Theorem: Gaussian Delta Sequence

thumb500pxThe 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:
:$\map {\delta_n} x := \dfrac n {\sqrt \pi} e^{- n^2 x^2}$
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
:$\map \delta x$ 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:
{{begin-eqn}}
{{eqn | l = \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x
      | r = \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
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | o = =:
      | r = I_1 + I_2 + I_3
}}
{{end-eqn}}
By definition, $\map \phi x$ is bounded.
Then:
{{begin-eqn}}
{{eqn | l = \size {\map g x}
      | r = \size {\map \phi x - \map \phi 0}
}}
{{eqn | o = < 
      | r = \size {\map \phi x} + \size {\map \phi 0}
      | c = Triangle Inequality for Real Numbers
}}
{{end-eqn}}
It follows that:
:$\exists M \in \R_{> 0} : \forall x \in \R : \size {\map g x} < M$ 
Then:
{{begin-eqn}}
{{eqn | l = \size {I_1}
      | r = \size {\int_{-\infty}^{-A} \sqrt {\frac n \pi} e^{-n x^2} \map g x \rd x}
}}
{{eqn | o = \le
      | r = M \size {\int_{-\infty}^{-A} \sqrt{\frac n \pi} e^{-n x^2} \rd x}
}}
{{eqn | r = \frac M {\sqrt \pi} \int_{-\infty}^{-A \sqrt n} e^{-y^2} \rd y
      | c = $y = \sqrt n x$, Integration by Substitution
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \to \infty} \size {I_1}
      | o = \le
      | r = \lim_{n \mathop \to \infty} \frac M {\sqrt \pi} \int_{-\infty}^{-A \sqrt n} e^{-y^2} \rd y
}}
{{eqn | r = \frac M {\sqrt \pi} \int_{-\infty}^{- \infty} e^{-y^2} \rd y
      | c = $A$ is a fixed strictly positive real number
}}
{{eqn | r = 0
}}
{{end-eqn}}
Analogously:
{{begin-eqn}}
{{eqn | l = \size {I_2}
      | r = \size {\int_A^\infty \sqrt {\frac n \pi} e^{-n x^2} \map g x \rd x}
}}
{{eqn | o = \le
      | r = M \size {\int_A^\infty \sqrt{\frac n \pi} e^{-n x^2} \rd x}
}}
{{eqn | r = \frac M {\sqrt \pi} \int_{A \sqrt n}^\infty e^{-y^2} \rd y
      | c = $y = \sqrt n x$, Integration by Substitution
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \to \infty} \size {I_2}
      | o = \le
      | r = \lim_{n \mathop \to \infty} \frac M {\sqrt \pi} \int_{A \sqrt n}^\infty e^{-y^2} \rd y
}}
{{eqn | r = \frac M {\sqrt \pi} \int_\infty^\infty e^{-y^2} \rd y
      | c = $A$ is a fixed strictly positive real number
}}
{{eqn | r = 0
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \size {I_3}
      | r = \size {\int_{-A}^A \sqrt {\frac n \pi} e^{-n x^2} \map g x \rd x}
}}
{{eqn | o = \le 
      | r = \int_{-A}^A \sqrt {\frac n \pi} e^{-n x^2} \size {\map g x} \rd x
}}
{{eqn | o = <
      | r = \epsilon \int_{-A}^A \sqrt {\frac n \pi} e^{-n x^2} \rd x
}}
{{eqn | r = \frac \epsilon {\sqrt \pi} \int_{- A \sqrt n}^{A \sqrt n} e^{-y^2} \rd y
      | c = $y = \sqrt n x$, Integration by Substitution
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \mathop \to \infty} \size {I_3}
      | o = <
      | r = \frac \epsilon {\sqrt \pi} \int_{-\infty}^{\infty} e^{-y^2} \rd y
      | c = $A$ is a fixed strictly positive real number
}}
{{eqn | r = \epsilon
      | c = Gaussian Integral
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} 
      | o = \le 
      | r = \size {I_1 + I_2 + I_3} 
}}
{{eqn | o = \le 
      | r = \size {I_1 + I_2} + \size {I_3} 
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | o = < 
      | r = \epsilon
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x
      | r = \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
      | c = Sum Rule for Limits of Real Functions
}}
{{eqn | r = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0
}}
{{eqn | r = 0
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x
      | r = \map \phi 0
}}
{{end-eqn}}
{{qed}}