# Power Law Delta Sequence

## Theorem

The graph of the $\ds \frac {\size x^{\frac 1 n - 1}} {2 n}$ 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:

$\ds \map {\delta_n} x := \frac {\size x^{\frac 1 n - 1}} {2 n}$

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 $\xi_\epsilon \ne 0$.

$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$

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.

$\blacksquare$