Open Access

Sobolev inequality of free boundary value problem for \((-1)^{M}(d/dx)^{2M}\)

  • Kazuo Takemura1Email author,
  • Atsushi Nagai2,
  • Yoshinori Kametaka3,
  • Kohtaro Watanabe4 and
  • Hiroyuki Yamagishi5
Journal of Inequalities and Applications20152015:54

https://doi.org/10.1186/s13660-015-0568-9

Received: 10 September 2014

Accepted: 19 January 2015

Published: 12 February 2015

Abstract

The Green function of the free boundary value problem for \((-1)^{M}(d/dx)^{2M}\) is found by using Whipple’s formula. The Green function is constructed through so-called symmetric orthogonalization method under a suitable solvability condition. Its Green function is a reproducing kernel for a suitable set of Hilbert space and an inner product. By using the fact, we compute the best constant (\(M=1,2,3,4,5\)) and a family of the best functions for a Sobolev inequality. It is possible for us to expect the best constant of the Sobolev inequality, but the proof has not been completed for \(M\geq6\) in the present paper.

Keywords

Green function Sobolev inequality best constant

MSC

34B27 46E35 41A44

1 Introduction

For \(M=1,2,3,\ldots \) , we introduce the Sobolev space
$$ H=H(M)= \biggl\{ u(x) \Bigm| u(x), u^{(M)}(x)\in L^{2}(-1,1), \int _{-1}^{1}u(x)x^{i} \,dx=0\ (0\le i\le M-1) \biggr\} $$
with the Sobolev inner product
$$ (u,v)_{M} =\int_{-1}^{1} u^{(M)}(x) \overline{v}^{(M)}(x) \,dx. $$
\((\cdot,\cdot)_{M}\) is proven to be an inner product of H in Section 4. H is the Hilbert space with the inner product \((\cdot,\cdot)_{M}\). The purpose of the present paper is to find a supremum of the Sobolev functional given by
$$ S(u)=S(M;u)= \Bigl(\sup_{|y|\le1} \bigl\vert u(y)\bigr\vert \Bigr)^{2}\big/\| u\|_{M}^{2},\quad \|u \|_{M}^{2}=\int_{-1}^{1}\bigl\vert u^{(M)}(x)\bigr\vert ^{2} \,dx. $$
The conclusion of the present paper is as follows.

Theorem 1.1

\(G(x,y)=G(M;x,y)\) is the Green function which is defined later in Theorem  3.1. \(\sup_{u \in H, u\not\equiv0}S(u)=C_{0}\) is given by
$$ C_{0}=C(M)=\max_{|y|\le1}G(y,y)=G(y_{0},y_{0}) \quad (M\geq1), $$
(1.1)
where \(y_{0}\) attains the maximum value of the diagonal of the Green function. The supremum \(C_{0}\) is attained by setting \(u(x)=G(x,y_{0})\). In particular, if \(M\leq5\), we have \(y_{0}=\pm1\), in other words,
$$ \max_{|y|\le1}G(y,y)=G(1,1)=G(-1,-1)\quad (M=1,2,3,4,5). $$
(1.2)
Explicit forms of \(G(1,1)=G(-1,-1)\) are given as
$$ G(1,1)=G(-1,-1)=\frac{2^{2M-1}\Gamma(2M-1)\Gamma(2M+1)}{ \Gamma(M)^{2}\Gamma(4M)}\quad (M\geq1). $$
(1.3)
Note that the equality
$$ \max_{|y|\le1}G(y,y)=G(\pm1,\pm1) $$
in (1.1) is proven only for \(M\leq5\) and is still open for \(M\geq6\). Here, we list explicit forms of \(C(M)\) (\(M=1,2,3,4,5\)),
$$\begin{aligned}& C(1)=\frac{2}{3},\qquad C(2)=\frac{8}{105},\qquad C(3)= \frac{4}{1\text{,}155}, \\& C(4)=\frac {32}{405\text{,}405},\qquad C(5)=\frac{4}{3\text{,}741\text{,}309}. \end{aligned}$$

Theorem 1.1 is equivalently rewritten as follows.

Theorem 1.2

For any \(u(x) \in H\), there exists a positive constant C such that the Sobolev inequality holds:
$$ \Bigl(\sup_{|y| \le1}\bigl\vert u(y)\bigr\vert \Bigr)^{2} \le C \int_{-1}^{1} \bigl\vert u^{(M)}(x)\bigr\vert ^{2} \,dx. $$
Among such C, the best constant \(C_{0}\) is the same as (1.1). If we replace C by \(C_{0}\), the equality holds for
$$ u(x)=c G(x,y_{0}),\quad c \in\mathbb{C}\backslash\{0\}\ (-1< x<1). $$
In particular, if \(M\leq5\), we have \(y_{0}=\pm1\).

Concerning the infimum of \(S(u)\), we have the following theorem.

Theorem 1.3

The infimum of the Sobolev functional is equal to zero,
$$ \inf_{u \in H, u\not\equiv0}S(u)=0. $$
(1.4)

Based on the previous research of Bliss [1], the best constant and a family of the best functions of the Sobolev inequality are first obtained independently by Aubin [2] and Talenti [3]. They mainly used the functional analysis technique to compute the best constant of the Sobolev inequality. On the other hand, we have computed the best constant of the Sobolev inequality by using the Green function corresponding to a boundary value problem. This is a new method quite different from that of the functional analysis. The Green function is the reproducing kernel for suitable set of Hilbert space and inner product. As an application, the best constant of the corresponding Sobolev inequality is expressed as the maximum value of the diagonal of the Green function. References [46] are related to the early studies based on these facts. The engineering meaning of the Sobolev inequality becomes that the square of maximum bending of a string (\(M=1\)) or a beam (\(M=2\)) is estimated from above by the constant multiple of the potential energy [7, 8]. We have already obtained the best constant of each Sobolev inequality which corresponds to clamped-free, Dirichlet and periodic boundary value problems for \((-1)^{M}(d/dx)^{2M}\) [911]. Further, in [12], we consider a time-periodic boundary value problem of nth order ordinary differential operator which appears typically in Heaviside cable and Thomson cable theory. In this problem, the physical meaning of a Sobolev type inequality becomes that the square of maximum of the absolute value of AC output voltage is estimated above by the constant multiple of the power of input voltage. The purpose of the present paper is to derive a Sobolev inequality which corresponds to free boundary value problem for \((-1)^{M}(d/dx)^{2M}\) and obtain the best constant by using some properties as the reproducing kernel of the Green function.

This paper is organized as follows. In Section 2, we construct a proto Green function that becomes the origin of the Green function from the corresponding eigenvalue problem. In Section 3, we construct the Green function from the proto Green function as to satisfy the properties of reproducing kernel. We call the technique symmetric orthogonalization method [13]. In the method, Whipple’s theorem concerning the hypergeometric series has an important role. In Section 4, we show that the Green function is the reproducing kernel for H and \((\cdot,\cdot)_{M}\). After deriving the Sobolev inequality, we give the proof of (1.1). Sections 5 and 6 are devoted to the proof of (1.2) (\(M=5\)) and (1.3) in the main Theorem 1.1. In Section 7, we prove Theorem 1.3.

2 Free boundary value problem and the proto Green function

We start with the following lemma concerning the eigenvalue problem.

Lemma 2.1

The eigenvalue problem
$$ \begin{cases} (-1)^{M} u^{(2M)}=\lambda u(x) & (-1< x<1), \\ u^{(i)}(\pm1)=0 &(M\le i\le2M-1) \end{cases} $$
has an eigenvalue \(\lambda=0\), and the corresponding eigenspace is M-dimensional. Its orthonormal base is
$$ \biggl\{ \varphi_{i}(x)=\sqrt{i+\frac{1}{2}}P_{i}(x) \Bigm| 0\le i\le M-1 \biggr\} , $$
where \(P_{i}(x)\) are Legendre polynomials.

We omit the proof of the lemma.

For convenience, we introduce the following monomials \(\{K_{j}(x)\}\):
$$ K_{j}(x)=K_{j}(M;x)= \begin{cases} \frac{x^{2M-1-j}}{\Gamma(2M-j)}& (0\le j\le2M-1), \\ 0 & (2M\le j). \end{cases} $$
For any bounded continuous function \(f(x)\) on \(-1< x<1\) satisfying the solvability condition
$$ \int_{-1}^{1}f(y) \varphi_{i}(y)\, dy=0\quad (0\le i\le M-1), $$
(2.1)
we consider the free boundary value problem
We first introduce the proto Green function \(G_{0}(x,y)\).

Lemma 2.2

Suppose that the boundary value problem (2.2) and (2.3) has a classical solution \(u(x)\). Then \(f(x)\) satisfies the solvability condition (2.1), and \(u(x)\) is expressed as
$$ u(x)=\sum_{i=0}^{M-1} \alpha_{i}\varphi_{i}(x)+\int_{-1}^{1}G_{0}(x,y)f(y) \, dy\quad (-1< x<1), $$
(2.5)
where \(\alpha_{0}, \alpha_{1}, \ldots,\alpha_{M-1}\) are appropriate constants. \(G_{0}(x,y)\) is a proto Green function or an equivalently fundamental solution for the differential operator \((-1)^{M}(d/dx)^{2M}\), which is defined by
$$ G_{0}(x,y)=\frac{(-1)^{M}}{2}K_{0}\bigl(\vert x-y\vert \bigr)\quad (-1< x,y<1). $$

Proof

We assume that (2.2) and (2.3) have a classical solution \(u(x)\). Introducing new functions
$$ \mathbf{u}={}^{t}( \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} u_{0}& u_{1}& \cdots& u_{2M-1} \end{array}),\quad u_{i}= u^{(i)}\ (0\le i\le2M-1) $$
and the \(2M\times2M\) nilpotent matrix
$$ \mathbf{N}= \begin{pmatrix} 0&1& & \\ &0&\ddots& \\ & &\ddots& 1 \\ & & & 0 \end{pmatrix}, $$
we can rewrite (2.2) and (2.3) as
$$\begin{aligned}& \mathbf{u}'=\mathbf{N}\mathbf{u}+{}^{t}( \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 &\cdots&0 &1 \end{array}) (-1)^{M}f(x)\quad (-1< x<1), \end{aligned}$$
(2.6)
$$\begin{aligned}& u_{i}(\pm1)=0\quad (M\le i\le2M-1). \end{aligned}$$
(2.7)
The fundamental solution \(\mathbf{E}(x)\) to the above initial value problem is expressed as
$$ \mathbf{E}(x)=\mathbf{K}(x)\mathbf{K}(0)^{-1}, $$
where
K ( x ) = ( K i + j ) ( x ) ( 0 i , j 2 M 1 ) , K ( 0 ) = ( 1 1 ) = K ( 0 ) 1 .
Solving (2.6), we have
$$\begin{aligned}& \mathbf{u}(x)=\mathbf{E}(x+1)\mathbf{u}(-1)+\int_{-1}^{x} \mathbf {E}(x-y){}^{t}( \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 &\cdots&0 &1 \end{array}) (-1)^{M}f(y)\, dy, \\& \mathbf{u}(x)=\mathbf{E}(x-1)\mathbf{u}(1)-\int_{x}^{1} \mathbf {E}(x-y){}^{t}( \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 &\cdots&0 &1 \end{array}) (-1)^{M}f(y)\, dy, \end{aligned}$$
or equivalently, for \(0\le i\le2M-1\),
$$\begin{aligned}& u_{i}(x)=\sum_{j=0}^{2M-1}K_{i+j}(x+1)u_{2M-1-j}(-1) +\int_{-1}^{x}(-1)^{M}K_{i}(x-y)f(y) \, dy, \\& u_{i}(x)=\sum_{j=0}^{2M-1}K_{i+j}(x-1)u_{2M-1-j}(1)- \int_{x}^{1}(-1)^{M}K_{i}(x-y)f(y) \, dy. \end{aligned}$$
Employing the boundary conditions (2.7), we have
$$\begin{aligned}& u_{i}(x)=\sum_{j=M}^{2M-1}K_{i+j}(x+1)u_{2M-1-j}(-1) +\int_{-1}^{x}(-1)^{M}K_{i}(x-y)f(y) \, dy, \end{aligned}$$
(2.8)
$$\begin{aligned}& u_{i}(x)=\sum_{j=M}^{2M-1}K_{i+j}(x-1)u_{2M-1-j}(1) -\int_{x}^{1}(-1)^{M}K_{i}(x-y)f(y) \, dy \end{aligned}$$
(2.9)
for \(0\le i\le2M-1\). If \(M\leq i\leq2M-1\), we have
$$\begin{aligned}& u_{i}(x)=\int_{-1}^{x}(-1)^{M}K_{i}(x-y)f(y) \, dy\quad (M\leq i\leq2M-1), \\& u_{i}(x)=-\int_{x}^{1}(-1)^{M}K_{i}(x-y)f(y) \, dy\quad (M\leq i \leq2M-1). \end{aligned}$$
Setting \(x=\pm1\) and employing the boundary conditions (2.7), we have
$$\begin{aligned}& 0=u_{i}(1)=\int_{-1}^{1}(-1)^{M}K_{i}(1-y)f(y) \, dy\quad (M\leq i\leq 2M-1), \end{aligned}$$
(2.10)
$$\begin{aligned}& 0=u_{i}(-1)=-\int_{-1}^{1}(-1)^{M}K_{i}(-1-y)f(y) \, dy \quad (M\leq i\leq 2M-1). \end{aligned}$$
(2.11)
From (2.10) and (2.11), it is shown that \(f(x)\) satisfies the condition
$$ \int_{-1}^{1}y^{i}f(y)\, dy=0 \quad (0 \le i\le M-1), $$
or equivalently
$$ \int_{-1}^{1}f(y)\varphi_{i}(y)\, dy=0 \quad (0\le i\le M-1), $$
which are necessary conditions for the existence of a classical solution to (2.2) and (2.3). Setting \(i=0\) in (2.8) and (2.9), we have
$$\begin{aligned}& u_{0}(x)=\sum_{j=M}^{2M-1}K_{j}(x+1)u_{2M-1-j}(-1) +\int_{-1}^{x}(-1)^{M}K_{0}(x-y)f(y) \, dy, \\& u_{0}(x)=\sum_{j=M}^{2M-1}K_{j}(x-1)u_{2M-1-j}(1) -\int_{x}^{1}(-1)^{M}K_{0}(x-y)f(y) \, dy. \end{aligned}$$
Taking the average of the above two equalities, we obtain an expression for the solution \(u(x)=u_{0}(x)\),
$$ u(x)=\sum_{j=0}^{M-1}\alpha_{j} \varphi_{j}(x) +\int_{-1}^{1}G_{0}(x,y)f(y) \, dy\quad (-1< x<1), $$
where \(\alpha_{j}\) (\(0\le j\le M-1\)) are suitable constants and
$$ G_{0}(x,y)=\frac{(-1)^{M}}{2}K_{0}\bigl(\vert x-y\vert \bigr)\quad (-1< x,y<1). $$
This shows (2.5) in Lemma 2.2. □

It is easy to prove the following lemma concerning the properties of the proto Green function \(G_{0}(x,y)\). We omit the proof of the lemma.

Lemma 2.3

The proto Green function \(G_{0}(x,y)\) satisfies the following properties.
$$\begin{aligned} (1)&\quad \partial_{x}^{2M}G_{0}(x,y)=0 \quad (-1< x,y<1, x\ne y), \\ (2)&\quad \partial_{x}^{i} G_{0}(x,y) |_{x=\pm1}= (\pm1)^{i}\frac{(-1)^{M}}{2}K_{i}(1\mp y) \\ &\qquad (0\le i\le2M-1, -1<y<1), \\ (3)&\quad \partial_{x}^{i} G_{0}(x,y) |_{y=x-0} - \partial_{x}^{i} G_{0}(x,y) |_{y=x+0} = \begin{cases} 0& (0 \le i \le2M-2), \\ (-1)^{M}& (i=2M-1) \end{cases} \quad (-1<x<1), \\ (4)&\quad \partial_{x}^{i} G_{0}(x,y) |_{x=y+0} - \partial_{x}^{i} G_{0}(x,y) |_{x=y-0} = \begin{cases} 0& (0 \le i \le2M-2), \\ (-1)^{M} & (i=2M-1) \end{cases} \quad (-1<y<1). \end{aligned}$$

3 Method of symmetric orthogonalization

Concerning the uniqueness and existence of the solution to \(\operatorname{BVP}(M)\), we have the following theorem.

Theorem 3.1

For any bounded continuous function \(f(x)\) on an interval \(-1< x<1\) satisfying (2.1), \(\operatorname{BVP}(M)\) has a unique classical solution \(u(x)\) that is expressed as
$$ u(x)=\int_{-1}^{1}G(x,y)f(y)\, dy\quad (-1< x<1). $$
The integral kernel \(G(x,y)=G(M;x,y)\) is given by
$$\begin{aligned}& G(x,y)=G_{0}(x,y)-\sum_{i=0}^{M-1} \bigl\{ \psi_{i}(x)\varphi_{i}(y)+\psi _{i}(y) \varphi_{i}(x) \bigr\} +\sum_{i,j=0}^{M-1} \gamma_{ij}\varphi_{i}(x)\varphi_{j}(y) \\& \quad (-1< x,y<1), \end{aligned}$$
(3.1)
where
$$\begin{aligned}& \psi_{i}(x)=\int_{-1}^{1}G_{0}(x,y) \varphi_{i}(y)\, dy\quad (0\le i\le M-1, -1< x<1), \end{aligned}$$
(3.2)
$$\begin{aligned}& \gamma_{ij}=\int_{-1}^{1} \psi_{i}(x)\varphi_{j}(x)\,dx=\int_{-1}^{1} \varphi _{i}(x)\psi_{j}(x)\,dx=\gamma_{ji}\quad (0 \le i,j\le M-1). \end{aligned}$$
(3.3)

The above procedure, in which the Green function \(G(x,y)\) is constructed from the proto Green function \(G_{0}(x,y)\), is called the symmetric orthogonalization method [5, 6, 13]. The proof of Theorem 3.1 is given in [13]. In the present paper, we give a more closed expression of \(G(x,y)\).

We have the following theorem.

Theorem 3.2

For \(j=0,1,\ldots, M-1\), the following equalities hold:
$$ \begin{aligned} &\psi_{j}(x)=\frac{(-1)^{M}}{2}\sqrt{j+ \frac{1}{2}} \bigl\{ Q_{j}(x)+(-1)^{j}Q_{j}(-x) \bigr\} , \\ &Q_{j}(x)=\sum^{j}_{k=0} \frac{2^{2M}(-1)^{j+k}\Gamma(j+k+1)}{\Gamma(2M+k+1)\Gamma (k+1)\Gamma(j-k+1)} \biggl(\frac{1+x}{2} \biggr)^{2M+k} \quad (-1< x<1). \end{aligned} $$
(3.4)

Proof

From (3.2) in Theorem 3.1, the functions \(\psi_{j}(x)\) are calculated as
$$\begin{aligned} \psi_{j}(x) =&\int^{1}_{-1}G_{0}(x,y) \varphi_{j}(y)\,dy= \frac{(-1)^{M}}{2}\sqrt{j+\frac{1}{2}}\int ^{1}_{-1}K_{0}\bigl(\vert x-y\vert \bigr)P_{j}(y)\,dy \\ =& \frac{(-1)^{M}}{2}\sqrt{j+\frac{1}{2}} \biggl\{ \int ^{x}_{-1}K_{0}(x-y)P_{j}(y) \,dy+\int^{1}_{x}K_{0}(y-x)P_{j}(y) \,dy \biggr\} \\ =& \frac{(-1)^{M}}{2}\sqrt{j+\frac{1}{2}} \biggl\{ \int ^{x}_{-1}K_{0}(x-y)P_{j}(y)\,dy +\int^{-x}_{-1}K_{0}(-y-x)P_{j}(-y) \,dy \biggr\} \\ =& \frac{(-1)^{M}}{2}\sqrt{j+\frac{1}{2}} \bigl\{ Q_{j}(x)+(-1)^{j}Q_{j}(-x) \bigr\} , \end{aligned}$$
(3.5)
where
$$ Q_{j}(x)=\int^{x}_{-1}K_{0}(x-y)P_{j}(y) \,dy\quad (-1< x<1). $$
It is shown through direct calculations that \(u=Q_{j}(x)\) satisfies the initial value problem:
The Legendre polynomial \(P_{j}(x)\) is expressed as
$$ P_{j}(x)=\sum^{j}_{k=0} \frac{(-1)^{j+k}\Gamma(j+k+1)}{\Gamma (k+1)^{2}\Gamma(j-k+1)} \biggl(\frac{1+x}{2} \biggr)^{k} \quad (-1< x<1), $$
and integrating (3.6) 2M times under the initial conditions (3.7), we obtain (3.4). This proves Theorem 3.2. □

Theorem 3.3

For \(M=1,2,3,\ldots \) , the coefficients \(\gamma_{ij}\) (\(0 \leq i,j \leq M-1\)) are expressed as
$$ \gamma_{ij}= \begin{cases} \frac{(-1)^{j+M}\pi\Gamma(2M+1)\sqrt{i+\frac{1}{2}}\sqrt{j+\frac{1}{2}}}{2^{2M}\Gamma(\frac{2M-i-j+1}{2})\Gamma(\frac{2M+i-j+2}{2}) \Gamma(\frac{2M-i+j+2}{2})\Gamma(\frac{2M+i+j+3}{2})}& (i-j\mbox{: } \textit{even}), \\ 0 & (i-j\mbox{: } \textit{odd}). \end{cases} $$

Proof

From (3.3) and (3.5), we have
$$\begin{aligned} \gamma_{ij} =&\int^{1}_{-1} \varphi_{i}(x)\psi_{j}(x)\,dx \\ =& \frac{(-1)^{M}}{2}\sqrt{ \biggl(i+\frac{1}{2} \biggr) \biggl(j+ \frac{1}{2} \biggr)} \biggl\{ \int^{1}_{-1}P_{i}(x)Q_{j}(x) \,dx +(-1)^{j}\int^{1}_{-1}P_{i}(x)Q_{j}(-x) \,dx \biggr\} \\ =& \frac{(-1)^{M}}{2}\sqrt{ \biggl(i+\frac{1}{2} \biggr) \biggl(j+ \frac{1}{2} \biggr)} \bigl\{ 1+(-1)^{i-j}\bigr\} \int ^{1}_{-1}P_{i}(x)Q_{j}(x)\,dx \\ =& \begin{cases} (-1)^{M} \sqrt{ (i+\frac{1}{2} ) (j+\frac{1}{2} )}\int^{1}_{-1}P_{i}(x)Q_{j}(x)\,dx& (i-j\mbox{: even}), \\ 0& (i-j\mbox{: odd}). \end{cases} \end{aligned}$$
Hereinafter, we investigate the case in which \(i-j\) is even. From the Rodrigues formula and (3.4), we have
1 1 P i ( x ) Q j ( x ) d x = 1 Γ ( i + 1 ) 2 i k = 0 j ( 1 ) j + k Γ ( j + k + 1 ) Γ ( 2 M + k + 1 ) Γ ( k + 1 ) Γ ( j k + 1 ) 2 k × 1 1 ( ( 1 ) i ( d d x ) i ( 1 x 2 ) i ) ( 1 + x ) 2 M + k d x = 1 Γ ( i + 1 ) 2 i k = 0 j ( 1 ) j + k Γ ( j + k + 1 ) Γ ( 2 M + k + 1 ) Γ ( k + 1 ) Γ ( j k + 1 ) 2 k × ( 2 M + k ) ( 2 M + k 1 ) ( 2 M + k i + 1 ) 1 1 ( 1 x 2 ) i ( 1 + x ) 2 M + k i d x = 1 Γ ( i + 1 ) 2 i k = 0 j ( 1 ) j + k Γ ( j + k + 1 ) Γ ( 2 M + k i + 1 ) Γ ( k + 1 ) Γ ( j k + 1 ) 2 k × 2 2 M + k + i + 1 B ( i + 1 , 2 M + k + 1 ) = ( 1 ) j 2 2 M + 1 k = 0 j ( 1 ) k Γ ( 2 M + k + 1 ) Γ ( j + k + 1 ) Γ ( 2 M + k i + 1 ) Γ ( 2 M + k + i + 2 ) Γ ( k + 1 ) Γ ( j k + 1 ) = ( 1 ) j 2 2 M + 1 Γ ( 2 M + 1 ) Γ ( 2 M i + 1 ) Γ ( 2 M + i + 2 ) k = 0 j ( j ) k ( j + 1 ) k ( 2 M + 1 ) k ( 2 M i + 1 ) k ( 2 M + i + 2 ) k 1 k ! = ( 1 ) j 2 2 M + 1 Γ ( 2 M + 1 ) Γ ( 2 M i + 1 ) Γ ( 2 M + i + 2 ) 3 F 2 ( j , j + 1 , 2 M + 1 2 M i + 1 , 2 M + i + 2 | 1 ) ,
(3.8)
where \((a)_{k}\) is Pochhammer’s symbol, defined by
$$ (a)_{k}=\frac{\Gamma(a+k)}{\Gamma(a)}\quad (a\ne0,-1,-2,\ldots). $$
Here, we present Whipple’s theorem concerning the hypergeometric series \({}_{3}F_{2}\).

Theorem 3.4

(Whipple’s theorem [14])

F 2 3 ( a , 1 a , b 1 + 2 b c , c | 1 ) = π 2 1 2 b Γ ( c ) Γ ( 1 + 2 b c ) Γ ( a + c 2 ) Γ ( a + 1 + 2 b c 2 ) Γ ( 1 a + c 2 ) Γ ( 2 + 2 b a c 2 ) .
Setting \(a=-j\), \(b=2M+1\), and \(c=2M-i+1\) in Theorem 3.4, we have
F 2 3 ( j , j + 1 , 2 M + 1 2 M i + 1 , 2 M + i + 2 | 1 ) = π Γ ( 2 M i + 1 ) Γ ( 2 M + i + 2 ) 2 4 M + 1 Γ ( 1 2 i 2 j 2 + M ) Γ ( M + i 2 j 2 + 1 ) Γ ( M i 2 + j 2 + 1 ) Γ ( M + i 2 + j 2 + 3 2 ) .
Substituting the above expression into (3.8), we obtain \(\gamma_{ij}\) in a closed form, as follows:
$$ \gamma_{ij} =\frac{(-1)^{j+M}\pi\Gamma(2M+1)\sqrt{i+\frac{1}{2}}\sqrt{j+\frac{1}{2}}}{ 2^{2M}\Gamma(\frac{2M-i-j+1}{2})\Gamma(\frac{2M+i-j+2}{2}) \Gamma(\frac{2M-i+j+2}{2})\Gamma(\frac{2M+i+j+3}{2})}\quad (i-j \mbox{: even}). $$
This proves Theorem 3.3.  □

The following statement follows from Theorems 3.2 and 3.3.

Corollary 3.1

\(\psi_{j}(x)\) is a polynomial of \((2M+j)\) th degree and satisfies \(\psi_{j}(-x)=(-1)^{j} \psi_{j}(x)\). Moreover, \(\psi_{j}(x)\) is expanded by \(\{\varphi_{i}(x)\}\), as follows:
$$ \psi_{2i}(x)=\sum_{j=0}^{M+i} \gamma_{2i,2j}\varphi_{2j}(x),\qquad \psi_{2i+1}(x)=\sum _{j=0}^{M+i}\gamma_{2i+1,2j+1} \varphi_{2j+1}(x). $$

Remark 3.1

It is often more convenient to express \(G(x,y)\) in the following equivalent form:
$$\begin{aligned} G(x,y) =& G_{0}(x,y)-\sum_{i=0}^{[\frac{M-1}{2}]} \sum_{j=0}^{M+i}\gamma_{2i,2j} \bigl[ \varphi_{2i}(x)\varphi_{2j}(y)+\varphi_{2i}(y) \varphi _{2j}(x) \bigr] \\ &{} -\sum_{i=0}^{[\frac{M-2}{2}]}\sum _{j=0}^{M+i}\gamma_{2i+1,2j+1} \bigl[ \varphi_{2i+1}(x)\varphi_{2j+1}(y)+ \varphi_{2i+1}(y) \varphi_{2j+1}(x) \bigr] \\ &{} +\sum_{i,j=0}^{[\frac{M-1}{2}]}\gamma_{2i,2j} \varphi _{2i}(x)\varphi_{2j}(y)+ \sum _{i,j=0}^{[\frac{M-2}{2}]}\gamma_{2i+1,2j+1}\varphi _{2i+1}(x)\varphi _{2j+1}(y) \\ & (-1< x,y<1), \end{aligned}$$
where
$$\begin{aligned}& \gamma_{2i,2j}= {\frac{(-1)^{M}\pi\Gamma(2M+1)\sqrt{(4i+1)(4j+1)}}{ 2^{2M+1}\Gamma(M+i-j+1)\Gamma(M-i+j+1)\Gamma(M+i+j+\frac{3}{2})\Gamma (M-i-j+\frac{1}{2})}} \\& \quad \bigl(0\leq i,j\leq\bigl[(M-1)/2\bigr]\bigr), \\& \gamma_{2i+1,2j+1}= {\frac{(-1)^{M+1}\pi\Gamma (2M+1)\sqrt{(4i+3)(4j+3)}}{ 2^{2M+1}\Gamma(M+i-j+1)\Gamma(M-i+j+1)\Gamma(M+i+j+\frac{5}{2})\Gamma (M-i-j-\frac{1}{2})}} \\& \quad \bigl(0\leq i,j\leq\bigl[(M-2)/2\bigr]\bigr). \end{aligned}$$

Theorem 3.1 is a direct consequence of the following theorem, which states that \(G(x,y)\) serves as the Green function of \(\operatorname{BVP}(M)\).

Theorem 3.5

The Green function \(G(x,y)\) satisfies the following conditions:
$$\begin{aligned} (1) &\quad (-1)^{M} \partial_{x}^{2M} G(x,y)=-\sum _{i=0}^{M-1}\varphi _{i}(x) \varphi_{i}(y)\quad (-1< x,y<1, x\ne y), \\ (2) &\quad \partial_{x}^{i} G(x,y) |_{x=\pm1}=0\quad (M\le i\le2M-1, -1<y<1), \\ (3) &\quad \partial_{x}^{i} G(x,y) |_{y=x-0} - \partial_{x}^{i} G(x,y) |_{y=x+0}= \begin{cases} 0 &(0 \le i \le2M-2), \\ (-1)^{M} &(i=2M-1) \end{cases} \quad (-1<x<1), \\ (4) &\quad \partial_{x}^{i} G(x,y) |_{x=y+0}- \partial_{x}^{i} G(x,y) |_{x=y-0}= \begin{cases} 0 &(0 \le i \le2M-2), \\ (-1)^{M} &(i=2M-1) \end{cases} \quad (-1<y<1), \\ (5) &\quad \int_{-1}^{1}\varphi_{i}(x)G(x,y) \,dx =0\quad (0\le i\le M-1, -1<y<1). \end{aligned}$$

The proof of the above theorem is given in [13].

4 Sobolev inequality

In this section, it is shown that the Green function \(G(x,y)\) is a reproducing kernel for a set of Hilbert space \(H=H(M)\) and its inner product \((\cdot,\cdot)_{M}\) introduced in Section 1. We also derive the Sobolev inequality from the reproducing relation.

Theorem 4.1

For any \(u(x) \in H\), we have the following reproducing relation:
$$ u(y)=\bigl(u(x),G(x,y)\bigr)_{M}= \int_{-1}^{1}u^{(M)}(x) \partial_{x}^{M} G(x,y)\,dx\quad (-1\le y\le1). $$
(4.1)
In particular, we have
$$ G(y,y)=\int_{-1}^{1}\bigl\vert \partial_{x}^{M}G(x,y) \bigr\vert ^{2}\,dx\quad (-1\leq y\leq1). $$
(4.2)

Proof

For functions \(u=u(x)\) and \(v=v(x)=G(x,y)\) with y arbitrarily fixed in \(-1\le y \le1\), we have
$$ u^{(M)}v^{(M)}- u(-1)^{M}v^{(2M)}= \Biggl( \sum_{j=0}^{M-1}(-1)^{M-1-j}u^{(j)}v^{(2M-1-j)} \Biggr)'. $$
Integrating with respect to x on intervals \(-1< x<y\) and \(y< x<1\) and using the properties of \(G(x,y)\) given in Theorem 3.5, we have
$$\begin{aligned} (u,v)_{M} =&\int_{-1}^{1} u^{(M)}(x)v^{(M)}(x)\,dx \\ =& \int_{-1}^{1} u(x) (-1)^{M}v^{(2M)}(x)\,dx \\ &{}+ \Biggl[\sum_{j=0}^{M-1} (-1)^{M-1-j}u^{(j)}(x)v^{(2M-1-j)}(x) \Biggr] \biggl\{ \biggl.\biggr|_{x=-1}^{x=y-0}+ \biggl.\biggr|_{x=y+0}^{x=1} \biggr\} \\ =&\sum_{j=0}^{M-1}(-1)^{M-1-j} \bigl[u^{(j)}(1)v^{(2M-1-j)}(1)- u^{(j)}(-1)v^{(2M-1-j)}(-1) \bigr] \\ &{}+\sum_{j=0}^{M-1}(-1)^{M-1-j}u^{(j)}(y) \bigl[v^{(2M-1-j)}(y-0)-v^{(2M-1-j)}(y+0) \bigr] \\ =& u(y). \end{aligned}$$
This proves (4.1). Equation (4.2) is shown by setting \(u(x)=G(x,y)\) in (4.1). □
Applying the Schwarz inequality to (4.1) and using (4.2), we have
$$ \bigl\vert u(y) \bigr\vert ^{2} \le\int_{-1}^{1} \bigl\vert \partial_{x}^{M}G(x,y) \bigr\vert ^{2}\,dx \int_{-1}^{1} \bigl\vert u^{(M)}(x) \bigr\vert ^{2}\,dx =G(y,y)\int _{-1}^{1} \bigl\vert u^{(M)}(x) \bigr\vert ^{2}\,dx. $$
Taking the supremum of the above inequality with respect to \(y\in[-1,1]\) and setting
$$ C_{0}= \max_{|y|\le1}G(y,y), $$
we obtain the following Sobolev inequality:
$$ \Bigl(\sup_{|y| \le1}\bigl\vert u(y)\bigr\vert \Bigr)^{2} \le C_{0}\int_{-1}^{1} \bigl\vert u^{(M)}(x)\bigr\vert ^{2} \,dx = C_{0}(u,u)_{M}. $$
(4.3)
This inequality shows that \((\cdot,\cdot)_{M}\) is positive definite. Note that, in order to prove (4.3), the Schwarz inequality is required but not the ‘positive definiteness’ of the inner product.
Let \(y_{0}\) be a value that attains the maximum of \(G(y,y)\). Applying this inequality to \(u(x)=G(x,y_{0})\in H\), we have
$$ \Bigl( \sup_{|y| \le1}\bigl\vert G(y,y_{0})\bigr\vert \Bigr)^{2} \le C_{0} \int_{-1}^{1} \bigl\vert \partial_{x}^{M}G(x,y_{0})\bigr\vert ^{2} \,dx=C_{0}^{2}. $$
Combining this expression and the trivial inequality:
$$ C_{0}^{2}= \bigl\{ G(y_{0},y_{0}) \bigr\} ^{2}\le \Bigl( \sup_{|y| \le1} \bigl\vert G(y,y_{0})\bigr\vert \Bigr)^{2}, $$
we have
$$ C_{0}^{2}\le \Bigl(\sup_{|y| \le1}\bigl\vert G(y,y_{0})\bigr\vert \Bigr)^{2} \le C_{0}\int _{-1}^{1} \bigl\vert \partial_{x}^{M}G(x,y_{0}) \bigr\vert ^{2} \,dx=C_{0}^{2}. $$
Hence, it is shown that the equality in (4.3) holds if \(u(x)=G(x,y_{0})\):
$$ \Bigl(\sup_{|y| \le1}\bigl\vert G(y,y_{0})\bigr\vert \Bigr)^{2} = C_{0}\int_{-1}^{1} \bigl\vert \partial_{x}^{M}G(x,y_{0})\bigr\vert ^{2} \,dx. $$

Remark 4.1

In the next section, it is shown that \(y_{0}=\pm1\) for \(M\leq5\), which is expected to hold also for \(M \geq6\).

5 Supremum of Sobolev functional

In this section, we prove (1.2) in Theorem 1.1 in the case of \(M=5\). The cases of \(M \leq4\) are comparatively simple, and so we omit their proofs. From Theorem 3.1, the diagonal value of the Green function \(G(x,y)=G(5;x,y)\) is given as
$$\begin{aligned} G(5;x,x) =& \frac{1}{429\text{,}083\text{,}246\text{,}592}\bigl( 3\text{,}693 + 1\text{,}750 \text{,}653x^{2} - 16\text{,}760\text{,}772x^{4} +56 \text{,}270\text{,}844x^{6} \\ &{} -78\text{,}971\text{,}346x^{8}+ 44\text{,}618 \text{,}574x^{10} - 7\text{,}910\text{,}916x^{12} + 1 \text{,}701\text{,}564x^{14} \\ &{} - 263\text{,}891x^{16} + 20\text{,}349x^{18}\bigr). \end{aligned}$$
Calculating \(G(5;1,1)-G(5;x,x)\), we have
$$\begin{aligned}& G(5;1,1)-G(5;x,x)=\frac{1}{429\text{,}083\text{,}246\text{,}592}\bigl(1-x^{2}\bigr)g \bigl(5;x^{2}\bigr), \\& g(5;x)=455\text{,}059 - 1\text{,}295\text{,}594x + 15\text{,}465 \text{,}178x^{2} - 40\text{,}805\text{,}666x^{3} + 38 \text{,}165\text{,}680x^{4} \\& \hphantom{g(5;x)=}{}- 6\text{,}452\text{,}894x^{5} + 1\text{,}458 \text{,}022x^{6} - 243\text{,}542x^{7} + 20 \text{,}349x^{8}\quad (0< x<1). \end{aligned}$$
It is sufficient to prove the following lemma.

Lemma 5.1

$$ g(x)=g(5;x)\ge0\quad (0\le x\le1). $$

Proof

We investigate the extremal values or zeros of \(g(x), g'(x),\ldots,g^{(4)}(x)\). Through straightforward calculations, we have
$$\begin{aligned}& g'''(0)=-244\text{,}833\text{,}996, g'''(1)=414\text{,}624\text{,}768, \\& g^{(4)}(x)=77\text{,}520\bigl( 11\text{,}816-9\text{,}989x+6 \text{,}771x^{2}-2\text{,}639x^{3}+441x^{4}\bigr) \\& \hphantom{g^{(4)}(x)}=77\text{,}520\bigl\{ 1\text{,}827 + 9\text{,}989(1-x) + 4 \text{,}132x^{2} + 2\text{,}639x^{2}(1-x) + 441x^{4} \bigr\} >0. \end{aligned}$$
Therefore, the equation \(g'''(x)=0\) (\(0< x<1\)) possesses a unique solution, \(x=x_{0} \in(0,1)\), and \(g''(x)\) takes its minimum at \(x=x_{0}\). Next, we prove
$$ g''(x_{0})< 0. $$
We can find \(\frac{30}{100}< x_{0}<\frac{31}{100}\) from the intermediate value theorem. Dividing \(g''(x)\) by \(g'''(x)\), we have
$$\begin{aligned}& g''(x)=\frac{1}{1\text{,}512}(252x -377)g'''(x)+ \frac{5}{126}r_{0}(x), \\& r_{0}(x)=r_{0}^{+}(x)-r_{0}^{-}(x), \\& r_{0}^{+}(x)=46\text{,}067\text{,}875x^{4} + 5\text{,}261 \text{,}460\text{,}050x^{2} + 613\text{,}870\text{,}628x, \\& r_{0}^{-}(x)=526\text{,}780\text{,}700x^{3} + 758\text{,}928 \text{,}637. \end{aligned}$$
Setting \(x=x_{0}\), we conclude \(g''(x_{0})<0\) as
$$\begin{aligned}& g''(x_{0})=\frac{5}{126}r_{0}(x_{0}), \\& r_{0}(x_{0})=r_{0}^{+}(x_{0})-r_{0}^{-}(x_{0}) < r_{0}^{+} \biggl(\frac{31}{100} \biggr)-r_{0}^{-} \biggl( \frac {30}{100} \biggr) \\& \hphantom{r_{0}(x_{0})}= -\frac{61\text{,}440\text{,}051\text{,}132\text{,}097}{800\text{,}000}<0. \end{aligned}$$
Together with the fact that \(g''(0), g''(1)>0\), there exist \(x_{1}\), \(x_{2}\) satisfying \(0< x_{1}<x_{0}<x_{2}<1\) and \(g''(x_{1})=g''(x_{2})=0\). In addition, \(g'(x)\) takes its maximal and minimal values at \(x=x_{1}\) and \(x=x_{2}\), respectively. Next, we show that
$$ g'(x_{2})>0. $$
Direct calculation shows that \(\frac{41}{100}< x_{2}<\frac{42}{100}\). Dividing \(g'(x)\) by \(g''(x)\), we have
$$\begin{aligned}& g'(x)=\frac{1}{1\text{,}764}(252x -377)g''(x)+ \frac{1}{441}r_{1}(x), \\& r_{1}(x)=r_{1}^{+}(x)-r_{1}^{-}(x), \\& r_{1}^{+}(x)=138\text{,}203\text{,}625x^{5} + 26\text{,}307 \text{,}300\text{,}250x^{3} + 4\text{,}604\text{,}029 \text{,}710x^{2} + 2\text{,}343\text{,}829\text{,}099, \\& r_{1}^{-}(x)=1\text{,}975\text{,}427\text{,}625x^{4} + 11 \text{,}383\text{,}929\text{,}555x. \end{aligned}$$
Setting \(x=x_{2}\), we have
$$\begin{aligned}& g'(x_{2})=\frac{1}{441}r_{1}(x_{2}), \\& r_{1}(x_{2})=r_{1}^{+}(x_{2})-r_{1}^{-}(x_{2})> r_{1}^{+} \biggl(\frac{41}{100} \biggr)-r_{1}^{-} \biggl( \frac {42}{100} \biggr) \\& \hphantom{r_{1}(x_{2})}= \frac{7\text{,}181\text{,}871\text{,}438\text{,}953\text{,}829}{80\text{,}000\text{,}000}>0, \end{aligned}$$
from which it is concluded that
$$ g'(x_{1})>g'(x_{2})>0. $$
Since we have \(g'(0)<0\), there exists a unique \(x_{3}\in(0,1)\) such that \(g'(x_{3})=0\), at which \(g(x)\) takes its minimum value. Finally, we prove that
$$ g(x_{3})>0. $$
Direct calculation and the intermediate value theorem show that \(\frac{5}{100}< x_{3}<\frac{6}{100}\). Dividing \(g(x)\) by \(g'(x)\), we have
$$\begin{aligned}& g(x)=\frac{1}{2\text{,}016}(252x -377)g'(x)+\frac{1}{1\text{,}008}r_{2}(x), \\& r_{2}(x)=r_{2}^{+}(x)-r_{2}^{-}(x), \\& r_{2}^{+}(x)=46\text{,}067\text{,}875x^{6}+13\text{,}153 \text{,}650\text{,}125x^{4}+3\text{,}069\text{,}353\text{,}140x^{3} \\& \hphantom{r_{2}^{+}(x)=}{}+4 \text{,}687\text{,}658\text{,}198x+214\text{,}480\text{,}003, \\& r_{2}^{-}(x)=790\text{,}171\text{,}050x^{5}+11\text{,}383 \text{,}929\text{,}555x^{2}. \end{aligned}$$
Using the same procedures, we obtain
$$\begin{aligned}& g(x_{3})=\frac{1}{1\text{,}008}r_{2}(x_{3}), \\& r_{2}(x_{3})=r_{2}^{+}(x_{3})-r_{2}^{-}(x_{3})>r_{2}^{+} \biggl(\frac{5}{100} \biggr)-r_{2}^{-} \biggl( \frac{6}{100} \biggr) \\& \hphantom{r_{2}(x_{3})}= \frac{653\text{,}353\text{,}651\text{,}584\text{,}933\text{,}307}{1\text{,}600\text{,}000\text{,}000} >0. \end{aligned}$$
This shows that
$$ g(x)\geq g(x_{3})>0\quad (0\le x\le1), $$
which completes the proof of Lemma 5.1. □

Although the proof is straightforward and simple, the proof for general M remains incomplete. Even if \(M\geq6\), if it is possible to carefully analyze the increase and decrease on the function \(G (M;x,x)\), we will be able to confirm that the maximum value is achieved on the boundary points \(x=\pm1\). However, a unified way to treat \(G(M;1,1)-G(M;x,x)\) has not been established and therefore the positivity remains to be unproved.

6 \(G(1,1)\)

The purpose of this section is to prove (1.3) in Theorem 1.1,
$$ G(1,1)=\frac{2^{2M-1}\Gamma(2M-1)\Gamma(2M+1)}{\Gamma(M)^{2}\Gamma(4M)}. $$
(6.1)
Setting \(y=1\) in (4.2), we have
$$ G(1,1)=\int_{-1}^{1} \bigl( \partial_{x}^{M} G(x,1) \bigr)^{2}\,dx. $$
(6.2)
We present the following key lemma.

Lemma 6.1

$$\begin{aligned} \partial_{x}^{M}G(x,1)&=\frac{\Gamma(M+1)}{2^{M}}K_{M}(1-x)K_{M-1}(1+x) \\ &=\frac{1}{2^{M}\Gamma(M)}(1-x)^{M-1}(1+x)^{M} \quad (-1< x<1). \end{aligned}$$
From the above lemma and (6.2), (6.1) is proven by
$$\begin{aligned} G(1,1)&=\frac{1}{2^{2M}\Gamma(M)^{2}}\int_{-1}^{1} (1-x)^{2M-2}(1+x)^{2M}\,dx \\ &=\frac{2^{2M-1}}{\Gamma(M)^{2}}B(2M-1,2M+1) =\frac{2^{2M-1}\Gamma(2M-1)\Gamma(2M+1)}{\Gamma(M)^{2}\Gamma(4M)}. \end{aligned}$$

Proof

It is easy to see that
$$ v(x)=\frac{\Gamma(M+1)}{2^{M}}K_{M}(1-x)K_{M-1}(1+x) $$
satisfies the boundary value problem:
Hence, it is sufficient to prove that
$$ u(x)=\partial_{x}^{M}G(x,1)\quad (-1< x<1) $$
satisfies the same boundary value problem (6.3)-(6.5).
From (3.1), we have
$$\begin{aligned} G(x,1) =& \frac{(-1)^{M}}{2}K_{0}(1-x)-\sum _{i=0}^{M-1} \bigl\{ \psi_{i}(x) \varphi_{i}(1)+\psi_{i}(1)\varphi_{i}(x) \bigr\} \\ &{} +\sum_{i,j=0}^{M-1}\gamma_{ij} \varphi_{i}(x)\varphi_{j}(1). \end{aligned}$$
Using \(\partial_{x}^{M}\varphi_{i}(x)=0\) (\(-1< x<1\), \(0\le i\le M-1\)), we have
$$ u(x)=\partial_{x}^{M}G(x,1)= \frac{1}{2}K_{M}(1-x)- \sum_{i=0}^{M-1} \varphi_{i}(1)\psi_{i}^{(M)}(x). $$
(6.6)
Recall the relation \((-1)^{M}\psi_{i}^{(2M)}(x)=\varphi_{i}(x)\) (\(-1< x<1\)), which follows from (3.2). We have
$$\begin{aligned} (-1)^{M-1}u^{(2M-1)}(x)&=(-1)^{M}\sum _{i=0}^{M-1}\varphi_{i}(1) \psi_{i}^{(3M-1)}(x) =\sum_{i=0}^{M-1} \varphi_{i}(1)\varphi_{i}^{(M-1)}(x) \\ &=\varphi_{M-1}(1)\varphi_{M-1}^{(M-1)}(x) = \biggl(M- \frac{1}{2} \biggr)P_{M-1}^{(M-1)}(x) \\ &=\frac{\Gamma(2M)}{2^{M}\Gamma(M)}. \end{aligned}$$
(6.7)
This shows that \(u(x)=\partial_{x}^{M} G(x,1)\) satisfies the differential equation (6.3).
Next, we prove that \(u(x)\) satisfies the same boundary conditions (6.4) and (6.5). From (6.6), we have
$$ u^{(k)}(x)=(-1)^{k}\frac{1}{2}K_{M+k}(1-x)- \sum_{i=0}^{M-1}\varphi_{i}(1) \psi_{i}^{(M+k)}(x). $$
We now prove the following lemma.

Lemma 6.2

For any k (\(0\le k\le M-2\)), we have
$$ (1)\quad u^{(k)}(1)=-\sum_{i=0}^{M-1} \varphi_{i}(1)\psi_{i}^{(M+k)}(1)=0. $$
For any k (\(0\le k\le M-1\)), we have
$$ (2)\quad u^{(k)}(-1)=(-1)^{k}\frac{1}{2}K_{M+k}(2)- \sum_{i=0}^{M-1}\varphi _{i}(1) \psi_{i}^{(M+k)}(-1)=0. $$

Proof

We first prove (1). Substituting
$$ \psi_{i}^{(M+k)}(1)=\int_{-1}^{1} \frac{(-1)^{M}}{2}K_{M+k}(1-y)\varphi_{i}(y)\,dy\quad (0\leq k \leq M-2) $$
into (1), we have
$$ \sum_{i=0}^{M-1}\varphi_{i}(1) \psi_{i}^{(M+k)}(1)= \int_{-1}^{1} \frac{(-1)^{M}}{2}K_{M+k}(1-y) \sum_{i=0}^{M-1} \varphi_{i}(1)\varphi_{i}(y)\,dy. $$
Since a vector space spanned by
$$ \bigl\{ \Gamma(M-k)K_{M+k}(1-y)=(1-y)^{M-1-k} | 0\le k\le M-2 \bigr\} $$
is the same as the vector space spanned by
$$ \bigl\{ \varphi_{j}(y)-\varphi_{j}(1) | 1\le j\le M-1 \bigr\} , $$
it is sufficient to show that
$$ \int_{-1}^{1} \bigl\{ \varphi_{j}(y)- \varphi_{j}(1) \bigr\} \sum_{i=0}^{M-1} \varphi_{i}(1)\varphi_{i}(y)\,dy=0\quad (1 \leq j \leq M-1). $$
This can be shown by
$$\begin{aligned} {(\mbox{l.h.s.})}&=\int_{-1}^{1} \biggl\{ \varphi_{j}(y)-\frac{\varphi _{j}(1)}{\varphi _{0}(1)}\varphi_{0}(y) \biggr\} \sum _{i=0}^{M-1}\varphi_{i}(1) \varphi_{i}(y)\,dy \\ &=\sum_{i=0}^{M-1} \biggl\{ \varphi_{i}(1)\int_{-1}^{1} \varphi _{j}(y)\varphi_{i}(y)\,dy- \frac{\varphi_{j}(1)\varphi_{i}(1)}{\varphi_{0}(1)}\int _{-1}^{1} \varphi _{0}(y)\varphi _{i}(y)\,dy \biggr\} =0, \end{aligned}$$
where we used the following relations:
$$ \varphi_{0}(y)=\varphi_{0}(1)=\frac{1}{\sqrt{2}} \quad \text{and}\quad \int_{-1}^{1}\varphi_{i}(y) \varphi_{j}(y)\,dy=\delta_{ij}. $$
We next show (2), that is,
$$ \sum_{i=0}^{M-1}\varphi_{i}(1) \psi_{i}^{(M+k)}(-1)=\frac{(-1)^{k}}{2}K_{M+k}(2)\quad (0\le k\le M-1). $$
Since
$$ \psi_{i}^{(M+k)}(-1)=\frac{(-1)^{k}}{2}\int _{-1}^{1}K_{M+k}(1+y)\varphi_{i}(y) \,dy\quad (0\le k\le M-1) $$
holds, it is sufficient to show that
$$ \int_{-1}^{1}K_{M+k}(1+y)\sum _{i=0}^{M-1}\varphi_{i}(1)\varphi _{i}(y)\,dy=K_{M+k}(2)\quad (0\le k\le M-1), $$
or equivalently
$$ \int_{-1}^{1} \bigl\{ K_{M+k}(2)-K_{M+k}(1+y) \bigr\} \sum_{i=0}^{M-1}\varphi_{i}(1) \varphi_{i}(y)\,dy=0\quad (0\le k\le M-1). $$
This is easily confirmed in the case in which \(k=M-1\), and therefore it is sufficient to prove this for \(0\le k\le M-2\). The vector space spanned by
$$\begin{aligned}& \bigl\{ \Gamma(M-k) \bigl(K_{M+k}(2)-K_{M+k}(1+y) \bigr)=2^{M-1-k}-(1+y)^{M-1-k} | \\& \quad 0\le k\le M-2 \bigr\} \end{aligned}$$
is the same as the vector space spanned by
$$ \bigl\{ \Gamma(M-k)K_{M+k}(1-y)=(1-y)^{M-1-k} | 0\le k\le M-2 \bigr\} . $$
Using the same procedure as in the proof of (1), we can prove (2). □

From (6.7) and Lemma 6.2, we have Lemma 6.1.  □

7 Infimum of Sobolev functional

In this section, we finally prove (1.4). It is sufficient to prove the following lemma.

Lemma 7.1

The Legendre polynomials \(P_{n}(x)\) (\(n=0,1,2,\ldots\)) satisfy the following properties:
$$\begin{aligned} (1)&\quad \sup_{|y|\le1}\bigl\vert P_{n}(y)\bigr\vert =1, \\ (2)&\quad \int_{-1}^{1}\bigl\vert P_{n}^{(M)}(x)\bigr\vert ^{2}\,dx \rightarrow \infty\quad (n\to\infty, M=1,2,\ldots). \end{aligned}$$

Proof

(1) follows from the Laplace-Mehler formula,
$$ P_{n}(\cos\theta)=\frac{1}{\pi}\int_{0}^{\pi} ( \cos\theta+\sqrt{-1}\sin\theta\cos\varphi)^{n}\, d\varphi, $$
and \(P_{n}(1)=1\). In order to prove (2), we present two lemmas.

Lemma 7.2

$$ P_{n}^{(M)}(1)=\frac{\Gamma(n+M+1)}{2^{M} \Gamma(M+1)\Gamma (n-M+1)} \quad (n\ge M). $$
(7.1)

Proof

\(u=P_{n}(x)\) satisfies the Legendre differential equation:
$$ \bigl(1-x^{2}\bigr)u''-2xu'+n(n+1)u=0. $$
By taking M times the derivative of the above equation, it is shown that \(v=v(x)=P_{n}^{(M)}(x)\) satisfies
$$ \bigl(1-x^{2}\bigr)v''-2(M+1)xv'+ \bigl\{ n(n+1)-M(M+1)\bigr\} v=0. $$
(7.2)
Setting \(x=1\) in (7.2), we have
$$ -2(M+1)P_{n}^{(M+1)}(1)+\bigl\{ n(n+1)-M(M+1)\bigr\} P_{n}^{(M)}(1)=0. $$
This is equivalently rewritten as
$$\begin{aligned} \frac{\Gamma(M+2)\Gamma(n-M)}{\Gamma(n+M+2)}2^{M+1}P_{n}^{(M+1)}(1) &= \frac{\Gamma(M+1)\Gamma(n-(M-1))}{\Gamma (n+M+1)}2^{M}P_{n}^{(M)}(1) \\ &=\frac{\Gamma(1)\Gamma(n+1)}{\Gamma(n+1)}P_{n}(1)=1. \end{aligned}$$
This proves Lemma 7.2. □

Lemma 7.3

For \(n\ge M\), we have
$$ \int_{-1}^{1}\bigl\vert P_{n}^{(M)}(x) \bigr\vert ^{2}\,dx \ge\frac{M}{n^{2}+n-M^{2}} \biggl(\frac{\Gamma(n+M+1)}{2^{M}\Gamma(M+1)\Gamma(n-M+1)} \biggr)^{2}. $$

Proof

Multiplying \(v=P_{n}^{(M)}(x)\) on both sides of (7.2), we have
$$ \bigl(1-x^{2}\bigr)v''v-2(M+1)xv'v+ \bigl\{ n(n+1)-M(M+1)\bigr\} v^{2}=0, $$
or equivalently
$$ \bigl(n^{2}+n-M^{2}\bigr)v^{2}= \bigl[Mxv^{2}-\bigl(1-x^{2}\bigr)v'v \bigr]'+\bigl(1-x^{2}\bigr) \bigl(v' \bigr)^{2}. $$
Integrating the above equality on the interval \(-1< x<1\), we have
$$\begin{aligned} \bigl(n^{2}+n-M^{2} \bigr)\int_{-1}^{1} \bigl\vert v(x)\bigr\vert ^{2}\,dx &=M \bigl(v(1)^{2}+v(-1)^{2} \bigr)+\int_{-1}^{1}\bigl(1-x^{2}\bigr) \bigl\vert v'(x) \bigr\vert ^{2}\,dx \\ &\geq M v(1)^{2}=M \bigl(P_{n}^{(M)}(1) \bigr)^{2}. \end{aligned}$$
Substituting (7.1) in the right-hand side of the above inequality, we obtain Lemma 7.3. □

Taking the limit \(n\to\infty\) in Lemma 7.3, we finally prove Lemma 7.1(2).  □

Proof of Theorem 1.3

From Lemma 7.1, we have
$$ S\bigl(M;P_{n}(x)\bigr)= \Bigl(\sup_{|y|\le1}\bigl\vert P_{n}(y)\bigr\vert \Bigr)^{2} \Big/ \int _{-1}^{1}\bigl\vert P_{n}^{(M)}(x) \bigr\vert ^{2}\,dx \rightarrow0 \quad (n\to\infty). $$
This proves (1.4). □

Declarations

Acknowledgments

This research was supported by J.S.P.S. Grant-in-Aid for Scientific Research (C) Nos. 24540199, 25400146, 25400210.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
College of Science and Technology, Nihon University
(2)
College of Industrial Technology, Nihon University
(3)
Graduate School of Engineering Science, Osaka University
(4)
Department of Computer Science, National Defense Academy
(5)
Tokyo Metropolitan College of Industrial Technology

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© Takemura et al.; licensee Springer. 2015