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Find all $d > 0$ that for any function $f$ such as $f(0) = f(1)$ for some $x_0 \in [0, 1 - d]$: $$ f(x_0) = f(x_0 + d), x_0 \in [0, 1 - d] $$

I suspect the answer is something like $(0, 0.5) \cup \{1\}$ as, obviously, for the function $f(x) = sin(2\pi x)$ $d$ cannot be greater than $0.5$ (except 1), but I can't prove that.

$f$ is continuous on $[0, 1]$.

Nikrom
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  • If you take $\sin(\pi x)$, or $x(1-x)$, I think you can restrict the answer even more. – Arthur Nov 20 '18 at 11:58
  • @Arthur, actually, for $f(x) = x(1 - x)$ and $f(x) = sin(\pi x)$ not restrictions are applied to $d$, as any $d \in [0, 1]$ satisfies the condition. – Nikrom Nov 20 '18 at 12:18
  • Ahh, I misunderstood $x_0$. I thought you meant "for all $x_0$", when really you meant "for some $x_0$" – Arthur Nov 20 '18 at 12:50

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