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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
Russian version:
Matematicheskii Sbornik, 2023, Volume 214, Number 3, DOI: https://doi.org/10.4213/sm9786 
Bibliographic databases:
    § 1. IntroductionThe following interpolation inequality holds on the sphere $\mathbb{S}^d$ (see [1] and also [2]):
$$
\begin{equation}
\frac{q-2}d\int_{\mathbb{S}^d}|\nabla\varphi|^2\,d\mu+ \int_{\mathbb{S}^d}|\varphi|^2\,d\mu\geqslant \biggl(\int_{\mathbb{S}^d}|\varphi|^q\,d\mu\biggr)^{2/q}.
\end{equation}
\tag{1.1}
$$
Here $d\mu$ is the normalized Lebesgue measure on $\mathbb{S}^d$:
$$
\begin{equation*}
d\mu=\frac{d\sigma}{\sigma_d}=\frac{d\sigma}{\frac{2\pi^{(d+1)/2}}{\Gamma((d+1)/2)}},
\end{equation*}
\notag
$$
so that $\mu(\mathbb{S}^d)=1$ (the gradient is calculated with respect to the natural metric). Next, $q\in[2, \infty)$ for $d=1,2$ and $q\in[2, 2d/(d-2)]$ for $d\geqslant 3$. The remarkable fact about (1.1) is that the constant $(q-2)/d$ is sharp for all admissible $q$. The inequality clearly degenerates and turns to equality on constants. The fact that the constant $(q-2)/d$ is sharp is verified by means of the minimizing sequence $\varphi_\varepsilon(s)=1+\varepsilon v(s)$, where $\varepsilon\to0$ and $v(s)$ is an eigenfunction of the Laplacian on $\mathbb{S}^d$, 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 $\varphi$ usually play the role of stream functions of divergence-free vector functions $u$, $u=\nabla^\perp\varphi$, and therefore without loss of generality $\varphi$ can be chosen to be orthogonal to constants. In this work we consider the two-dimensional sphere $\mathbb{S}^2$ only and are interested in writing the Sobolev embedding $H^1(\mathbb{S}^2)\hookrightarrow L^q(\mathbb{S}^2)$ as a multiplicative inequality of Gagliardo-Nirenberg type involving the $L^2$-norms of $\varphi$ and $\nabla\varphi$ on the right-hand side: $\|\varphi\|_{L^2(\mathbb{S}^2)}=:\|\varphi\|$ and $\|\nabla \varphi\|_{L^2(\mathbb{S}^2)}=:\|\nabla\varphi\|$. It is also well known that in the case of $\mathbb{R}^d$ 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 $x\to mx$) 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 $H^1$ with norm and inner product
$$
\begin{equation*}
\|\varphi\|_{H^1}^2:=m^2\|\varphi\|^2+\|\nabla\varphi\|^2\quad\text{and} \quad (\varphi_1,\varphi_2)_{H^1}:=m^2(\varphi_1,\varphi_2)+ (\nabla\varphi_1,\nabla\varphi_2),
\end{equation*}
\notag
$$
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 $H^1$-orthonormal families proved in [4]. We can now state and discuss our main result. Theorem 1. Let a family of zero-mean functions $\{\varphi_j\}_{j=1}^n\in\dot{H}^1(\mathbb{S}^2)$ be orthonormal with respect to the inner product Then for $1\leqslant p<\infty$ the function satisfies the inequality where These inequalities were proved in the case of $\mathbb{R}^d$ in [4] for ${p=\infty}$ and $d=1$, for ${1\leqslant p<\infty}$ and $d=2$ and for the critical $p=d/(d-2)$ and $d\geqslant3$. 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 $\mathrm{B}_2\leqslant(4\pi)^{-1/2}$ for $\mathbb{T}^2$, $\mathbb{S}^2$ and $\mathbb{R}^2$; this was based on the following two inequalities for a lattice sum over $\mathbb{Z}^2_0=\mathbb{Z}^2\setminus\{0,0\}$ and a series with respect to the spectrum of the Laplacian on $\mathbb{S}^2$, which were proved there in the special case when $p=2$:
$$
\begin{equation}
J_p(m):= \frac{(p-1)m^{2(p-1)}}\pi\sum_{n\in{\mathbb Z}_0^2}\frac1{(m^2+|n|^2)^p}<1,
\end{equation}
\tag{1.5}
$$
$$
\begin{equation}
I_p(m):=m^{2(p-1)}(p-1)\sum_{n=1}^\infty\frac{2n+1}{\bigl(m^2+n(n+1)\bigr)^p}<1.
\end{equation}
\tag{1.6}
$$
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 $\mathbb{T}^2$ and $\mathbb{S}^2$ once we have at our disposal inequalities (1.5) and (1.6) for all $1< p< \infty$. Inequality (1.5) and Theorem 1 for the torus $\mathbb{T}^2$ 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 $\mathbb{R}^2$, instead of (1.5) and (1.6), we have the equality
$$
\begin{equation}
\frac{(p-1)m^{2(p-1)}}\pi\int_{\mathbb {R}^2}\frac{dx}{(m^2+|x|^2)^p}=1.
\end{equation}
\tag{1.7}
$$
For one function (that is, for $n=1$) Theorem 1 turns to Sobolev’s inequality with parameter for the embedding $H^1\hookrightarrow L^q$, $q=2p\in [2,\infty)$, which can equivalently be written as the Gagliardo-Nirenberg inequality
$$
\begin{equation}
\|f\|_{L^q}\leqslant\biggl(\frac{1}{4\pi}\biggr)^{(q-2)/(2q)} \biggl(\frac q2\biggr)^{1/2}\|f\|^{2/q}\|\nabla f\|^{1-2/q},
\end{equation}
\tag{1.8}
$$
which holds for $\mathbb{R}^2$, $\mathbb{T}^2$ and $\mathbb{S}^2$, see the corollary. For the torus $\mathbb{T}^2$ 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 $\mathbb{R}^2$ 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 $\mathbb{R}^2$ with the best to date estimate for the constant (see [11]):
$$
\begin{equation}
\|\varphi\|_{L^q(\mathbb{R}^2)}\leqslant\biggl(\frac1{4\pi}\biggr)^{(q-2)/(2q)} \frac{q^{(q-2)/q}}{(q-1)^{(q-1)/q}} \biggl(\frac q2\biggr)^{1/2} \|\varphi\|^{2/q}\|\nabla\varphi\|^{1-2/q}, \quad q\geqslant2;
\end{equation}
\tag{1.9}
$$
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 $\mathbb{R}^2$ and inequality (2.4) for $\mathbb{T}^2$ are special cases of the Gagliardo-Nirenberg inequality. For $\mathbb{R}^2$ the best constant is known for every $q\geqslant2$ 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 $\mathbb{T}^2$ was recently obtained in [8]. As far as the case of the sphere $\mathbb{S}^2$ 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 $\mathbb{T}^2$ is that the orthonormal spherical functions are not uniformly bounded in $L^\infty$. Our approach makes it possible to prove similar inequalities in the vector case. Namely, we show that for $u\in\mathbf{H}^1_0(\Omega)\cap\{\operatorname{div} u=0\}$
$$
\begin{equation*}
\|u\|_{L_q(\mathbb{S}^2)}\leqslant\biggl(\frac{1}{4\pi}\biggr)^{(q-2)/(2q)} \biggl(\frac q2\biggr)^{1/2}\|u\|^{2/q}\|\operatorname{rot} u\|^{1-2/q}.
\end{equation*}
\notag
$$
Here $\Omega\subseteq\mathbb{S}^2$ is an arbitrary domain on $\mathbb{S}^2$. This inequality looks very similar to (1.8), the important difference being that, unlike the scalar case, the vector Laplacian on $\mathbb{S}^2$ is positive definite, and we can freely use the extension by zero for ${\Omega\varsubsetneq\mathbb{S}^2}$. 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 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]): 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 $s\in\mathbb{S}^{2}$ 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. 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 $\mathbf{H}^1$: Then for $1\leqslant p<\infty$ the function satisfies the inequality where 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: 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 $\mathbb{S}^2$ 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 § 3. Proof of estimate (1.6) 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 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 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, 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
To complete the proof, we only need to prove the inequality 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 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 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 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
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 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 $J_p(\infty)=1$. 
Citation in format AMSBIB
\Bibitem{ZelIly23} |
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