How can I prove this property for delta function? ($a$ is a root of $f$)

$$\delta(f(x)-f(a)) = \frac{\delta(x-a)}{|f ' (a)|}$$

I tried to prove it by expanding $f$ using Taylor series, and inserting in Fourier integral, but I could not prove it. I would greatly appreciate if someone could please provide a proof.

  • $\begingroup$ Try a change of variable. And use the definition of the delta function in terms of what you get when you integrate it multiplied by a test function. $\endgroup$
    – user14972
    Oct 5 '14 at 12:23
  • $\begingroup$ tnx for your comment. something like f(x)=u ? $\endgroup$
    – Mr. Nobody
    Oct 5 '14 at 12:24
  • $\begingroup$ i did not get any result... $\endgroup$
    – Mr. Nobody
    Oct 5 '14 at 12:44
  • $\begingroup$ Yes, that substitution should be useful. $\endgroup$
    – user14972
    Oct 5 '14 at 12:44

To start, I will list three other identities that will be used later. $$1) \quad \int_{a}^{b} \delta(x)g(x) dx = \begin{cases} 0, & 0 \notin (a,b) \\ g(0), & 0 \in (a,b) \end{cases} $$ $$2) \quad \int \delta(px)g(x) dx = \frac{1}{|p|} \int \delta(x)g(x) dx $$ $$3) \quad \int_{a}^{b} \delta(x-p)g(x) dx = \begin{cases} 0, & p \notin (a,b) \\ g(p), & p \in (a,b) \end{cases} $$

I will neglect the proofs to these, but they are all fairly straightforward.

To prove your property

$$\delta(f(x))= \frac{\delta(x-x_0)}{|f'(x_0)|} $$

We will multiply both sides by some function $g(x)$, integrate from $a$ to $b$ with respect to $x$, and use property $(3)$ on the right hand side to get the expression

$$\int_{a}^{b} \delta(f(x))g(x)dx = \frac{g(x_0)}{|f'(x_0)|} $$

Where $f(x_0)=0$ and $x_0 \in (a,b)$. Therefore, proving this identity is equivalent to proving the property you want.

Now, can split the integral on the left side into

$$\int_{a}^{b} \delta(f(x))g(x)dx = \int_{a}^{x_0-\epsilon} \delta(f(x))g(x)dx + \int_{x_0-\epsilon}^{x_0+\epsilon} \delta(f(x))g(x)dx + \int_{x_0+\epsilon}^{b} \delta(f(x))g(x)dx $$

For some arbitrarily small $\epsilon$. Now, we know that $f(x)$ does not have a zero in the interval $(a,x_0-\epsilon)$ or $(x_0+\epsilon,b)$, so we know from property $(1)$ that these two integrals equal $0$. We now have

$$\int_{a}^{b} \delta(f(x))g(x)dx = \int_{x_0-\epsilon}^{x_0+\epsilon} \delta(f(x))g(x)dx$$

Taking the Taylor series of $f(x)$ centered at $x_0$ up to the first order yields

$$\begin{aligned} \int_{x_0-\epsilon}^{x_0+\epsilon} \delta(f(x))g(x)dx =& \int_{x_0-\epsilon}^{x_0+\epsilon} \delta \big(f(x_0)+f'(x_0)(x-x_0) \big)g(x)dx \\ =& \int_{x_0-\epsilon}^{x_0+\epsilon} \delta \big(f'(x_0)(x-x_0) \big)g(x)dx \\ =& \frac{1}{|f'(x_0)|}\int_{x_0-\epsilon}^{x_0+\epsilon} \delta (x-x_0)g(x)dx \end{aligned}$$

From property (2). Also, since $0 \in (-\epsilon, \epsilon)$, we can use property (3) to conclude that

$$ \int_{x_0-\epsilon}^{x_0+\epsilon} \delta(f(x))g(x)dx = \frac{1}{|f'(x_0)|}g(x_0) $$


$$\int_{a}^{b} \delta(f(x))g(x)dx = \frac{g(x_0)}{|f'(x_0)|}$$

Which concludes our proof.


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