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 ${\sum }_k\Vert T{e}_k{\Vert }^2<\infty$ for a Hilbertian basis $(e_k)$. Let $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$, extend $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis, and let the linear functional $\varphi$ equal one on $v$ and zero on every other basis vector. Then $Tx=\varphi (x)e_1$ has $T{e}_k=0$ for every $k$ but $Tv=e_1$. For $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k$, $v-v_N\to 0$ 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.