Suppose $\{f_{n}\}_{n \ge 1} \subset L^{1}(\Omega)$ is a uniformly bounded sequence and $\Omega = (0,1)$. Can we say that $f_{n}$ converges in $\mathcal{D}'(\Omega)$ (up to a subsequence) to a limit $f$?
My thoughts:
We can embed $L^{1}(\Omega)$ into the space $\mathcal{M}_{b}(\Omega)$ of finite measures. Since $\mathcal{M}_{b}(\Omega)$ is the dual space of $C_{b}(\Omega)$ and $f_{n}$ is also bounded in $\mathcal{M}_{b}(\Omega)$ we have that (up to a subsequence) $f_{n}$ converges weak-* to a limit $f$ in $\mathcal{M}_{b}(\Omega)$. Now $f$ being a measure induces a distribution via the mapping $$ \phi \mapsto \int_{\Omega}\phi ~df. $$ But I am not sure if this argument will imply convergence in $\mathcal{D}'$. Sure the limit is a distribution but I am not sure if the convergence will actually take place in the distributional sense. I would appreciate any help.