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?
Attempt
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?
