Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Is $X$ a Banach space? – saz Dec 28 '12 at 10:58
up vote 1 down vote accepted

($\Leftarrow$): 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.

share|cite|improve this answer
thanks! was the other direction correct? So we can not use closed graph here? – Johan Dec 28 '12 at 13:17
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.