Continuous Inverse theorem proof.

Question

Question Statement:

Let $f:[0,1]\rightarrow [0,1]$ be continious such that $f(0)=0$ and $f(1)=1$ and $f(f(x))=x, \forall x\in [0,1]$. Prove that $f(x)=x, \forall x\in [0,1]$.

My attempt:

Consider $f(f(x))=x$. Since the RHS is injective then $f(f(x))$ must be injective. Hence $f(x)$ must be injective. Since $f(x)$ is continuous and injective, then by the Continious Inverse theorem, $f^{-1}$ is well defined as well as continuous and surjective. Since $f^{-1}$ is defined, this implies that $f$ is unique. And since $f(x)=x$ satisfies the conditions, it must be the unique solution, hence $f(x)=x$.

However, I am not sure of the bold part. I am not really sure how I would prove that.

I don't see how you can get the bold part. If you have $f(f(x)) = x$, you've already argued that $f$ is injective. And clearly $f$ is a surjection, and so $f$ is bijective and is invertible. This holds true without any continuity assumption. There are plenty of non-continuous solutions to your equation. So your justification of uniqueness has to use something about the continuity, which I don't see used. — Willie Wong, Apr 20 '17 at 13:33
Incidentally, when are you using the continuous inverse theorem? The usual proof of the proposition you stated uses only monotonicity. — Willie Wong, Apr 20 '17 at 13:34
given an inverse, the function is unique. But the existence of an inverse cannot show that the function is unique. — Lelouch, Apr 20 '17 at 13:36
@WillieWong In the hints, the question said to use the Continuous Inverse Theorem or the Continuous Injection Theorem . I didn't see how I would use the later so I expected the proof would use some form of the continuous inverse theorem. How would I use monotonicity here? — AspiringMat, Apr 20 '17 at 13:37
Possible duplicate of Continuous involutions on $\mathbb R$ with at least two fixed points — mlc, Apr 20 '17 at 13:40
You should use the continuous injection theorem instead (which states that a continuous injection of a interval is strictly monotone). Then prove by contradiction: suppose $f(x) < x$ for some $x \in [0,1]$, show that this implies $f(f(x)) < x$. — Willie Wong, Apr 20 '17 at 13:41
@WillieWong Thanks. I think I got it. So if it is monotonic. Then two case. If $\exists x$ such that $f(x)<x$. Then we have $f(f(x))<f(x)$ by monotonicity since $f(x)<x$ and hence $f(f(x))<x$ which contradicts $f(f(x))=x$. Similarly if there is an $x$ where $f(x)>x$, then again it contradicts $f(f(x))=x$. Since there are no such $x$ where $f(x)<x$ or $f(x)>x$ then $f(x)=x$ for all the $x$'s — AspiringMat, Apr 20 '17 at 13:55
@AspiringMat: looks good. Now copy that to an answer below and earn your self-learner badge. — Willie Wong, Apr 20 '17 at 14:29

score 1 · Answer 1 · answered Apr 21 '17 at 20:15

Here is the answer for completion. Consider $f(f(x))=x$. Since the RHS is injective then $f(f(x))$ must be injective. Hence $f(x)$ must be injective. Since $f(x)$ is continuous and injective, then by the Continious Inverse theorem, $f(x)$ is monotone.

Now for the sake of contradiction assume $f(x)\neq x$. Then $\exists x$ such that $f(x)<x$ or $f(x)>x$ (otherwise $f(x)=x$ for all $x$'s). If $\exists x$ where $f(x)<x$ then $f(f(x))<f(x)$ by monotonicity and $f(x)<x$ and hence $f(f(x))<x$ which is a contradiction to $f(f(x))=x$. Similarly if $\exists x$ where $f(x)>x$ then $f(f(x))>f(x)$ by monoticity and $f(x)>x$ contradicting $f(f(x))=x$. Hence, in both cases, there is a contradiction and hence $f(x)$ must equal $x$.

Continuous Inverse theorem proof.

1 Answers1