Could someone verify whether my solution to the following exercise is correct? The reason I am a bit in doubt is because the chapter of which this Exercise is part consists of the Banach-Steinhaus Theorem and Closed Graph Theorem, so I would expect I have to use those theorems. However, I guess my direct approach works.
Let $X$ be a Banach space and let $T : X \to \ell^{\infty} (\mathbb{N})$ 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.
My solution is the following:
Suppose $T$ is bounded. Then there exists a $M > 0$ such that $\| Tx \|_{\infty} \leq M \| x \|_X$ whenever $x \in X$. Therefore, it holds for all $x \in X$ that $$ |f_n(x)| = |(Tx)_n| \leq \sup_{n \in \mathbb{N}} |(Tx)_n| \leq M \|x\|_X, $$ and hence $f_n$ is bounded.
Suppose $f_n$ is bounded. Then, for each $n \in \mathbb{N}$, there exists an $M_n > 0$ such that $|f_n(x)| \leq M_n \| x \|_X$. Let $$ K := \sup_{n \in \mathbb{N}} \frac{|f_n (x)|}{\|x\|_X} $$ Then, for all $x \in X$, $$ \| Tx\|_{\infty} = \sup_{n \in \mathbb{N}} |f_n (x)| \leq K \| x \|_X ,$$ and hence $T$ is bounded.
The next exercise is the same as the one above, but with $\ell^{\infty}$ replaced with $\ell^{1}$. I guess a similar approach works there?
What you did for the first part is correct.
For the converse, however, there is a problem. You have to prove that the sequence $\left(M_n\right)_{n\geqslant 1}$ is bounded. Your $K$ depends on $X$, and the task is to prove that $K$ is finite.
You can solve the question using the uniform boundedness principle: since $Tx$ belongs to $\ell^\infty$ for each $x\in X$, we have that $\sup_{n\in\mathbb N}\left|f_n(x)\right|\lt+\infty$.