For which \(\sigma\in\mathbb R\) does there exist \(C_\sigma<\infty\) such that \[ \left|\sum_{j,k=1}^\infty (1+|j-k|)^\sigma a_jb_k\right| \le C_\sigma\|a\|_{\ell^2}\|b\|_{\ell^2} \] for all \(a,b\in\ell^2\)?
Solution
The estimate holds exactly when \[ \boxed{\sigma<-1}. \]
1. Sufficiency for \(\sigma<-1\).
Set \[ K_{jk}=(1+|j-k|)^\sigma. \] If \(\sigma<-1\), then \[ M:=1+2\sum_{m=1}^\infty(1+m)^\sigma<\infty. \] For every \(j\), \[ \sum_{k=1}^\infty K_{jk}\le M, \] and similarly every column sum is at most \(M\). By Schur’s test, the operator \[ (Tb)_j=\sum_{k=1}^\infty K_{jk}b_k \] is bounded on \(\ell^2\) with \(\|T\|\le M\). Therefore \[ \left|\sum_{j,k}K_{jk}a_jb_k\right| =|\langle a,Tb\rangle| \le M\|a\|_2\|b\|_2. \]
2. Necessity.
For \(N\ge1\), let \[ a_j=b_j=\begin{cases}N^{-1/2},&1\le j\le N,\\0,&j>N.\end{cases} \] Then \(\|a\|_2=\|b\|_2=1\), while the bilinear form equals \[ S_N=\frac1N\sum_{j,k=1}^N(1+|j-k|)^\sigma =1+\frac2N\sum_{m=1}^{N-1}(N-m)(1+m)^\sigma. \] For \(1\le m\le N/2\), one has \(N-m\ge N/2\), hence \[ S_N\ge \sum_{m=1}^{\lfloor N/2\rfloor}(1+m)^\sigma. \] If \(\sigma=-1\), the right-hand side grows like \(\log N\). If \(\sigma>-1\), it grows on the order of \(N^{\sigma+1}\). Thus \(S_N\to\infty\) whenever \(\sigma\ge-1\). No uniform constant \(C_\sigma\) can then exist.
Hence the estimate holds exactly for \(\sigma<-1\).