Power Law Delta Sequence
Theorem
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
- $\ds \map {\delta_n} x := \frac {\size x^{\frac 1 n - 1}} {2 n}$
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 $a \in \R_{> 0}$.
Then:
| \(\ds \int_{-a}^a \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | \(=\) | \(\ds \int_{-a}^0 \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x + \int_0^a \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_{-a}^0 \frac {\paren {-x}^{\frac 1 n - 1} } {2 n} \rd x + \int_0^a \frac {x^{\frac 1 n - 1} } {2 n} \rd x\) | Definition of Absolute Value | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^a \frac {x^{\frac 1 n - 1} } n \rd x\) | Integration by Substitution | |||||||||||
| \(\ds \) | \(=\) | \(\ds \bigintlimits {x^{\frac 1 n} } 0 a\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a^{\frac 1 n}\) |
Furthermore:
- $\ds \forall a \in \R_{> 0} : \lim_{n \mathop \to \infty} a^{\frac 1 n} = 1$
Suppose $a, b \in \R_{> 0} : 0 < a < b$.
Then:
| \(\ds \lim_{n \mathop \to \infty} \int_a^b \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\int_0^b \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x - \int_0^a \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x}\) | Sum of Integrals on Adjacent Intervals for Integrable Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\frac {b^{\frac 1 n} } 2- \frac {a^{\frac 1 n} } 2 }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 - \frac 1 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Analogously, suppose $a < b < 0$.
Then:
| \(\ds \lim_{n \mathop \to \infty} \int_a^b \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_a^0 \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x - \lim_{n \mathop \to \infty} \int_b^0 \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | Sum of Integrals on Adjacent Intervals for Integrable Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_a^0 \frac {\paren {-x}^{\frac 1 n - 1} } {2 n} \rd x - \lim_{n \mathop \to \infty} \int_b^0 \frac {\paren {-x}^{\frac 1 n - 1} } {2 n} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\frac {\paren {-a}^{\frac 1 n} } 2- \frac {\paren {-b}^{\frac 1 n} } 2 }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 - \frac 1 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Let $\epsilon, a \in \R_{> 0}$ such that $\epsilon < a$.
Then:
| \(\ds \lim_{n \mathop \to \infty} \int_{-a}^a \map \phi x \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \int_{-a}^{-\epsilon} \map \phi x \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x + \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \map \phi x \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x + \lim_{n \mathop \to \infty} \int_\epsilon^a \map \phi x \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi {\xi_-} \lim_{n \mathop \to \infty} \int_{-a}^{-\epsilon} \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x + \map \phi {\xi_+} \lim_{n \mathop \to \infty} \int_{\epsilon}^a \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x\) | Mean value theorem for integrals, $\xi_\epsilon \in \closedint {-\epsilon} \epsilon$, $\xi_- \in \hointl {-a} {-\epsilon}$, $\xi_+ \in \hointr \epsilon a$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0 + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-\epsilon}^{\epsilon} \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x + 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi {\xi_\epsilon}\) |
$\epsilon$ 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 $\xi_{+'} \in \hointr {\epsilon'} a$ and $\xi_{-'} \in \hointl {-a} {-\epsilon'}$.
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 $\xi_\epsilon = 0$.
Taking the limit $a \to \infty$ in the above expressions yields the desired result.
 | Further research is required in order to fill out the details. In particular: The above results depdends on the order of limits wrt a and n. It would be nice to find exactly, in which sense the above is an example for a delta sequence You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by finding out more. 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 {{Research}} from the code. |
$\blacksquare$
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