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Could anyone give some idea about the following problem? Many thanks！

Suppose that $f,g: \mathbb{R}\to\mathbb{R}$ are two periodic functions such that $\lim_{x\to\infty}[f(x)-g(x)]=0$. Show that $f(x)=g(x)$ for all $x\in\mathbb{R}$.

OnoL
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  • Or a weaker one is actually enough: Suppose the conditions above are satisfied. Show that if $f(x+T)=f(x)$ for some $T\in\mathbb{R}$, then $g(x+T)=g(x)$. – OnoL Sep 21 '13 at 23:59

2 Answers2

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Hint: Suppose $f(x_0)\neq g(x_0)$ for some $x_0\in\mathbb{R}$ and let $d=f(x_0)-g(x_0)>0$. Given that $f$ and $g$ are periodic can you reach a contradiction of the assumption that $\lim_{x\to\infty}|f(x)-g(x)|=0$?

Dan Rust
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  • Would please be more concrete in how to deal with the problem that we have no idea about whether $f$ and $g$ have common cycle? Many thanks! – OnoL Sep 22 '13 at 04:57
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Let's assume $\exists\ x\ \ni\ {\rm f}\left(x\right) - {\rm g}\left(x\right) = s \not=0$. Given $0 < \epsilon < \left\vert s\right\vert$ and given $\underline{any}$ $N$, we can always find a ( T: period ) $\tilde{x} \equiv x + nT > N\ \ni\ \left\vert{\rm f}\left(\tilde{x}\right) - {\rm g}\left(\tilde{x}\right)\right\vert = \left\vert s\right\vert > \epsilon$.

That means $\lim_{x \to \infty} \left[ {\rm f}\left(\tilde{x}\right) - {\rm g}\left(\tilde{x}\right)\right] = 0$ isn't true. Then, there isn't any $x$ which satisfies the condition mentioned above.

Felix Marin
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