Find all triples such that a quadratic bound implies a constant function

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Find all triples such that a quadratic bound implies a constant function

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Source: 2019 Philippine IMO TST2 Problem 3

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#1 h May 4, 2022, 4:38 PM

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Determine all ordered triples $(a, b, c)$ of real numbers such that whenever a function $f : \mathbb{R} \to \mathbb{R}$ satisfies $|f(x) - f(y)| \le a(x - y)^2 + b(x - y) + c$ for all real numbers $x$ and $y$ , then $f$ must be a constant function.

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#2 h May 4, 2022, 6:44 PM

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All triples are $a>0$ and $b^2=4ac$ , or $a=b=c=0$ .
Let $g(t)=at^2+bt+c$ . Then:
1) $g(t)\geq0$ holds for all $t\in\mathbb{R}$ . Otherwise if $g(t_0)<0$ for some $t_0$ , we get $|f(x+t_0)-f(x)|\leq g(t_0)<0$ , which is impossible)
2) $\min g(t)=0$ . Otherwise if $\min g(t)=m>0$ , take $f(x)=\begin{cases}{m,}&{x=0,}\\{0,}&{x\neq0.}\end{cases}$ We get $|f(x)-f(y)|\leq m\leq g(x-y)$ .
From 1)2) we know that $a>0$ and $b^2=4ac$ , or $a=b=c=0$ .
It remains to prove that $f$ has to be constant under this condition.
The $a=0$ case is trivial. Now let $a>0$ and we know
$|f(x)-f(y)|\leq a(x-y-r)^2$ for some $r\in\mathbb{R}$ . If $r\neq 0$ , let $x=y+r$ and then we know that $f$ is periodic with period $r$ .
Hence
$|f(x)-f(y)|=|f(x+r)-f(y)|\leq a(x+r-y-r)^2=a(x-y)^2.$ So we only need to consider the case $r=0$ . This time we have
$\left|\dfrac{f(x)-f(y)}{x-y}\right|\leq a|x-y|.$ By taking the limit $y\rightarrow x$ , we know that $f'(x)=0$ for all $x\in\mathbb{R}$ , which means that $f$ is constant.

This post has been edited 1 time. Last edited by Andrew_Sakura, May 4, 2022, 6:51 PM

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#3 h May 4, 2022, 7:36 PM

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I think we can not use derivative here.
But we can end solution in next way
We want to prove that $|f(x)-f(y)| \leq a(x-y)^2$ with $a>0$ has only const solution
Let $x-y=nd$ where $n$ is natural
$|f(x)-f(y)| =|f(y+nd)-f(y)| \leq |f(y+nd)-f(y+(n-1)d)|+|f(y+(n-1)d)-f(y+(n-2)d)|+...+|f(y+d)-f(y)| \leq and^2=\frac{a(x-y)^2}{n}$
So by taking limit $n\to +\infty$ we get $f(x)=f(y)$

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