Is it true that if $f(f(x))$ is continuous and strictly decreasing then $f$ is continuous$?$

First of all, how can $f(f(x))$ be strictly decreasing. If $f(x)$ is increasing then $f(f(x))$ is also increasing and if $f(x)$ is decreasing, then $f(f(x))$ is increasing again.

I think this statement is trivially true. If we use Boolean algebra, if $p$ is not true then $p \Rightarrow q$ is always true.

Am I thinking correctly$?$


1 Answer 1


What if $f(x)$ is neither increasing nor decreasing.

The answer is NO. Indeed, here is a counterexample.

First, partition the interval $(0,\infty)$ into pairs of two elements.

This can be done for example, by paring $x \in (0, \infty)$ with $\frac{1}{x}$, as long as they are both not integers, and pairing $n$ with $\frac{1}{n+1}$ (the last one to avoid the fact that $x=\frac{1}{x}$ for $x=1$).

This way, we get a family $\{ A_i\}_{i \in I}$ of sets, with the following properties:

  • $|A_i|=2$ for each $i$.
  • $A_i \cap A_j= \emptyset$ for all $i \neq j$.
  • $\bigcup_{i \in I} A_i =(0,\infty)$.

Now, define $f: \mathbb R \to \mathbb R$ the following way: First set $f(0)=0$.

Next, for each $i \in I$, if $A_i=\{ a,b\}$ define $$f(a)=b \\ f(b)=-a \\ f(-a)=-b \\ f(-b)=-a$$

Then $f \circ f(x)=-x$ for all $x \in \mathbb R$.

By chosing the right $A_i$ you can make $f$ discontinuous, but $f \circ f$ is continuous and decreasing.

P.S. A simpler way to pair $(0, \infty)$ is by pairing the intervals $(2n, 2n+1]$ and $(2n+1, 2n+2]$ via the $x \to x+1$.

Then, if I didn't make a mistake, here is your function $$f(x)= \left\{ \begin{array}{lc} f(0)=0 & \\ f(x)=x+1 & \mbox{ if } \lceil x \rceil \mbox{ is positive and odd } \\ f(x)=-x+1 & \mbox{ if } \lceil x \rceil \mbox{ is positive and even} \\ f(x)=x-1 & \mbox{ if } \lfloor x \rfloor \mbox{ is negative and odd} \\ f(x)=-x-1 & \mbox{ if } \lfloor x \rfloor \mbox{ is negative and odd} \\ \end{array} \right. $$

Here $\lceil x \rceil$ and $ \lfloor x \rfloor$ are the ceiling and floor functions.

Added Note that actually you can prove something simpler, and probably this is what the question asks:

Lemma If $f \circ f$ is strictly decreasing, then $f$ cannot be continuous on $\mathbb R$.

Proof: Assume by contradiction that $f$ is continuous. Since $f\circ f$ is one to one, $f$ must be a one-to-one function.

Since $f$ is one-to-one and continuous, by a simple application of the Intermediate value Theorem it is monotonic.

But then, by the argument you made $f$ is increasing, contradiction.

  • $\begingroup$ I am confused isn't by this rule for instance $\{1/2, 2\}$ and $\{2, 1/3\}$ happening? Since $1/2$ is no integer and then because 2 is? Also you really want $f(b) = -a$? Because I don't think that is well-defined. $\endgroup$
    – hal4math
    Sep 24, 2019 at 3:18
  • $\begingroup$ @hal4math For the first point, there was a typo, I fixed it. I mean when $x$ and $\frac{1}{x}$ are not integers. For the second, remember that I am defining a function from $\mathbb R$ to $\mathbb R$. $\endgroup$
    – N. S.
    Sep 24, 2019 at 4:02
  • $\begingroup$ Ah, okay. But lets say I have now $x\in\mathbb{R}$, $ x > 0$ and I find $A$ with $x$ in it and so is say $y$. Is $f(x) = y$ or $f(x) = - y$? I think I can't tell from your definition. $\endgroup$
    – hal4math
    Sep 24, 2019 at 4:23
  • $\begingroup$ @hal4math You have to make a choice, which is irrelevant, but needs to be made. Note that if you chose $f(x)=y$ then $f(y)=-x$ while if you make the other choice $x, y$ get interchanged, i.e $f(x)=-y$ and $f(y)=x$...... $\endgroup$
    – N. S.
    Sep 24, 2019 at 5:33
  • 2
    $\begingroup$ @hal4math If you what to make a consistent choice, you can simply say: order $a <b$ and then use the four cycle $a \to b \to -a \to -b$ and back to $a$, i.e. smaller goes to the other, and larger goes to negative. $\endgroup$
    – N. S.
    Sep 24, 2019 at 5:36

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