I'm trying to prove a result mentioned in this thread, i.e.,
Let $\Omega$ be an open subset of $\mathbb R^d$. Then $\mathcal C_c^\infty (\Omega)$ is dense in $L^\infty (\Omega)$ w.r.t. the weak topology $\sigma(L^\infty, L^1)$.
Could you confirm if my below attempt is correct?
Proof We have $L^\infty =(L^1)^*$. Let $T: L^\infty \to \mathbb R$ be a linear function that is continuous w.r.t. the weak topology of $\sigma(L^\infty, L^1)$. By Proposition 3.14. in Brezis's Functional Analysis, there is $u \in L^1$ such that $$ T (f) = \langle f, u \rangle = \int_\Omega u f \quad \forall f \in L^\infty (\Omega). $$
Assume $T = 0$ on $\mathcal C_c^\infty (\Omega)$. By Corollary 1.8. in the same textbook, it suffices to prove that $T =0$ on $L^\infty (\Omega)$.
Corollary 4.24 Let $\Omega$ be an open subset of $\mathbb R^d$. Let $u \in L_{\text{loc}}^1 (\Omega)$ such that $$ \int_{\Omega} uf =0 \quad \forall f \in \mathcal C_c^\infty (\Omega). $$ Then $u=0$ a.e. on $\Omega$.
It follows from above corollary that $u=0$ a.e. on $\Omega$ and thus $T \equiv 0$. This completes the proof.
