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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. –  joriki Nov 13 '12 at 10:39
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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. –  Christopher A. Wong Nov 13 '12 at 10:57
    
@ChristopherA.Wong what is the proper notation for the continuous dual space? –  Jimmy R Nov 13 '12 at 11:00
    
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. –  Christopher A. Wong Nov 13 '12 at 11:05

3 Answers 3

The problem is that you would have to check the condition for all $f \in C[0,1]$ with $\|f\|_2 < \delta$, not just $f=0$.

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For a linear functional continuity at a point implies continuity and vice versa. –  Jimmy R Nov 13 '12 at 10:48
    
Yes, but he didn't check continuity at the single point 0, because he only showed that one point of his $\delta$-neighbourhood is mapped to the $\epsilon$-neighbourhood of $f(0)$, which is trivial. –  Thomas Nov 13 '12 at 11:23

Consider $g_n$ such that $f_n(0)=1$, $g_n(n^{-1})=0$, $g_n(1)=0$ and $f_n$ is piecewise linear. Then take $f_n:=\sqrt{g_n}$^to see that $L$ is norm continuous if we endow $C[0,1]$ with the $L^2$ norm.

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I didn't get this, isn't $L^2$ a norm for sequences? –  Jimmy R Nov 13 '12 at 10:55
    
I meant the norm on the space of continuous functions given by $||f||^2=\int_0^1f(t)^2dt$. –  Davide Giraudo Nov 13 '12 at 10:57
    
It's exactly what I wrote, isn't it? –  Davide Giraudo Nov 13 '12 at 11:03

Let $f_n(x):=(1-n x)^+:=\max\{1-nx,0\}$. These $f_n$ are continuous, and $L(f_n)=1$ for all $n\geq 1$. From $$\|f_n\|_2^2=\int_0^{1/n}(1-2n x+ n^2 x^2)\ dx={1\over 3n}$$ it follows that $${L(f_n)\over \|f_n\|_2}=\sqrt{3n}\to\infty\qquad(n\to\infty)\ .$$ This shows that there is no $C>0$ with $$|L(f)|\leq C\|f\|_2\qquad \forall f\in C\bigl([0,1]\bigr)\ .$$ Therefore $L$ is not continuous with respect to the $L^2$-norm on $C\bigl([0,1]\bigr)$.

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When finding the norm, why do you integrate with $1/n$ as upper bound? –  Jimmy R Nov 13 '12 at 11:10
    
@Jimmy R: By definition $a^+$ is $=a$ if $a\geq0$ and $=0$ if $a<0$. –  Christian Blatter Nov 13 '12 at 11:35

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