Tuning Fork Delta Sequence
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Theorem
The graph of the tuning fork delta sequence. As $n$ grows, the rectangles becomes thinner and longer. The area of each shape is equal to $1$, where the area under the axis contributes negatively. Note that at $x = 0$ the graph is always negative, and its value here approaches $-\infty$. Only taking the entire graph we can ensure the proper behaviour for $n$ approaching $\infty$,
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
- $\map {\delta_n} x := \begin{cases}
-n & : \size x < \frac 1 {2n} \\ 2n & : \frac 1 {2n} \le \size x \le \frac 1 n \\ 0 & : \size x > \frac 1 n \end{cases}$
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
Let $\phi \in \map \DD \R$ be a test function.
Then:
| \(\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\int_{- \infty}^{-\frac 1 n } \map \phi x \map {\delta_n} x \rd x + \int_{- \frac 1 n }^{- \frac 1 {2n} } \map \phi x \map {\delta_n} x \rd x + \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \map \phi x \map {\delta_n} x \rd x + \int_{\frac 1 {2n} }^{\frac 1 n} \map \phi x \map {\delta_n} x \rd x + \int_{\frac 1 n}^{\infty} \map \phi x \map {\delta_n} x \rd x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {2n \int_{- \frac 1 n }^{- \frac 1 {2n} } \map \phi x \rd x - n \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \map \phi x \rd x + 2n \int_{\frac 1 {2n} }^{\frac 1 n} \map \phi x \rd x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} n \paren {2 \map \phi {\xi_1} \paren {- \frac 1 {2n} - \paren {- \frac 1 n} } - \map \phi {\xi_2} \paren {\frac 1 {2n} - \paren {- \frac 1 {2n} } } + 2 \map \phi {\xi_3} \paren {\frac 1 n - \frac 1 {2n} } }\) | Mean Value Theorem for Integrals, $\xi_1 \in \closedint {-\frac 1 n } {-\frac 1 {2n} }$, $\xi_2 \in \closedint {-\frac 1 {2n} } {\frac 1 {2n} }$, $\xi_3 \in \closedint {\frac 1 {2n} } {\frac 1 n }$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren { \map \phi {\xi_1} - \map \phi {\xi_2} + \map \phi {\xi_3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi {\lim_{n \mathop \to \infty} \xi_1} - \map \phi {\lim_{n \mathop \to \infty} \xi_2} + \map \phi {\lim_{n \mathop \to \infty} \xi_3}\) | Limit of Image of Sequence on Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi 0 - \map \phi 0 + \map \phi 0\) | $\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$, Squeeze Theorem for Real Sequences | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \delta \phi\) | Definition of Dirac Delta Distribution |
Furthermore:
| \(\ds \int_{-\infty}^\infty \map {\delta_n} x \rd x\) | \(=\) | \(\ds 2n \int_{- \frac 1 n }^{- \frac 1 {2n} } \rd x - n \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \rd x + 2n \int_{\frac 1 {2n} }^{\frac 1 n} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n \paren {2 \paren {- \frac 1 {2n} - \paren {- \frac 1 n} } - \paren {\frac 1 {2n} - \paren {- \frac 1 {2n} } } + 2 \paren {\frac 1 n - \frac 1 {2n} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n \paren {\frac 2 {2n} - \frac 2 {2n} + \frac 2 {2n} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Proof 2
We have that:
| \(\ds \int_{-\infty}^\infty \map {\delta_n} x \rd x\) | \(=\) | \(\ds 2n \int_{- \frac 1 n }^{- \frac 1 {2n} } \rd x - n \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \rd x + 2n \int_{\frac 1 {2n} }^{\frac 1 n} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n \paren {2 \paren {- \frac 1 {2n} - \paren {- \frac 1 n} } - \paren {\frac 1 {2n} - \paren {- \frac 1 {2n} } } + 2 \paren {\frac 1 n - \frac 1 {2n} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n \paren {\frac 2 {2n} - \frac 2 {2n} + \frac 2 {2n} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
Let $\map g x = \map \phi x - \map \phi 0$
Then:
| \(\ds \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x\) | \(=\) | \(\ds \map \phi 0 \int_{-\infty}^\infty \map {\delta_n} x \rd x + \int_{-\infty}^\infty \map {\delta_n} x \map g x \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi 0 + \int_{-\infty}^\infty \map {\delta_n} x \map g x \rd x\) |
Furthermore:
| \(\ds \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x\) | \(=\) | \(\ds 2n \int_{- \frac 1 n}^{- \frac 1 {2n} } \map g x \rd x - n \int_{- \frac 1 {2 n} }^{\frac 1 {2n} } \map g x \rd x + 2n \int_{\frac 1 {2 n} }^{\frac 1 n} \map g x \rd x\) |
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 $\ds \delta = \frac 1 N$ where $N \in \R_{> 0}$.
Let $\ds \epsilon' = \frac \epsilon 3$ where $\epsilon \in \R_{> 0}$.
Then
- $\ds \forall \epsilon > 0 : \exists N \in \R_{> 0} : \size x < \frac 1 N \implies \size {\map g x} < \frac \epsilon 3$
Suppose:
- $\ds n \in \N : \size x \le \frac 1 n < \frac 1 N$
That is, suppose:
- $\ds n \in \N : \frac 1 {\size {x} } \ge n > N$
Then:
| \(\ds \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x\) | \(\le\) | \(\ds 2n \int_{- \frac 1 n}^{- \frac 1 {2 n} } \size {\map g x} \rd x + n \int_{- \frac 1 {2n} }^{\frac 1 {2 n} } \size {\map g x} \rd x + 2n \int_{\frac 1 {2 n} }^{\frac 1 n} \size {\map g x} \rd x\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \frac \epsilon 3 \paren {2 n \paren {- \frac 1 {2 n} + \frac 1 n} + n \paren {\frac 1 {2 n} + \frac 1 {2 n} } + 2 n \paren {\frac 1 n - \frac 1 {2 n} } }\) | $\ds \frac 1 {2n} < \frac 1 n < \frac 1 N$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
Since $n$ was arbitrary:
- $\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x\) | \(=\) | \(\ds \map \phi 0\) |
$\blacksquare$