Gaussian Delta Sequence
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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 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 $\xi_\epsilon \ne 0$.
By Real Numbers are Densely Ordered:
- $\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$
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 $\xi_\epsilon = 0$.
 | This article needs proofreading. In particular: Check the rigour of the last few lines. Make the argument stricter. If you believe all issues are dealt with, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
$\blacksquare$
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x\) | \(=\) | \(\ds \map \phi 0\) |
$\blacksquare$