Gaussian Delta Sequence/Proof 2
Theorem
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 Arfken, Hans J. Weber and Frank E. Harris: Mathematical Methods for Physicists (7th ed.): Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function