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.
Best 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.