Question about $\delta\big(f(x)\big) = \frac{\delta(x)}{\left|f'(x) \right|}$ (that is-non discrete zeros in f(x))

Question

The following holds for Dirac delta function where $f(x)$ has discrete zeros at $a_i$

$$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}$$

See: Dirac Delta Function of a Function

My question is:

But what about $f(x)$ where the zeros are not discrete, does

$\delta\big(f(x)\big) = \frac{\delta(x)}{\left|f'(x) \right|}$ , where $f'$ is not zero,

hold? If yes how is it derived?

(I would also appreciate references for the non discrete zeros case .)

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Re. defining $\delta(f(x))$ when the zeros are not discrete :

Here is a reference that discusses $\delta(P)$ where $P$ is a hypersurface (eg. spherical shell).

I.M. Gelfand et al "Generalized Functions, volume 1", Academic Press 1964

rough highlights:

p.209: hypersurface definition $P(x1,...xn) = 0$

p.210: require no singular points on $P=0$

p.220: definition of the form $\omega$

(?analogous to Euclidian differential combined with the Jacobian ?)

p.222: $\delta(P)$ definition:

$(\delta(P),\phi )= \int \delta(P)\phi dx)= \int_{P=0} \phi(x)\omega $

So my limited understanding of this is that if the argument of the delta is the equation

of a surface, the integral of the delta with a test function $\phi$ is the surface

integral of $\phi$ over the surface $P$. It seems to me like this could be generalized to volumes, ect..

But I do not see how any of this could lead to $\delta\big(f(x)\big) = \frac{\delta(x)}{\left|f'(x) \right|}$

Answer 1

You need to define $\delta(f(x))$. The obvious definition is for $f$ continuous and $\phi$ smooth supported on some finite interval $[a,b]$

$$\int_{-\infty}^\infty \delta(f(x))\phi(x)dx = \lim_{n \to \infty}\frac{n}{2} \int_{-\infty}^\infty 1_{|f(x)| < 1/n}\ \phi(x)dx$$ whenever the limit converges

That is $\delta(f(x)) = \lim_{n \to \infty} \frac{n}{2}1_{|f(x)| < 1/n}$ with the limit taken in the sense of distributions, assuming it does converge in the sense of distributions.

If $f$ is smooth and $Z(f) = \{c\in \Bbb{R}, f(c) = 0\}$ is discrete and for $c \in Z(f),f' \ne 0$ then the sequence does converge in the sense of distributions, to $$\delta(f(x))= \sum_{c \in Z(f)} \frac{1}{|f'(c)|} \delta(x-c)$$

Note $f$ doesn't need to be smooth as $\delta(f(x)) = \delta(|f(x)|)$.

Let $g(x) = 1 - |x| 1_{|x| < 1}$ and $$f_K(x) = \min_{k=1}^K \min_{m=1}^{2^{k-1}} g( 2^{2^k} (x-\frac{2m-1}{2^k})), \qquad f(x) = \lim_{K \to \infty} f_K(x)$$ Where $\min_{j=1}^l u_j(x) $ means the result of the sequence $v_1(x)=1, v_{j+1}(x) = \min(v_j(x),u_j(x))$.

Then $$\delta(f_K(x)) = \sum_{k=1}^k\sum_{m=1}^{2^{k-1}} \frac{1}{2^{2^k}} \delta(x-\frac{2m-1}{2^k}), \qquad \delta(f(x)) = \sum_{k=1}^\infty\sum_{m=1}^{2^{k-1}} \frac{1}{2^{2^k}} \delta(x-\frac{2m-1}{2^k})$$ which is well-defined even if $f$ has infinitely many zeros.