= Finite column sum does not imply boundedness
An everywhere-defined <linear operator> $T$ on an infinite-dimensional <separable Hilbert space> need not be bounded merely because $\sum_k\|Te_k\|^2<\infty$ for a <Hilbertian basis> $(e_k)$. Let $v=\sum_{k\geq1}2^{-k}e_k$, extend $\{e_k:k\geq1\}\cup\{v\}$ to a <Hamel basis>, and let the <linear functional> $\phi$ equal one on $v$ and zero on every other basis vector. Then $Tx=\phi(x)e_1$ has $Te_k=0$ for every $k$ but $Tv=e_1$. For $v_N=\sum_{k=1}^N2^{-k}e_k$, $v-v_N\to0$ while $T(v-v_N)=e_1$, proving discontinuity. Thus the definition of a <Hilbert-Schmidt operator> must require boundedness, or first define the operator on finite <linear combinations> and take its unique continuous extension.
Back to article page
