Functional inequality $f(x)\cdot \cos( x-y)\le f(y)$.

Asked Aug 07 '23 at 17:49

Active Aug 08 '23 at 09:18

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I am doing special manual of functional equations. While i was searching problems for my book, i have found this: Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x) \cdot \cos(x-y)\le f(y)$.

The answer is easy, but i don't know how to prove it. $f(x)=c, c\ge 0$

It's not difficult to prove, that $f(x+2\pi)=f(x)$, but i don't know the next move of solution. The only way i see how to solve this is to prove that $f'(x)\ge 0$

$\displaystyle f'(x)=\lim_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x} \ge \lim_{\Delta x \to 0} \dfrac{f(x)(\cos \Delta x - 1)}{\Delta x}=0$. But in the task not given, that f is differentiable.

edited Aug 08 '23 at 09:18

asked Aug 07 '23 at 17:49

Dmitry Ch

It is periodic so bounded (assuming continuous), so, if not constant it would have a maximum. But if $x_0$ is a local maximum then there are larger values around $x_0$ based on the inequality, – Salcio Aug 07 '23 at 18:05
Why do larger values around $x_0$ exist? – Dmitry Ch Aug 07 '23 at 18:07
because $f(y) >= 0$ (easy to prove) and the inequality. That is take $x_0$ and $y$ close to $x_0$ since $\cos t < 1$ for $t != 0$ you get the claim. – Salcio Aug 07 '23 at 18:11
I don't understand. We can have $f(x_0)\cdot\cos (x_0-y) \le f(y)$. But $\cos (x_0-y)$ can equal, for example $\frac{1}{2}$. And we have $f(x_0) \le 2 f(y)$. It doesn't do anything. if we switch $x_0$ and $y$ it doesn't do anything too.
Sorry for my poor english. I am from another country
– Dmitry Ch Aug 07 '23 at 18:20
I am talking about local max. Have you taken a course in Analysis 1? – Salcio Aug 07 '23 at 18:46
I am studying at school. Yes, local maximum $f'(x)=0$. But there is nothing in task about derivative – Dmitry Ch Aug 07 '23 at 18:48
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Related: AoPS thread, found using Approach$0$. – Bruno B Aug 07 '23 at 19:42
Oh, that was so easy... Thanks a lot!! – Dmitry Ch Aug 07 '23 at 19:47
It isn't a functional equation (with an equal sign) as you write it in your title but a functional inequality. – Jean Marie Aug 07 '23 at 22:38
Thanks for correcting me! – Dmitry Ch Aug 08 '23 at 09:19

Functional inequality $f(x)\cdot \cos( x-y)\le f(y)$.

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