Limit $\lim_{\epsilon \rightarrow0} \frac{1}{\sqrt{\epsilon}}\exp\left({-\frac{x^2}{\epsilon}}\right)$ in the sense of distributions

Question

Compute the following limit:

$$\lim_{\epsilon \rightarrow0} \frac{1}{\sqrt{\epsilon}}\exp\left({-\frac{x^2}{\epsilon}}\right)$$

I do not know how the "sense of distributions" is applied in this example, could someone explain this?

Answer 1

This goes to $0$, except if $x=0$.

In the sense of distribution, however, the associated distributions converge to the Dirac delta. You have to multiply your gaussians by a smooth test function $\varphi$, integrate and prove that the limit is always $\varphi(0)$ to prove this result.

Answer 2

If you set $\frac{1}{\varepsilon} = t $ you get $$ \lim_{t \to \infty} \frac{\sqrt{t}}{e^{tx^2 }} = 0 $$ But what you are asking for is convergence in distrubution, which means $$ \lim_{n \to \infty}F_n(X_n) = F(X) $$ where $F_n$ is the cdf of $X_n$ and $F$ is the cdf of $X$. An example of it is CLT.

Answer 3

I would say that "in the sense of distributions" means: "which element of the dual space of integrable functions does the set of functionals $f_\epsilon (g) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{\epsilon}} \exp \left(-\frac{x^2}{\epsilon} \right) \, g(x) \, dx$ converge to as $\epsilon$ goes to $0$ from above?"

As others have pointed out, $f_{\epsilon}$ as defined above converges to the dirac delta "function", which is defined as the map which takes an integral function $g$ to its value at $0$, $g(0)$.

In other words, you need to show $$ \delta(g) = g(0) = \lim_{\epsilon \to 0^+} f_{\epsilon} (g) = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty \frac{1}{\sqrt{\epsilon}} \exp\left(-\frac{x^2}{\epsilon} \right) \, g(x) \, dx $$ for all integrable functions $g$, as this is precisely what it means to show $\lim_{\epsilon \to 0} f_{\epsilon} = \delta$.

You may have noticed that I defined the $f_\epsilon$ not as functions, but as functionals, i.e. maps which take functions in and spit out real numbers. Your teacher is abusing notation by asking for the convergence of the $f_\epsilon$ defined as given, since that set of functions does not converge. This abuse of notation is justified since we may often identify functions $f$ with their corresponding functionals $g \mapsto \int f g$.

N.B.: the set of functions that you have to "test" the $f_\epsilon$ against is often known (creatively enough) as the set of "test functions." I only guessed that the set of test functions in this case was the set of integrable functions on $\mathbb{R}$, but your teacher may have had another set of test functions in mind (for example all smooth, compactly supported functions).

Answer 4

A sequence of functions $f_n$ converges to $f$ in the sense of distributions if for every $\varphi \in C_c^\infty$ (compactly supported smooth functions) the sequence $\int f_n(x) \, \varphi(x) \, dx$ converges to $\int f(x) \, \varphi(x) \, dx.$

So let $\varphi \in C_c^\infty.$ Then, $$ \int \frac{1}{\sqrt{\epsilon}}\exp\left({-\frac{x^2}{\epsilon}}\right) \, \varphi(x) \, dx = \{ x = \sqrt{\epsilon}y \} = \int \exp\left(-y^2\right) \, \varphi(\sqrt{\epsilon}y) \, dy \to \int \exp\left(-y^2\right) \, \varphi(0) \, dy = \left( \int \exp\left(-y^2\right) dy \right) \varphi(0) = \sqrt{\pi}\varphi(0) = \int \sqrt{\pi}\delta(x) \, \varphi(x) \, dx. $$

Thus, $\frac{1}{\sqrt{\epsilon}}\exp\left({-\frac{x^2}{\epsilon}}\right) \to \sqrt{\pi}\delta(x)$ as $\epsilon \to 0.$