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$$\Large-\textbf{Problem}-$$ Is there at least a continuous mapping from an arbitrary interval $[a,b]$ to $[0,1]$ with a periodic point of period $3$?

$$\Large-\textbf{Thoughts and Ideas}-$$ Let $$T_{\lambda}(\theta) = \theta + \dfrac{2\pi p}{3}$$ where $(p,3) = 1$, $\lambda = p/3$ and $p$ is an integer. Denote this by $T:S^1\rightarrow S^1$, where $S^1$ is the set of values modulo $2\pi$ radians. I was thinking a bit about the translation of the circle with periodic points of period $3$. However, I can't find a continuous mapping $f:[a,b] \rightarrow [0,1]$ from here. So I find another approach.

For a mapping to have periodic point with period $3$, it needs not to be a homeomorphism; that is because a homeomorphism can have no periodic points with prime period greater than $2$. Then, I need to find a continuous mapping that is either 1-1 or onto or (possibly) not both. This type of approach, to me, is fine, but it doesn't help me a lot to find such mapping.

Any suggestions or comments?

NasuSama
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2 Answers2

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You can rescale $\sin(x)$ or $\cos(x)$ . You know $\sin(x)$ has period $2\pi$. So, rescale the argument $x$ so that either $\sin(x)$ or $\cos(x)$ can hit every value three times in the interval $[a,b]$. This gives a continuous mapping from an arbitrary interval $[a,b] \to [0,1]$ with a periodic point of period $3$.

guero
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  • I suspect you're not using "periodic" in the desired sense. A periodic point $x$ of period $p$ for a function $f$ is one such that, with $x_0 = x$ and $x_{i+1} = f(x_i)$, $x_0, x_1, \ldots, x_{p-1}$ are distinct but $x_p = x$ – Robert Israel Mar 08 '14 at 01:08
  • Now that you mention it, since the tag is Dynamical Systems. I guess I should have been more careful. – guero Mar 08 '14 at 01:12
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Let's construct a function $f$ from $[0,1]$ to $[0,1]$ such that some point $x_0$ is periodic with period $3$. That is, $f(x_0) = x_1$, $f(x_1) = x_2$, $f(x_2) = x_0$, for some distinct $x_0$, $x_1$ and $x_2$ in $[0,1]$. In fact, we can take any three distinct points. For example, try $x_0 = 0$, $x_1 = 1$, $x_2 = 1/2$, and interpolate linearly: $$ f(x) = \cases{ 1 - 2 x & for $0 \le x \le 1/2$\cr -1/2 + x & for $1/2 \le x \le 1$\cr}$$

If you want $f: [a,b] \to [0,1]$ with $[a,b]$ arbitrary, you'll have trouble if $[0,1]$ and $[a,b]$ don't overlap, because $f \circ f$ won't be defined. If $[0,1] \cap [a,b] = [c,d]$ is an interval of positive length, then you could define $f$ similar to the above on $[c,d]$ and extend to $[a,b]$ so that the values stay in $[c,d]$.

Robert Israel
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