Is the map $L^{1}_{loc}\left( \mathbb{R}^{n} \right) \hookrightarrow \mathcal{D}^{\prime}\left( \mathbb{R}^{n} \right)$ a topological embedding?

Question

Let $d$ be a positive integer, and denote by $L^{1}_{loc}\left( \mathbb{R}^{d} \right)$ the space of locally integrable functions on $\mathbb{R}^{d}$ and by $\mathcal{D}^{\prime}\left( \mathbb{R}^{d} \right)$ the space of distributions on $\mathbb{R}^{d}$ that we endow with their usual topologies. It is well known that the map $I \colon L^{1}_{loc}\left( \mathbb{R}^{d} \right) \rightarrow \mathcal{D}^{\prime}\left( \mathbb{R}^{d} \right)$ given by $$\forall \varphi \in \mathcal{D}\left( \mathbb{R}^{d} \right), \, \left\langle I(f), \varphi \right\rangle = \int_{\mathbb{R}^{d}} f \varphi \, d\lambda$$ is a continuous injective linear map, where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^{d}$ and $\mathcal{D}\left( \mathbb{R}^{d} \right)$ denotes the space of smooth functions on $\mathbb{R}^{d}$ with compact support. Is this map a topological embedding? Equivalently, if $\left( f_{n} \right)_{n \geq 0}$ is a sequence of elements of $L^{1}_{loc}\left( \mathbb{R}^{d} \right)$ such that $$\forall \varphi \in \mathcal{D}\left( \mathbb{R}^{d} \right), \, \lim_{n \rightarrow +\infty} \int_{\mathbb{R}^{d}} f_{n} \varphi \, d \lambda = 0 \, \text{,}$$ then do we have necessarily $$\forall K \subset \mathbb{R}^{d} \text{ compact}, \, \lim_{n \rightarrow +\infty} \int_{K} \left\lvert f_{n} \right\rvert \, d\lambda = 0 \, \text{?}$$ I could not find the answer in the literature. I would be surprised if this question has a positive answer, but I have not been able to construct a counterexample.

Answer 1

Let's try this.
In $\mathbb R$ . For $k \in \mathbb N$, let $$ f_k = k\mathbf1_{(0,1/k)} \qquad\text{and}\qquad g_k = k\mathbf1_{(-1/k,0)} $$ so that as $k \to \infty$ we have $f_k \to \delta$ and $g_k \to \delta$ in $\mathcal D'$.
Then consider $h_k = f_k-g_k$. So $h_n \to 0$ in $\mathcal D'$. But for $K = [-1,1]$, we have $\int_K |h_k| = 2 \not\to 0$.