# Dirac's delta composition with function

Reading this I see that the statement:

$\delta \left( f(x) \right) = \sum_i \dfrac{\delta(x - a_i)}{|f'(a_i)|}$

is equivalent to showing that:

$\int_{-\infty}^{\infty} g(x)\delta \left( f(x) \right) = \sum_i \dfrac{g(a_i)}{|f'(a_i)|}$

Where $f(a_i) = 0 \:\: \forall i$

Can somebody explain to me why these two statements are equivalent? Thanks in advance

(I understand some of the proofs in that post, I just don't get why the statements are the same)

• You misquoted the statement, but if you understood even a single one of those proofs, you understood why it's true (that's the meaning of "proof"). If you didn't understand them, why don't you ask their authors, in that thread? – Professor Vector Aug 19 '17 at 14:57
• math.stackexchange.com/questions/2389100/… – md2perpe Aug 19 '17 at 17:02

This is the definition of the Dirac delta : $$\int_{-\infty}^\infty g(x) \delta(x)dx = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty g(x) \frac{1_{|x| < \epsilon}}{2 \epsilon}dx=g(0)$$ whenever $g$ is continuous, which means $\delta(x)$ is the limit in the sense of distributions of $\frac{1_{|x| < \epsilon}}{2 \epsilon}$ as $\epsilon \to 0^+$.
Thus it is natural to define $\delta(f(x))$ as the limit in the sense of distributions of $\frac{1_{|f(x)| < \epsilon}}{2 \epsilon}$ as $\epsilon \to 0^+$, which means $$\int_{-\infty}^{\infty} g(x)\delta \left( f(x) \right)dx =\lim_{\epsilon \to 0^+}\int_{-\infty}^{\infty} g(x)\frac{1_{|f(x)| < \epsilon}}{2 \epsilon}dx= \sum_{f(\alpha)=0} \dfrac{g(\alpha)}{|f'(\alpha)|}\\=\int_{-\infty}^\infty g(x)\sum_{f(\alpha)=0} \dfrac{\delta(x-\alpha)}{|f'(\alpha)|} dx$$ whenever $g$ is continuous and $f$ is $C^1$ and has finitely many zeros.