# Gaussian Delta Sequence/Proof 2

## Theorem

The graph of the Gaussian delta sequence. As $n$ grows, the graph becomes thinner and taller. The area of under each Gaussian is equal to $1$.

Let $\sequence {\map {\delta_n} x}$ be a sequence such that:

$\ds \map {\delta_n} x := \frac n {\sqrt \pi} e^{- n^2 x^2}$

Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.

That is, in the distributional sense it holds that:

$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$

or

$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$

where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.

## Proof

Let $\map g x = \map \phi x - \map \phi 0$.

Then:

$\ds \int_{- \infty}^\infty \map \phi x \map {\delta_n} x \rd x = \map \phi 0 + \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$

Let $A \in \R_{> 0}$.

Then:

 $\ds \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$ $=$ $\ds \int_{- \infty}^{- A} \map g x \map {\delta_n} x \rd x + \int_A^\infty \map g x \map {\delta_n} x \rd x + \int_{- A}^A \map g x \map {\delta_n} x \rd x$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=:$ $\ds I_1 + I_2 + I_3$

By definition, $\map \phi x$ is bounded.

Then:

 $\ds \size {\map g x}$ $=$ $\ds \size {\map \phi x - \map \phi 0}$ $\ds$ $<$ $\ds \size {\map \phi x} + \size {\map \phi 0}$ Triangle Inequality for Real Numbers

It follows that:

$\exists M \in \R_{> 0} : \forall x \in \R : \size {\map g x} < M$

Then:

 $\ds \size {I_1}$ $=$ $\ds \size {\int_{-\infty}^{-A} \sqrt {\frac n \pi} e^{-n x^2} \map g x \rd x}$ $\ds$ $\le$ $\ds M \size {\int_{-\infty}^{-A} \sqrt{\frac n \pi} e^{-n x^2} \rd x}$ $\ds$ $=$ $\ds \frac M {\sqrt \pi} \int_{-\infty}^{-A \sqrt n} e^{-y^2} \rd y$ $y = \sqrt n x$, Integration by Substitution $\ds \leadsto \ \$ $\ds \lim_{n \to \infty} \size {I_1}$ $\le$ $\ds \lim_{n \mathop \to \infty} \frac M {\sqrt \pi} \int_{-\infty}^{-A \sqrt n} e^{-y^2} \rd y$ $\ds$ $=$ $\ds \frac M {\sqrt \pi} \int_{-\infty}^{- \infty} e^{-y^2} \rd y$ $A$ is a fixed strictly positive real number $\ds$ $=$ $\ds 0$

Analogously:

 $\ds \size {I_2}$ $=$ $\ds \size {\int_A^\infty \sqrt {\frac n \pi} e^{-n x^2} \map g x \rd x}$ $\ds$ $\le$ $\ds M \size {\int_A^\infty \sqrt{\frac n \pi} e^{-n x^2} \rd x}$ $\ds$ $=$ $\ds \frac M {\sqrt \pi} \int_{A \sqrt n}^\infty e^{-y^2} \rd y$ $y = \sqrt n x$, Integration by Substitution $\ds \leadsto \ \$ $\ds \lim_{n \to \infty} \size {I_2}$ $\le$ $\ds \lim_{n \mathop \to \infty} \frac M {\sqrt \pi} \int_{A \sqrt n}^\infty e^{-y^2} \rd y$ $\ds$ $=$ $\ds \frac M {\sqrt \pi} \int_\infty^\infty e^{-y^2} \rd y$ $A$ is a fixed strictly positive real number $\ds$ $=$ $\ds 0$

We have that $\map g 0 = 0$.

By definition, $\phi$ is a smooth real function on $\R$.

By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$.

Then:

$\ds \forall \epsilon \in \R_{> 0} : \exists \delta \in \R_{> 0} : 0 < \size x < \delta \implies \size {\map g x} < \epsilon$

Let $A$ be such that $A < \delta$.

Then:

 $\ds \size {I_3}$ $=$ $\ds \size {\int_{-A}^A \sqrt {\frac n \pi} e^{-n x^2} \map g x \rd x}$ $\ds$ $\le$ $\ds \int_{-A}^A \sqrt {\frac n \pi} e^{-n x^2} \size {\map g x} \rd x$ $\ds$ $<$ $\ds \epsilon \int_{-A}^A \sqrt {\frac n \pi} e^{-n x^2} \rd x$ $\ds$ $=$ $\ds \frac \epsilon {\sqrt \pi} \int_{- A \sqrt n}^{A \sqrt n} e^{-y^2} \rd y$ $y = \sqrt n x$, Integration by Substitution $\ds \leadsto \ \$ $\ds \lim_{n \mathop \to \infty} \size {I_3}$ $<$ $\ds \frac \epsilon {\sqrt \pi} \int_{-\infty}^{\infty} e^{-y^2} \rd y$ $A$ is a fixed strictly positive real number $\ds$ $=$ $\ds \epsilon$ Gaussian Integral

Then:

 $\ds \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x}$ $\le$ $\ds \size {I_1 + I_2 + I_3}$ $\ds$ $\le$ $\ds \size {I_1 + I_2} + \size {I_3}$ Triangle Inequality for Real Numbers $\ds$ $<$ $\ds \epsilon$

To sum up:

$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : \forall n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$

By definition of the limit of a real sequence:

$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$

However:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0 \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \rd x$ Sum Rule for Limits of Real Functions $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0$ $\ds$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x$ $=$ $\ds \map \phi 0$

$\blacksquare$

## Sources

2013: George ArfkenHans J. Weber and Frank E. Harris: Mathematical Methods for Physicists (7th ed.): Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function