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I'm struggling to solve this question:

Let $f:\mathbb R^m\to \mathbb R^m$ be a diffeomorphism and $B$ be the closed unitary ball of $\mathbb R^m$ such that $f(B)\subset B$. Suppose $|\det f'(x)|\lt 1$ for every $x\in B$. Prove for every continuous function $g:B\to \mathbb R$ we have $$\lim_{n\to \infty}\int_{f^n(B)}g(x)dx=0$$

My attempt

Changing the variables we have for the case $n=2$:

\begin{align} & \int_{f^2(B)}g(y)dy\\[10pt] = {} & \int_{f(B)}g(f(x))|\det f'(x)|dx \\[10pt] = {} & \int_B g(f^2(x))|\det f'(f(x))||\det f'(x)|dx \\[10pt] \end{align}

By induction we have:

$$\int_{f^n}g(y)dy=\int_B g(f^n(x))|\det f'(f^{n-1}(x))||\det f'(x)|^ndx$$

Using the fact $|\det f'(x)|\lt 1$ we have $\displaystyle\int_Bg^n(x)dx\lt \int_B1\cdot dx=\operatorname{vol} B$

So I don't know if my reasoning is right and how to do with this information I got:

$$\int_Bg^n(x)dx\lt \operatorname{vol} B$$

I need help

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    At the n-th step you have $|det f'(x)|^n$ – 2017-01-28
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    @Legoman I didn't understand why. For example in the $n=2$ case, I'm applying the change of variables to the function: $h(x)=f(x)\det f'(x)$ – 2017-01-28
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    Why? You need to change the coordinates to the function $g(w)=g(f(x))$ and $|\det f'(x)|$ appear another time so you find $|\det f'(x)|^2$ – 2017-01-28
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    @Legoman of course, I edited the question. Thank you again – 2017-01-28
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    I see now your comment. I post an answer to explain what I mean. :) – 2017-01-28

3 Answers 3

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By induction, the change of variables is $$ \begin{align} \int_{f^n(B)}g(x)\,\mathrm{d}x &=\int_{f^{n-1}(B)}g(f(x))\left|\det\!\left(f'(x)\right)\right|\,\mathrm{d}x\\ &=\int_{f^{n-2}(B)}g\!\left(f^2(x)\right)\left|\det\!\left(f'(f(x))\right)\right|\,\left|\det\!\left(f'(x)\right)\right|\,\mathrm{d}x\\ &=\int_{f^{n-3}(B)}g\!\left(f^3(x)\right)\left|\det\!\left(f'\!\left(f^2(x)\right)\right)\right|\,\left|\det\!\left(f'(f(x))\right)\right|\,\left|\det\!\left(f'(x)\right)\right|\,\mathrm{d}x\\ &=\int_Bg\!\left(f^n(x)\right)\prod_{k=0}^{n-1}\left|\det\!\left(f'\!\left(f^k(x)\right)\right)\right|\,\mathrm{d}x\tag{1} \end{align} $$ Since $f$ is a diffeomorphism on $\mathbb{R}^m$ and $B$ is compact, $\left|\det\!\left(f'(x)\right)\right|$ attains is maximum, $\lambda\lt1$, on $B$. $|g(x)|$ also attains its maximum, $\gamma$. Therefore, $$ \left|\int_{f^n(B)}g(x)\,\mathrm{d}x\right| \le\gamma\,\lambda^n\int_B1\,\mathrm{d}x\tag{2} $$ and since $\lambda\lt1$, $(2)$ tends to $0$ as $n\to\infty$.

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This is to answer to comments above. The function $f:\Bbb R^m\rightarrow \Bbb R^m$ is a global diffeomorphism, so we can use it to change the variables on integrals.

Now:

$$\int_{f^n(B)}g(y)dy=\int_{f^{n-1}(B)}g(f(x))|\det f'(x)|dx=\dots =\int_Bg(f^n(x))\Big (\prod_{k=0}^{n-1}|\det f'|(f^k(x))\Big)dx$$

Since $|\det f'(x)|<1$ then $|\det f'(x)|\rightarrow 0$ for $n\rightarrow \infty$.

Essentially applying the change of variable $n$ times, you obtain $|\det f'(x)|$ inside the integral.

So $$\lim_{n\to \infty}\int_{f^n(B)}g(x)dx=0$$

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    Thank you for your answer. In order that this integral is zero, shouldn't $g(f^n(x))$ be bounded? – 2017-01-28
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    $g : B \to \mathbb{R}$ in continuous, $B$ is compact, so $g(B)$ is compact, en thus $g$ is uniformly bounded. – 2017-01-28
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    Thank you again. One last question, I didn't understand yet why the absolute values remain $|\det f'(x)||$ untouchable. for the case $n=2$ for example, don't we have to redefine another function $h(x)=g(f(x))|\det f'(x)|$? Notice that $|\det f'(x)|$ is not constant, we can't put it outside the integral. – 2017-01-28
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    To be nitpicking, I guess the correct formula is $$ \int_{f^n(B)} g(x) \, dx = \int_{B} g(f^n(x)) \left( \prod_{k=0}^{n-1} |\det f'|(f^k(x))\right) \, dx. $$ At each step, all the Jacobians pupped up in previous steps should be transformed as well. – 2017-01-28
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    And @user42912, It seems to me that you need an extra assumption that $f(B) \subset B$, otherwise there is no guarantee that $g$ is defined on $f^n(B)$ and $|\det f'| < 1$ on each $f^k(B)$ ($k = 1, 2, \cdots $). Consider $f(x) = 2017 + x/2$, which satisfies $|\det f'| = 1/2^m$ but still $f(B_1(0)) \nsubseteq B_1(0)$. – 2017-01-28
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    @SangchulLee thank you, I edited my question – 2017-01-28
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    @user42912, Thank you, now everything looks fine. :) – 2017-01-28
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    @SangchulLee you right! thanks – 2017-01-28
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    @SangchulLee As I asked Legoman, For example in the n=2 case, I'm applying the change of variables to the function: $h(x)=f(x)| \det f′(x)|$ so we have to have $\int_Bg(f(f(x)))|\det f'(f(x))||\det f'(x)|$ – 2017-01-28
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    @user42912, You are correct. And I guess we are talking about the same thing. In my previous comment, I intended $|\det f'|(f^k(x))$ as the Jacobian $|\det f'|$ evaluated at $f^k(x)$. (That is, I used $|\det f'|$ as a name of the Jacobian of $f$.) – 2017-01-28
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    @SangchulLee did you mean $f^{k-1}(x)$? – 2017-01-28
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    @user42912, My product runs from $k = 0$ to $n-1$, with the convention that $f^0 = \mathrm{id}_{\Bbb{R}^m}$. Expanding the product will give $$|\det f'(f^{n-1}(x))| \cdots |\det f'(f(x))| \cdot |\det f'(x)|.$$ I am pretty sure this is also what you are thinking of. – 2017-01-28
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    Now everything seems ok – 2017-01-28
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    @SangchulLee and Legoman yes, thank you very much to you both – 2017-01-28
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    I think it is essential to use that $B$ is compact and $\left|\det\!\left(f'(x)\right)\right|$ is continuous on $B$ to get that $\left|\det\!\left(f'(x)\right)\right|\le\lambda\lt1$ for all $x\in B$. – 2017-01-28
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Assuming that you have access to the dominated convergence theorem, the following solution is also available:

Proof. By the assumption, there is $\lambda \in [0, 1)$ such that

$$|f(x) - f(y)| \leq \lambda|x - y|, \qquad \forall x, y \in B.$$

(Indeed, you may take $\lambda = \sup_{x \in B} |\det f'(x)|$. By the continuity of $f'$ and the compactness of $B$, we have $\lambda < 1$. Then the claim follows from the mean value inequality.) Then by the contraction mapping theorem, there is a unique fixed point $x_0$. Now $$ \max_{x \in B} |f^n(x) - x_0| \leq \lambda^n \max_{x \in B} |x - x_0| $$ and thus $f^n (B) \to \{x_0\}$ in the sense that $\mathbf{1}_{f^n(B)}(x) \to \mathbf{1}_{\{x_0\}}(x)$ for each $x \in B$. Since $g$ is integrable on $B$, by the dominated convergence theorem we have

$$ \int_{f^n(B)} g = \int_{B} g \mathbf{1}_{f^n(B)} \xrightarrow[n\to\infty]{} \int_{B} g \mathbf{1}_{\{x_0\}} = 0. $$

Remark. This argument allows us to weaken the assumption on $f$. For instance, it suffices to assume that $f : B \to B$ is continuous and satisfies $|f(x) - f(y)| < |x - y|$ for all distinct $x, y \in B$.