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So I have the following statement to prove

Let $L:C\:[0,1]\to \mathbb{C}$ be a linear functional defined by $$ Lf=f(0)$$ Show that $L\notin(C[0,1],||\cdot||_2)^*$, where $||\cdot||_2$ is the usual 2-norm.

This functional is definitely linear, so I guess I need to show that it is not continuous, I understand that if I show that it is not continuous at one point, then it's not continuous anywhere.

But suppose I take $f=0\in C\:[0,1]$, then if I get $\varepsilon>0$ by taking $\delta=\epsilon$ I obtain $$ ||0||_2<\delta \: \Rightarrow\: |0|<\epsilon $$ So it seems that $L$ is in this dual space, where is my reasoning wrong and what would be the correct way of proving this statement?

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    Problems are solved, statements are proved.2012-11-13
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    Just a comment about terminology: you're trying to show that $L$ isn't in the *continuous* dual space. $L$ *does*, however, belong to the algebraic dual space, which is just the set of all (possibly unbounded) linear functionals.2012-11-13
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    @ChristopherA.Wong what is the proper notation for the continuous dual space?2012-11-13
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    The notation you used is fine. Almost all textbooks on analysis concern themselves only with the continuous dual space, for obvious reasons. I just was pointing out something that you might want to specify in the future.2012-11-13

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