4
$\begingroup$

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.

  • 2
    Lebesgue integral is not defined for many bounded funtions.2012-12-09

1 Answers 1

2

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 $[-1,1]$.