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Can someone help me to construct a linear functional in $\mathcal{C}([0,1])$ that does not attain its norm?

Actually, I want to prove that $\mathcal{C}([0,1])$ is not reflexive Banach space. Is it sufficient to construct that kind of functional?

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2 Answers 2

Try $\varphi(f) = \int_{0}^{1/2} f(x)\,dx - \int_{1/2}^{1} f(x)\,dx$.

As for the second question, yes. For every $x \in X$ there is by Hahn-Banach a functional $\varphi \in X^{\ast}$ with $\|\varphi\| = 1$ such that $\varphi(x) = \|x\|$. Now apply this to $X^{\ast}$ and use reflexivity.

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Sir , I am trying to see why $||\phi||=1$ . Can you give me some hints . I would like to work it out. –  Theorem Nov 27 '12 at 20:55
    
$\leq$ is trivial, for $\geq$ take $f$ piecewise linear, constant equal to $1$ on $[0,1/2 - \varepsilon]$ and constant equal to $-1$ on $[1/2 +\varepsilon, 1]$. –  t.b. Nov 29 '12 at 14:19
    
i don't get it why $\le$ is trivial, can u give me more details . –  Theorem Nov 29 '12 at 14:39
    
triangle inequality + $\lvert \int f \rvert \leq \int \rvert f\lvert$. –  t.b. Nov 29 '12 at 14:42
    
yes , got it . But are you sure for $\ge$ the function u gave will work ? –  Theorem Nov 29 '12 at 17:25

Knowing some deeper theorems will help to see what to do.

The Riesz representation theorem says that the continuous linear functionals on $C([0,1])$ are precisely the signed Radon measures on $[0,1]$. It's not hard to see that any bounded measurable function is a continuous linear functional on the space of signed measures, i.e. is an element of $C([0,1])^{**}$. So the existence of bounded discontinuous functions on $[0,1]$ shows that $C([0,1])$ cannot be reflexive.

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