Composition of Dirac Delta Distribution with Function with Simple Zero

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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

Dirac delta distribution is even:

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

which can be abreviated 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:

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

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:

$\ds T_{\map H x}' = \delta_0 = \delta_{x}$

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

$\blacksquare$