ R(p)\biggr)=:G(m,p), \end{aligned} \end{equation} \tag{3.6}
where
\begin{equation*} R(p):=\sum_{n=2}^\infty\frac{2n+1}{(n^2+n)^p}. \end{equation*} \notag

To complete the proof, we only need to prove the inequality

\begin{equation*} G(m,p)<1 \end{equation*} \notag
for all p\in[1,2] and all m\in[0,m_0].

Again we apply the Euler-Maclaurin formula to the series R(p) (taking into account that now summation starts with n=2). Setting

\begin{equation*} f(n):=\frac{2n+1}{(n^2+n)^p}, \end{equation*} \notag
we have
\begin{equation*} \int_1^\infty f(x)\,dx=\frac{2^{1-p}}{p-1}, \qquad f(1)=\frac3{2^p}, \qquad f'(1)=\frac1{2^p}\biggl(2-\frac{9p}2\biggr) \end{equation*} \notag
and
\begin{equation*} \begin{aligned} \, f''''(n) &=\frac{(2n+1)p(p+1)}{(n^2+n)^{p+4}}\bigl((16p^2-4)n^4+(32p^2-8)n^3 \\ &\qquad+(24p^2+20p+4)n^2+(8p^2+20p+8)n+p^2+5p+6\bigr). \end{aligned} \end{equation*} \notag
Since clearly f''''(n)>0, it follows from (3.3) that the last two terms in the Euler-Maclaurin formula add up to zero:
\begin{equation*} \frac1{720}f'''(1)+R_4\leqslant\frac1{720}\biggl(f'''(1)+\int_1^\infty f''''(x)\,dx\biggr)=0. \end{equation*} \notag
Therefore,
\begin{equation*} R(p)\leqslant\int_1^\infty f(x)-\frac12f(1)-\frac1{12}f'(1)= \frac1{2^p(p-1)}\,\frac{9p^2-49p+88}{24}. \end{equation*} \notag
We substitute this into (3.6) and set z:=m^2/2. We also assume that z\leqslant1. Then, since e^{-x}\leqslant1/(1+x), x\geqslant0, and taking into account that \ln z\leqslant0 we have
\begin{equation*} z^{p-1}=e^{(p-1)\ln z}<\frac1{1-(p-1)\ln z}. \end{equation*} \notag
Using this and Bernoulli’s inequality (1+z)^p>1+pz we obtain
\begin{equation} \begin{aligned} \, \notag G(m,p)-1 &=\frac12z^{p-1}\biggl(\frac{3(p-1)}{(1+z)^p}+2^p(p-1)R(p)\biggr)-1 \\ \notag &<\frac12\frac1{(1-(p-1)\ln z)}\biggl(\frac{3(p-1)}{1+pz}+\frac{9p^2-49p+88}{24}\biggr)-1 \\ \notag &=\frac{p-1}{48(pz+1)(1-(p-1)\ln z)} \\ &\qquad\times\bigl(9zp^2+(48 z\ln z-40 z+9)p+48 \ln z+32\bigr) \notag \\ &=:A(z,p)\phi(z,p). \end{aligned} \end{equation} \tag{3.7}
We point out that (3.7) holds for all m\leqslant\sqrt{2} (so that z\leqslant1) and all p>1 and in this case A(z,p)>0. Therefore, the sign of G(m,p)-1 coincides with that of the quadratic polynomial
\begin{equation*} \phi(z,p)=9zp^2+(48 z\ln z-40 z+9)p+48 \ln z+32. \end{equation*} \notag
For fixed p the function \phi(z,p) is monotone increasing with respect to z. In fact, since p\ln z+1/z\geqslant p(1-\ln p), for p>1 we have
\begin{equation} \partial_z\phi(z,p)=9p^2+8p+48\biggl(p\ln z+\frac1z\biggr)\geqslant 9p^2+56p-48p\ln p>0. \end{equation} \tag{3.8}

Now returning to p\in (1,2] we observe that for z=m_0^2/2=0.6504<1 we have \ln z<0, and therefore

\begin{equation*} \phi\biggl(\frac{m_0^2}2,p\biggr)=5.854p^2-30.446p+11.358<0 \quad \text{for } p\in[1,2]. \end{equation*} \notag
Hence
\begin{equation*} \phi\biggl(\frac{m^2}2,p\biggr)<0 \quad\text{for all } m\in[0,m_0]\quad\text{and} \quad p\in[1,2]. \end{equation*} \notag
This completes the proof of inequality (3.1) for p\in (1,2].

We are now ready to verify (3.1) for p>2 as well. The key idea here is to use the fact that I_p(m) is monotone decreasing with respect to p for 0\leqslant m\leqslant m_1(p), where m_1(p) is given below. Indeed, let

\begin{equation*} f(n)=f(m,n,p)=\frac{m^{2(p-1)}(p-1)(2n+1)}{(m^2+n^2+n)^p}. \end{equation*} \notag
Then
\begin{equation*} \partial_pf(n)=\frac{m^{2(p-1)}(2n+1)}{(m^2+n^2+n)^p} \biggl(1+(p-1)\ln\frac{m^2}{m^2+n^2+n}\biggr), \end{equation*} \notag
and the derivative is negative for all n\in\mathbb N if
\begin{equation*} m<m_1(p):=\frac{\sqrt{2}}{\sqrt{e^{1/(p-1)}-1}}. \end{equation*} \notag

Now let p>2. Two cases are possible

\begin{equation*} m_0(p)<m_1(p)\quad\text{and} \quad m_0(p)\geqslant m_1(p). \end{equation*} \notag

In the first case inequality (3.1) holds for all m, since if m>m_0(p), then it holds in view of (3.4) and (3.5), while if m<m_0(p)<m_1(p), then it holds in view of the established monotonicity with respect to p and the fact that (3.1) holds for p=2.

In the second case, first we find the interval with respect to p where the inequality m_1(p)\leqslant m_0(p) actually holds. Namely, it holds for

\begin{equation*} p\in[2,p_*], \quad\text{where } p_*=2.10915\dots; \end{equation*} \notag
see Figure 1, where the unique value of p_* is found numerically.

Thus, inequality (3.1) holds for p>p_* and we only need to look at the interval p\in[2,p_*]. Furthermore, since m_0(p) in (3.5) is monotone increasing, we must only check (3.1) for

\begin{equation*} p\in[2,p_*]\quad\text{and} \quad m\in[0,m_*], \quad m_*=m_0(p_*)=1.169\dotsc\,. \end{equation*} \notag
In view of (3.6)(3.8) and the remark after (3.7) we have the following sequence of implications and equivalences:
\begin{equation*} \begin{aligned} \, \bigl\{I_p(m)<1\bigr\} &\Longleftarrow\bigl\{G(m,p)-1<0\bigr\}\Longleftarrow \bigl\{A(z,p)\phi(z,p)<0\bigr\} \\ &\Longleftrightarrow\bigl\{\phi(z,p)<0\bigr\}\Longleftarrow\bigl\{\phi(z_*,p)<0\bigr\} \\ &\Longleftrightarrow\bigl\{6.1495p^2-30.8222p+13.7197<0,\ p\in[2,p_*]\bigr\}=\{\text{`true'}\}, \end{aligned} \end{equation*} \notag
where z=m^2/2\leqslant z_*=m_*^2/2=0.6832<1 and m_*=1.169<\sqrt{2}.

Inequality (3.1) is now proved for the whole range of parameters.

The proposition is proved.

Remark 3. The case p=2, which is important for applications, was treated by more elementary means in [7].

Remark 4. Calculations show that for each p tested the function I_p(m) is monotone increasing with respect to m. We are not able to prove this rigorously at the moment. However, it was shown in [8] that the lattice sum J_p(m) in (1.5) is monotone increasing in m, which obviously implies (1.5) since J_p(\infty)=1.


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Citation: S. V. Zelik, A. A. Ilyin, “On a class of interpolation inequalities on the 2D sphere”, Sb. Math., 214:3 (2023), 396–410
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\by S.~V.~Zelik, A.~A.~Ilyin
\paper On a~class of interpolation inequalities on the 2D sphere
\jour Sb. Math.
\yr 2023
\vol 214
\issue 3
\pages 396--410
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\crossref{https://doi.org/10.4213/sm9786e}
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