Application of Banach-Steinhaus theorem to judge on continuity. Let $X$ be a Banach space and $T: X \rightarrow l^1$ linear map $T(x) = (T_n(x))_{n=1}^\infty$. Also for all sequences $a_n \in \{-1, 1\}$ funcional:
$$F: X \ni x \rightarrow\sum_{n=1}^\infty a_n T_n(x) \in R$$
is continuous. I want to prove that then $T$ also has to be continuous.
My work so far
The hint I was given is to use $\textbf{Banach Steinhaus}$ theorem i.e.:
Let $X$ be a Banach space and $Y$ normed space. $(T_n)$ is a sequence of linear and continuous maps $X \rightarrow Y$ for which $\sup_n \|T_n(x)\| < \infty \;\;\;\forall_{x \in X}$. Then $\sup_n\|T_n\| <\infty$.
So I was trying to apply this theorem to sequence $a_nT_n$ (which we know is linear and continuous).
$$\|a_nT_n(x)\|_1 = |\sum_{n=1}^\infty a_nT_n| \le \sum_{n=1}^\infty|a_nT_n| = \sum_{n=1}^\infty|T_n|<\infty \;\;\;\;((T_n(x)) \in l^1)$$
So it means that $\sup_n||a_nT_n(x)\|_1 < \infty$ and by Banach Steinhaus theorem we have that $\sup_n\|a_nT_n\| < \infty$.
But now I don't know that I should do to prove linearity of $T$. Could you please give me a hint how to process this proof further?
 A: So you have a Banach space $X$ and an operator $T:X\to\ell^1$, where $Tx=(T_nx)_{n=1}^\infty$, where $T_n:X\to\mathbb{R}$ are linear maps. We also know that, if $a_n\in\{-1,1\}$ for all $n$, then the functional $\phi_{\{a_n\}}:X\to\mathbb{R}$ given by
$$\phi_{\{a_n\}}(x)=\sum_{n=1}^\infty a_n T_n(x)$$
is well-defined and bounded. The goal is to show that $T$ is bounded.
Since $\phi_{\{a_n\}}$ is bounded we have that $$\bigg|\sum_{n=1}^\infty a_nT_n(x)\bigg|\leq C_{\{a_n\}}\cdot\|x\|$$
where $C_{\{a_n\}}=\|\phi_{\{a_n\}}\|>0$. Note that, by the triangle inequality, if $x\in X$, then $$\sup_{\{a_n\}\in\{-1,1\}^\mathbb{N}}|\phi_{\{a_n\}}(x)|\leq\sum_{n=1}^\infty|T_n(x)|=\|Tx\|_{\ell^1}<\infty$$
(Note that we do know that $T:X\to\ell^1$ is a well-defined linear map so the above makes sense).
By applying the principle of uniform boundedness (aka Banach-Steinhaus) we conclude that
$$\sup_{\{a_n\}\in\{-1,1\}^\mathbb{N}}\|\phi_{\{a_n\}}\|=M<\infty$$
and therefore
$$\bigg|\sum_{n=1}^\infty a_n T_n(x)\bigg|\leq M\|x\|$$
is true for all $x\in X$ and all $a_n\in\{-1,1\}$. Fix a $x\in X$ and choose $a_n=\frac{|T_n(x)|}{T_n(x)}$ when $T_n(x)\neq0$ and choose $a_n$ arbitrarily when $T_n(x)=0$. You will obtain
$$\sum_{n=1}^\infty|T_n(x)|\leq M\|x\|$$
and this is true for all $x$; but the left hand side is equal to $\|Tx\|_{\ell^1}$, which is what we wanted to prove.
