Exercise 23 from chapter 4 (“Hilbert Spaces: An Introduction”) of Stein & Shakarchi's “Real Analysis”

Question

Consider exercise 23 from chapter 4 ("Hilbert Spaces: An Introduction") of [1] (p. 198). Any help will be much appreciated. Thank you in advance.

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.]

Note: I realize why - as the hint suggests - the ranges of the $T_k$'s are mutually orthogonal. Indeed, let $k, j \in \mathbb{Z}$ be distinct, and let $h_k, h_j \in \mathcal{H}$. Then $$ \left<T_j h_j, T_k h_k\right> = \left<T_k^* T_j h_j, h_k\right> = \left<0, h_k\right> = 0 $$

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References

[1] Stein, Elias M. and Shakarchi, Rami. Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press (2005)

Answer 1

Let $P_{k}$ be the orthogonal projection onto the closure of the range of $T_{k}$. Then $P_{k}P_{k'}=P_{k'}P_{k}=0$ for $k\ne k'$ because $(T_{k}x,T_{k'}y)=(T_{k'}^{\star}T_{k}x,y)=0$ for all $x,y \in H$. Similarly, if $Q_{k}$ is the orthogonal projection onto the closure of the range of $T_{k}^{\star}$, then $Q_{k}Q_{k'}=0$ for $k\ne k'$. Furthermore, $$ (T_{k}x,y)=(T_{k}x,P_{k}y)=(x,T_{k}^{\star}P_{k}y)=(Q_{k}x,T_{k}^{\star}P_{k}y)= (T_{k}Q_{k}x,P_{k}y). $$ If $F$ is any finite subset of the integers $\mathbb{Z}$, then $\|T\| \le 1$, the Cauchy-Schwarz inequality, and Bessel's inequality together yield $$ \begin{align} |\sum_{k\in F}(T_{k}x,y)| & \le \sum_{k\in F}|(T_{k}Q_{k}x,P_{k}y)| \\ & \le \sum_{k\in F}\|T_{k}Q_{k}x\|\|P_{k}y\| \\ & \le \sum_{k\in F}\|Q_{k}x\|\|P_{k}y\| \\ & \le \left(\sum_{k\in F}\|Q_{k}x\|^{2}\right)^{1/2}\left(\sum_{k\in F}\|P_{k}y\|^{2}\right)^{1/2} \\ & \le \|x\|\|y\|. \end{align} $$ It follows that, regardless of the finite subset $F$, one has the inequality $$ \|\sum_{k\in F}T_{k}x\| \le \|x\|,\;\;\; x \in \mathcal{H}. $$ Because $\{ T_{k}x\}_{k\in\mathbb{Z}}$ is a mutually orthogonal set of vectors, then $$ \sum_{k\in F}\|T_{k}x\|^{2} = \|\sum_{k\in F}T_{k}x\|^{2} \le \|x\|^{2}. $$ Therefore, $\sum_{k\in\mathbb{Z}}\|T_{k}x\|^{2} \le \|x\|^{2}$. Because $s_{N}=\sum_{k=-N}^{N}\|T_{k}x\|^{2}<\infty$ is a convergent sequence, then it is a Cauchy sequence. So, $\sum_{k=-N}^{N}T_{k}x$ is a Cauchy sequence in $\mathcal{H}$ because, for $0 < N < M$, $$ \begin{align} \|\sum_{k=-M}^{M}T_{k}x-\sum_{k=-N}^{N}T_{k}x\|^{2}& =\sum_{N < |k| \le M}\|T_{k}x\|^{2} \\ & =\sum_{k=-M}^{M}\|T_{k}x\|^{2}-\sum_{k=-N}^{N}\|T_{k}x\|^{2}. \end{align} $$ Knowing that $\mathcal{H}$ is a Hilbert space allows one to conclude that $$ Tx = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}T_{k}x $$ exists for all $x\in \mathcal{H}$. Furthermore, by what was shown, $\|Tx\|\le \|x\|$ because the same is true for all finite sums $\sum_{k=-N}^{N}T_{k}x$.

Answer 2

First observation: For all $u,v\in\mathcal H$ and $j\ne k$ $$ \langle T_ju,T_k v\rangle=\langle T_j^*u,T_k^* v\rangle=0. $$ Thus the linear subspaces $T_j^*\mathcal H$, $j\in\mathbb Z$, are perpendicular to each other, and so are their closures. Set $$ Y=\bigoplus_{j\in\mathbb Z}\overline{T_j^*\mathcal H}. $$ This infinite direct sum contains elements of the form $t=\sum_{j\in\mathbb Z}t_j$, where $\sum_{j\in\mathbb Z}\|t_j\|^2<\infty$, and hence $Y$ is a closed subspace of $\mathcal H$.

Then $\mathcal H=Y\oplus Y^\perp$, and thus every $h\in \mathcal H$ can be writen (uniquely) as $$ h=z+\sum_{j\in\mathbb Z}t_j, \quad z\in Y^\perp,\,t_j\in\overline{T_j^*\mathcal H},\qquad(\star) $$ and clearly $$ \|h\|^2=\|z\|^2+\sum_{j\in\mathcal Z}\|t_j\|^2. \qquad (\star\star) $$ Now, if $z\in Y^\perp$, then $\langle z,T_j^*u\rangle=0$, for all $u\in\mathcal H$, and thus $\langle T_jz,u\rangle=0$, for all $u\in\mathcal H$, which means that $T_jz=0$. Also, if $t_j\in \overline{T_j^*\mathcal H}$, then clearly $T_kt_j=0$, for all $k\ne j$.

Hence, using representation $(\star)$ we have $$ S_Nh=\sum_{|j|\le N}T_j\Big(z+\sum_{k\in\mathbb Z}t_k\Big)=\sum_{|j|\le N}T_jt_j. $$ In particular, $$ \|S_Mh-S_Nh\|^2=\sum_{N<|j|\le M}\|T_jt_j\|^2\le \sum_{N<|j|\le M}\|t_j\|^2\to 0, $$ as $M,N\to\infty$, since the series $(\star\star)$ converges. Thus $$ Th=\lim_{N\to\infty}S_Nh, $$ is well-defined, and $$ \|Th\|^2=\sum_{j\in\mathbb Z} \|T_jt_j\|^2\le \sum_{j\in\mathbb Z} \|t_j\|^2\le \|h\|^2, $$ and hence $\|T\|\le 1$.