# weak convergence to delta function

Let $f_\epsilon\in L^1(\mathbb{R}^n)$ be a function which depends on a parameter $\epsilon\in(0,1)$, and is such that

1. $\operatorname{supp}{f_\epsilon}\subset\{|x|\leq\epsilon\}$,
2. the total integral of $f_\epsilon$ is $1$, and
3. $\displaystyle\int_{\mathbb{R}^n}{|f_\epsilon(x)|\,dx}\leq\mu \lt \infty$ for $\epsilon\in(0,1)$.

How do I show that $f_\epsilon\rightarrow\delta$ (in the space of tempered distributions on $\mathbb{R}^n$) as $\epsilon\rightarrow0^+$,

i.e. how do I show $$\int_{\mathbb{R}^n}{f_\epsilon(x)\,\phi(x)\,dx}=\int_{|x|\leq\epsilon}{f_\epsilon(x)\,\phi(x)\,dx}\;\xrightarrow{\varepsilon \to 0^+}\;\phi(0)$$ for all test functions $\phi$?

-

Let $t \gt 0$. Since $\phi$ is continuous, you can choose $d$ so small that $|x| \leq d$ implies $|\phi(x)-\phi(0)| \lt \dfrac{t}{\mu}$.
Thus, for $\varepsilon \leq d$ we have
$$\left| \phi(0) - \int_{\mathbb R^n} f_\varepsilon \cdot \phi \right| = \left| \int_{\mathbb R^n} f_\varepsilon \cdot (\phi(0) - \phi ) \right| = \left| \int_{\{x \leq \varepsilon\}} f_\varepsilon \cdot (\phi(0) - \phi) \right| \leq \|f_\varepsilon\|_1 \cdot \frac{t}{\mu} \leq t$$ and the result follows, as $t \gt 0$ was arbitrary.