Let X be a Banach space.Prove that a linear map $M:X \rightarrow C([0,1])$ is continuous iff for every $t\in [0,1]$, the rule $x \mapsto (Mx)(t)$ definies a continuous linear functional on X.
My try: $\ell_t(x) = (Mx)(t)$
$(\Rightarrow)$ $\lim_{n\rightarrow \infty} \ell_t(x_n) = \lim_{n\rightarrow \infty}(Mx_n)(t) = (Mx)(t) = \ell_t(x)$ $(\Leftarrow)$ By closed graph theorem: $\lim_{n\rightarrow \infty} x_n = x$ and $\lim_{n\rightarrow \infty} Mx_n = y$ we want to show that $Mx = y$. $\|Mx- y\| =\sup_t|\ell_t(x) - y(t)| = \sup_t |\lim_{n\rightarrow \infty} \ell_t(x_n) - y(t)| \leq \epsilon $ where the second continuity is because $\ell_t(x)$ is continuous. I'm not at all sure about this and I think I missed something with uniformed boundedness, please correct me.