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 $\phi \in \map \DD \R$ 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:


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

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

Then:

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

Then:

Altogether: