How to define the dirac delta as the limit of an arbitrary probability distribution?

Question

Is it true that, for an arbitrary probability density, $\rho(x)$, that, $$\lim_{\alpha\to0^{+}}\frac{1}{\alpha}\rho(x/\alpha)=\delta(x),$$ where $\delta(x)$ is the dirac delta?

If this is not true please can you provide some additional minimal conditions under which it becomes so.

(for clarity by probability density I mean a function which satisfies, $\rho(x)\ge0$ $\forall x$ and $\int_{-\infty}^{\infty}\rho(x)\mathrm{d}x=1$)

Answer 1

By definition, $\frac{1}{\alpha} \rho(x/\alpha) \to \delta$ as $\alpha \to 0^+$ if $\int \frac{1}{\alpha} \rho(x/\alpha) \, \varphi(x) \, dx \to \varphi(0)$ for every $\varphi \in C_c^\infty.$

Therefore, let $\varphi \in C_c^\infty.$ Then, $$ \int \frac{1}{\alpha} \rho(x/\alpha) \, \varphi(x) \, dx = \{ \text{ substitution $x = \alpha y$ } \} = \int \rho(y) \, \varphi(\alpha y) \, dy\ . $$

Now, since $|\rho(y)\,\varphi(\alpha y)| \leq |\rho(y)|\,\sup|\varphi| \in L^1$ we can apply the dominated convergence theorem and get the limit $$ \int \rho(y) \, \varphi(0) \, dy = \int \rho(y) \, dy \ \varphi(0) = \varphi(0) . $$ Thus, $\frac{1}{\alpha} \rho(x/\alpha) \to \delta.$

Answer 2

The question is ambiguous, especially given the use of the tag "distribution-theory". Convergence can be understood in the sense of Schwartz's theory of distribution or in the sense of probability theory. The answer by md2perpe addressed the first interpretation, so let me consider the second. If $\rho$ is density for the probability law of a random variable $X$, then $\rho(\cdot/\alpha)/\alpha$ is the density for the law of $\alpha X$. For a bounded continuous function $F$, $\mathbb{E}[F(\alpha X)]$ converges to $F(0)$, because of the dominated convergence theorem and the sequential criterion for convergence. This, by definition means that $\alpha X$ converges in distribution to a random variable almost-surely equal to zero, i.e., a random variable with law given by the unit Dirac mass at the origin.