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An extremal problem for positive definite functions with support in a ball A. D. Manov St Petersburg State University, St Petersburg, Russia
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     § 1. IntroductionWe fix some notation: $|\cdot|$ is the Euclidean norm in $\mathbb{R}^n$, $\mathbb{B}_r:=\{x\in \mathbb{R}^n\colon |x|<r\}$ is the open ball of radius $r>0$ centred at the origin, $\overline{\mathbb{B}_r}$ is the closure of this ball, $\widetilde{f}(x):=\overline{f(-x)}$, $(f\ast g)(x)=\displaystyle\int_{\mathbb{R}^n}f(x-t)g(t)\,dt$, and $L_\infty^{\mathrm{loc}}(\mathbb{R}^n)$ is the space of functions locally bounded almost everywhere on $\mathbb{R}^n$. A complex-valued function $f\colon \mathbb{R}^n\to\mathbb{C}$ is said to be positive definite on $\mathbb{R}^n$ ($f\in\Phi(\mathbb{R}^n)$) if for each $m\in\mathbb{N}$, any $\{x_i\}_{i=1}^m\subset \mathbb{R}^n$ and each set of complex numbers $\{c_i\}_{i=1}^m\subset\mathbb{C}$ we have the inequality
$$
\begin{equation}
\sum_{i,j=1}^m c_i\overline{c_j}f(x_i-x_j)\geqslant0.
\end{equation}
\tag{1.1}
$$
If $f\in\Phi(\mathbb{R}^n)$, then from (1.1), for $m=2$ we obtain $|f(x)|\leqslant f(0)$ for $x\in \mathbb{R}^n$, and $f$ is an Hermitian function, that is, $f=\widetilde{f}$. In this paper we are interested in the following convex subset of positive definite functions. Let $r>0$. Then we denote by $\mathfrak{F}_r(\mathbb{R}^n)$ the set of functions
$$
\begin{equation*}
{\varphi\in\Phi(\mathbb{R}^n)\cap C(\mathbb{R}^n) \quad \text{such that}\ \ \varphi(0)=1 \quad \text{and}\ \ \operatorname{supp}\varphi\subset\overline{\mathbb{B}_r}.}
\end{equation*}
\notag
$$
It is obvious that the class $\mathfrak{F}_r(\mathbb{R}^n)$ is nonempty. For example, if $u\in L_2(\mathbb{R}^n)$ is a function such that $u(x)=0$ for $|x|\geqslant r/2$ and $\|u\|_2=1$, then the following function belongs to $\mathfrak{F}_r(\mathbb{R}^n)$:
$$
\begin{equation}
\varphi(x)=(u\ast\widetilde{u})(x)=\int_{\mathbb{R}^n}u(x-t)\widetilde{u}(t)\,dt, \qquad x\in\mathbb{R}^n.
\end{equation}
\tag{1.2}
$$
In fact, we have $\varphi\in C(\mathbb{R}^n)$ and $ \operatorname{supp}\varphi\subset\overline{\mathbb{B}_{r/2}}+\overline{\mathbb{B}_{r/2}} =\overline{\mathbb{B}_r}$. That $\varphi$ is positive definite is straightforward. Note that for $n=1$ it follows from the Boas–Kac–Krein theorem (for instance, see [1], Theorem 3.10.2) that each function $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$ can be represented in the form (1.2). Generally speaking, this is not true for $n\geqslant2$. In this paper we consider the following extremal problem for positive definite functions $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$. Problem. Let $r>0$, and let $\rho\in L_\infty^{\mathrm{loc}}(\mathbb{R}^n)$ be a radial real function. Find the quantity For $\rho(x)\equiv1$ the quantity $M(n,\rho,r)$ was found by Siegel [2] in 1935 and, independently, by Boas and Kac (see [3], Theorem 5) in 1945 for $n=1$. They showed that
$$
\begin{equation*}
M(n,\rho,r)=\operatorname{vol}(\mathbb{B}_{r/2})=\frac{\pi^{n/2}r^n}{2^n\Gamma(n/2+1)},
\end{equation*}
\notag
$$
where $\operatorname{vol}(\,\cdot\,)$ is the Lebesgue measure in $\mathbb{R}^n$. In this case the extremal function is the convolution of the characteristic function of the ball $\mathbb{B}_{r/2}$ with itself:
$$
\begin{equation*}
\varphi(x)=\frac1{\operatorname{vol}(\mathbb{B}_{r/2})}(\chi_{\mathbb{B}_{r/2}}\ast\chi_{\mathbb{B}_{r/2}})(x),\qquad x\in\mathbb{R}^n.
\end{equation*}
\notag
$$
Siegel’s result was also re-discovered by Gorbachev [4] in 2001, who used other methods. It can also be mentioned that another method of the proof of Siegel’s results can be found in the recent paper [5]. Note that for $\rho(x)\equiv1$ the problem under consideration is an extremal problem of Turán type. In this class of problems one must find the supremum of the values of the integral $\displaystyle\int_{\mathbb{R}^n}\varphi(x)\,dx$ over all functions $\varphi\in\Phi(\mathbb{R}^n)\cap C(\mathbb{R}^n)$ with fixed value at the origin whose support lies in a fixed centrally symmetric convex body. To this date, apart from the ball case, solutions of the Turán problem are only known for polytopes tiling the space (see Arestov and Berdysheva [6]) and for spectral bodies (see Kolountzakis and Révész [7]). It is also worth pointing out that similar problems arise in a natural way in various fields of mathematics, for instance, convex analysis (see [8]). Applications to function theory can be found in [9]. Also note Efimov’s paper [10], where he considered a version of Turán problem for a ball. For further information about the history, versions and applications of this type of problem the reader can consult Révész’s paper [11]. For $n=1$ and weaker assumptions about the function $\rho$ an analogue of the problem in question was considered by this author in [12]. Here we establish the following theorem, which solves the problem for $n\neq2$. Theorem 1. Let $n\neq2$, $r>0$, and let $\rho\in L_\infty^{\mathrm{loc}}(\mathbb{R}^n)$ be a radial real function. Let $A_\rho\colon L_2(\mathbb{B}_{r/2})\to L_2(\mathbb{B}_{r/2})$ be the operator defined by Then $A_\rho$ is a compact selfadjoint operator in $L_2(\mathbb{B}_{r/2})$ and where $\|A_\rho\|$ is the norm of $A_\rho$ in $L_2(\mathbb{B}_{r/2})$.Remark 1. It follows from Theorem 1 that the solution of the problem under consideration reduces to finding the eigenvalue of $A_\rho$ that is greatest in modulus. Note also that Theorem 1 is an analogue of Szász’s theorem for nonnegative trigonometric polynomials (see [13], Theorem IV). Remark 2. The proof of Theorem 1 is based on the Rudin–Efimov representation (see Theorem 5 below) of functions $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$. In general, it has been established for $n\neq2$. On the other hand, if $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$ is infinitely smooth, then this representation also holds for $n=2$. In some cases, in the above problem it is sufficient to consider infinitely smooth functions from $\mathfrak{F}_r(\mathbb{R}^n)$. For instance, this is true when $\rho$ is continuous. In this case Theorem 1 also holds for $n=2$. If $\rho(x)$ is a polynomial, then our problem is connected with sharp bounds for entire functions of exponential spherical type $\leqslant r$. Recall that an entire function $f\colon \mathbb{C}^n\to\mathbb{C}$ is said to have an exponential spherical type $\leqslant r$ if for each $\varepsilon>0$ there exists a positive constant $A_\varepsilon$ such that
$$
\begin{equation*}
|f(z)|\leqslant A_\varepsilon e^{(r+\varepsilon)|z|}, \qquad z\in\mathbb{C}^n, \quad\text{where } |z|=\biggl(\sum_{k=1}^n|z_k|^2\biggr)^{1/2}.
\end{equation*}
\notag
$$
We let $W_{p,r}(\mathbb{R}^n)$ denote the set of entire functions of exponential spherical type $\leqslant r$ such that their restrictions to $\mathbb{R}^n$ belong to $L_p(\mathbb{R}^n)$, $p\geqslant1$, and let $W_{p,r}^{+}(\mathbb{R}^n)$ denote the subset of functions in $W_{p,r}(\mathbb{R}^n)$ that are nonnegative on $\mathbb{R}^n$. Then the following result holds. Theorem 2. Let $n,m\in\mathbb{N}$, $r>0$, let $\Delta$ be the Laplace operator, $L$ be a linear differential operator of the form and let Then the following sharp inequality holds for $f\in W_{1,r}^{+}(\mathbb{R}^n)$: The paper is organized as follows. In § 2 we present some auxiliary facts and statements. In §§ 3 and 4 we prove Theorems 1 and 2, respectively. In § 5 we present for example the solution of the problem under consideration for $\rho(x)\equiv1$. Also, in § 5 we obtain solutions for $\rho(x)=|x|^2$ and $n\neq2$, and for $\rho(x)=x^{2m}$ and $n=1$, $m\in\mathbb{N}$. As a consequence, we deduce sharp inequalities of Bernstein–Nikol’skii type for functions in $W_{1,r}^{+}(\mathbb{R}^n)$. § 2. Auxiliary facts and statementsWe point out the following properties of functions in $\Phi(\mathbb{R}^n)$. Let $f,f_{i}\in\Phi(\mathbb{R}^n)$. Then: (1) $|f(x+y)-f(x)|^2\leqslant 2f(0)(f(0)-\operatorname{Re}f(y))$, $x,y\in \mathbb{R}^n$; (2) $\lambda_{1}f_{1}+\lambda_{2}f_{2}$, $\overline{f}$, $\operatorname{Re}f$, $f_1f_2\in\Phi(\mathbb{R}^n)$ for $\lambda_i\geqslant 0$; (3) if a finite limit $\lim_{n\to\infty} f_{n}(x)=:g(x)$ exists for all $x\in \mathbb{R}$, then $g\in\Phi(\mathbb{R}^n)$. Properties (1)–(3) are well known (for instance, see [14], [1] and [15]). In 1932 Bochner and, independently, Khintchine proved the following criterion for a function to be positive definite. Theorem 3 (Bochner–Khintchine theorem). A function $f$ belongs to $\Phi(\mathbb{R}^n)\cap C(\mathbb{R}^n)$ if and only if there exists a finite nonnegative Borel measure $\mu$ on $\mathbb{R}^n$ such that The proof can be found, for instance, in [14], [1] or [15]. As a direct consequence (for instance, see [1], Theorem 1.8.7), we obtain the following criterion of positive definiteness in terms of the Fourier transform. Theorem 4. If $f\in C(\mathbb{R}^n)\cap L_1(\mathbb{R}^n)$, then where and in this case $\widehat f\in L_1(\mathbb{R}^n)$. The following result is important for the proof of Theorem 1. Theorem 5 (Rudin and Efimov). Let $n\neq2$, $r>0$, and let $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$ be a radial function. Then $\varphi$ can be represented by a uniformly convergent series: where $u_k\in L_2(\mathbb{R}^n)$ and $u_k(x)=0$ as $|x|\geqslant r/2$. Remark 3. For $n=1$ Theorem 5 is a corollary to the Boas–Kac–Krein theorem. For ${n\in\mathbb{N}}$ Theorem 5 was proved by Rudin under the assumption that $\varphi$ is infinitely differentiable (see [16] and [1], Theorem 3.10.4). Ehm, Gneiting and Richards [17] noted without proof that in Rudin’s theorem it is sufficient to assume merely that $\varphi$ is continuous. In [18] Efimov proved Theorem 5 for $n\geqslant3$. Remark 4. Representation (2.1) for radial infinitely smooth functions in $\mathfrak{F}_r(\mathbb{R}^n)$ was used by Rudin in [16] to show that each positive definite radial function in a ball in $\mathbb{R}^n$ can be extended to a positive definite function on the whole space. This generalizes a theorem of M. Krein on the extension of positive definite functions from an interval to the whole line $\mathbb{R}$ (see [19]). Problems of the extension of positive definite radial functions to spaces of higher dimension were considered in [20]. Since a positive definite function is Hermitian, in the problem under consideration we can assume that functions $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$ are even. It follows from the lemma below that they can also be assumed to be radial. Lemma. Let $n\neq1$ and $r>0$, let $\rho\in L_\infty^{\mathrm{loc}}(\mathbb{R}^n)$ be a radial function, let $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$, and let $\varphi_{\mathrm{rad}}$ be the radialization of $\varphi$: where $\operatorname{SO}(n)$ is the special orthogonal group and $d\tau$ is the normalized Haar measure on $\operatorname{SO}(n)$. Then $\varphi_{\mathrm{rad}}\in\mathfrak{F}_r(\mathbb{R}^n)$ and Proof. The fact that $\varphi_{\mathrm{rad}}$ belongs to $\mathfrak{F}_r(\mathbb{R}^n)$ is verified directly. We prove (2.2):
$$
\begin{equation*}
\begin{aligned} \, \int_{\mathbb{R}^n}\varphi_{\mathrm{rad}}(x)\rho(x)\,dx &=\int_{\mathbb{R}^n}\biggl(\int_{\operatorname{SO}(n)}\varphi(\tau x)\,d\tau\biggr)\rho(x)\,dx \\ &=\int_{\operatorname{SO}(n)}\biggl(\int_{\mathbb{R}^n}\varphi(\tau x)\rho(x)\,dx\biggr)d\tau \\ &=\int_{\operatorname{SO}(n)}\biggl(\int_{\mathbb{R}^n}\varphi(x)\rho(\tau^{-1} x)\,dx\biggr)d\tau \\ &=\int_{\operatorname{SO}(n)}\biggl(\int_{\mathbb{R}^n}\varphi(x)\rho(x)\,dx\biggr)d\tau \\ &=\int_{\operatorname{SO}(n)}d\tau\cdot\int_{\mathbb{R}^n}\varphi(x)\rho(x)\,dx =\int_{\mathbb{R}^n}\varphi(x)\rho(x)\,dx. \end{aligned}
\end{equation*}
\notag
$$
The lemma is proved. § 3. Proof of Theorem 1Step 1. The operator $A_\rho$ is bounded and compact in $L_2(\mathbb{B}_{r/2})$ because ${\rho\in L_\infty^{\mathrm{loc}}(\mathbb{R}^n)}$. It is selfadjoint because $\rho$ is real and radial. For convenience, below we identify $L_2(\mathbb{B}_{r/2})$ with the subspace of functions ${u\in L_2(\mathbb{R}^n)}$ such that $u(x)=0$ almost everywhere for $|x|\geqslant r/2$. Let $u\in L_2(\mathbb{R}^n)$ and $u(x)=0$ for $|x|\geqslant r/2$. Then
$$
\begin{equation*}
\int_{\mathbb{R}^n}(u\ast\widetilde{u})(x)\rho(x)\,dx=(A_\rho u,u),
\end{equation*}
\notag
$$
where $(\,\cdot\,{,}\,\cdot\,)$ is the inner product in $L_2(\mathbb{B}_{r/2})$. In fact,
$$
\begin{equation*}
\begin{aligned} \, \int_{\mathbb{R}^n}(u\ast\widetilde{u})(x)\rho(x)\,dx &=\int_{\mathbb{R}^n}\biggl(\int_{\mathbb{R}^n}u(x-t)\widetilde{u}(t)\,dt\biggr)\rho(x)\,dx \\ &=\int_{\mathbb{R}^n}\biggl(\int_{\mathbb{R}^n}u(x+t)\overline{u(t)}\,dt\biggr)\rho(x)\,dx \\ &=\int_{\mathbb{R}^n}\biggl(\int_{\mathbb{R}^n}u(x+t)\rho(x)\,dx\biggr)\overline{u(t)}\,dt \\ &=\int_{\mathbb{R}^n}\biggl(\int_{\mathbb{R}^n}\rho(x-t)u(x)\,dx\biggr)\overline{u(t)}\,dt \\ &=\int_{\mathbb{B}_{r/2}}\biggl(\int_{\mathbb{B}_{r/2}}\rho(t-x)u(x)\,dx\biggr)\overline{u(t)}\,dt =(A_\rho u,u). \end{aligned}
\end{equation*}
\notag
$$
Step 2. We show that $M(n,\rho,r)\leqslant \|A_\rho\|$. It follows from the above lemma that we can assume in the problem under consideration that the functions $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$ are radial. Let $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$, and let $\varphi$ be a radial function. It follows from Theorem 5 that
$$
\begin{equation*}
\varphi(x)=\sum_{k=1}^\infty (u_k\ast\widetilde{u_k})(x), \qquad x\in\mathbb{R}^n,
\end{equation*}
\notag
$$
where the series is uniformly convergent, $u_k\in L_2(\mathbb{R}^n)$ and $u_k(x)=0$ for $|x|\geqslant r/2$. In addition, since $\varphi(0)=1$, we have $\sum_{k=1}^\infty\|u_k\|_2^2=1$ and therefore
$$
\begin{equation*}
\begin{aligned} \, \biggl|\int_{\mathbb{R}^n}\varphi(x)\rho(x)\,dx\biggr| &=\biggl|\int_{\mathbb{R}^n}\sum_{k=1}^\infty (u_k\ast\widetilde{u_k})(x)\rho(x)\,dx\biggr| =\biggl|\sum_{k=1}^\infty\int_{\mathbb{R}^n} (u_k\ast\widetilde{u_k})(x)\rho(x)\,dx\biggr| \\ &=\biggl|\sum_{k=1}^\infty(A_\rho u_k,u_k)\biggr| \leqslant \sum_{k=1}^\infty|(A_\rho u_k,u_k)| \leqslant \|A_\rho\|\sum_{k=1}^\infty|(u_k,u_k)| \\ &=\|A_\rho\|\sum_{k=1}^\infty\|u_k\|_2^2=\|A_\rho\|. \end{aligned}
\end{equation*}
\notag
$$
Hence Step 3. We show that $M(n,\rho,r)\geqslant \|A_\rho\|$. Let $u\in L_2(\mathbb{R}^n)$, $u(x)=0$ for $|x|\geqslant r/2$ and $\|u\|_2=1$. Then $u\ast\widetilde{u}\in\mathfrak{F}_r(\mathbb{R}^n)$ and we have
$$
\begin{equation*}
|(A_\rho u,u)| =\biggl|\int_{\mathbb{R}^n}(u\ast\widetilde{u})(x)\rho(x)\,dx\biggr|\leqslant M(n,\rho,r).
\end{equation*}
\notag
$$
As $A_\rho$ is selfadjoint, it follows that
$$
\begin{equation*}
\|A_\rho\|=\sup_{\|u\|_2=1}|(A_\rho u,u)|\leqslant M(n,\rho,r).
\end{equation*}
\notag
$$
The proof is complete. § 4. Proof of Theorem 2First we note that by multiplying each function $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$ by $e^{i(t,\cdot)}$, $t\in\mathbb{R}^n$, we map $\mathfrak{F}_r(\mathbb{R}^n)$ to itself bijectively, so that
$$
\begin{equation}
M(n,\rho,r)=\sup\biggl\{\biggl|\int_{\mathbb{R}^n}\varphi(x)e^{i(t,x)}\rho(x)\,dx\biggr|\colon \varphi\in\mathfrak{F}_r(\mathbb{R}^n)\biggr\} \quad\text{for all } t\in\mathbb{R}^n.
\end{equation}
\tag{4.1}
$$
Let $f\in W_{1,r}^{+}(\mathbb{R}^n)$ and $f(x)\not\equiv0$. Since $L$ is a linear operator, we can assume without loss of generality that $\|f\|_1=(2\pi)^n$. By the Paley–Wiener theorem (for instance, see [15], § 3.4.9, or [21], § 3.2.6) and Theorem 4 the Fourier transform establishes a bijection between $W_{1,r}^{+}(\mathbb{R}^n)$ and the set of functions $\varphi\in\Phi(\mathbb{R}^n)\cap C(\mathbb{R}^n)$ such that $\operatorname{supp}\varphi\subset\overline{\mathbb{B}_{r}}$. Thus,
$$
\begin{equation}
f(t)=\int_{\mathbb{R}^n}\varphi(x)e^{i(t,x)}\,dx, \qquad t\in\mathbb{R}^n,
\end{equation}
\tag{4.2}
$$
where $\varphi\in\Phi(\mathbb{R}^n)\cap C(\mathbb{R}^n)$ and $ \operatorname{supp}\varphi\subset\overline{\mathbb{B}_{r}}$. By the formula for the inverse Fourier transform
$$
\begin{equation*}
\varphi(x)=\frac1{(2\pi)^n}\int_{\mathbb{R}^n}f(t)e^{-i(t,x)}\,dt, \qquad x\in\mathbb{R}^n,
\end{equation*}
\notag
$$
so that $\varphi(0)=1$. Thus, $\varphi\in\mathfrak{F}_r(\mathbb{R}^n)$. Applying the differential operator $L$ to both sides of (4.2) we obtain
$$
\begin{equation*}
Lf(t)=\int_{\mathbb{R}^n} \varphi(x)\rho(x)e^{i(t,x)}\,dx, \qquad t\in\mathbb{R}^n.
\end{equation*}
\notag
$$
It follows from (4.1) that Theorem 2 is proved. § 5. Some examples5.1. Example 1Let $n\neq2$, $r>0$ and $\rho(x)\equiv1$. Then the operator $A_\rho$ is finite-dimensional and has the form
$$
\begin{equation*}
(A_\rho u)(t)=\int_{\mathbb{B}_{r/2}}u(x)\,dx, \qquad t\in\mathbb{B}_{r/2}.
\end{equation*}
\notag
$$
We find all $\lambda\neq0$ for which the equation $A_\rho u=\lambda u$ has nontrivial solutions. Let ${u(t)=a}$ for $t\in\mathbb{B}_{r/2}$, where $a\in\mathbb{C}\setminus\{0\}$. Then $a\cdot\operatorname{vol}(\mathbb{B}_{r/2})=\lambda a$, and therefore $\lambda=\operatorname{vol}(\mathbb{B}_{r/2})$. Thus,
$$
\begin{equation*}
M(n,\rho,r)=\operatorname{vol}(\mathbb{B}_{r/2})=\frac{\pi^{n/2}r^n}{2^n\Gamma(n/2+1)}.
\end{equation*}
\notag
$$
5.2. Example 2Let $n\neq2$, $r>0$ and $\rho(x)=|x|^2$, $x\in\mathbb{R}^n$. It is easy to verify that in this case
$$
\begin{equation}
M(n,\rho,r_1)=\biggl(\frac{r_1}{r_2}\biggr)^{n+2}M(n,\rho,r_2), \qquad r_1,r_2>0.
\end{equation}
\tag{5.1}
$$
Let $r=2$. Then $A_\rho$ is a finite-dimensional operator of the following form:
$$
\begin{equation*}
\begin{aligned} \, &(A_\rho u)(t)=\int_{\mathbb{B}_1}|t-x|^2u(x)\,dx \\ &\ =\biggl(\sum_{k=1}^n t_k^2\biggr)\int_{\mathbb{B}_1}u(x)\,dx -2\sum_{k=1}^n t_k\int_{\mathbb{B}_1}x_k u(x)\,dx +\sum_{k=1}^n \int_{\mathbb{B}_1}x_k^2 u(x)\,dx, \qquad t\in\mathbb{B}_1. \end{aligned}
\end{equation*}
\notag
$$
We find all $\lambda\neq0$ for which the equation
$$
\begin{equation}
(A_\rho u)(t)=\lambda u(t), \qquad t\in\mathbb{B}_1,
\end{equation}
\tag{5.2}
$$
has nontrivial solutions. Let $u(x)=a|x|^2+(b,x)+c$ for $x\in\mathbb{B}_1$, where $a,c\in\mathbb{C}$ and $b=(b_1,\dots,b_n)\in\mathbb{C}^n$. Here $(\,\cdot\,{,}\,\cdot\,)$ is the scalar product in $\mathbb{C}^n$. Then we have the equalities
$$
\begin{equation*}
\int_{\mathbb{B}_1}u(x)\,dx=a\int_{\mathbb{B}_1}|x|^2\,dx+c\int_{\mathbb{B}_1}dx, \qquad \int_{\mathbb{B}_1}x_k u(x)\,dx=b_k\int_{\mathbb{B}_1}x_1^2\,dx
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sum_{k=1}^n\int_{\mathbb{B}_1}x_k^2 u(x)\,dx=a\int_{\mathbb{B}_1}|x|^4\,dx+c\int_{\mathbb{B}_1}|x|^2\,dx.
\end{equation*}
\notag
$$
Now for convenience we introduce the following notation:
$$
\begin{equation*}
\begin{gathered} \, \xi:=\int_{\mathbb{B}_1}|x|^2\,dx=\frac{2\pi^{n/2}}{(n+2)\Gamma(n/2)}, \qquad \eta:=\int_{\mathbb{B}_1}dx=\frac{2\pi^{n/2}}{n\Gamma(n/2)}, \\ \tau:=\int_{\mathbb{B}_1}|x|^4\,dx=\frac{2\pi^{n/2}}{(n+4)\Gamma(n/2)}, \qquad \theta:=-2\int_{\mathbb{B}_1}x_1^2\,dx=\frac{-4\pi^{n/2}}{n(n+2)\Gamma(n/2)}. \end{gathered}
\end{equation*}
\notag
$$
Equating the coefficients in (5.2) we obtain
$$
\begin{equation*}
\begin{pmatrix} \xi & 0 & \dots & 0 & \eta \\ 0 & \theta & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & \theta & 0 \\ \tau & 0 & \dots & 0 & \xi \end{pmatrix} \begin{pmatrix} a \\ b_1 \\ \vdots \\ b_n \\ c \end{pmatrix} =\lambda \begin{pmatrix} a \\ b_1 \\ \vdots \\ b_n \\ c \end{pmatrix}.
\end{equation*}
\notag
$$
Solving the equation
$$
\begin{equation*}
\begin{vmatrix} \xi-\lambda & 0 & \dots & 0 & \eta \\ 0 & \theta-\lambda & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & \theta-\lambda & 0 \\ \tau & 0 & \dots & 0 & \xi-\lambda \end{vmatrix} =(\theta-\lambda)^n(\lambda^2-2\xi\lambda+\xi^2-\tau\eta)=0
\end{equation*}
\notag
$$
we obtain
$$
\begin{equation*}
\begin{gathered} \, \lambda_1=\theta=\frac{-4\pi^{n/2}}{n(n+2)\Gamma(n/2)}, \\ \lambda_2=\xi-\sqrt{\tau\eta}=\frac{2\pi^{n/2}}{\Gamma(n/2)} \biggl(\frac1{n+2}-\frac{\sqrt{n(n+4)}}{n(n+4)}\biggr) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\lambda_3=\xi+\sqrt{\tau\eta}=\frac{2\pi^{n/2}}{\Gamma(n/2)} \biggl(\frac1{n+2}+\frac{\sqrt{n(n+4)}}{n(n+4)}\biggr).
\end{equation*}
\notag
$$
Clearly, $|\lambda_3|>|\lambda_2|$. For $n=1$ we have $|\lambda_3|-|\lambda_1|=(-10+6\sqrt{5})/15>0$. If $n\geqslant2$, then
$$
\begin{equation*}
|\lambda_3|-|\lambda_1|=\frac{2\pi^{n/2}}{\Gamma(n/2)}\biggl(\frac{n-2}{n(n+2)}+\frac{\sqrt{n(n+4)}}{n(n+4)}\biggr)>0.
\end{equation*}
\notag
$$
Thus, $M(n,\rho,2)=\lambda_3$ for $n\neq2$. One eigenfunction corresponding to $\lambda_3$ is, for example,
$$
\begin{equation}
u(t)=|t|^2+\sqrt{\frac{\tau}{\eta}}=|t|^2+\sqrt{\frac{n}{n+4}}, \qquad t\in\mathbb{B}^n.
\end{equation}
\tag{5.3}
$$
In the general case it follows from (5.1) that
$$
\begin{equation}
M(n,\rho,r)=\frac{r^{n+2}\pi^{n/2}}{2^{n+1}\Gamma(n/2)} \biggl(\frac1{n+2}+\frac{\sqrt{n(n+4)}}{n(n+4)}\biggr), \qquad n\neq2, \quad r>0.
\end{equation}
\tag{5.4}
$$
5.3. Example 3Let $n=1$, $r>0$ and $\rho(x)=x^{2m}$, where $m\in\mathbb{N}$. It is easy to verify that then
$$
\begin{equation}
M(n,\rho,r_1)=\biggl(\frac{r_1}{r_2}\biggr)^{2m+1}M(n,\rho,r_2), \qquad r_1,r_2>0.
\end{equation}
\tag{5.5}
$$
Let $r=2$. Then $A_\rho$ is a finite-dimensional operator of the following form:
$$
\begin{equation*}
(A_\rho u)(t)=\int_{-1}^1(t-x)^{2m}u(x)\,dx =\sum_{i=1}^{2m+1}(-1)^{i+1}C_{2m}^{i-1}t^{2m+1-i}s_{i-1},
\end{equation*}
\notag
$$
where We find all $\lambda\neq0$ for which the equation
$$
\begin{equation}
(A_\rho u)(t)=\lambda u(t), \qquad -1\leqslant t\leqslant1,
\end{equation}
\tag{5.6}
$$
has nontrivial solutions. We seek $u(x)$ in the following form:
$$
\begin{equation*}
u(x)=\sum_{p=0}^{2m} a_p x^p, \qquad a_p\in\mathbb{C}.
\end{equation*}
\notag
$$
In this case
$$
\begin{equation*}
s_k=\sum_{j=0}^{2m}\frac{1+(-1)^{k-j}}{2m+k-j+1}a_{2m-j}, \qquad k=0,\dots,2m.
\end{equation*}
\notag
$$
Equating the coefficients in (5.6) we obtain the system $A_m\cdot a=\lambda a$, where $a:=(a_0,\dots,a_{2m})$ and
$$
\begin{equation}
A_m:=\biggl(\frac{(-1)^{i+1}C_{2m}^{i-1}(1+(-1)^{i-j})}{2m+i-j+1}\biggr)_{i,j=1}^{2m+1}.
\end{equation}
\tag{5.7}
$$
Thus,
$$
\begin{equation}
M(1,\rho,r)=\biggl(\frac{r}{2}\biggr)^{2m+1}|\lambda_{\max}|,
\end{equation}
\tag{5.8}
$$
where $\lambda_{\mathrm{max}}$ is the eigenvalue of the matrix (5.7) with the greatest modulus. In particular, for $m=1$ the matrix $A_m$ has the following form:
$$
\begin{equation*}
A_1= \begin{pmatrix} \dfrac23 & 0 & 2 \\ 0 & -\dfrac43 & 0 \\ \dfrac25 & 0 & \dfrac23 \end{pmatrix}.
\end{equation*}
\notag
$$
The eigenvalues of $A_1$ are
$$
\begin{equation*}
\lambda_1=-\frac43, \quad \lambda_2=\frac{10-6\sqrt5}{15}\quad\text{and} \quad \lambda_3=\frac{10+6\sqrt5}{15},
\end{equation*}
\notag
$$
which corresponds to Example 2. 5.4. Remarks Remark 5. It follows from Theorem 2 and Example 2 that for $n\neq2$, for each function $f\in W_{1,r}^{+}(\mathbb{R}^n)$ we have the sharp inequality Remark 6. It follows from Theorem 2 and Example 3 that for $m\in\mathbb{N}$ and each $f\in W_{1,r}^{+}(\mathbb{R})$ we have the sharp inequality Remark 7. This author proved in [12] that the following sharp inequality holds for ${f\in W_{1,r}^{+}(\mathbb{R})}$: Remark 8. Ibragimov proved in 1959 (see [22], Corollary 2) that for (not necessarily nonnegative) functions $f\in W_{1,r}(\mathbb{R})$ we have the inequalities Inequalities of type (5.9), (5.10) and (5.12) belong to the class of Bernstein–Nikol’skii type inequalities. For more information about inequalities of this type, see [24] by Gorbachev. 
Citation in format AMSBIB
\Bibitem{Man24} |
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