= Solution
Here the intended space must consist of \b[bounded] <linear operators>, namely <Hilbert-Schmidt operators>. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic <linear span> of $(e_k)$, for example $v=\sum_{k\geq1}2^{-k}e_k$. Extend the linearly independent family $\{e_k:k\geq1\}\cup\{v\}$ to a <Hamel basis>. Define a <linear functional> $\phi$ on this <Hamel basis> by $\phi(v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined <linear operator> $Rx=\phi(x)e_1$ satisfies
$$
Re_k=0\quad(k\geq1),\qquad Rv=e_1\ne0.
$$
Its printed column sum is zero although $R\ne0$. More explicitly, $v_N=\sum_{k=1}^N2^{-k}e_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined <adjoint operator> of the bounded kind used in the question. This proves <finite column sum does not imply boundedness>. We use $T\in\mathcal B(X)$ in the intended argument below.
By the <Parseval identity for a Hilbertian basis> for $(e_j)$, followed by the <adjoint operator> identity and then the <Parseval identity for a Hilbertian basis> for $(h_k)$,
$$
\begin{aligned}
\sum_k\|Th_k\|^2
&=\sum_k\sum_j|\langle Th_k,e_j\rangle|^2\\
&=\sum_j\sum_k|\langle h_k,T^*e_j\rangle|^2\\
&=\sum_j\|T^*e_j\|^2.
\end{aligned}
$$
The interchange is valid by <Tonelli's theorem> for nonnegative double <series>, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals $\sum_k\|Te_k\|^2$. Hence
$$
\boxed{\sum_k\|Th_k\|^2=\sum_j\|T^*e_j\|^2=\sum_k\|Te_k\|^2.}
$$
In particular the <Hilbert-Schmidt norm> is independent of the <Hilbertian basis>, and $\|T^*\|_2=\|T\|_2$.
Back to article page
