Weakly-periodic functions compared with lines

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Weakly-periodic functions compared with lines

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Source: KoMaL A. 895

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#1 h Jan 13, 2025, 2:20 PM

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Let's call a function $f:\mathbb R\to\mathbb R$ weakly periodic if it is continuous and $f(x+1)=f(f(x))+1$ for all $x\in\mathbb R$ .
a) Does there exist a weakly periodic function such that $f(x)>x$ for all $x\in\mathbb R$ ?
b) Does there exist a weakly periodic function such that $f(x)<x$ for all $x\in\mathbb R$ ?

Proposed by: András Imolay, Budapest

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internationalnick123456

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internationalnick123456

#2 h Jan 13, 2025, 4:22 PM • 1 Y

Y by Miquel-point

Solution for part a)
We will prove by contradiction that there does not exist a weakly periodic function $f$ satisfying $f(x) > x$ for all $x \in \mathbb{R}$ .
First, observe that $f(x) \neq x + 1$ for all $x \in \mathbb{R}$ . By the continuity of $f$ , we conclude that either $f(x) > x + 1$ for all $x \in \mathbb{R}$ , or $f(x) < x + 1$ for all $x \in \mathbb{R}$ .
Case 1. $f(x) > x + 1$ for all $x \in \mathbb{R}$
We will use induction to show that $f(x) > x + n$ for all $x \in \mathbb{R}$ and for all $n \in \mathbb{N}$ . The base case is true for $n = 1$ , as $f(x) > x + 1$ by assumption.
Now, assume the statement holds for some $n \in \mathbb{N}$ , i.e., $f(x) > x + n$ for all $x \in \mathbb{R}$ . We will show that $f(x) > x + n + 1$ for all $x \in \mathbb{R}$ .
Consider $f(x + 1) = f(f(x)) + 1$ . By the inductive hypothesis, we know $f(x) > x + n$ , so:
$f(x + 1) = f(f(x)) + 1 > f(x) + 1 > x + n + 1.$ Thus, $f(x + 1) > x + n + 1$ , and by induction, $f(x) > x + n$ for all $n \in \mathbb{N}$ .
Taking the limit as $n \to \infty$ , we obtain a contradiction because $f(x)$ cannot grow without bound while remaining bounded by $x$ . Thus, we conclude that the assumption $f(x) > x + 1$ for all $x \in \mathbb{R}$ leads to a contradiction.
Case 2. $f(x) < x + 1$ for all $x \in \mathbb{R}$
In this case, define $g(x) = f(x) - x$ for all $x \in \mathbb{R}$ . Then, we have $0 < g(x) < 1$ for all $x \in \mathbb{R}$ , and the equation $g(x + 1) = g(x) + g(g(x) + x)$ holds for all $x \in \mathbb{R}$ .
By the Min-Max Theorem for continuous functions, there exists a minimum value of $g(x)$ over the interval $[0,1]$ . Let this minimum be $a$ , so $a = \min\limits_{x \in [0,1]} g(x)$ . Since $g(x + 1) > g(x)$ for all $x \in \mathbb{R}$ , it follows that $g(x) \geq a$ for all $x \geq 0$ .
Now consider $g(x + g(x)) \geq a$ for all $x \geq 0$ . This implies:
$g(x + 1) \geq g(x) + a \quad \text{for all} \quad x \geq 0.$ Thus, we have: $g(n) \geq na \quad \text{for all} \quad n \in \mathbb{N}.$ However, this leads to a contradiction, since $g(x) < 1$ for all $x \in \mathbb{R}$ , and thus $g(n)$ cannot exceed 1 for large $n$ . Therefore, the assumption that $f(x) < x + 1$ for all $x \in \mathbb{R}$ also leads to a contradiction.
Since both cases lead to contradictions, the assumption that such a weakly periodic function $f$ exists is false. $\boxed{\text{qed}}$

This post has been edited 1 time. Last edited by internationalnick123456, Jan 13, 2025, 4:23 PM

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#3 h Apr 15, 2025, 9:15 AM • 1 Y

Y by Pirkuliyev Rovsen

a) Answer: No. Suppose such function $f$ does exist. We have
$f(x+1)>f(x)+1>f(x)\eqno(1).$ Further,
$f(x+1)-(x+1)=[f(x)-x]+[f(f(x))-f(x)]>f(x)-x\eqno(2).$ Denote $m=\min_{x\in[0,1]}(f(x)-x)>0$ . From (2) we get $f(x)-x\ge m, \forall x\ge 0$ . Again, by (2), $f(x+1)-(x+1)\ge f(x)-x+m$ , which means that $f(x)-x$ is not bounded as $x\to\infty$ . Since it is continuous, there exists $N\in\mathbb{N}$ and $x_0>0$ , such that $f(x_0)-x_0=N$ . Hence,
$f(x_0+1)=f(f(x_0))+1=f(x_0+N)+1>f(x_0+N),$ which contradicts (1).

b) Answer: Yes. If $f(x)$ is defined consistently for all $x\le 0$ , it can be extended for $x\ge 0$ using the recurrent relation. So, we define first $f(x)$ appropriately over $[-1,0]$ and then we extend it over each interval $[-N,-N+1]$ .

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