On a dual Hardy-Hilbert's inequality and its generalization (Q2568535)

The paper deals with a dual Hardy-Hilbert's inequality with a best constant factor involving the beta function, which is an extension of the Hilbert's inequality with \((p,q)\)-parameter. The main results are the following: If \(p>1\), \(1/p+1/q=1\), \(2-\min\{p,q\}<\lambda \leq 2\), and \(a_n,b_n\geq 0\), such that \[ 0<\sum^\infty_{n=0}\left(n+\frac 12\right)^{(p-1)(2-\lambda)-1} a^p_n<\infty,\quad 0<\sum^\infty_{n=0}\left(n+\frac 12 \right)^{(q-1)(2-\lambda)-1}b^q_n<\infty, \] then \[ \begin{aligned} & \sum^\infty_{m=0}\, \sum^\infty_{n=0} \frac{a_mb_n}{(m+n+1)^\lambda}\\ &\quad <k_\lambda (p)\left(\sum^\infty_{n=0}\left(n+\frac 12\right)^{(p-1)(2-\lambda)-1}a^p_n\right)^{1/p} \left(\sum^\infty_{n=0}\left(n+\frac 12\right)^{(q-i)(2-\lambda)-1}b^q_n\right)^{1/q}, \end{aligned} \] where the constant factor \(k_\lambda(p)=B\big(\frac{p+\lambda-2}{p},\frac{q+\lambda-2}{q}\big)\) is the best possible. In particular, for \(\lambda=1\), \[ \sum^\infty_{m=0}\, \sum^\infty_{n=0}\frac{a_m b_n}{m+n+1} <\frac{ \pi}{\sin(\pi/p)}\left(\sum^\infty_{n=0}\left(n+\frac 12\right)^{p-2} a^p_n\right)^{1/p} \left(\sum^\infty_{n=0}\left(n+\frac 12\right)^{q-2} b^q_n\right)^{1/q},\tag{1} \] where the constant factor \(\pi/\sin(\pi/p)\) is also the best possible. If \(p>1\), \(1/p+1/q=1\), \(2-\min\{p,q\}<\lambda\leq 2\), \(a_n\geq 0\), and \[ 0<\sum^\infty_{n=0}\left(n+\frac 12\right)^{(p-1)(2-\lambda)-1}a^p_n<\infty, \] then \[ \sum^\infty_{n=0}\left(n+\frac 12 \right)^{p+\lambda-3}\left(\sum^\infty_{m=0}\frac{a_m}{(m+n+1)^\lambda}\right)^p <(k_\lambda(p))^p\left(\sum^\infty_{n=0}\left(n+\frac 12\right)^{(p-1)(2-\lambda)-1}a^p_n\right)^{1/p}, \tag{2} \] where the constant factor \(k_\lambda(p)^p=B\big(\frac{p+\lambda-2}p,\frac{q+\lambda-2}q\big)^p\) is the best possible. Inequality (2) is equivalent to (1). In particular, for \(\lambda=1,\) \[ \sum^\infty_{n=0}\left(n+\frac 12\right)^{p-2}\left(\sum^\infty_{m=0}\frac{a_m}{m+n+1}\right)^p <\left(\frac \pi{\sin(\pi/p)}\right)^p\left(\sum^\infty_{n=0}\left(n+\frac 12\right)^{p-2}a^p_n\right)^{1/p}, \] where the constant factor \((\pi/\sin(\pi/p))^p\) is still best possible.