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It's easy to show $C[0,1]$ is not dense in $L^{\infty}[0,1]$ in the norm topology, but $C[0,1]$ is dense in $L^{\infty}[0,1]$ in the weak*-topology when take $L^{\infty}$ as the dual of $L^{1}$. how to prove it?

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up vote 6 down vote accepted

Given $f\in L^\infty$, apply Lusin's theorem infinitely many times to get a sequence $(f_n)_n$ in $C[0,1]$ uniformly bounded by $\|f\|_\infty$ such that the measure of the set where $f$ and $f_n$ are unequal is less than $\frac{1}{n}$. If $g\in L^1$, then given $\varepsilon\gt 0$ there exists $N$ such that $\int_A|g|<\varepsilon$ whenever the measure of $A$ is less than $\frac{1}{N}$. Then $|\int_0^1(f-f_k)g|\leq 2\|f\|_\infty\varepsilon$ for all $k\geq N$. This implies that $f_n\to f$ in the weak-$*$ topology, because $\int_0^1 f_ng\to\int_0^1fg$ for all $g\in L^1$.

Alternatively, you can show that the sequence of Cesàro means of the Fourier series of $f$ converges to $f$ in the weak-$*$ topology.

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Let $\psi_n$ be a sequence of standard mollifiers, symmetric about 0. If $f \in L^\infty$, $g \in L^1$, it is a quick application of Fubini's theorem to see that $\int (f * \psi_n) g = \int f (\psi_n * g)$, where $*$ denotes convolution. Since $f * \psi_n$ is continuous and $\psi_n * g \to g$ in $L^1$, we are done.

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I know that $\int_\mathbb{R} (f \star \psi_n) g = \int_\mathbb{R} f (g\star \psi_n)$ but it is still true if we replace $\mathbb{R}$ by $[0,1]$? – user37238 Dec 18 '14 at 12:24

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