Convergence of the series $\sum_{i=1}^{\infty}{w_iT_i}$ in a Hilbert space

Question

I asked this question before (link: Proving that the sequence of partial sums of an infinite series of operators is a Cauchy sequence) and I am writing it again in order to get some help.
Let $H$ be a Hilbert space and let $(T_i)_{i\in{I}}$ be a countable family of firmly nonexpansive operators from $H$ to $H$. Denote by $T:=\sum_{i=1}^{\infty}{w_iT_i}$, where for every $i\in I$, $w_i\in (0,1)$ and $\sum_{i=1}^{\infty}{w_i}=1$.
Is $T$ a well defined operator? Namely, how can I show that for every $x\in H$, the series $Tx=\sum_{i=1}^{\infty}{w_iT_i(x)}$ converges? (Note: This question is due to the reading of Patrick L. Combettes's paper, "Construction d’un point fixe commun `a une famille de contractions fermes").

Answer 1

Let $B(H)$ denote the normed vector space of bounded linear maps from $H$ to itself and note that it is a Banach space. Let us define the partial sums \begin{equation} S_n = \sum\limits_{k=1}^n w_i T_i. \end{equation} Clearly $S_n \in B(H)$. Then for any $1 \leq n \leq m$ it holds that \begin{align} ||S_m - S_n|| &= ||\sum\limits_{k=n+1}^m w_k T_k || \\ &\leq \sum\limits_{k=n+1}^m |w_k| \end{align} where we used the triangle inequality as well as that the $T_k$ are (firmly) nonexpansive so that $||T_k|| \leq 1$ for all $k$. I see from your previous question that $w_i$ are assumed to be positive. Since the infinite sum \begin{equation} \sum\limits_{k \geq 1} w_k = \sum\limits_{k \geq 1} |w_k| = 1 \end{equation} converges, it implies that its partial sums are Cauchy. Thus, by the above display the sequence of partial sums $(S_k)_{k \geq 1}$ is also Cauchy with respect to the operator norm. But as we noted before $B(H)$ is s Banach space i.e. complete with respect to the operator norm which implies that Cauchy sequences converge.