Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

With results: "For convex subsets of a locally convex space, a, originally( strongly) closed equals weakly closed, and b, originally (strongly dense equals weakly dense." Could you help me solve this problem?

$L^\infty[0,1]$ has its norm topology coming from the essential supremum norm $$\|f\|_\infty=\operatorname{ess sup}|f|=\inf\{a\in\mathbb{R}\mid \mu(\{x\in X\mid |f(x)|>a\})=0\}$$ and its weak*-topology as the dual of $L^1$. Prove that C, the space of all continuous functions on $[0,1]$, is dense in $L^\infty$ in one of these topologies but not in the other. (Compare with the above result). Shoe the same with "closed " in place of "dense."

Thanks in advance.

share|improve this question
    
What is the source of the problem? –  Jonas Meyer Jan 20 '13 at 5:13
1  
Thank Bro @JonasMeyer, this problem from problem 7, chap3, functional analysis, W.Rudin (p.87) –  user52523 Jan 20 '13 at 5:33
add comment

1 Answer

As seen in On the density of $C[0,1]$ in the space $L^{\infty}[0,1]$, $C[0,1]$ is weak-* dense in $L^\infty[0,1]$. Because a uniform limit of continuous functions is continuous, $C[0,1]$ is norm closed in $L^\infty[0,1]$. Because $C[0,1]\neq L^\infty[0,1]$, these facts imply respectively that $C[0,1]$ is not weak-* closed and not norm dense.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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