= Solution
As in the preceding part, take $T$ to be a <bounded linear operator>. For $x\in X$, the <Hilbertian basis> expansion and continuity of $T$ give
$$
Tx=\lim_{N\to\infty}\sum_{k=1}^N\langle x,e_k\rangle Te_k.
$$
The <triangle inequality>, the <Cauchy-Schwarz inequality> for finite sums, and the <Parseval identity for a Hilbertian basis> give
$$
\begin{aligned}
\left\|\sum_{k=1}^N\langle x,e_k\rangle Te_k\right\|
&\leq\sum_{k=1}^N|\langle x,e_k\rangle|\,\|Te_k\|\\
&\leq\left(\sum_{k=1}^N|\langle x,e_k\rangle|^2\right)^{1/2}
\left(\sum_{k=1}^N\|Te_k\|^2\right)^{1/2}
\leq\|x\|\|T\|_2.
\end{aligned}
$$
Passing to the limit and taking the supremum over the <unit ball> proves
$$
\boxed{\|T\|\leq\|T\|_2.}
$$
Thus the <Hilbert-Schmidt norm> controls the <operator norm>. Without the boundedness interpretation, the nonzero $R$ constructed in the preceding part has zero printed column sum and makes the requested inequality false.
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