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I found this question where a nice formula is given for the composition $\delta(f(x))$. Is there a similar general formula for $\delta'(f(x))$? In other words, is there a nice way to express the integral

$$ \displaystyle\int_{\mathbb{R}^n} \delta'(f(x))g(x) dx,$$

where $f,g:\mathbb{R}^n\to \mathbb{R}$?

cav
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  • With $\delta'(f(x)$ do you mean $\left.\frac{d}{dx}\delta(x)\right|_{x = f(\cdot)}$ or $\frac{d}{dx}[\delta(f(x))]$? In any case see https://en.wikipedia.org/wiki/Dirac_delta_function#Distributional_derivatives In the latter case you have to use the chain-rule first. – Winther Aug 10 '17 at 13:41
  • I mean $\left.\frac{d}{dx}\delta(x)\right|_{x = f(\cdot)}$ – cav Aug 10 '17 at 13:44
  • Start with a rigorous definition of the Dirac delta $$\int_{-\infty}^\infty \delta(x) \phi(x)dx \overset{def}= \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty \frac{1_{|x| < \epsilon}}{2 \epsilon} \phi(x)dx$$ To make sense to $\delta'(x)$ you can replace $\frac{1_{|x| < \epsilon}}{2 \epsilon}$ by a smoothed version $\varphi_\epsilon(x) = \frac{1}{\epsilon} e^{-\pi x^2/\epsilon^2}$ – reuns Aug 10 '17 at 13:48
  • The standard rigorous definition of $\delta$ is as a linear functional acting on a "nice" function $\phi \in C_c^\infty(\mathbb R^n)$ in such way that $\langle \delta, \phi \rangle = \phi(0)$. The derivative is defined by $\langle \delta', \phi \rangle = - \langle \delta, \phi' \rangle$. – md2perpe Aug 10 '17 at 15:05
  • @md2perpe What is $\delta(f(x))$ then ? Starting from $\delta(x) = \lim_{\epsilon \to 0^+} \frac{1}{\epsilon} e^{-\pi x^2/\epsilon^2}$ has many advantages. In particular $(\delta(f(x))' = \delta'(f(x)) f'(x)$ and $\langle \delta'(f(x)),\phi \rangle = -\langle \delta(f(x),(\frac{\phi(x)}{ f'(x)})' \rangle$ – reuns Aug 10 '17 at 17:04
  • Motivated by what happens for ordinary functions, for a distribution $u$ I would define $u \circ f,$ where $f \in C^\infty$ with $f' > 0$, through $$\langle u \circ f, \phi \rangle = \langle u, \frac{\phi \circ f^{-1}}{f' \circ f^{-1}} \rangle$$ The definition can be extended to cover some other cases where $f' \not > 0$. – md2perpe Aug 10 '17 at 18:42
  • With $f' \not > 0$ I meant that $f' > 0$ isn't globally satisfied, not that we globally have $f' \leq 0$. – md2perpe Aug 10 '17 at 18:49
  • @md2perpe What you wrote is the best way to see people saying $\delta(x) = \delta(2x)$. Why hide that distributions are just limits of functions sequences (in the topology induced by the pairing) ? – reuns Aug 10 '17 at 20:50
  • How do you get $\delta(x) = \delta(2x)$ from my definition of $u \circ f$? And why say that distributions are limits of functions when there is a more precise definition? – md2perpe Aug 10 '17 at 20:57

1 Answers1

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We start by looking at a derivation of the given formula for $\delta(f(x))$ in one dimension:

Let $I$, $J$ be intervals on $\mathbb R$ and assume that $f : I \to J$ is differentiable and invertible. If $0 \in J$ we then formally have $$ \int_I \delta(f(x)) \, \phi(x) \, dx = \{ x = f^{-1}(y) \} = \int_J \delta(y) \, \phi(f^{-1}(y)) \, \frac{dy}{|f'(f^{-1}(y))|} \\ = \phi(f^{-1}(0)) \, \frac{1}{|f'(f^{-1}(0))|} = \phi(x_0) \, \frac{1}{|f'(x_0)|} \\ $$ where $x_0 = f^{-1}(0)$ is the zero of $f$.

Generally, if $f : \mathbb R \to \mathbb R$ is differentiable and have isolated zeros so that we can take intervals $I_n$ containing only one zero each, say $x_n$, then we get $$ \int_{-\infty}^{\infty} \delta(f)(x)) \, \phi(x) \, dx = \sum_n \int_{I_n} \delta((f|I_n)(x)) \, \phi(x) \, dx = \sum_n \phi(x_n) \frac{1}{|f'(x_n)|} \\ = \sum_n \int_{-\infty}^{\infty} \frac{1}{|f'(x_n)|} \delta(x-x_n) \, \phi(x) \, dx = \int_{-\infty}^{\infty} \sum_n \frac{1}{|f'(x_n)|} \delta(x-x_n) \, \phi(x) \, dx. $$ Since this is valid for any "nice" $\phi$ we have $$\delta(f(x)) = \sum_n \frac{1}{|f'(x_n)|} \delta(x-x_n).$$

Now we instead consider $\delta'(f(x))$, still in one dimension:

Here we need to know that $$\int_{-\infty}^{\infty} \delta'(t) \, g(t) \, dt = -g'(0)$$

In this case the first calculation becomes $$\begin{align} \int_I \delta'(f(x)) \, \phi(x) \, dx & = \left. -\frac{d}{dy} \left( \phi(f^{-1}(y)) \, \frac{1}{|f'(f^{-1}(y))|} \right) \right|_{y=0} \\ & = \left. - \left( \phi'(f^{-1}(y)) \frac{1}{f'(f^{-1}(y))} \cdot \frac{1}{|f'(f^{-1}(y))|} + \phi(f^{-1}(y)) \cdot \frac{-f''(f^{-1}(y))}{|f'(f^{-1}(y))|^3} \right) \right|_{y=0} \\ & = - \left( \phi'(x_0) \frac{1}{f'(x_0)} \cdot \frac{1}{|f'(x_0)|} + \phi(x_0) \cdot \frac{-f''(x_0)}{|f'(x_0)|^3} \right) \\ & = \frac{f''(x_0)}{|f'(x_0)|^3} \phi(x_0) - \frac{f'(x_0)}{|f'(x_0)|^3} \phi'(x_0) \end{align}$$

Therefore our final formula becomes $$\delta'(f(x)) = \sum_n \frac{f''(x_n) \delta(x-x_n) + f'(x_n) \delta'(x-x_n)}{|f'(x_n)|^3}$$

Enrico M.
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md2perpe
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  • Thank you - do you know if a similar argument could be used in higher dimensions? – cav Aug 11 '17 at 12:03
  • I think so. The change of variable will however involve spaces of different dimensions which might complicate things. – md2perpe Aug 11 '17 at 15:26