# Composition of Dirac Delta Distribution with Function with Simple Zero

## Theorem

Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.

Let $\sequence {\map {\delta_n} x}_{n \mathop \in \N}$ be a delta sequence.

Let $f : \R \to \R$ be a real function with a simple zero at $x_0$.

Let $f$ be strictly monotone.

Let $\phi \in \map \DD \R$ be a test function.

Then in the distributional sense it holds that:

$\ds \map \delta {\map f x} = \frac {\map \delta {x - x_0}}{\size {\map {f'} {x_0}} }$

which can be interpreted as:

$\ds \int_{-\infty}^\infty \map \delta {\map f x} \map \phi x \rd x = \frac {\map \phi {x_0}}{\size {\map {f'} {x_0}} }$

which more strictly means that:

$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} {\map f x} \map \phi x \rd x = \frac {\map \phi {x_0}}{\size {\map {f'} {x_0}} }$

### Corollary

$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} {-x} \map \phi x \rd x = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x$

which can be abbreviated to:

$\map \delta {-x} = \map \delta x$

## Proof 1

Suppose $\map {f'} {x_0} > 0$.

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{\infty} \map {\delta_n} {\map f x} \map \phi x \rd x$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{\infty} \map {\delta_n} y \frac {\map \phi {\map x y} } {\map {f'} {\map x y} } \rd y$ Derivative of Inverse Function, Integration by Substitution, $y = \map f x$ $\ds$ $=$ $\ds \frac {\map \phi {x_0} }{\map {f'} {x_0} }$ Definition of Delta Sequence, $\map x 0 = x_0$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \frac {\map \phi x \map {\delta_n} {x - x_0} } {\map {f'} {x_0} } \rd x$

Suppose $\map {f'} {x_0} < 0$.

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{\infty} \map {\delta_n} {\map f x} \map \phi x \rd x$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{\infty}^{-\infty} \map {\delta_n} y \frac {\map \phi {\map x y} } {\map {f'} {\map x y} } \rd y$ Derivative of Inverse Function, Integration by Substitution, $y = \map f x$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{\infty} \map {\delta_n} y \frac {\map \phi {\map x y} } {-\map {f'} {\map x y} } \rd y$ $\ds$ $=$ $\ds \frac {\map \phi {x_0} }{- \map {f'} {x_0} }$ Definition of Delta Sequence, $\map x 0 = x_0$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \frac {\map \phi x \map {\delta_n} {x - x_0} } {-\map {f'} {x_0} } \rd x$

Altogether:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{\infty} \map {\delta_n} {\map f x} \map \phi x \rd x$ $=$ $\ds \frac {\map \phi {x_0} } {\size { \map {f'} {x_0} } }$ Definition of Delta Sequence, $\map x 0 = x_0$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \frac {\map \phi x \map {\delta_n} {x - x_0} } {\size {\map {f'} {x_0 } } } \rd x$

$\blacksquare$

## Proof 2

Let $H$ be the Heaviside step function.

Let $T \in \map \DD \R$ be a distribution associated with $\map H {\map f x}$:

$\ds T = T_{\map H {\map f x}}$

We have that:

 $\ds \map H {\map f x}$ $=$ $\ds \begin{cases} \map H {x - x_0} & : \forall x \in \R : \map {f'} x > 0 \\ 1 - \map H {x - x_0} & : \forall x \in \R : \map {f'} x < 0 \end{cases}$

Taking the derivative of the left hand side yields:

 $\ds \dfrac \d {\d x} \map H {\map f x}$ $=$ $\ds \dfrac {\d \map f x} {\d x} \dfrac {\d \map H {\map f x} } {\d \map f x}$ $\ds$ $=$ $\ds \map {f'} x \dfrac {\d \map H {\map f x} } {\d \map f x}$

Taking the derivative of the right hand side yields:

$\forall x \ne x_0 : \dfrac \d {\d x} \map H {\map f x} = 0$

Furthermore:

$\forall x \in \R : \map {f'} x > 0 : \paren {\map f {x_0^+} = 1} \land \paren {\map f {x_0^-} = 0}$
$\forall x \in \R : \map {f'} x < 0 : \paren {\map f {x_0^+} = 0} \land \paren {\map f {x_0^-} = 1}$

By Jump Rule the right hand side reads:

 $\ds T_{ \map H {\map f x} }'$ $=$ $\ds \begin{cases} \delta_{x_0} & : \forall x \in \R : \map {f'} x > 0 \\ -\delta_{x_0} & : \forall x \in \R : \map {f'} x < 0 \end{cases}$

Define the composite Dirac delta distribution according to Distributional Derivative of Heaviside Step Function.

We have that:

$T_{\map H x}' = \delta_0 = \delta_x$

Then:

 $\ds T_{\map H {\map f x} }'$ $=$ $\ds T_{\map {f'} x \frac {\d \map H {\map f x} } {\d \map f x} }$ Derivative of Composite Function $\ds$ $=$ $\ds \map {f'} x \delta_{\map f x}$ Multiplication of Distribution induced by Locally Integrable Function by Smooth Function

Hence:

 $\ds \forall x \in \R : \map {f'} x$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds \size {\map {f'} x} \delta_{\map f x}$ $=$ $\ds \delta_{x_0}$ $\ds \leadsto \ \$ $\ds \delta_{\map f x}$ $=$ $\ds \frac {\delta_{x_0} } {\size {\map {f'} x} }$ $\ds$ $=$ $\ds \frac {\delta_{x_0} } {\size {\map {f'} {x_0} } }$ Product of Smooth Function and Dirac Delta Distribution

and:

 $\ds \forall x \in \R : \map {f'} x$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds -\size {\map {f'} x} \delta_{\map f x}$ $=$ $\ds -\delta_{x_0}$ $\ds \leadsto \ \$ $\ds \delta_{\map f x}$ $=$ $\ds \frac {\delta_{x_0} } {\size {\map {f'} x} }$ $\ds$ $=$ $\ds \frac {\delta_{x_0} } {\size {\map {f'} {x_0} } }$ Product of Smooth Function and Dirac Delta Distribution

$\blacksquare$