Consider the Gaussian function:

$$ f_{N(\mu, \sigma^2)}(t) = \frac{1}{\sigma \sqrt{\pi}} \exp \left[ {-\frac{1}{2}\left( \frac{t - \mu}{\sigma} \right)^2} \right] $$

I have seen in some texts showing that:

$$ \lim_{\sigma \to 0} f_{N(\mu, \sigma^2)}(t) = \delta(t-\mu) $$

Now, I know that the Dirac's Delta has been the subject of many studies and has been defined in many ways. One of the early definitions is that such a function can be modeled as a impulse whose base's length is a function of $R^{-1}$ and height is a function of $R$ (for a given parameter $R \in \mathbb{R}$).

Distribution theory

However, modern Distribution Theory given this function a full characterization as a distribution (aka generalized function) defined as:

$$ \langle \delta, \phi \rangle = \phi(0) $$

Having $\phi \in \mathcal{D}$ a test function.

How can we prove, in the distribution framework, that the Gaussian function converges to a Dirac Delta distribution when the variance vanishes?


I have tried setting a proof. Since we need to prove that a function converges to a distribution, we first need to work in the distribution space by considering the generalized function of the Gaussian function.

I know from Distribution Theory that a generic function $f:\mathbb{R} \mapsto \mathbb{R}$ can be transformed into a distribution $T_f$ by:

$$ \langle T_f, \phi \rangle = \int_{-\infty}^{+\infty} f(t)\phi(t)dt $$

So our Gaussian distribution would be the following functional:

$$ \langle T_{N(\mu, \sigma^2)}, \phi \rangle = \frac{1}{\sigma \sqrt{\pi}} \int_{-\infty}^{+\infty} \exp \left[ {-\frac{1}{2}\left( \frac{t - \mu}{\sigma} \right)^2} \right] \phi(t) dt $$

From here I am a little insecure how to move on. My goal would be to extract the limit of the distribution and show a convergence to the Delta distribution. I remember that generalized functions have a well defined definition of functional convergence which is based on a convergence model defined on the test function space. How to proceed?


1 Answer 1


Make the substitution $s=\frac {t-\mu} {\sigma}$. You get $\frac 1 {\sqrt {2\pi}}\int_{\mathbb R} e^{-s^{2}/2} \phi (\mu+\sigma s) ds$. It is easy to justify taking the limit inside (by DCT) so the expression tends to $\phi (\mu) \frac 1 {\sqrt {2\pi}}\int_{\mathbb R} e^{-s^{2}/2} ds=\phi (\mu)$.


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