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I stumbled upon a question and I can't seem to find the answer. Here it is:

Suppose $f$ a continuous function (from $\Bbb R$ to $\Bbb R$) with 2 attractive fixed points. Let's call them $a$ and $b$. The basin of attraction of $a$ is $(-\infty,p)$ and the basin of attraction of $b$ is $(p, \infty)$ for $p$ in $\Bbb R$. What can you say about the point $p$? My intuition would be to say that the function $f$ is not derivable in $p$, but I don't really know why.

Also, what would be an example of a function that has these properties?

Thank you in advance for your answers!

Daniele Tampieri
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A. Smith
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2 Answers2

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If $f(p)>p$, then there is a neighborhood of $p$ on which $f(x)>p$. Therefore, that neighborhood would belong to the attractive basin of $b$. This is a contradiction, since all points $<p$ belong to the attractive basin of $a$.

Similarly, if $f(p)<p$ then a neighborhood of $p$ would belong to the attractive basin of $a$, which also cannot be since all points $>p$ must belong to the basin of $b$.

Therefore, $f(p)=p$, i.e. $p$ is also a fixed point.

The information given doesn't tell you anything about the differentiability at $p$. You were given the example of $\frac{4}{\pi}\arctan(x)$ with $p=0$, which is differentiable at $p$. You can bend it a bit to produce an example that is not differentiable at $p$:

$$f(x)=\begin{cases}\frac{4}{\pi}\arctan(x),&x\geq0\\\frac{4(1+\epsilon)}{\pi}\arctan\left(x/(1+\epsilon)\right),&x>0\end{cases}$$

user647486
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$$ f(x) = \frac{4}{\pi} \; \arctan x $$

Will Jagy
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