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I just did one exercise stating: Prove that the linear map $M: X \rightarrow C([0,1])$, is continuous iff for every $t\in[0,1]$, the rule $x\rightarrow (Mx)(t)$ defines a continuous linear functional on X. the next exercise stated: State, and prove a similar continuity criterion for linear maps $M:X\rightarrow Y$ where Y is an arbitrary Banach space.

Is there some theorem which states that $M$ is continuous iff $x\mapsto \ell(Mx)$ is continous for all linear functionals $\ell:Y\rightarrow \mathbb{K}$ in $Y'$? or what does it mean?

I posted a new try to a proof, can someone please confirm it or post another one?

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up vote 3 down vote accepted

Your answer seems to be circling around the right ideas, but I'm not sure if you are using them in the right way.

Fix $x\in X$. Consider the map $T_x:Y'\to\mathbb K$, given by $T_x(\ell)=\ell(Mx)$. Note that $$ \|T_x\|=\sup\{|T_x(\ell)|:\ \|\ell\|=1\}=\sup\{|\ell(Mx)|:\ \|\ell\|=1\}=\|Mx\| $$ where the last equality is a classical application of the Hahn-Banach Theorem.

Now apply the Uniform Boundedness Theorem over the unit ball of $X$ (so we need $X$ to be Banach): for a fixed $\ell$, $$ \sup\{|T_x(\ell)|:\ \|x\|\leq1\}=\sup\{|\ell(Mx)|:\ \|x\|\leq1\}=\|\ell\circ M\|<\infty. $$ By the UBP, there exists $c>0$ such that $\|T_x\|\leq c$ for all $x$ with $\|x\|\leq1$. So $$ \|Mx\|=\|T_x\|\leq c,\ \text{ if }\|x\|\leq1. $$ So $M$ is bounded in the the unit ball, and is thus continuous.

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Great explanation, thanks – user25470 Jan 2 '13 at 21:07

Operators $\ell:\: X \rightarrow Y$ are not linear functionals. Think about how the linear functionals where defined in the example of $Y = C([0,1])$ as a composition of something with $M$ and try to generalize that. Use the representation of the norm in $X$ via linear functionals, and the uniform boundedness principle to prove the statement.

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thanks, It is changed now – Johan Dec 10 '12 at 9:22
Your interpretation is right if you replace $\ell(Mx)$ by $x \mapsto \ell(Mx)$ or simply $\ell \circ M$. – Thomas Dec 10 '12 at 9:24

The map $x\mapsto \hat{\ell}(Mx)$ is a linear map $\ell \colon X \mapsto \mathbb{K}$. By the continuity of $\hat{\ell}$ and uniformed boundedness we have $\sup_{\ell} \|\ell\| = c$.

\begin{equation} \begin{split} \|M\| =& \sup_{\|x\| = 1} \|Mx\| = \sup_{\|\hat{\ell}\| \leq 1}\sup_{\|x\| = 1} \|\hat{\ell}(Mx)\| \leq \sup_{\hat{\ell}} \sup_{\|x\| = 1} \|\hat{\ell}(Mx)\| \\ =& \sup_{\ell , \|x\| = 1} \|\ell x\| \leq \|\ell\|= c \end{split} \end{equation} Is this correct?

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