Show that $C_{c}^{\infty}(\Omega)$ is dense in $L^{\infty}(\Omega)$ with respect to the topology $\sigma(L^{\infty},L^{1})$, where $\Omega$ is an open subset of $\mathbb{R^n}$.

My try: Let $$\Omega_n=\{x \in \Omega: d(x,\partial \Omega) \gt \frac{2}{n},||x|| \lt n\}$$

For $u \in L^{\infty}(\Omega)$, define $\bar{u}$ by extending $u$ by $0$ outside $\Omega$. Let $\xi_n=\chi_{\Omega_n},\xi_=\chi_{\Omega}$, where $\chi$ is the usual characteristic function. Then $\Omega_{n} \subset \Omega_{n+1},\forall n$ and $\xi_n \to \xi$ in $\mathbb{R^n}$. Let $\rho_n$ be a sequence of mollifiers with $\rho_n \ge 0$ and $\int \rho_n=1$.

Define $v_n=\rho_n *(\xi_n\bar{u}), v=\xi \bar{u}$. It is then clear that $v_n \in C_c^{\infty}(\Omega), v_n \rightharpoonup v$ in $L^{\infty}(\Omega)$ weak$^* \sigma(L^{\infty},L^1)$ and $\int_{B}|v_n-v| \to 0$ for every ball $B$. From here I can extract a subsequence of $v_n$ which converges to $v$ almost everywhere in $B$.

I need to do diagonalization argument to get hold of a subsequence which converges to $v$ a.e in $\Omega$ which I am unable to do. Choosing the open sets $\Omega_n$ should do the job but I am little in doubt.

Thanks for the help!!


1 Answer 1


Since the dual of $(L^\infty,\sigma(L^\infty,L^1))$ is $L^1$ (more precisely, every element of the dual is of the form $f\mapsto \int fg$ for some $g\in L^1$) the Hahn-Banach theorem tells you what to show: every $g\in L^1$ with $\int fg =0$ for all $f\in C_c^\infty$ is $0$. This can be done by approximating (w.r.t the $L^1$-norm) the sign $|g|/g$ by smooth functions.

  • $\begingroup$ I don't understand how the approximation w.r.t. the $L^1$-norm works. Would you mind adding some details, please? $\endgroup$
    – Andromeda
    Commented May 8, 2023 at 8:24
  • $\begingroup$ A standard procedure is convolution (of the extension by $0$ outside $\Omega$) with smooth functions with very small support and integral $1$. $\endgroup$
    – Jochen
    Commented May 8, 2023 at 19:05

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