# -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 $\sum_{-\infty}^{\infty}{a_{n}}=A < \infty$? Is is true that the convergence preserves and $||T|| \leq A$?

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Hint: let $P_j$ be the self-adjoint projection on $\text{Ran}(T_j)$. – Robert Israel Apr 16 '12 at 7:28
More details please? :) – Anna Apr 16 '12 at 7:59
More hints: $\text{Ran}(T_j)$ are orthogonal to each other, and $S_N(P_j f) = T_j P_j f$ if $|j| \le N$. – Robert Israel Apr 16 '12 at 18:23
Can anyone provide a proof with more details? I'm not sure I understand after all.. – Anna Apr 19 '12 at 2:33
Oops, that wasn't quite right. $P_j$ should be the projection on $\overline{\text{Ran}(T_j^*)} = (\text{Ker}(T_j))^\perp$. I'll write out an answer. – Robert Israel Apr 19 '12 at 5:09

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 $M > N$, $\|S_M f - S_N f\|^2 \le \sum_{N < |i|\le M} \|P_i f\|^2$. For any $f$ and any $\epsilon > 0$, we can take $K$ large enough that $\sum_{|i|>K} \|P_i f\|^2 < \epsilon$, and so $\|S_M f - S_N f\|^2 < \epsilon$ 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 $\|T\| \le 1$.