# Decreasing function and period $2$ points

I'm trying to prove the following result:

Let $$h: [a,b] \rightarrow [a,b]$$ be a monotonic decreasing function and given $$x \in [a,b]$$ we define the sequence \begin{align*} y_{0,x} & = x\newline y_{n+1,x} & = h(y_{n,x}), \forall n>1 \end{align*} Then, $$\{y_{n,x}\}_{n=0}^\infty$$ is convergent for every $$x \in [a,b]$$ iff every point of period $$2$$ of $$h$$ is, in fact, a fixed point.

I've managed to prove that if every sequence of that type is convergent, then every point of period $$2$$ is a fixed point (in fact, you don't even need to assume that $$h$$ is decreasing). My problem is with the reciprocal.

I've proven that for any $$x \in [a,b]$$ the subsequences $$x_n=y_{2n,x}$$ and $$z_n=y_{2n+1,x}$$ are always increasing and decreasing respectively (or vice versa). As they are bounded, because they are contained in the interval $$[a,b]$$, this proves that both of them are convergent.

Now, if $$h$$ was continuous, then it is easy to see (using now that every point of period $$2$$ of $$h$$ is fixed) that both limits are equal, and so, that the sequence $$y_{n,x}$$ converges.

My doubt is then if this result is still true without the condition that $$h$$ is continuous. I cannot prove without this condition, but I also cannot give any counterexample, so any help would be appreciated.

• Is $h$ supposed to be injective? In other words, is $h$ required to be strictly decreasing? Feb 1 at 22:26
• It looks people are divided on the definition of "monotonic decreasing". Wolfram says it means "strictly decreasing". Wikipedia says it means "weakly decreasing". Check this post. Feb 1 at 23:38

It is not true that if every point of period $$2$$ of $$h$$ is a fixed point, then $$\{y_{n,x}\}_{n=0}^\infty$$ is convergent for every $$x \in [a,b]$$.

Here is a counterexample.

$$\quad$$

Let $$f:[0,1]\to[0,1]$$, $$\begin{array}{crcl} f: &[0,1] &\to &[0,1]\\ &x &\mapsto &1-\frac x2 &\text{if } x<\frac12\\ &x &\mapsto &\frac 23 &\text{if } \frac12\le x\le\frac34\\ &x &\mapsto &\frac54-x &\text{if } \frac34< x\\ \end{array}$$

The only periodical point of $$f(x)$$ is $$x=\frac23$$ with $$f(\frac23)=\frac23$$.

If $$x=y_{0,x}=0$$, then $$y_{2n+1,x}=\frac34+\frac1{2^{n+2}}$$ and $$y_{2n,x}=\frac12-\frac1{2^{n+1}}$$ for all $$n\ge0$$. $$\lim_{n\to\infty}y_{2n,x}=\frac12\not=\frac34=\lim_{n\to\infty}y_{2n+1,x}$$

Hence $$f$$ is a counterexample.

Let $$g(x):[0,1]\to[0,1]$$ be the same as $$f(x)$$ except for $$\frac12\le x\le\frac34$$, $$g(x)=\frac23-\frac13(x-\frac23)=\frac89-\frac x3$$.

What have been said above about $$f$$ also holds for $$g$$. In particular, $$g$$ is also a counterexample. (As mentioned by OP, neither $$f$$ nor $$g$$ can be continuous. In fact, both are discontinuous at $$x=\frac12$$ and $$x=\frac34$$.)

Note that while $$f$$ is weakly decreasing, $$g$$ is strictly decreasing.

• It was so simple and yet, I couldn't find it. Thanks! Feb 2 at 11:32