I am having a problem with the following exercise:
Show that for every bounded borelian function $\varphi : \mathbb{R} \rightarrow \mathbb{R}$,
$\underset{n}{lim} \frac{n}{\sqrt{2\pi}} \underset{\mathbb{R}}{\int} \varphi(x) e^\frac{-n^2(x-a)^2}{2} dx = \varphi(a)$, where $a \in \mathbb{R}$.
Somebody suggested me to consider a sequence of measurable simple functions $(\varphi_n)$ so that $\varphi_n(x) \nearrow \varphi(x)$ $\forall x$. But then?
Thank you!
Hint: LHS can be written as convolution $\varphi*\psi_n(a)$, $\psi_n(y):=(2\pi)^{-1/2}n\exp(-n^2 y^2 /2)$...
Show that $(\psi_n)$ converges to $\delta_0$ (which is "identity" for convolution) in some (weak) sense...
The $\psi_n$ are Gaussian distributions with mean $0$ and standard deviation $1/n$. Need convergence $\psi_n\to \delta_0$ as distribution, hence $\int \psi_n f dx\to f(0)$ for any $f$ smooth, compactly supported. With $y=z/n$ have $$\int \psi_n f dy = \int (2\pi)^{-1/2}n\exp(-n^2 y^2 /2) f(y) dy = \int (2\pi)^{-1/2}\exp(- z^2 /2) f(z/n) dz,$$ where $f(z/n)\to f(0)$ point-wisely, and dominated convergence gives $$\int (2\pi)^{-1/2}\exp(- z^2 /2) f(z/n) dz \to \int (2\pi)^{-1/2}\exp(- z^2 /2) f(0) dz=f(0).$$ Hence $\psi_n\to \delta_0$ as distribution, and for any $a$, $\varphi*\psi_n(a)\to \delta_0*\varphi (a)= \varphi(a)$.