5
$\begingroup$

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$?$

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 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 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 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 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 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.