I'm trying to solve the following exercise $23$ of chapter $4$ of Stein's Real Analysis.
Exercise 23
Suppose $\{T_k\}$ is a collection of bounded operators on a Hilbert space $\mathcal{H}$, with $\|T_k\|\leq1$ for all $k$. Suppose also that $$ T_kT_j^* = T_k^*T_j = 0\hspace{1cm}\mbox{for all } k \neq j. $$ Let $S_N = \sum_{k = -N}^N T_k$.
Show that $S_N(f)$ converges as $N \rightarrow \infty$, for every $f \in \mathcal{H}$. If $T(f)$ denotes the limit, prove that $\|T\| \leq 1$.
[Hint: Consider first the case when only finitely many of the $T_k$ are non-zero, and note that the ranges of the $T_k$ are mutually orthogonal.]
Now, I'm studying this page to understand the exercise, and there are some points that I can't fully understand.
The first point is the following sentence;
It follows that, regardless of the finite subset $F$, one has the inequality $$\Vert \sum_{k\in F}T_{k}x\Vert \le \Vert x\Vert ,\qquad x \in \mathcal{H}.$$
I've tried to connect this with the previous inequality $|\sum_{k\in F}(T_{k}x,y)| \le \Vert x \Vert \Vert y \Vert$, but I couldn't find the relationship btw these. Applying $y = x$ doesn't lead to any useful result.
The second is the following sentence
Because $s_{N}=\sum_{k=-N}^{N}\|T_{k}x\|^{2}<\infty$ is a convergent sequence,
Right before this sentence, this proved the following statement $\sum_{k\in F}\Vert T_{k}x\Vert ^{2} = \Vert \sum_{k\in F}T_{k}x\Vert ^{2} \le \Vert x\Vert ^{2}$. However, $\Vert x \Vert$ is not always finite, so $\sum_{k\in F}\Vert T_{k}x\Vert ^{2}$ is not always finite. This is what I've understood. Then, why $s_N$ is convergent??
Finally, I want to make this statement more clearly.
Furthermore, by what was shown, $\Vert Tx\Vert \le \Vert x\Vert$
Assuming the all of the previous statements are correct, $\Vert \sum_{k\in F}T_{k}x\Vert ^{2} \le \Vert x\Vert ^{2}$ holds for all $F \subset \mathbb N$. Then, $\Vert \sum_{k = -N}^N T_{k}x\Vert ^{2} \le \Vert x\Vert ^{2}$ holds for all $N \in \mathbb N$. So, $\lim_{N\to \infty} \Vert \sum_{k = -N}^N T_{k}x\Vert \le \Vert x\Vert$. This is what I've understood. Then, how could I extend to the result that we want?
Thank you.
Let $v= \sum_{k\in F}T_{k}x$. if $v=0$ it is trivial that $\|v\|\le\|x\|$. If $v\ne 0$, let $y=\frac{v}{\|v\|}$. Then $\|y\| = 1$ and $$|\langle v, y\rangle| \le \|x\| \|y\| = \|x\|$$ but $$|\langle v, y\rangle| = |\langle v, \frac{v}{\|v\|}\rangle| = \frac{\|v\|^2}{\|v\|} = \|v\|$$ So $\|v\|\le \|x\|$.
Second question - $\|x\|$ is always finite because $x$ is a assumed to be a member of $\mathcal{H}$.
Third question - Let $v_N = \sum_{k=-N}^N T_{k}x$ and $v = \lim_{N\rightarrow\infty} v_N$. For all $\varepsilon > 0$ there exists an $m$ such that $\|v - v_k\| \le \varepsilon$ for all $k > m$. But $\|v_k\| \le \|x\|$ for all $k>m$, so by the triangle inequality $\|v\| \le \|x\| + \varepsilon$. Since $\varepsilon$ was any positive number it follows that $\|v\|\le \|x\|$.