Does convergence in $L^1_{\text{loc}}(\mathbb{R}^n)$ implies convergence in $\mathcal{D}'(\mathbb{R}^n)$?

Question

Every function $f\in L^1_{\text{loc}}(\mathbb{R}^n)$ defines an element $\Lambda_f$ of $\mathcal{D}'(\mathbb{R}^n)$.

If $f_n\to f$ in $L^1_{\text{loc}}(\mathbb{R}^n)$, does that mean that $\Lambda_{f_n}\to \Lambda_f$ in $\mathcal{D}'(\mathbb{R}^n)$?

Answer 1

Yes. The point is that convergence in $L^1_{\mathrm{loc}}(\mathbb R^n)$ implies convergence in each $L^1(\Omega)$ for every bounded domain $\Omega \subset \mathbb R^n,$ and hence weak $L^1$ convergence in $\Omega.$ But $L^1(\Omega)^* \cong L^{\infty}(\Omega),$ which contains $C^{\infty}_c(\Omega).$

Going along these lines, we can check this directly. Suppose $f_n \rightarrow f$ in $L^1_{\mathrm{loc}}(\mathbb R^n)$ and let $\varphi \in C^{\infty}_c(\mathbb R^n).$ Pick a bounded domain $\Omega \subset \mathbb R^n$ such that $\operatorname{supp} \varphi \subset \Omega$ and note that $f_n \rightarrow f$ in $L^1(\Omega).$ Then we get, $$ \Lambda_{f_n}(\varphi) - \Lambda_f(\varphi) = \int_{\mathbb R^n} (f_n - f)\varphi \,\mathrm{d}x = \int_{\Omega} (f_n - f)\varphi \,\mathrm{d}x \leq \lVert f_n - f\rVert_{L^1(\Omega)} \lVert \varphi \rVert_{L^{\infty}_{\Omega}} \longrightarrow 0 $$ as $n \rightarrow \infty.$ Since $\varphi$ was arbitrary, $\Lambda_{f_n} \rightarrow \Lambda_f$ in $\mathcal{D}'(\mathbb R^n).$