0
$\begingroup$

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!!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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