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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.

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  • $\begingroup$ What is the source of the problem? $\endgroup$ – Jonas Meyer Jan 20 '13 at 5:13
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    $\begingroup$ Thank Bro @JonasMeyer, this problem from problem 7, chap3, functional analysis, W.Rudin (p.87) $\endgroup$ – user52523 Jan 20 '13 at 5:33
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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.

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