A relation between two properties of sequences of operators We have $(T_l)_l$ a sequence of bounded linear operators from $\ell^2$ to $\ell^2$.
$\bullet$ We say $(T_l)_l$ satisfies the property "A" if $\sup_{||x||_{\ell^2}=1}\sum_{l=1}^\infty||T_l(x)||^2<\infty$.
$\bullet$ And we say that it satisfies the property "B" if $\sup_{N\in\mathbb{N}}\int_0^{2\pi} ||\sum_{l=1}^{N}e^{ilt} \cdot T_l||^2dt<\infty$.
$$ $$
It is very easy to see that if $(T_l)_l$ satisfies $B$, it also satisfies $A$, because
$$\sum_{l=1}^N||T_l(x)||^2=\int_0^{2\pi}||\sum_{l=1}^N e^{ilt} \cdot T_l(x)||^2dt\leq \int_0^{2\pi} ||\sum_{l=1}^N e^{ilt} \cdot T_l||^2||x||^2dt.$$
My question is, if $(T_l)_l$ satisfies $A$, can it not satisfy "B? Or these two properties are really the same?
Thanks in advance for any help.
 A: Let us consider the family of operators
$$
T_\ell ((x_n)_n) = x_\ell \cdot e_1,
$$
where $e_1 = (1, 0,0,\dots) \in \ell^2$ is the first basis vector.
This family satisfies the first property, since for $x = (x_n)_n \in \ell^2$ with $\| x\| = 1$, we have
$$
\sum_{\ell=1}^\infty \|T_\ell x\|^2 = \sum_\ell |x_\ell|^2 = \|x\|^2 \leq 1.
$$
But it is not too hard to see (see below) that
$$
\bigg\| \sum_{\ell=1}^N \alpha_\ell T_\ell \bigg\| = \|(\alpha_\ell)_{\ell =1 , \dots,N}\|_{\ell^2}
$$
for arbitrary coefficients $\alpha_\ell \in \Bbb{C}$.
This easily implies that your property (B) does not hold in this case.
To justify the above calculation, note
\begin{align*}
\bigg\| \sum_{\ell=1}^N \alpha_\ell T_\ell \bigg\|_{B(\ell^2)}
&= \sup_{\|x\| = 1} \bigg\| \sum_{\ell=1}^N \alpha_\ell T_\ell(x) \bigg\|_{\ell^2} \\
&= \sup_{\|x\| = 1} \bigg\| \sum_{\ell=1}^N \alpha_\ell x_\ell e_1\bigg\|_{\ell^2} \\
&= \sup_{\|x\|=1} \sum_{\ell=1}^N |\alpha_\ell x_\ell| \\
&= \| (\alpha_\ell)_{\ell=1, \dots,N}\|_{\ell^2} 
\end{align*}
