Equivalent statements about unconditionally convergent series Problem
I want to show the following equivalence:

Problem. Let $T\subseteq\mathbb R$ be a bounded set with at least two elements. A series $\sum_k a_k$ in a banach space $E$ is perfectly convergent (see definition below) iff $\sum_k \theta_k a_k$ converges for every sequence $(\theta_k)$ in $T$.

Here, perfectly convergent means:

Definition. A series $\sum_k a_k$ in a banach space $E$ is perfectly convergent iff for every sequence $(\alpha_k)$ with $\alpha_k\in\{-1,1\}$ for $k\in\mathbb N$ the series $\sum_k \alpha_k a_k$ converges.

So my problem is to show that perfect convergence with sequences in $\{-1,1\}$ can be generalized to sequences in an arbitrary bounded set $T$ with $|T|\geq 2$.
What I already know
Perfect convergence is an important property for series, because it is equivalent to unconditional convergence:

Definition. A series $\sum_k a_k$ in a banach space $E$ is said to be unconditionally convergent iff every rearrangement converges, i.e. for each permutation $\pi\colon\mathbb N\rightarrow\mathbb N$  the series $\sum_k a_{\pi(k)}$ converges.

One can also show, that a series $\sum_k a_k$ in $E$ is perfectly convergent iff every subseries converges, i.e. $\sum_k a_{\sigma(k)}$ converges for every strictly increasing mapping $\sigma\colon\mathbb N\rightarrow\mathbb N$.
All equivalences are shown in "Kadets, V. (2011, November 11). Series in Banach Spaces: Conditional and Unconditional Convergence (Operator Theory: Advances and Applications, 94) (1997th ed.). Birkhäuser."
Except for the first equivalence, it is listed as an exercise in the book.
My approach
I list some ideas for both directions of the proof:

*

*$\theta_k$ is bounded, hence there is a convergent subsequence with limit in the closure of $T$. But I don't know how to implement this fact in my proof.

*We can assume that $0\notin T$ without loss of generality, because otherwise we could reduce to subseries. Hence $\sum_k \frac{\theta_k}{|\theta_k|}a_k$ convergences if we assume perfect convergence. But I can't deduce the convergence of $\sum_k \theta_k a_k$ from it, although I know that $|\theta_k|\leq \theta$ for some $\theta>0$.

*Perhaps I can use methods from functional analysis, because the series $(\theta_k)$ lies in $\ell^\infty$. But I haven't found anything in the literature on this yet.

 A: We want to show that the following two statements are equivalent:
$\def\abajo{\\[0.3cm]}$ $\def\x\{E}$ $\def\Re{\operatorname{Re}}$ $\def\ds{\displaystyle}$ $\def\NN{\mathbb N}$

*

*$\ds\sum_k\theta_kx_k$ converges for all $\theta\in\{-1,1\}^\NN$;


*$\ds\sum_k\theta_kx_k$ converges for all $\theta\in T^\NN$
1$\implies$2. The proof relies on the following equality: given $x_1,\ldots,x_n\in\x$,
\begin{equation}\tag1
\max\Big\{\Big\|\sum_{k=1}^n \theta_kx_k\Big\|:\ \theta_k\in[-1,1]\Big\}
=\max\Big\{\Big\|\sum_{k=1}^n \theta_kx_k\Big\|:\ \theta_k=\pm1\Big\}
\end{equation}
(one would initially write $\sup$ instead of $\max$, but it is easy to check that all  sets are continuous images of compact domains; in any case, the proof does not change if one writes sup instead of max).
Having equality $(1)$, we get that $\sum_k\theta_kx_k$ converges for all $\theta\in\ell^\infty(\NN)$ (since any $\theta\in\ell^\infty(\NN)$ can be rescaled to the unit disk, and real and imaginary parts can be considered separately).
2$\implies$1. Let $t_1,t_2\in T$, with $t_1\ne t_2$. Let $\{n_k\}\subset\NN$ be a sequence. Write $G=\{n_k\}$ and let
$$
\theta_k=\begin{cases} t_1,&\ k\in \NN\setminus G\abajo t_2,&\ k\in G\end{cases}
$$By hypothesis, the series
$$
\sum_k t_1x_k\qquad\text{ and } \qquad \sum_k \theta_k\,x_k
$$
converge. Then their difference also converges, and this is
$$
\sum_kt_1x_k-\sum_k\theta_kx_k=\sum_{k\in G}(t_1-t_2)x_k=(t_1-t_2)\sum_k x_{n_k}.
$$
This shows that $\sum_kx_{n_k}$ converges for any increasing sequence $\{n_k\}$, and hence the series converges unconditionally.

Proof of  $(1)$. One proves
\begin{aligned}
\max\Big\{\Big\|\sum_{k=1}^n \theta_kx_k\Big\|:\ \theta_k\in[-1,1]\Big\}
&=\max\Big\{\sum_{k=1}^n|\Re g(x_k)|:\ g\in\x^*,\ \|g\|=1\Big\}\abajo
&=\max\Big\{\Big\|\sum_{k=1}^n \theta_kx_k\Big\|:\ \theta_k=\pm1\Big\}
\end{aligned}
Let $g\in\x^*$ with $\|g\|=1$ and $\ds g\Big(\sum_{k=1}^n\theta_kx_k\Big)=\Big\|\sum_{k=1}^n\theta_kx_k\Big\|$ (this is an easy consequence of Hahn-Banach). Then
\begin{align*}
\Big\|\sum_{k=1}^n\theta_kx_k\Big\|
&=g\Big(\sum_{k=1}^n\theta_kx_k\Big)=\Re g\Big(\sum_{k=1}^n\theta_kx_k\Big)=\sum_{k=1}^n\theta_k \Re g(x_k)\abajo
&\leq\sum_{k=1}^n|\theta_k|\,|\Re g(x_k)|\leq\sum_{k=1}^n|\Re g(x_k)|.
\end{align*}
Conversely, given $g\in\x^*$ with $\|g\|=1$ choose $\theta_k\in\{1,-1\}$ so that $\theta_k \Re g(x_k)=|\Re g(x_k)|$. Then
$$
\sum_{k=1}^n |\Re g(x_k)|=\sum_{k=1}^n \theta_k\Re g(x_k)=\Re g\Big(\sum_{k=1}^n \theta_kx_k\Big)\leq\Big\|\sum_{k=1}^n \theta_kx_k\Big\|, 
$$
completing the proof of the first line. The equality with the last line follows easily from the fact that in the proof of the first line we chose all $\theta_k$ with $\theta_k=\pm1$.
