Unfortunately I do not succeed in completing the following exercise:
Let $X$ be a Banach space and let $T : X \to \ell^{1} (\mathbb{N})$ be a linear operator. For each $n \in \mathbb{N}$, let $(Tx)_n$ denote the $n^{th}$ entry of $Tx$ and let $f_n$ be a linear functional on $X$ given by $x \mapsto (Tx)_n$. Prove that $T$ is bounded if and only if each $f_n$ is bounded.
I have shown that each $f_n$ is bounded whenever $T$ is bounded. However, I have difficulties in showing the converse. My attempt until now is the following:
Suppose that each $f_n$ is bounded. Since $$\sum_{n \in \mathbb{N}} |f_n (x)| < \infty$$ for all $x \in X$, it follows that $\| f_n (x) \|_{\infty} < \infty$, and hence $\sup_{n \in \mathbb{N}} \|f_n \|_{op} < \infty$ by the principle of uniform boundedness.
However, I am not sure how to proceed from here.
Any help is appreciated. Thanks in advance.
Suppose each $f_n$ is bounded, i.e. for each $n$, we have $\|f_n\|=\sup_{\|x\|=1}|f_n(x)|<\infty$. Define $T_n:X\to \ell^1(\mathbb N)$ by $x\mapsto \sum_{i=1}^n f_i(x)e_i$, where $\{e_i\}$ is the canonical basis for $\ell^1(\mathbb N)$. If $\|x\|=1$ then $$\sup_n\|T_n x\|=\sum_{i=1}^\infty|f_i(x)|<\infty, $$ so by Banach-Steinhaus we have $$\sup_n\sup_{\|x\|=1}\|T_n x\|=\sup_{\|x\|=1}\sum_{i=1}^\infty |(Tx)_n|=\|T\|<\infty. $$