Given a linear operator $T$ and a linear functional $\phi_n(x)=(T(x))(n)$, show that $T$ is continuous iff $\phi \in X^*$

Question

Given $X$ a Banach space, and $T:X \rightarrow l_p$ a linear operator, with $1 \leq p \leq \infty$, for all $n \in \mathbb{N}$ consider the linear functional $\phi_n:X \rightarrow \mathbb{K}$, defined by:

$\phi_n(x)=(T(x))(n)\,$ $\,\forall \, x \in X$.

Show that $T$ is continuous if and only if $\phi_n(x) \, \in \, X^*$, for all $n \in \mathbb {N}$.

$\mathbb{K}=\mathbb{R},\mathbb{C}$. Any suggestion?
There are different ways to see the continuity of a linear operator (for example, the continuity in zero, the acotation in the unit circle,...), right? But how can I get this "iff" with the given functional?

Answer 1

I think the 'if' part matters to you. To prove that $T$ is continuous, show that $$ x_k \to x,\;Tx_k \to y \Rightarrow Tx= y. $$ Then by closed graph theorem, we know that $T$ is continuous.

Now, since $Tx_k \to y$, we have $$ Tx_k(n) \to y(n) $$ for all $n$. Also by the assumption that $\phi_n$ is continuous, we also have $\phi_n(x_k) =Tx_k(n) \to \phi_n(x) = Tx(n)$ for each $n$. This establishes $Tx(n) = y(n)$ for all $n$, and therefore the claim follows.