Exercise 7 page 93 Functional Analysis book of Conway The following is Exercise 7 page 93 in Functional Analysis book of Conway:

Let $1 \le p \le \infty$ and suppose $(a_{ij})$ is a matrix such that $(Af)(i) = \sum_{j=1}^{\infty} a_{ij} f(j)$ defines an element $Af$ of $\ell^p$ for every $f$ in $\ell^p$. Show that $A \in \mathcal{B}(\ell^p)$.

Case $p=\infty$: If $f(j)$ is such that $||f(j)||_{\infty} < \infty$ and $\sum_{j=1}^{\infty} a_{ij} f(j) < \infty $ how to show that $(Af)(i)$ is bounded so that A maps $\ell^{\infty}$ to $\ell^{\infty}$?
Case $1 \le p < \infty$ : For this case same approach I think but this time we use Holder's inequality I suppose?
NOTE : I looking for a solution based on the Uniform Boundedness Principle, a solution without using the Closed Graph Theorem. The answer in here has just one sentence mentioning that the result could be proved using Baire Category Theorem, but such mention is not really an answer. The answer itself is based on applying the Closed Graph Theorem.
 A: For $p \in [1, +\infty]$, let $q \in [1, +\infty]$, such that $\frac1p +\frac1q =1$, (with the usual convention that $\frac1\infty=0$).
It is immediate that $A$ is linear. So it remains to be proved that $A$ is bounded.
For all $i$ and $k$ positive integers, let us define $\phi_{i k}$ on $\ell^p$ by, for all $f \in \ell^p$,
$$ \phi_{i k}(f) = \sum_{j=1}^k a_{ij}f(j) $$
It is clear that $\phi_{i k}$ is linear. Note that, for each $i$ and $k$, we have that $(a_{ij})_{j=1}^k \in \ell^q$. So $\phi_{i k}$ is bounded linear and $\| \phi_{i k} \| = \| (a_{ij})_{j=1}^k \|_q$ (see Remark).
Now, note that, for all $f \in \ell^p$, $\sup_{i, k} | \phi_{i k}(f) | < \infty$. So, by the Uniform Boundness Principle,
$$\sup_{i, k} \| \phi_{i k} \| < \infty$$
that is,
$\sup_{i, k} \| (a_{ij})_{j=1}^k \|_q < \infty$. Let $M =\sup_{i, k} \| (a_{ij})_{j=1}^k \|_q < \infty$.
It follows that
$$ \sup_{i} \| (a_{ij})_{j=1}^\infty \|_q \leq M < \infty$$
Now,, for each $i$ positive integer, let us define $\phi_i$ on $\ell^p$ by, for all $f \in \ell^p$,
$$ \phi_i(f) = \sum_{j=1}^\infty a_{ij}f(j) $$
It is clear that $\phi_i$ is linear and, for all $f \in \ell^p$,
$$ |\phi_i(f)| \leq \| (a_{ij})_{j=1}^\infty \|_q \|f\|_p \leq M \|f\|_p $$
So $\phi_i$ is bounded.
For each $i$  positive integer, let $e_i \in \ell^p$ be such that $e_i = (e_{i,j})_{j=1}^\infty$ and $e_{i,i} = 1$ and $e_{i,j}=0$ if $i \ne j$.
Now, for each $r$ positive integer, let us define $\psi_r$ on $\ell^p$ by, for all $f \in \ell^p$,
$$ \psi_r(f) = \sum_{i=1}^r \phi_i(f) e_i \in  \ell^p $$
It is clear that $\psi_r$ is linear and bounded. Moreover,
$$ \sup_r \|\psi_r(f) \|_p \leq \|Af\|_p < \infty$$
So, by the Uniform Boundness Principle, we have that
$$ \sup_r \|\psi_r \|_{\mathcal{B}(\ell^p, \ell^p)} < \infty$$
Let $K = \sup_r \|\psi_r \|_{\mathcal{B}(\ell^p, \ell^p)} < \infty$.
So, we have, for all $r$ and all $f \in \ell^p$,
$$\|\psi_r(f)\|_p \leq K \|f\|_p \tag{1} $$
Note that, for all $f \in \ell^p$, $\psi_r(f)$ converges to $Af$ in $\ell^p$, so we have that   $\|\psi_r(f)\|_p$ converges to $\|Af\|_p$. Form $(1)$, we have that
$$ \|Af\|_p \leq K \|f\|_p $$
So, $A$ is bounded.
Remark:  Since $(a_{ij})_{j=1}^k \in \ell^q$  and, for all $f \in \ell^p$,
$ \phi_{i k}(f) = \sum_{j=1}^k a_{ij}f(j) $, we have that $\| \phi_{i k} \| = \| (a_{ij})_{j=1}^k \|_q$. It is a standard result on the duality of $\ell^p$ and $\ell^q$.
