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How does one prove that $C^0([0;1],\mathbb{R})$ equipped with the sup norm is not reflexive?

I don't understand how to show that the $J$ mapping is not surjective.

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You are missing "not" in your first sentence? Do you know what the dual of $C^0([0,1],\mathbb R)$ is? –  Jonas Meyer Jun 3 '12 at 22:17
David - I am not too sure of the space that has been given in the is. I know that C^infinity([0:1],R) is the set of infinitely differentiable functions but I couldn't find the definition on the internet. Also, what is the norm that the linear functional x' is equipped with? –  nada Jun 3 '12 at 22:39
Okay, yes, I get the solution now. Thank you. :) –  nada Jun 3 '12 at 22:56

1 Answer 1

up vote 5 down vote accepted

As we have seen in a previous question, if $E$ is a reflexive Banach space then each linear continuous functional attains its norm. So, in order to show that $E:=C^0([0,1],\Bbb R)$ endowed with the supremum norm is not reflexive, it's enough to find a linear functional which is not norm attaining. We can define $$x'(f):=\int_0^{1/2}f(t)dt-\int_{1/2}^1f(t)dt;$$ $x'$ is linear and $|x'(f)|\leq \lVert f\rVert_{\infty}$, hence $x'$ is continuous and its norm is $\leq 1$.

To see that the norm is indeed $1$, for $n$ integer, consider a function which is $1$ on $[0,1/2-1/n)$ and $(1/2+1/n,1)$, and linear on $(1/2-1/n,1/2+1/n)$. We can see that $\lVert f_n\rVert=1$ and $x'(f_n)=1-2/n$.

Now, we have to show that we can't find $f\in E$ such that $x'(f)=1$ and $\lVert f\rVert=1$. If $f$ is continuous of norm $1$, for a fixed $\varepsilon>0$ we can find $\delta>0$ such that $|f(t)-f(1/2)|\leq \varepsilon$ whenever $|t-1/2|\leq \delta$. We have $$x'(f)=\int_0^{1/2-\delta}f(t)dt+2\delta\varepsilon-\int_{1/2+\delta}^1f(t)dt$$ hence $|x'(f)|\leq 1-2\delta+2\delta\varepsilon=1-2\delta(1-\varepsilon)$. Taking $\varepsilon=1/2$, we get that $|x'(f)|\leq 1-\delta<1$.

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