The number of decreasing function f:R--R such that f(f(x))=x+1
for all x€R

Question

The number of decreasing function f:R-->R such that f(f(x))=x+1 
for all x€R

parthkohli · Answer

Let $f(x)$ be $ax+b$ where $a<0$. Then $f(f(x)) = a(ax+b)+b = a^2 x+ab+b $. 
The only possible value for $a$ then has to be $-1$ by comparing coefficients. But then there'd be no way to get the constant term as +1 'cuz $ab+b = -1\cdot b + b = 0$.

bobo-i-bo · Answer

@ParthKohli  That only says that f(x) isn't in the form ax+b. Doesn't mean there isn't another function which satisfies the conditions. I have a hypothesis which I don't know whether it's true, but I'm guessing it is, and it might help with the Q:
If f:R->R is a function such that f(f(x)) is continuous, then f(x) is continuous.

bobo-i-bo · Answer

Maybe the condition that f(f(x)) is not only continuous but also bijective is necessary for my hypothesis.

kainui · Answer

Ahh ok I think I figured it out, the question specifically states decreasing, and nothing about continuous functions. So check this out:

$$f(f(f(x))) = f(x)+1 = f(x+1)$$

So since $f(x+1)-f(x) >0$ all choice of f(x) are increasing. So none satisfy it.

bobo-i-bo · Answer

Nice!

bobo-i-bo · Answer

I'd still like to find out whether my hypothesis is true though :P

kainui · Answer

I didn't come across this so nicely as this, I had taken the derivative, looking for some kinda thing to do with it decreasing: $f'(f(x))f'(x) = 1$ and recognized if I plugged in f(x) I'd get $f'(x+1)f'(f(x))=1$ and have $f'(x)=f'(x+1)$ and that's when I backpedaled haha...

Alright let's see I'm trying to see if I can find a counterexample to your hypothesis first before I try to prove it haha >:D

samigupta8 · Answer

Well did! Kainui sir ..
Thanks.

parthkohli · Answer

@Bobo-i-bo That's exactly what it means. Do give an example of a function which isn't of that form and still satisfies $f(f(x))=x+1$

bobo-i-bo · Answer

@ParthKohli Can't think of one! I don't know if one exists. :P But your argument does not prove that there does not exist a function which isn't of that form and still satisfies f(f(x))=x+1!

kainui · Answer

I had found a counterexample to the nonbijective case, but I think you sorta had that idea in mind and my computer died right before I hit post. I think my example was $f(x) = \lfloor \sin(\pi x) \rfloor$ but yeah, clearly not bijective so I'm still wondering about this too. I feel like the definition of continuous maps as I learned it in real analysis has the answer in it something about composition of continuous maps etc etc but I'm too scared to look right now haha.