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Let $f:(0,1)\rightarrow\mathbb{R}$ be a function with nonconstant derivative $f'$ on $(0,1)$. Is it possible that there exists a real number $t$ such that $f'(x)=f'(x+t)$ for all $x$ such that $x,x+t\in(0,1)$, but $f(x)\neq f(x+t)$ for some $x$ such that $x,x+t\in(0,1)$?

If the question were the other way round (i.e., $f(x)=f(x+t)$ for all $x$ but $f'(x)=f'(x+t)$ for some $x$), but it is not possible by definition of derivative. But for the direction in question, it's much less clear. Also, this question is different because it assumes that $f$ is periodic.

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python55
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1 Answers1

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Take $f(x) = 40\pi x+\cos(20\pi x)$. Its derivative $f^\prime$ is periodic with period $1/10$, yet $f$ is injective (it is strictly increasing).

Clement C.
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  • More generally, by writing for some $x_0\in(0,1)$ such that $x_0+t\in(0,1)$ that $$ f(x) - f(x_0) = \int_{x_0}^x f^\prime(u)du = \int_{x_0}^x f^\prime(u+t)du = \int_{x_0+t}^{x+t} f^\prime(u)du = f(x+t)-f(x_0+t) $$ you get $$ f(x+t) = f(x) + \left(f(x_0+t) - f(x_0)\right) $$ and the last term has to be zero for the function to be $t$-periodic. – Clement C. Mar 20 '15 at 19:37