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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

2 Answers 2

up vote 7 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 is enough to find a linear functional which is not norm attaining. We can define $$x'(f):=\int_0^{1/2}f(t)\mathrm dt-\int_{1/2}^1f(t)\mathrm dt;$$ $x'$ is linear and $|x'(f)|\leqslant \lVert f\rVert_{\infty}$, hence $x'$ is continuous and its norm is $\leqslant 1$.

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 cannot find $f\in E$ such that $x'(f)=1$ and $\lVert f\rVert=1$. Let $f$ be continuous of norm $1$. We have to show that $x'(f)\neq 1$, for a fixed $\varepsilon>0$ we can find $\delta>0$ such that $|f(t)-f(1/2)|\leqslant \varepsilon$ whenever $|t-1/2|\leqslant \delta$. Since $$\tag{*} x'(f)=\int_0^{1/2-\delta}f(t)\mathrm dt+\int_{1/2-\delta}^{1/2+\delta}f(t)\mathrm dt -\int_{1/2+\delta}^1f(t)\mathrm dt,$$ we can assume that $f(x)=1$ on $[0,1/2-\delta]$ and $f(x)=-1$ on $[1/2+\delta,1]$, otherwise it is clear that $x'(f)\neq 1$.

By (*), it follows that $|x'(f)|\leqslant 1-2\delta+2\delta\varepsilon+\delta| f(1/2)|$. Taking $\varepsilon\lt 1-|f(1/2)|$, we get that $|x'(f)|\lt 1$.

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Why does $2 \delta \varepsilon$ appear? I obtained $$ \underset{1/2 - \delta}{\overset{1/2}{\int}} f(t) dt - \underset{1/2}{\overset{1/2 + \delta}{\int}} f(t) dt \leq 2 \delta f(1/2) - \frac{1}{2} f(1/2) + \varepsilon / 2 $$ Thank you! –  g.pomegranate Mar 25 at 19:14
    
@rodie You are right. Thank you for spotting an inaccuracy. I have edited. –  Davide Giraudo Mar 25 at 20:23
    
But why does $2 \delta \varepsilon$ appear? I obtained $$x'(f) \leq 2 \delta (f(\frac{1}{2}) + \varepsilon) + 1 - 2 \delta.$$ Is this because $f(\frac{1}{2}) < \infty$ and then we can consider $\varepsilon : = f(\frac{1}{2}) + \varepsilon $? Thak you! –  g.pomegranate Mar 27 at 11:46
    
You are right, I have edited again. –  Davide Giraudo Mar 27 at 12:11

you could also use the following facts:

  • $BV[a,b]$ is not separable,
  • $C[a,b]$ is separable,
  • if $X$ is a Banach Space and $X^*$ is its dual and is separable, then $X$ is separable.

Since $BV[a,b]$ is the dual of $C[a,b]$, reflexivity would imply $BV[a,b]$ is separable.

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