First let me recap some definitions. A set of functions $\{K_\delta\}$ is said to be an approximation of the identity as $\delta \to 0$ if
- $\int_{-\infty}^{\infty} K_\delta(x)dx = 1,$
- $\int_{-\infty}^{\infty} |K_\delta(x)|dx \leq M$,
- For every $\eta > 0$, we have $\int_{|x|> \eta} |K_\delta(x)|dx \to 0$ as $\delta \to 0$.
With this, let $K_\delta(x) = \delta^{-1/2} e^{-\pi x^2/\delta}$. I want to show that $\{K_\delta\}$ is an approximation of the identity.
For that, the first two items of the definition are immediate. The first is a result of the gaussian integral
$$\int_{-\infty}^{\infty}\delta^{-1/2} e^{-\pi x^2/\delta}dx = \delta^{-1/2} \sqrt{\dfrac{\pi}{\pi/\delta}}=\dfrac{1}{\sqrt{\delta}}\sqrt{\delta}=1.$$
The second property holds with $M = 1$ because $K_\delta > 0$ for each $\delta$ and so
$$\int_{-\infty}^{\infty} |K_\delta(x)|dx = \int_{-\infty}^{\infty} K_\delta(x)dx=1\leq 1.$$
The third property is the one giving me trouble. I mean, let $\eta > 0$ we have
$$\int_{|x|> \eta}\delta^{-1/2}e^{-\pi x^2/\delta}dx = \int_{-\infty}^{-\eta}+\int_{\eta}^\infty \delta^{-1/2} e^{-\pi x^2/\delta}dx,$$
which changing variables to $y = \delta^{-1/2} x$ we have $x^2/\delta = y^2$, and then
$$\int_{|x|> \eta}\delta^{-1/2}e^{-\pi x^2/\delta}dx =\int_{-\infty}^{-\eta/\delta^{1/2}}+\int_{\eta/\delta^{1/2}}^\infty e^{-\pi y^2}dy.$$
Now, I can't see how to prove that this goes to zero as $\delta \to 0$. I think I understand the idea, as $\delta \to 0$ the "part of the line" we are integrating over gets smaller, because $\pm \eta/\delta^{1/2}\to \pm \infty$.
But I don't understand how to make this rigorous and show that this is indeed the case.
How do I prove this part of the definition of approximation of the identity for the gaussians? How do I finish this proof?
If $r>1$ then $$ \int_r^\infty e^{-y^2}\,dy<\int_r^\infty e^{-y}\,dy=e^{-r}. $$ You can also argue as follows: if $\int_0^\infty f(y)\,dy$ is convergent, then $\lim_{r\to\infty}\int_r^\infty f(y)\,dy=0$.