A baby version of the Stein-Cotlar almost-orthogonality lemma The following is an exercise from Stein and Shakarchi's Real Analysis.
Suppose $\{T_k\}$ is a collection of bounded operators on a Hilbert space $H$, each with norm at most $1$. Suppose also that 
$$T_kT^*_j = T^*_k T_j =0$$
for $k\neq j$. Let $S_n(f)=\sum_{k=1}^n T_k(f)$. The problem asks to show that $\lim_{N\rightarrow \infty} S_n(f)$ converges for every $f$ and that the the resulting operator $S$ has norm at most $1$.
Is the sketch written in the spoiler box below correct? I am mostly worried about the construction in the second and third sentences. Any alternative slick solutions are also welcome. 

 Consider first the case of a finite number of operators. Since the operators have mutually orthogonal ranges, the closures of their ranges are also mutually orthogonal. This allows us to decompose the space into $B\oplus_1^n V_k$, where the $V_k$ are the closures of the ranges and $B$ is what is left over after taking the direct sum of all the $V_k$. Let $f=\sum v_k$ denote the decomposition of a function $f$ into these spaces. Then, recalling the ranges are orthogonal, $\|S_n(f)\|\le \sum_1^n \|T_k(f)\|\le \sum |T_k (v_k)|\le \sum_1^n |v_k|\le |f|.$ This shows the sequence is absolutely convergent, so it is convergent, since we are in a complete space. It also immediately shows the limit is an operator and that this operator has norm at most 1.

 A: *

*There's a problem with $\sum_1^n \|T_k(f)\|\le \sum |T_k (v_k)|$: the decomposition $f=\sum v_k$ has to do with the ranges of $T_k$, but when you plug things into $T_k$, it's the domain, not the range, that matters. 

*Also a problem with $\sum_1^n |v_k|\le |f|$: needs squares to make it right. 


My solution. First, note that not only $T_k$, but also their adjoints $T_k^*$ have orthogonal ranges. Since $(\ker T_k)^\perp $ is the closure of $\operatorname{ran} T_k^* $ (see below), it follows that the spaces $(\ker T_k)^\perp $ are orthogonal. Let $v_k$ be the orthogonal projection of $v$ onto $(\ker T_k)^\perp $. Observe that $T_kv=T_kv_k$, because $v-v_k\in\ker T_k$.
Now we are ready to roll: 
$$\Big\|\sum_k T_k(v)\Big\|^2 = \sum_k \|T_k v\|^2 =  \sum_k \|T_k v_k\|^2 \le \sum_k \| v_k\|^2 \tag1$$
I did not put the limits of summation in (1), because we should do two things with it:


*

*sum over the tail $M\le k\le N$ to make sure that $\sum_{k=M}^N T_k(v)\to 0$ as $M,N\to \infty$; 

*sum the whole thing to estimate it by $\|v\|^2$ and thus get the norm bound.



The equality $(\ker T)^\perp =\overline{\operatorname{ran} T^*}$ follows from $\ker T = (\operatorname{ran} T^*)^\perp$. The latter is proved thus:
$$Tx=0 \iff \langle Tx,y\rangle=0 \ \ \forall y \iff \langle x,T^*y\rangle=0\ \ \forall y $$
