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lolm2k

#1 h Mar 24, 2018, 5:50 PM • 2 Y

Y by Adventure10, Mango247

Let $f: \mathbb R \rightarrow \mathbb R$ be a continuous function such that $f(f(f(x))) = x^2+1, \forall x \in \mathbb R$ show that $f$ is even.

This post has been edited 1 time. Last edited by lolm2k, Mar 24, 2018, 5:52 PM

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#2 h Mar 24, 2018, 6:25 PM • 4 Y

Y by matinyousefi, lolm2k, Adventure10, Mango247

Substituting $x=f(x)$ we get that $f(x^{2}+1)=f(x)^{2}+1$ . Now setting $\pm x$ we get that $f(x)^{2}=f(-x)^{2}$ . Since $f$ is continious we get that $f$ is either even or odd for all x. Indeed, if we assume that for some x $f(x)=f(-x)$ then for all $y$ very close to $x$ we must have that $f(x)$ is very close to $f(y)$ which means that $f(-y)$ will be very close to $f(-x)=f(x)$ but this will be impossible if $f(-y)=-f(y)$ . Analoguously, for $f(x)=-f(-x)$ . If $f$ is odd then $f(0)=0$ . Setting $x=0$ in the equation we get $0=1$ . Whence, a contradiction.

This post has been edited 3 times. Last edited by Lamp909, Mar 24, 2018, 7:04 PM

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#4 h Mar 24, 2018, 6:32 PM • 2 Y

Y by Adventure10, Mango247

The above arguments are valid only for $x \neq 0$ but when we get that the function is odd for nonzero $x$ it is then easy to prove it for $x=0$ by taking $x$ to converge to 0.

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lolm2k

#5 h Mar 24, 2018, 7:35 PM • 2 Y

Y by Adventure10, Mango247

yep, your argument can be made for $x=1$ since $f(1)$ cannot be zero. Then extend for all reals, good solution!

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lolm2k

#6 h Mar 24, 2018, 8:02 PM • 1 Y

Y by Adventure10

actually now that I thought about it nothing garantees we can extend our result to all reals, all we proved is that if f(x) = f(-x) for some number there is a ball centered in x such that everyone in there also satisfies it. I would like to see a rigorous extension of this result, which seems to be true.

Please tell me if this is ok:
you proved if $f(x) = f(-x)$ there is this open ball around $x$ such that this equality happens.
Let $A = \{ x \in \mathbb R, f(x) = f(-x) \}$ then $A$ contains and is contained in the union of balls around each element of $A$ such that $f(x) = f(-x)$ . So $A$ can be thought as an union of open sets and therefore it is open.
So is $B = \{x \in \mathbb R , f(x) = -f(-x) \}$ .
So we would have a contradition of the fact that $\mathbb R$ is conected. Therefore either A or B must be empty.

This post has been edited 2 times. Last edited by lolm2k, Mar 24, 2018, 8:31 PM
Reason: typo

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#7 h Mar 24, 2018, 8:39 PM • 4 Y

Y by lolm2k, MNJ2357, Adventure10, Mango247

Lamp909 wrote:

we get that $f(x)^{2}=f(-x)^{2}$ . Since $f$ is continious we get that $f$ is either even or odd for all x.

This is false. Consider the function defined $f(x)=1+x$ for $x < 0$
$= 1 -x$ for $x$ between $0$ and $1$
$= x-1$ for $x>1$
Assume there exists a such as $f(-a)=-f(a)$ different from $0$ . Then there exists $b$ such as $f(b)=f(-b)=0$ . Now suppose $f(x)=0$ for some $x$ then $x^{2}+1=f(f(0))$ so $x$ is at most $2$ values hence $f(x)=0$ iff $x=b$ or $x= -b$ . This means if $|x| > b$ , $f(x)=-f(-x)$ and $f(x)=f(-x)$ otherwise. Since $f$ is continuous, we can easily prove that it's surjective, but then $f^{3}$ would be surjective as well, contradiction.

This post has been edited 2 times. Last edited by mkhayech, Mar 28, 2018, 6:20 PM

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lolm2k

#8 h Mar 24, 2018, 9:02 PM • 2 Y

Y by Adventure10, Mango247

why there exists b such as f(b)=f(-b)=0 ?

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#9 h Mar 24, 2018, 9:29 PM • 3 Y

Y by lolm2k, Adventure10, Mango247

Because f(-a)=-f(a) so one of them is negative and the other is postive, f is continous.

I am on mobile so didnt elaborate some parts, but if you try to draw what the graph of f can look like you will get the idea

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lolm2k

#10 h Mar 24, 2018, 10:15 PM • 2 Y

Y by Adventure10, Mango247

i believe in most you sayed, indeed b cannot be zero so there is b and -b. and it feels pretty natural that |x| > b leads to f(x) = -f(-x) and in order to avoid f(0) =0 we must have f(x) = f(-x) inside |x|<b. I'm having trouble seeing f must be surjective tho

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#11 h Mar 24, 2018, 11:10 PM • 2 Y

Y by lolm2k, Adventure10

The fact that $f(a)=f(b)$ then $a^2=b^2$ (...#) finishes the problem.
If $f(0) > 0$ , there exists a positive real number $\epsilon$ such that $|x|<\epsilon \Longrightarrow f(x)>0 \Longrightarrow f(x)=f(-x)$ , then $f(x)$ is even for all real numbers $x$ (by #)
We can use the same arguement for $f(0)<0$ .

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lolm2k

#12 h Mar 24, 2018, 11:40 PM • 2 Y

Y by Adventure10, Mango247

^ you're wrong, read mkhayech first answer

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#13 h Mar 25, 2018, 1:11 AM • 3 Y

Y by lolm2k, Adventure10, Mango247

$f$ is surjective because: $f(f(f(x)))= x^{2} + 1$ so every value bigger than $1$ is in the range. $f(b)=0$ and $f$ iscontinous so every positive number is in the range. Now if $c < 0$ then we can find $x< -b$ such that $f(x)=-f(-x)=-(-c)=c$ (since $-c$ is positive so it's in the range).

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#14 h Mar 25, 2018, 5:37 AM • 2 Y

Y by Adventure10, Mango247

lolm2k wrote:

^ you're wrong, read mkhayech first answer

That is why I mentioned $f(a)=f(b) \text{ then } a^2=b^2$ If this doesn't answer your question, could you kindly tell me which part is wrong?

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lolm2k

#15 h Mar 26, 2018, 8:51 PM • 2 Y

Y by Adventure10, Mango247

the big mistake is that even if there is an interval (which it does exist) in which $f(x)=f(-x)$ this does NOT implies in a obvious manner that the function is even for all reals.

if $f(0)=0$ then the functio does not exist so i'm not sure why you considered this case, not that this is too relevant it is just weird you tried to prove f is even when this happens

This post has been edited 2 times. Last edited by lolm2k, Mar 27, 2018, 12:18 AM
Reason: NOT

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#16 h Mar 26, 2018, 9:33 PM • 1 Y

Y by Adventure10

When substituting by x=f(x); aren't you assuming that the function has a fixed point? (I might be mistaken tho)

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lolm2k

#17 h Mar 26, 2018, 11:35 PM • 2 Y

Y by Adventure10, Mango247

never did that particular substitution, only when i assumed $f(0) = 0$ to prove this leads to contradiction

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#18 h Mar 27, 2018, 2:09 AM • 3 Y

Y by lolm2k, Adventure10, Mango247

lolm2k wrote:

the big mistake is that even if there is an interval (which it does exist) in which $f(x)=f(-x)$ this does not implies in a obvious manner that the function is even for all reals.

if $f(0)=0$ then the function does not exist so i'm not sure why you considered this case, not that this is too relevant it is just weird you tried to prove f is even when this happens

I didn't realize that there are no functions $f$ such that $f(0)=0$ . Fixed!!

And if $f$ increases for $x \in (0,\epsilon)$ , $f$ increases for all $x>0$ , or else because of continuity, there will be two positive reals $a,b$ such that $f(a)=f(b)$ . Do the same thing for $x<0$ , and we're done.
If my post isn't clear, here is another solution:

Since there are no solutions for $f(x) \in \{x,-x\}$
(if then $x^2+1=f(f(f(x))) \in \{x,-x\}$ )
If $f(0)>0 \Longrightarrow f(x)>|x| \forall x\in \mathbb{R}$ so $f$ is even.
same for $f(0)<0$ , so we are done!!

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hung9A

#19 h Jun 6, 2025, 4:33 PM

Y by

Note that $f(x^2 + 1) = f(f(f(f(x)))) = f(x)^2 + 1, \forall x \in \mathbb{R}$ , so $f(x)^2 = f(-x)^2, \forall x \in \mathbb{R}$ . In addition, if $f(a) = f(b)$ for some $a, b \in \mathbb{R}$ then $a^2 + 1 = f(f(f(a))) = f(f(f(b))) = b^2 + 1$ , so $a^2 = b^2$ .

Consider the two sets
$A = \{x \in \mathbb{R}: f(x) \neq f(-x)\}, B = \{x \in \mathbb{R}: f(x) \neq -f(-x)\}.$ Assume that both sets are non-empty. Then there exists $a, b \geq 0$ such that $a \in A, b \in B$ . Let $I$ be the closed segment with endpoints $a, b$ , and $J$ be the closed segment with endpoints $-a, -b$ . Then $f$ is injective on $I$ and injective on $J$ , so $f$ is monotonic on $I$ and on $J$ .
Now if $f(a)f(b) > 0$ then $f(-a)f(-b) < 0$ . This means that $f$ has a zero on $J$ but not on $I$ , but this cannot happen since $f(x)^2 = f(-x)^2, \forall x \in \mathbb{R}$ . We get a similar contradiction if $f(a)f(b) < 0$ .

Therefore either $A$ or $B$ is non-empty, so $f$ is either odd or even. Since $f(f(f(x))) = x^2 + 1$ is an even function, $f$ is also an even function.

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