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In a comment on Math.SE, I came today across the remark that the expression $\delta(f(x))$ has no meaning. Being a physicist, I have been surprised but I then remembered Schwartz distribution theory that only defines $\delta$ using the bracket $\langle \delta,\,\varphi\rangle=\varphi(0)$ for a "test-function" $\varphi$ (infinitely differentiable). In physics, we often use $\delta$ functions in order to express constraints without applying to a test-function.

On Math.SE, I found only one question related to calculus using $\delta$'s, but the references in there are actually not addressing the composition of $\delta$ with a function...

However, things seem pretty natural. There are only two simple and acceptable rules to state : extend the translation rule valid for regular functions $f$ $$\langle f,\,T_a\varphi\rangle=\int f(x)\varphi(x+a)\mathrm d x=\int f(x'-a)\varphi(x')\mathrm dx'=\langle T_{-a}f,\,\varphi\rangle$$ to $\delta$, and do the same for the dilation rule $$\langle f,\,D_\lambda\varphi\rangle=\int f(x)\varphi(\lambda x)\mathrm d x=\frac{1}{|\lambda|}\int f\left(-\frac{x'}{|\lambda|}\right)\varphi(x')\mathrm dx'=\frac{1}{|\lambda|}\langle D_{1/|\lambda|}f,\,\varphi\rangle.$$

Considering these points, it is well founded I think to write an expression like $\delta(ax+b)$ for $a\neq0$ and define it by $$\int\delta(ax+b)\varphi(x)\mathrm dx=\frac{1}{|a|}\varphi\left(-\frac ba\right).\tag1$$ This rule (1) is applied in several physics papers, but seems to be absent from the mathematics textbooks. It is sometimes very useful, even if one needs to be careful. So here comes at last my question : are there any mathematical textbooks addressing these points ? If yes which ones and if not, why ?

Notee added in edit The implication of being able to write $\delta(ax+b)$ is that one can write expressions like, for instance, $$\int\delta(x_1-x_2\xi)\delta(y_1-y_2\xi)\mathrm d\xi=\delta\left(\left\lvert\begin{smallmatrix}x_1&x_2\\y_1&y_2\end{smallmatrix}\right\rvert\right), \quad (y_1,\,y_2\neq0)$$ that encodes the colinearity constraint for vectors in $\mathbb R^2$.

Tom-Tom
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    Of course $\delta(f(x))$ has a meaning. For example through the useful formula recalled here (https://math.stackexchange.com/questions/662226/how-to-write-delta-fx-in-terms-of-delta-x). Some "pure spirits" in their ivory tower will sent to hell people using it, but it is a very handy way to express complicated though meaningful situations. But (there is a "but"), one should be very careful when using it in computations (the floor can be very slippery...) – Jean Marie Dec 12 '17 at 22:43
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    You can compose distributions with diffeomorphisms, so $(1)$ is fine. For a locally integrable function $f\colon U \to \mathbb{C}$ and a diffeomorphism $h \colon V \to U$, the composition $f\circ h$ is locally integrable, so for $\varphi \in \mathscr{D}(V)$ we can look at $$\int_V f(h(y))\varphi(y),dy,.$$ Per change-of-variable, $y = h^{-1}(x)$, that becomes $$\int_U f(x)\varphi(h^{-1}(x))\lvert J_{h^{-1}}(x)\rvert,dx,.$$ Since $x \mapsto \varphi(h^{-1}(x))\cdot \lvert J_{h^{-1}}(x)\rvert \in \mathscr{D}(U)$, this shows that – Daniel Fischer Dec 12 '17 at 22:45
  • $$(T\circ h)[\varphi] = T[(\varphi \circ h^{-1})\cdot \lvert J_{h^{-1}}\rvert]$$ is well-defined. One verifies that this defines a distribution on $V$ if $T\in \mathscr{D}'(U)$. One can generalise that further, but it makes sense only for nice enough functions. Things go totally haywire if one tries to compose distributions with arbitrary functions. – Daniel Fischer Dec 12 '17 at 22:45
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    That last formula is nonsense though, you can't let $\delta$ act on itself (or multiply them), at least not in the setting of distribution theory. Since $\delta$ is a compactly supported distribution of order $0$, you may pair it with functions that are continuous, but not in general with other distributions. – mrf Dec 13 '17 at 22:21
  • The product of two distributions is a whole 'nother can o' worms. – Daniel Fischer Dec 13 '17 at 22:22
  • @mrf. Nope. It is absolutely correct ! – Tom-Tom Dec 13 '17 at 22:26
  • @mrf. The expression above is simply a convolution by a delta, that is namely, as written above, a translation and a dilation. And the translation and dilation of a delta is a delta. – Tom-Tom Dec 13 '17 at 22:29
  • I repeat myself in all those threads : pick a concrete definition of $\delta$, for example : $\int_{-\infty}^\infty \delta(x) \varphi(x)dx = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty \frac{1_{|x| < \epsilon}}{2\epsilon} \varphi(x)dx$. The limit exists and is $\varphi(0)$ whenever $\varphi$ is continuous. Of course you can extend the definition to $\int_{-\infty}^\infty \delta(f(x))\varphi(x)dx=\lim_{\epsilon \to 0^+}\int_{-\infty}^\infty \frac{1_{|f(x)| <\epsilon}}{2\epsilon}\varphi(x)dx$, and show if it defines a distribution, then it is compatible with the change of variable – reuns Dec 13 '17 at 22:37
  • @JeanMarie: The people in the "ivory tower" have no problem with $\delta(f(x))$. The problem is not that the "floor is slippery" -- the problem is that people never bother saying where the floor actually is. And honestly, I doubt the floor is very slippery at all once you actually know where it is; I expect it only seems slippery to those who are never taught where to tread and are only taught to mimic someone else's footsteps. Which is most people, I think. –  Dec 14 '17 at 03:51

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