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.

  • $\begingroup$ 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? $\endgroup$ – Chris Eagle Dec 9 '12 at 1:03
  • 1
    $\begingroup$ 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. $\endgroup$ – GEdgar Dec 9 '12 at 1:04
  • $\begingroup$ @GEdgar But, the lebesgue integral would be one such extension, right? $\endgroup$ – RKD Dec 9 '12 at 1:10
  • 2
    $\begingroup$ Lebesgue integral is not defined for many bounded funtions. $\endgroup$ – GEdgar Dec 9 '12 at 1:14

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 $[-1,1]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.