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I am having trouble with Rudin's proof of the change of variable theorem for multiple integrals. The theorem is for 1-1 $\mathscr{C'}$ mappings from $R^k$ into $R^k$. In theorem 10.7 just before the change of variable theorem, he proves that if $\mathbf{F}(\mathbf{x})$ is a $\mathscr{C'}$ mapping of an open set $E\subset{R^k}$ into $R^k$ with $0\in{E}$, with $\mathbf{F}(\mathbf{0})=0$ and $\mathbf{F'}(0)$ invertible, then there is a neighborhood of $\mathbf{0}$ in which the representation $$\mathbf{F}(\mathbf{x})=B_1\cdots B_{n-1}\mathbf{G}_n\circ \cdots \mathbf{G}_1(\mathbf{x})$$ is valid, with each $\mathbf{G}_i(\mathbf{x})$ being a primitive $\mathscr{C'}$ mapping in some neighborhood of zero, $\mathbf{G}_i(\mathbf{0})=0$, and $\mathbf{G'}_i(0)$ is invertible, and each $B_i$ is either a flip or the identity operator. In the change of variable theorem, he claims that we can write $T(\mathbf{x})$, our 1-1 $\mathscr{C'}$ on $R^k$ mapping, as $$\mathbf{T}(\mathbf{x})=\mathbf{T}(\mathbf{a})+B_1\cdots B_{k-1}\mathbf{G}_k\circ \cdots \mathbf{G}_1(\mathbf{x-a})$$ If $\mathbf{T}(\mathbf{x})$ is linear, I understand how theorem 10.7 applies, because $\mathbf{T}(\mathbf{0})=\mathbf{0}$ for all linear transformations and we can apply the theorem to $\mathbf{T}(\mathbf{x-a})$. But if T is not linear, how does he arrive at this equation?

Secondly, even if the equation does hold, clearly $\mathbf{T}(\mathbf{x-a})$ is composition of primitive $\mathscr{C'}$ mappings and flips, but why is $\mathbf{T}(\mathbf{x})$. Doesn't the addition of the constant term $\mathbf{T}(\mathbf{x})$ change things?

nickodel
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  • Consider the map $U \colon \mathbf{y} \mapsto \mathbf{T}(\mathbf{a}+\mathbf{y}) - \mathbf{T}(\mathbf{a})$. – Daniel Fischer Aug 07 '14 at 18:31
  • @DanielFischer Thank you for the comment. I see how this makes U(0)=0. However, I am still unclear on a few points. Theorem 10.7 mentioned above, requires that 0 be an element of the set on which the map is defined, which does not seem to be guaranteed anywhere for this U transformation. Also, there is this strange difference in the variables, T is a function of x but the composition function is a function of (x-a). And the addition of T(a) seems to make it such that the right side isn't simply a composition of the right kinds of functions. Any hints? – nickodel Aug 07 '14 at 20:53

1 Answers1

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$$\mathbf{T}(\mathbf{x}) = \mathbf{T}(\mathbf{a}) + B_1\dotsb B_{k-1} \mathbf{G}_{k}\circ \dotsb \circ \mathbf{G}_1(\mathbf{x} - \mathbf{a})\tag{1}$$

represents $\mathbf{T}$ as the composition of two translations, $k$ primitive mappings, and at most $k-1$ flips. We obtain that representation by looking at the function

$$\mathbf{U}\colon \mathbf{y} \mapsto \mathbf{T}(\mathbf{a} + \mathbf{y}) - \mathbf{T}(\mathbf{a})\tag{2}$$

defined on the open set

$$E - \mathbf{a} = \{ \mathbf{x} - \mathbf{a} : \mathbf{x} \in E\}.$$

Since $\mathbf{U}(\mathbf{0}) = \mathbf{0}$, we have the representation

$$\mathbf{U}(\mathbf{y}) = B_1\dotsb B_{k-1} \mathbf{G}_{k}\circ \dotsb \circ \mathbf{G}_1(\mathbf{y})\tag{3}$$

from theorem 10.7, and since

$$\mathbf{T}(\mathbf{x}) = \mathbf{T}(\mathbf{a}) + \bigl(\mathbf{T}(\mathbf{a} + (\mathbf{x}-\mathbf{a})) - \mathbf{T}(\mathbf{a})\bigr) = \mathbf{T}(\mathbf{a}) + \mathbf{U}(\mathbf{x} - \mathbf{a}),$$

$(1)$ is a direct consequence of $(3)$.

What Rudin does not explicitly say in the proof of the change-of-variables formula is that the formula also holds for translations (a translation is, by the way, a composition of at most $k$ primitive $\mathscr{C}'$ maps, which then shows again that the transformation formula holds for translations). So $(1)$ is a representation of $\mathbf{T}$ as a composition of maps for which the transformation formula holds, and hence it also holds for $\mathbf{T}$.

Daniel Fischer
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  • wow fantastic, thank you so much! I literally never would have figured that out :) – nickodel Aug 07 '14 at 20:57
  • Dear Daniel. I have one question: Why translation is a composition of at most $k$ primitive $C'$ mappings? – RFZ Apr 03 '16 at 13:44
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    @RFZ The map $T_m \colon (x_1,\dotsc,x_k) \mapsto (x_1,\dotsc, x_{m-1},x_m + a_m, x_{m+1},\dotsc, x_k)$ is a primitive $\mathscr{C}'$ map for every $m$. The translation $x \mapsto x+a$ is the composition $T_1 \circ \dotsc \circ T_k$ - the order doesn't matter, since these maps commute with each other. We can omit those maps where $a_m = 0$, since the corresponding primitive map is the identity, hence "at most". – Daniel Fischer Apr 03 '16 at 13:58
  • Dear Daniel! Thank you very much for answer. Very nice. Just now i am stuck on this theorem and working with its proof about week and have some questions. If i create post can you help me with it please? I would be very grateful! – RFZ Apr 03 '16 at 14:05
  • Somebody will certainly be able to help you @RFZ. Whether it's going to be me or somebody else … – Daniel Fischer Apr 03 '16 at 14:12
  • Dear Daniel, I have a question too. I dont see how this part connects with the proposition of the theorem... Could you please explain? It must be a very elementary question and sorry to bother you. – Keitaro Ninomiya Jun 21 '20 at 09:43