# Gaussian Delta Sequence

## 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 1

 $\ds \int_0^\infty \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x$ $=$ $\ds \int_0^\infty \frac 1 {\sqrt \pi} e^{- \paren {n x}^2} \rd \paren {n x}$ $\ds$ $=$ $\ds \int_0^\infty \frac 1 {\sqrt \pi} e^{- y^2} \rd y$ $n x = y$, Integration by Substitution $\ds$ $=$ $\ds \frac 1 2$ Integral to Infinity of Exponential of -t^2

Furthermore:

 $\ds \int_{-\infty}^\infty \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x$ $=$ $\ds 1$

Let $a,b \in \R$.

Then:

 $\ds \int_a^b \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x$ $=$ $\ds \int_a^b \frac 1 {\sqrt \pi} e^{- \paren {n x}^2 } \rd \paren {n x}$ $\ds$ $=$ $\ds \int_{n a}^{n b} \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ $n x = y$, Integration by Substitution

Suppose $0 < a < b$.

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ $=$ $\ds \lim_{n \mathop \to \infty} \int_0^{b n} \frac 1 {\sqrt \pi} e^{- y^2 } \rd y - \lim_{n \mathop \to \infty} \int_0^{a n} \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=$ $\ds \int_0^\infty \frac 1 {\sqrt \pi} e^{- y^2 } \rd y - \int_0^\infty \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ $\ds$ $=$ $\ds \frac 1 2 - \frac 1 2$ $\ds$ $=$ $\ds 0$

Analogously, suppose $a < b < 0$.

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{a n}^0 \frac 1 {\sqrt \pi} e^{- y^2 } \rd y - \lim_{n \mathop \to \infty} \int_{b n}^0 \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=$ $\ds \int_{-\infty}^0 \frac 1 {\sqrt \pi} e^{- y^2 } \rd y - \int_{-\infty}^0 \frac 1 {\sqrt \pi} e^{- y^2 } \rd y$ $\ds$ $=$ $\ds \frac 1 2 - \frac 1 2$ $\ds$ $=$ $\ds 0$

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

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \map \phi x \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x + \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \map \phi x \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x + \lim_{n \mathop \to \infty} \int_\epsilon^\infty \map \phi x \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x$ $\ds$ $=$ $\ds \map \phi {\xi_-} \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x + \map \phi {\xi_+} \lim_{n \mathop \to \infty} \int_{\epsilon}^\infty \frac n {\sqrt \pi} e^{- n^2 x^2} \rd x$ Mean value theorem for integrals, $\xi_\epsilon \in \closedint {-\epsilon} \epsilon$, $\xi_- \in \hointl {-\infty} {-\epsilon}$, $\xi_+ \in \hointr \epsilon \infty$ $\ds$ $=$ $\ds 0 + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-n \epsilon}^{n \epsilon} \frac 1 {\sqrt \pi} e^{- y^2 } \rd y + 0$ $\ds$ $=$ $\ds \map \phi {\xi_\epsilon}$

$\epsilon$ is an arbitrary positive real number.

Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.

Suppose $\xi_\epsilon \ne 0$.

$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$

Then with respect to $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.

But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.

Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.

$\blacksquare$

## Proof 2

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

$\blacksquare$