4

Given that the function $f$ is continuous and has the property $f(f(x))=1-x$ for all $x\in[0,1]$. Find $J=\int_{0}^{1}f(x)dx$.

My try:

I did this problem without finding the function $f(x)$, but I am interested in finding the function $f(x)$. So what I did is I replaced $x$ with $f^{-1}(x)$ in the given equation (I assumed $f^{-1}$ exists.):

$$\tag{1} f(x)=1-f^{-1}(x).$$

Also, by applying $f^{-1}$ on both sides of given equation

$$f(x)=f^{-1}(1-x).$$

Replacing $x$ with $1-x$,

$$f(1-x)=f^{-1}(x).$$

Substituting in $(1)$ I got

$$f(x)+f(1-x)=1.$$

This actually seemed to be a easier equation but I couldn't solve it. Particularly I think that it might be a piecewise function.

Gary
  • 31,845
Learner
  • 327
  • 1
    I don't think you will need to solve this, your effort has been well directed. Indeed, it seems that integrating the last equation from $0$ to $1$ and using a simple change of variable to show $\int_0^1 f(1-x)dx = \int_0^1 f(x)dx$ should be giving the answer. If this works out, then do respond, but your effort was well directed. One small thing is, why is $f$ invertible? The point is that $f \circ f$ is bijective on $[0,1]$ so this is the clincher, although as the other comment points out you don't even need this! – Sarvesh Ravichandran Iyer Apr 17 '21 at 14:15
  • 1
    Note that $1 - f(f(x)) = x$. Substituting $x=f(x)$ gives $1 - f(f(f(x))) = f(x)$. But $f(f(x))=1-x$, thus $1-f(1-x)=f(x)$, i.e., $f(x)+f(1-x)=1$. You do not need to assume invertibility. – Gary Apr 17 '21 at 14:17
  • 1
    @user20672 But your $f$ does not satisfy $f(f(x))=1-x$. – Gary Apr 17 '21 at 14:19
  • @Gary That's explained by the same logic that is used for numbers with the whole division by zero business , it also occurs in trigonometric equations. Basically, implications can very much add extraneous solutions. One of the steps is adding extraneous solutions, bonus for figuring it out. – Sarvesh Ravichandran Iyer Apr 17 '21 at 14:23
  • 1
    @Teresa Lisbon it actually worked,and also as I mentioned I figured out the value of $J$ .I am interested in finding $f(x)$. – Learner Apr 17 '21 at 14:28
  • @Learner So you want $f$ such that $f(x)+f(1-x) = 1$, or you want $f$ such that $f(f(x)) = 1-x$? – Sarvesh Ravichandran Iyer Apr 17 '21 at 14:31
  • @Teresa Lisbon according to what I did,both are actually same? – Learner Apr 17 '21 at 14:33
  • @Learner In fact, this is not true. As mentioned, $f(x) = -x$ satisfies $f(x) + f(1-x) = 1$, but not $f(f(x)) = 1-x$. The point is, that implications add solutions : there should be a step which you can't "reverse" and so you end up adding solutions which aren't solutions of the original equation. – Sarvesh Ravichandran Iyer Apr 17 '21 at 14:35
  • @Teresa Lisbon yes but what I assumed is that $f^{-1}(x)$ exists and as $f(x)=-x$ has a inverse the function should satisfy it? I mean all other implications I did are true if inverse exists? – Learner Apr 17 '21 at 14:39
  • Implications add solutions is a weird way of putting it. More generally the proof above has shown that $f(f(x)) = 1 - x \Rightarrow f(x) + f(1-x) = 1$. So any solution of $f(f(x)) = 1 - x$ will be a solution of $f(x) + f(1-x) = 1$, but has you haven't proved that (and infact it isn't true) $f(x) + f(1-x) = 1 \Rightarrow f(f(x)) = 1 - x$, the other way around isn't true. In common terminology, this is called necessary and sufficient conditions. The new equation is a necessary condition but not sufficient. – Anon Apr 17 '21 at 14:40
  • 1
    @Kaind Maybe in this context, yes I could have been more precise. But to be exact now : the implication $f(x) + f(1-x) = 1 \implies f^{-1}(1-x) = f(x)$ or $f(x) + f(1-x) = 1 \implies f(1-x) = f^{-1}(x)$ doesn't go through, as the counterexample shows. Anyway, the context of implications adding solutions is probably better reserved for numbers and trigonometric equations (extraneous solutions), but I just brought the same thing forward for functions, where extraneous solutions are created from following irreversible steps (not a bad thing, of course, provided you can throw the bad apples out). – Sarvesh Ravichandran Iyer Apr 17 '21 at 14:42
  • 1
    @Learner I showed in my comment above that you do not need to assume that the inverse exists to conclude $f(x)+f(1-x)=1$. – Gary Apr 17 '21 at 14:46
  • @Gary You can show that $f$ is monotone, and hence inverse on some subset of $[0,1]$ exists. – Anon Apr 17 '21 at 14:47
  • @Teresa Lisbon $f(x)=-x$ Is not satisfying because it is not satisfying eq($1$) which means that our function must satisfy both eq($1$) and last one – Learner Apr 17 '21 at 14:48
  • 1
    Let me just put it out there: The solutions for $f(x) + f(1-x) = 1$ are precisely all solutions which satisfy $f(x) = g(x) , 0.5 < x \le 1 ; 1/2 , x = 1/2 , 1 - g(1-x) , 0 \le x < 0.5$, where $g(x)$ is any continuous function on $[0.5,1]$ satisfying $g(0.5) = 0.5$. However this is not the solution set for $f(f(x)) = 1-x$. – Anon Apr 17 '21 at 14:50
  • @Learner So do you want to find a function that satisfies eq (1)? That's related to the functional square root of a function, which talks about : given $g(x)$, when does there exist an $f$ such that $f(f(x)) = g(x)$? For that, this is useful. Of course, the square root is not going to be continuous, however : if $f(f(x)) = 1-x$ then $f$ is injective and hence monotone, but then $f(f(x))$ has to be monotone increasing. – Sarvesh Ravichandran Iyer Apr 17 '21 at 14:52
  • @TeresaLisbon can you explain what does the answer in the link meant by "there are an even number of cycles of f of any given even length" – IITM Apr 17 '21 at 15:03
  • duplicate? https://math.stackexchange.com/q/2820310/886804 – hgmath Apr 17 '21 at 15:37
  • @IITM Perhaps you would like to come to my chatroom, since this place may not be right to discuss it? The chatroom link is here, and we can discuss that answer in length there. – Sarvesh Ravichandran Iyer Apr 17 '21 at 17:16

1 Answers1

1

In absence of continuity there are infinitely many such function. Pick g a HAMEL function (for which g(x+y)=g(x)+g(y) ) Then $f(x)=\frac{g(x)}{g(1)}$ satisfies $f(x)+f(1-x)=\frac{g(x)+g(1-x)}{g(1)}=\frac{g(x+1-x)}{g(1)}=1$

Petre Ciobanu
  • 11
  • 1