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Let $p\in\mathbb{N}$. Let $f:I\to\mathbb{R}$ differentiable in the closed interval $I$ (bounded or not), with $f(I) \subset I$, and let $g = f\circ f\circ \cdots \circ f = f^p$, where $\circ$ means composition. If $\lvert g'(x) \rvert \leq c < 1, \forall x\in I$ where $c$ is a constant, show that there is only one $a\in I$ such that $f(a) = a$.

Some thoughts

I proved it for $p=1$, i.e. $g = f$ and $\lvert g'(x) \rvert = \lvert f'(x) \rvert \leq c < 1$. Notice that $I$ can be unbounded so the existence of the fixed point is not trivial.

For the task, I use two results:

  1. Let $f:I\to\mathbb{R}$ differentiable in the interval $I$. The function $f$ satisfies $\lvert f(x) - f(y) \rvert \leq c\lvert x - y \rvert$ for all $x,y\in I$ if and only if $\lvert f'(x) \rvert \leq c$ for all $x\in I$.

  2. Let $0\leq c < 1$. If the sequence $(x_n)$ is such that $\lvert x_{n+2} - x_{n+1}\rvert \leq c\lvert x_{n+1} - x_n\rvert $ for all $n\in \mathbb{N}$, then the sequence $(x_n)$ converges.

Now take any point $x_0\in I$ and build the following sequence: $x_1 = g(x_0), x_2 = g(x_1), \cdots, x_n = g(x_{n-1}), \cdots$. Since $\lvert g'(x) \rvert \leq c$, from result 1 we have $\lvert g(x_{n+1}) - g(x_{n}) \rvert \leq c\lvert x_{n+1} - x_{n} \rvert$, then $\lvert x_{n+2} - x_{n+1} \rvert \leq c\lvert x_{n+1} - x_{n} \rvert$. Since $0\leq c < 1$, from result 2 we have $(x_n)\to a$. Since $I$ is closed, $a\in I$. Finally, since $x_n = g(x_{n-1})$, this means $g(a) = a$. The uniqueness comes from supposing another $b$ and applying IVT to get contradiction.

I am having trouble trying to prove for $p > 1$.

Update: I marked Matthew's suggestion because it contained the key idea for the solution. The remaining bits for the solution are in the comments to his suggestion.

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  • $\begingroup$ You asked the same question 13 hours ago, and did a substantial edit one hour ago. You should've flagged that one for reopening instead of making a new one that's basically a copy. Actually, as of now it's officially reopened, but it's deleted instead. $\endgroup$ – Arthur Jul 20 '16 at 13:29
  • $\begingroup$ I'm sorry, I didn't know that. I deleted it instead. $\endgroup$ – Michael Jul 20 '16 at 13:30
  • $\begingroup$ Fair enough, you seem fairly new here. Just keep it in mind for next time. $\endgroup$ – Arthur Jul 20 '16 at 13:31
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Assuming your proof of the $p=1$ case is valid, you can apply it to $g=f^p$. You know $g$ has a unique fixed point $a$. You can show that $f(a)$ is a fixed point of $g$, too. But this means $a=f(a)$.

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  • $\begingroup$ Thanks for the idea. This was the key. $\endgroup$ – Michael Jul 22 '16 at 3:04
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    $\begingroup$ Extending from the original post, let now $g = f\circ f\circ \cdots \circ f = f^p$. We know there is a unique $a$ such that $g(a)=a$. We want to show that such $a$ is also the fixed point of $f$. Now, $g(a) = f^{p}(a) = a$, then $g(f(a)) = f^{p+1}(a) = f(a)$, therefore $f(a)$ is also fixed point of $g$, hence $a=f(a)$ by uniqueness. $\endgroup$ – Michael Jul 22 '16 at 3:10

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