# Continuous mapping between $X$ and $C([0,1])$

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.

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Is $X$ a Banach space? – saz Dec 28 '12 at 10:58

($\Leftarrow$): It's not that obvious why $\sup (\ldots) \leq \varepsilon$ since you only know that $x \mapsto \ell_t(x)$ is continuous for arbritary (but fixed) $t \in [0,1]$.
You could use the following proof: We have $$\sup_{t \in [0,1]} |\ell_t(x)| = \sup_{t \in [0,1]} |(Mx)(t)| \leq \|Mx\|_{\infty} < \infty$$ for all $x \in X$ (since $Mx \in C[0,1]$) and therefore (by uniform boundedness theorem) we conclude $$M :=\sup_{t \in [0,1]} \|\ell_t\| < \infty$$ Hence $$\|Mx\|_{\infty} = \sup_{t \in [0,1]} |(Mx)(t)| \leq \underbrace{\sup_{t \in [0,1]} \|\ell_t\|}_{M} \cdot \|x\|$$ which means that $M$ is continuous.
Yes, the other one is correct. Probably you could use somehow closed graph theorem to prove $\Leftarrow$, but as you already recognized you need the uniform boundedness. (And if you apply the theorem of uniform boundedness, you are already done as you can see above.) – saz Dec 28 '12 at 15:50