-Almost- self-adjoint bounded operators on Hilbert spaces I ran into the following question about bounded operators on Hilbert spaces; I could really use some help.
It goes like this: Suppose that $\left\{T_{k}\right\}$ is a collection of bounded operators on a Hilbert space $\mathcal{H}$, with $||T_{k}|| \leq 1$ for all $k$. Suppose also that $T_{k}T_{j}^{*}=T_{k}^{*}T_{j}=0$ for all $k \neq j$. Let $S_{N}=\sum_{k=-N}^{N}{T_{k}}$. Then, we have that $S_{N}(f)$ converges as $N \rightarrow \infty$, for every $f \in \mathcal{H}$. Moreover, if $T(f)$ denotes the limit, then $||T|| \leq 1$.
What if the condition $T_{k}T_{j}^{*}=T_{k}^{*}T_{j}=0$ is replaced by $||T_{k}T_{j}^{*}|| \leq a_{k-j}^{2}$ and $||T_{k}^{*}T_{j}|| \leq a_{k-j}^{2}$ for positive constants $a_{n}$ with the property $\sum_{-\infty}^{\infty}{a_{n}}=A &lt \infty$? Is is true that the convergence preserves and $||T|| \leq A$?  
 A: In the first version, let $P_j$ be the orthogonal projection of $\cal H$ on $\overline{\text{Ran}(T_j^*)} = (\text{Ker}(T_j))^\perp$.  Note that these are orthogonal to each other since $T_i T_j^* = 0$ for $i \ne j$: we have $P_i P_j = T_j P_i = 0$, while $T_i P_i = T_i$. 
 Thus for any $f \in \cal H$ and any $N$, we can write
$f = g_N + \sum_{j=-N}^N P_j f$ where $g_N = (I - \sum_{j=-N}^N P_j) f$ and
$\|f\|^2 = \|g_N\|^2 + \sum_{j=-N}^N \|P_j f\|^2$.  In particular, $\sum_{j=-\infty}^\infty \|P_j f\|^2 \le \|f\|^2$.  
Now $T_i f = T_i P_i f$ so $S_N f = \sum_{i=-N}^N T_i P_i f$ with
$$\eqalign{\|S_N f\|^2 &= \sum_{i=-N}^N \sum_{j=-N}^N \langle P_i f, T_i^* T_j P_j f \rangle = \sum_{i=-N}^N \langle P_i f, T_i^* T_i P_i f \rangle\cr & = \sum_{i=-N}^N \|T_i P_i f\|^2
\le \sum_{i=-N}^N \|P_i f\|^2 \le \|f\|^2\cr}$$
Similarly, for $M > N$, $\|S_M f - S_N f\|^2 \le \sum_{N &lt |i|\le M} \|P_i f\|^2$.
For any $f$ and any $\epsilon > 0$, we can take $K$ large enough that $\sum_{|i|>K} \|P_i f\|^2 &lt \epsilon$, and so $\|S_M f - S_N f\|^2 &lt \epsilon$ for all $M, N > K$.  Thus
$\lim_N S_N f$ exists.  Of course the limit is linear in $f$, so $T$ exists, and our bound on $\|S_N f\|^2$ shows that $\|T\| \le 1$.
I'm not sure what happens in the generalized version.
