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Source: 239 MO 2024 S1

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#1 h May 22, 2024, 8:15 PM • 1 Y

Y by astralAce1729

Let $f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ be a continuous function such that $f(0)=0$ and $f(x)+f(f(x))+f(f(f(x)))=3x$ for all $x>0$ . Show that $f(x)=x$ for all $x>0$ .

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#2 h May 23, 2024, 3:16 AM • 1 Y

Y by pco

Clearly $f$ is injective and a well known result states that a continuous and injective function is monotone. Suppose $f$ is increasing, if $f(x)>x$ then $f(f(x))\ge f(x) > x$ and $f(f(f(x)))\ge f(f(x))>x$ so their sum is greater than $3x$ which is a contradiction therefore $x \ge f(x)$ , meaning $x \ge f(x)\ge f(f(x))$ and $x\ge f(x) \ge f(f(x)) \ge f(f(f(x)))$ so their sum is $\le 3x$ and equality must hold everywhere giving $f(x)=x$ . The case when $f$ is decreasing is not possible because $x>0 \implies f(x) < f(0) = 0$ .
@below fixed

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pco

#3 h May 23, 2024, 4:28 AM

Y by

MrOreoJuice wrote:

The case when $f$ is decreasing is similar.

No, since then $f(f(x))$ and $f(x)$ are not monotonous in the same direction.

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#4 h May 28, 2024, 11:16 AM • 1 Y

Y by Khuatlam-VietNam

In fact, we need neither the condition $f(0)=0$ nor the continuity:
Fix $x$ and consider the sequence $a_n=f^{(n)}(x)$ so that $a_{n+3}+a_{n+2}+a_{n+1}=3a_n$ .
But from the explicit formula of such a recurrence sequence it is easy to see that such a sequence is either constant or negative at some point.
Since the latter is excluded, we find that $a_n$ is constant and in particular $f(x)=a_1=a_0=x$ . Since $x$ was arbitrary, the claim follows.

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#5 h May 28, 2024, 11:47 AM • 1 Y

Y by k12byda5h

Same idea as this P3:
https://www4352.vu.lt/matematikos-olimpiados/wp-content/uploads/2015/11/mifmo2015.pdf

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#6 h May 28, 2024, 11:51 AM

Y by

R8kt wrote:

Same idea as this P3:
https://www4352.vu.lt/matematikos-olimpiados/wp-content/uploads/2015/11/mifmo2015.pdf

Well, of course the trick is much older (for instance see this problem from the IMO Shortlist 1992).

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#7 h Aug 13, 2024, 5:18 AM

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R8kt wrote:

Same idea as this P3:
here

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