Theorem: Power Law Delta Sequence

thumb600pxThe graph of the $\dfrac {\size x^{\frac 1 n - 1} } {2 n}$ delta sequence. As $n$ grows, the graph becomes steeper and thinner. The area under each graph is infinite unless the range of integration is a finite interval say, $\closedint {-a} a$. However, in the limit $n \to \infty$, this area approaches $1$ regardless of $a$.
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\map {\delta_n} x := \dfrac {\size x^{\frac 1 n - 1} } {2 n}$
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
:$\map \delta x$ 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:
{{begin-eqn}}
{{eqn | l = \int_{-a}^a \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x
      | r = \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
}}
{{eqn | r = \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
      | c = {{Defof|Absolute Value|index = 1}}
}}
{{eqn | r = \int_0^a \frac {x^{\frac 1 n - 1} } n \rd x
      | c = Integration by Substitution
}}
{{eqn | r = \bigintlimits {x^{\frac 1 n} } 0 a
}}
{{eqn | r = a^{\frac 1 n}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_a^b \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x
      | r = \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}
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\frac {b^{\frac 1 n} } 2- \frac {a^{\frac 1 n} } 2}
}}
{{eqn | r = \frac 1 2 - \frac 1 2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Analogously, suppose $a < b < 0$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_a^b \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x 
      | r = \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
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \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
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\frac {\paren {-a}^{\frac 1 n} } 2 - \frac {\paren {-b}^{\frac 1 n} } 2}
}}
{{eqn | r = \frac 1 2 - \frac 1 2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Let $\epsilon, a \in \R_{> 0}$ such that $\epsilon < a$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-a}^a \map \phi x \frac {\size x^{\frac 1 n - 1} } {2 n} \rd x
      | r = \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
}}
{{eqn | r = \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
      | c = Mean value theorem for integrals, $\xi_\epsilon \in \closedint {-\epsilon} \epsilon$, $\xi_- \in \hointl {-a} {-\epsilon}$, $\xi_+ \in \hointr \epsilon a$
}}
{{eqn | r = 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
}}
{{eqn | r = \map \phi {\xi_\epsilon}
}}
{{end-eqn}}
We have {{hypothesis}} that $\epsilon$ 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 {{WRT}} $\epsilon'$ we have that:
:$\xi_\epsilon = \xi_{+'}$
or:
:$\xi_\epsilon = \xi_{-'}$
where:
:$\xi_{+'} \in \hointr {\epsilon'} a$
:$\xi_{-'} \in \hointl {-a} {-\epsilon'}$
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.
{{Research|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}}
{{qed}}
{{NoSources}}
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
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