Solution verification: Let $f$ be a function from $(0,1)$ to $(0,1)$ such that $f(x)<x$ for every $0<x<1$. Must $f^n(x)\rightarrow 0$ for some $x$?

Question

Let $f$ be a function (need not be continuous) from $(0,1)$ to $(0,1)$ such that $f(x)<x$ for every $0<x<1$.

Must it be the case that for some $0<x<1$ we have $f^n(x)\rightarrow 0$ as $n\rightarrow\infty$?

My attempt:

Let the interval $(2^{-k},2^{-k+1}]=I_k,$ $(I_1=(\frac12, 1))$ then take for $x\in I_k$ $$f(x)=\frac{x-2^{-k}}2+2^{-k}=\frac12(x+2^{-k}).$$

We can verify that $f(x)<x\space\forall x\in(0,1)$ and that if $x\in I_k$ then $f(x)\in I_k$ and so we conclude that the answer is 'no' as $\forall x\in(0,1)\space\exists\epsilon>0$ such that $f^n(x)>\epsilon\space \forall n\ge1$. In particular if $x\in I_k$ then $f^n(x)>2^{-k}\space\forall n$.

Is this correct? I was really expecting the answer to be 'yes' and this sort of solution is not typical for the questions I am doing. I also have a knack of making lots of mistakes with these types of questions.

(I am also looking to improve my solution writing so if this is correct feel free to criticise how I have written it up.)

Thanks in advance

Answer 1

Your solution is good.

You can notice that on each $I_k=(a,b]$, what $f$ does is return the midpoint between $x$ and $a$. This guarantees that $a<f(x)<x$, so that the iterates of $f$ approach $a$ but never quite get there. So what's really important in your solution is that the collection of intervals $I_k$ that partition $(0,1)$ never contain an $I_j$ with $a=0$.

The takeaway here, I guess, is that general functions can be weird. And the function you've pieced is quite on the nicer end of things -- it is smooth almost everywhere! There's only a handful of discontinuities, in the transitions between each $I_k$.