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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?

2 Answers 2

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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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    Try again: f_\varepsilon(x) = \begin{cases} 1 & 0 \leq x \leq 1/2 - \varepsilon \\ \text{linear} & \text{on }1/2 - \varepsilon \lt x \lt 1/2+\varepsilon \\ -1 & 1/2 + \varepsilon \leq x \leq 1.\end{cases} You should get $\varphi(f_\varepsilon) = 1 - \varepsilon$ while $\lVert f_\varepsilon \rVert = 1$, now let $\varepsilon \to 0$. Yes, no reflexivity needed: see [here](http://math.stackexchange.com/questions/129751/) and [here](http://math.stackexchange.com/q/80773/).2012-11-29
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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.