Let $\mathcal{H}$ be a Hilbert space and $\{T_n\}_{n\in \Bbb N}$ is a sequence of bounded linear operators on $\mathcal{H}$. Define the diagonal operator $M:\oplus_n \mathcal{H}^{(n)}\to \mathcal{H}^{(n)}$ such that \begin{bmatrix}T_1 & 0 & 0 & ...\\0 & T_2 & 0 & ...\\0 & 0 & T_3 & ...\\ \vdots &\vdots&\vdots& \ddots\end{bmatrix} Is $M$ bounded? I think if $\{T_n\}_{n\in \Bbb N}$ is uniformly bounded then $M$ is bounded. Is there any condition?
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I think that $M$ is not necessarily bounded. Suppose $H$ is $\mathbb{R}$. Let $T_{n}x = n^{2}x$. Each $T_{n}$ is bounded. Then $M$ is clearly not bounded. Take $x_{n}(m) = \frac{1}{m}$ if $m \leq n$ and $0$ otherwise. Since $\sum \frac{1}{m^{2}}$ converges, each $x_{n}$ is a valid element and $\{x_{n}\}$ converges to $x(m) = \frac{1}{m}$. Next, $\sum_{m= 1}^{\infty}|M(x_{n}(m))|^{2} > |M(x_{n}(n))|^{2} = n^{4}/n^{2} = n^{2}\to \infty$ so $M$ is not bounded.
If you think about this argument, I think you should see why the uniformly bounded condition fixes the problem.
