# Proving the Delta Function Composition Identity

Everywhere I've looked online, I see the proof for $$\tag{1}\delta(f(x))=\frac{\delta(x-x_0)}{|f'(x_0)|},$$ where $x_0$ is a root of $f$. The problem I'm working on is slightly different and I have an answer for it, but wanted to confirm it since it is not matching the final answer I'm supposed to get.

Question: Prove $$\tag{2}\delta[f(x')-f(x)]=\frac{\delta(x'-x)}{f'(x')}, \quad x' \mathrm{are\ roots\ of\ f}$$ My attempt: Let $u=f(x')-f(x)$, $du=-f'(x)dx$, $f^{-1}(u)=x'-x$. Then, $$\int \delta[f(x')-f(x)]g(x)dx=-\int \delta(u)g(x'-f^{-1}(u)) \frac{du}{f'(x'-f^{-1}(u)}=-\frac{g(x')}{f'(x')}=-\int \frac{\delta(x'-x)}{f'(x')}g(x)dx.$$ This holds for any $g(x)$, thus we get $$\delta[f(x')-f(x)]=-\frac{\delta(x'-x)}{f'(x')}.$$

Here, I have used $f^{-1}(0)=0$. In conclusion, here you don't end up with the absolute value, and I think you shouldn't because when we do a $u$ sub for solving Eq. (1), $du$ is positive, but here the nature of the problem gives a negative $du$. So I think my answer is right, but it is different than the expected one, Eq. (2). I wanted to seek some confirmation on it.

• The dirac delta function is a distribution, and thus takes a function for it's input. I am not sure what you are doing here. Also, how do you obtain $f^{-1}(u)=x^\prime -x$? Are you assuming that $f$ is linear? – Quoka Mar 18 '18 at 18:16
• I'm not too good in terms of analysis, so my steps can be wrong. – Ptheguy Mar 18 '18 at 18:23
• But the statement (equation (1)) doesn't make sense either. Could you clarify? – Quoka Mar 18 '18 at 18:25
• Isn't that just a property of delta function of a function? Here it is on Wikipedia en.wikipedia.org/wiki/Dirac_delta_function, under composition with a function. – Ptheguy Mar 18 '18 at 18:33
• Okay sorry. I wasn't aware of how the convolution of the dirac delta with a smooth function was defined. Let me know if the answer I posted is what you were looking for – Quoka Mar 18 '18 at 19:36

The composition of the dirac delta function with a smooth function $f$ is defined to be the function $\delta\circ f$ for which the following holds;
For all compactly supported test functions $g$ we have $$\int_{\mathbb{R}^n} [\delta\circ f](x)\cdot g(f(x))\cdot\lvert f^\prime(x)\rvert \,dx =\int_{g(\mathbb{R}^n)}\delta(u)g(u)\,du = g(0)$$
Now, if $x_0$ is a root of $f$ then note that the function $$\frac{\delta(x-x_0)}{|f'(x_0)|}$$ satisfies the above property. Indeed, $$\int_{\mathbb{R}^n} \frac{\delta(x-x_0)}{|f'(x_0)|}g(f(x))\lvert f^\prime(x)\rvert \,dx =\frac{g(f(x_0))\lvert f^\prime(x_0)\rvert}{\lvert f^\prime(x_0)\rvert} = g(0)$$
• The problem with what you are doing is that you set $u = f(x^\prime)-f(x)$ and then write $f^{-1}(u)=x^\prime-x$, which is not usually true for many reasons. First, you don't know that the inverse function of $f$ exists. Even assuming this, you would have $f^{-1}(u)= f^{-1}(f(x^\prime)-f(x)) \neq x^\prime-x$ – Quoka Mar 18 '18 at 23:28
• Ok yeah you're right. I realized I was assuming way too much by my $u$ sub. I figured a way (although it's nor rigorous) for proving what I needed. Thank you – Ptheguy Mar 19 '18 at 3:32