Let \(c_{00}\subset\ell^2\) be the finitely supported sequences, with standard basis \((e_m)\). Suppose a linear operator \(T:c_{00}\to\ell^2\) has matrix coefficients \[ a_{mn}=\langle Te_m,e_n\rangle. \]
Assume \[ \sum_{m,n=1}^\infty |a_{mn}|^2<\infty. \] Show that \(T\) is bounded on \(c_{00}\) and therefore extends uniquely to a bounded operator on \(\ell^2\).
If instead only \[ \sup_{m,n}|a_{mn}|<\infty, \] must \(T\) extend boundedly to \(\ell^2\)? Explain.
Erratum: the source states the problem for an arbitrary linear map \(T:\ell^2\to\ell^2\). For such a map part (a) fails: an everywhere-defined discontinuous linear map can vanish on every \(e_m\) and still be nonzero off the dense subspace \(c_{00}\), so all its matrix coefficients vanish while the map is unbounded. The statement on this card defines \(T\) on \(c_{00}\).
Solution
1. Square-summable matrix coefficients give a bounded operator.
Let \(x=\sum_m x_m e_m\in c_{00}\). For each \(n\), \[ (Tx)_n=\sum_m a_{mn}x_m. \] Hence, by Cauchy–Schwarz in the \(m\)-sum, \[ |(Tx)_n|^2 \le \left(\sum_m|a_{mn}|^2\right) \left(\sum_m|x_m|^2\right). \] Summing over \(n\) gives \[ \begin{aligned} \|Tx\|_2^2 &=\sum_n|(Tx)_n|^2\\ &\le \left(\sum_{m,n}|a_{mn}|^2\right)\|x\|_2^2. \end{aligned} \] Thus \[ \|Tx\|_2\le \left(\sum_{m,n}|a_{mn}|^2\right)^{1/2}\|x\|_2. \] Since \(c_{00}\) is dense in \(\ell^2\), \(T\) extends uniquely by continuity to a bounded operator on \(\ell^2\).
2. Uniformly bounded entries do not suffice.
Define \(T:c_{00}\to\ell^2\) on the basis by \[ Te_m=e_1+e_2+\cdots+e_m. \] Then \[ a_{mn}=\langle Te_m,e_n\rangle = \begin{cases} 1,&n\le m,\\ 0,&n>m, \end{cases} \] so \[ \sup_{m,n}|a_{mn}|=1. \] However, \[ \|Te_m\|_2=\sqrt m, \qquad \|e_m\|_2=1. \] Therefore \(T\) is unbounded on \(c_{00}\) and cannot extend to a bounded operator on \(\ell^2\).