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Let $$G = \left\{ f : [0,1] \to (0,\infty)\text{ Lebesgue measurable and }\int_{[0,1]} fd\mu=1\right\}$$

Question: is $$f \odot g = f\left(\int_0^x g(y)dy\right) g(x)$$ a group law on $G$ with identity $1$ and inverse $\frac1{f(F^{-1}(x))}, F(x)=\int_0^x f(y)dy$ ?

ie. when integrating the elements of $G$ we'd obtain a subgroup of the group $C$ of bijective strictly increasing continuous functions $[0,1] \to [0,1]$ with group law given by composition. The map $G \to C$ is not surjective because of those kind of things constructed from the Cantor function.

Both $f\left(\int_0^x g(y)dy\right) g(x)$ and $\frac1{f(F^{-1}(x))}$ are measurable so the problem is to find if they integrate to $1$ and trying to produce a counter-example if they don't.

reuns
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  • I wonder whether $f\odot f$ is in $L^1$ when we pick some $f\in L^1\setminus L^2$, for example $f(x) = \frac 12x^{-1/2}$. – amsmath Sep 07 '19 at 17:04
  • I checked it for $f(x) = cx^{-1+\epsilon}$. They're all fine. – amsmath Sep 07 '19 at 17:15
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    The inverse is $\frac 1{f\circ h^{-1}}$, where $h(x) = \int_0^xf(y),dy$. So, the one given there is false. – amsmath Sep 07 '19 at 17:25
  • sure just a typo sorry – reuns Sep 07 '19 at 17:39
  • @reuns not sure where you got this problem from (I hope it is not a homework question), but everything becomes pretty obvious when you realize that $f\odot g=(I_f\circ I_g)'$ where $I_f,I_g$ denote the integrals. See my answer for details. – pre-kidney Sep 07 '19 at 17:46
  • @amsmath It is easy to see the functions continuous on $[0,1]$ minus finitely many points are a subgroup of $G$ (because in that case improper Riemann integral =Lebesgue integral). Usually to solve $L^1$ problems I take a sequence of continuous functions converging to it but here everything is not quite continuous – reuns Sep 07 '19 at 17:55
  • @pre-kidney I didn't mention that morally we should have $I_{f\odot g}=I_f\circ I_g$ because I was thinking it was obvious, sorry (see the part about composition of continuous functions) since the beginning my question was if the chain rule works for the distributional derivative of $I_f \circ I_g$ which isn't elementary (see the part about Cantor's function) and r9m's link showing $V\circ U$ is not absolutely continuous with $U(x)=x^2 \sin^2 1/x, V(x)=x^{1/2}$. – reuns Sep 07 '19 at 19:53
  • @amsmath Tks a lot for your research, I didn't write an answer yet because I'm searching for a simple argument showing that the map $\int : L^1 \to C^0$ has continuous inverse when we restrict to subsets satisfying a $\sum_j y_j-x_j < \delta(\epsilon) \implies \sum_j |I_f(y_j)-I_f(x_j)| < r\epsilon$ condition, as well as a simple argument for the AC of $I_f^{-1}$ – reuns Sep 08 '19 at 17:52

1 Answers1

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The first thing to check is that the operation $\odot $ is closed, i.e. that if $f,g$ are positive and measurable with $\|f\|_{L^1([0,1])}=\|g\|_{L^1([0,1])}=1$ then the same holds for $f\odot g$. Let $I_f(x)=\int_0^x f(y)\ dy$ and similarly for $I_g$. Then since $I_f,I_g$ are increasing functions on $[0,1]$ with $I_f(0)=I_g(0)=0$ and $I_f(1)=I_g(1)=1$, $$ \int_0^1 (f\odot g)(x)\ dx=\int_0^1 f(I_g(x))I_g'(x)\ dx=[I_f(I_g(x))]_0^1=1. $$ So the group operation is closed.

Showing that the constant function $1$ is the identity is easy, since $ 1\odot g=g $ is immediate and $(f\odot 1)(x)=f(\int_0^x 1\ dy)=f(x)$.

Before computing the inverse, let me first point out that the group law can be equivalently written as $f\odot g=(I_f\circ I_g)'$. Thus if we want $(f\odot g)(x)=1$, it is equivalent to $I_f\circ I_g=x$, i.e. $I_g(x)=I_f^{-1}(x)$. Therefore $$ g(x)=I_g'(x)=\frac{d}{dx}I_f^{-1}(x)=\frac{1}{f(I_f^{-1}(x))} $$ is the inverse, and it is seen that $f\odot g=g\odot f=1$.

pre-kidney
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  • I believe "closed" is the wrong term. You're checking that it's well-defined (i.e., that the image of the domain actually lies within the specified codomain). You seem to have skipped over the part about the inverse. – John Hughes Sep 07 '19 at 17:25
  • @JohnHughes The term "closed" is fine in this context. However, the invese is indeed missing. In the comments I provided the correct inverse. – amsmath Sep 07 '19 at 17:27
  • @JohnHughes I am using the term "closed" in the same manner as here https://en.wikipedia.org/wiki/Group_(mathematics)#Definition – pre-kidney Sep 07 '19 at 17:28
  • Yeah ... Wikipedia's not a compelling answer to me for this one. I would say, for instance, that the set of even integers is closed under the operation "plus" defined on the integers. In other words, a subset is closed under a particular operation defined on some set. But for that to make sense, the operation must be well-defined. So I'd claim that it's wrong to say that the "reals aren't closed under the operation $x \mapsto 1/x$", but one should instead say that the operation $x \mapsto 1/x$ is not defined for all reals. Presumably this is a potato/potahto issue, so I'll leave it at that – John Hughes Sep 07 '19 at 17:34
  • In this case the group operation is defined as a binary operation on the larger space $L^1([0,1])$, and we are showing that it restricts to a well-defined on operation on the subspace of positive elements with norm $1$. – pre-kidney Sep 07 '19 at 17:38
  • @amsmath I have finished writing the portion of my answer regarding the inverse – pre-kidney Sep 07 '19 at 17:44
  • How do you know $F(G(x))' = F'(G(x))G'(x)$ works for $F,G$ absolutely continuous (equivalently (?) how do you know $F(G(x))$ is absolutely continuous ?) – reuns Sep 07 '19 at 17:46
  • By the fundamental theorem of Lebesgue integral calculus, as described here https://en.wikipedia.org/wiki/Absolute_continuity#Equivalent_definitions – pre-kidney Sep 07 '19 at 17:48
  • ??? It is not about composition. The word "composition" doesn't appear in https://en.wikipedia.org/wiki/Absolute_continuity#Equivalent_definitions – reuns Sep 07 '19 at 17:53
  • That page shows you that a function is AC if and only if it is the Lebesgue integral of a measurable function. If you are happy with that, then the chain rule claim is a basic property of the Radon-Nikodym derivative, which is covered in any measure theoretic real analysis textbook (for instance Folland's book) and also blogged about online in places like here https://unapologetic.wordpress.com/2010/07/12/the-radon-nikodym-chain-rule/ – pre-kidney Sep 07 '19 at 18:00
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    Anyway, it is clear that I answered your initial question and you have been moving the goalposts on your question with slight modifications after seeing my answer (and other comments as well). If your actual question is about the chain rule, you should ask that in a separate post - but this one is answered and should be marked as such... – pre-kidney Sep 07 '19 at 18:02
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    This seems to provide a counter-example for this context. Of course we can choose a positive $h$ (in the linked post) to have the same effect. – r9m Sep 07 '19 at 18:04
  • @pre-kidney ??? Not at all, this is a difficult problem that I created. If you think the Radon-Nikodym derivative (equivalently the distributional derivative) helps then show us – reuns Sep 07 '19 at 18:04
  • @r9m I think non-negativity of the derivative changes everything – reuns Sep 07 '19 at 18:11
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    @reuns I think the absolute continuity of $I_f\circ I_g$ follows by definition of abs. continuity (and the fact that $I_g$ is non-decreasing).: $\sum|I_f(I_g(a_{n+1}))-I_f(I_g(a_n))|<\epsilon$ if $\sum(I_g(a_{n+1})-I_g(a_n)) < \delta$ if $\sum(a_{n+1}-a_n)<\delta'$. – amsmath Sep 07 '19 at 20:09
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    @pre-kidney Where did you show that the inverse is contained in $G$? – amsmath Sep 07 '19 at 20:15
  • @reuns $I_g$ is absolutely continuous. So, of course it does imply that by definition. Here is the definition: https://en.wikipedia.org/wiki/Absolute_continuity#Definition – amsmath Sep 07 '19 at 20:32
  • @amsmath Great I think it works tks, so $U \circ V$ is AC whenever $U,V$ are AC and $V' \ge 0$ – reuns Sep 07 '19 at 20:40
  • @reuns Exactly. You only need non-negativity for the derivative. Now, for the inverse you only need to show that $V^{-1}$ is also $AC$ (if $V'>0$ a.e.) This seems to be harder to me, though. – amsmath Sep 07 '19 at 20:46
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    @reuns For the latter see here: https://math.stackexchange.com/questions/1544618/is-the-inverse-of-a-absolutely-continuous-function-with-almost-everywhere-positi – amsmath Sep 07 '19 at 21:26