5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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

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

Therefore

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

Which concludes our proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.