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Let $\Phi: C[0,1]\longrightarrow\mathbb{R}$ the linear functional defined by $\Phi(f)=\int_0^1 f(x)dx$. Let $\tilde{\Phi}$ an extension of $\Phi$ to the normed space $(B[0,1]$ (of bounded functions on $[0,1]$, with the $\sup$ norm) such that $\|\Phi\|=\|\tilde{\Phi}\|$. Such an extension is guaranteed by the Hahn-Banach theorem. Let $h(x)$ be as follows: $h(x)=1$ if $x\leq 1/2$ and $h(x)=-1$ if $x>1/2$.

How to calculate $\tilde{\Phi}(h)$? My difficulty is that it is impossible to approximate $h$ uniformly by continuous function.

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It doesn't make sense to calculate $\tilde\Phi$ until you've chosen a specific extension of $\Phi$, of which there are infinitely many. So which extension are you using? –  Chris Eagle Dec 9 '12 at 1:03
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That's the thing about Axiom of Choice. you get the existence of a whole lot of items, but not a recipe to compute any one of them. –  GEdgar Dec 9 '12 at 1:04
    
@GEdgar But, the lebesgue integral would be one such extension, right? –  Ravi Donepudi Dec 9 '12 at 1:10
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Lebesgue integral is not defined for many bounded funtions. –  GEdgar Dec 9 '12 at 1:14
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up vote 2 down vote accepted

For your particular example.

Consider the extension obtained by adding just one more function. Namely adding your function $h$ to $C[0,1]$. If you are reading the correct proof of the HB theorem, it shows what the possible values of the extension are. There is a certain interval, and we may choose any number in that interval as the value.

Here, the interval we compute is $[-1,1]$.

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