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I came across the following problem:

Let $g \colon [0,1] \to [0,1]$ be a continuous map and consider the iteration $x_{n+1}=g(x_n)$.Then Which of the following maps will yield a fixed point for $g$?

The options are as follows:
a. $g(x)=\frac{x^2}{4}$,
b. $g(x)=\frac{x^2}{32}$,
c. $g(x)=\frac{x^2}{8}$,
d. $g(x)=\frac{x^2}{16}$.

Can someone point me in the right direction? Thanks in advance for your time.

Community
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learner
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2 Answers2

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The answer is certainly all of them, according to a very general and powerful theorem called the Brower fixed point theorem. You'll probably get some funny looks if you turn that in, though, as I think it's quite a bit easier than that.

Don't all the maps have an obvious common fixed point? Is that really the whole problem?

Mark McClure
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  • The question is not about the existence of a fixed point, which as you say is obvious, but whether the iteration of $g$ gets to it, I guess from any starting point. Perhaps you mean Banach's fixed-point theorem. – lhf May 15 '13 at 13:08
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    @lhf Perhaps, that's what the question is supposed to be. – Mark McClure May 15 '13 at 13:41
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Graph each of these functions in a square $[0,1]\times[0,1]$. A fixed point is exactly where the graph crosses the diagonal of the square. Where does this happen for each function?

Alternately, in the vein of using too-powerful technology, since $[0,1]$ is a complete metric space, you can use the contraction mapping theorem (bearing in mind that differentiable functions on compact spaces are globally Lipschitz, with global Lipschitz constant the maximum of the derivative.)

Neal
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