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Suppose $f_n(x)$ is continuous and periodic, $f_n(x+T_n)=f_n(x)$.

If we know that $f_n(x) \to f(x)$ uniformly and $T_n \to T$,

can we conclude that $f(x+T)=f(x)$ ?

Arrow
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Takanashi
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  • Have you checked does the standard trick $$\vert f(x+T) - f(x) \vert = \vert \left ( f(x+T) - f_n(x+T) \right ) + \left( f_n(x+T)- f_n(x+T_n) \right ) + \left ( f_n(x+T_n) - f(x) \right \vert $$ work here? – Evgeny Aug 31 '16 at 08:20
  • This trick (especially, the second term in the sum) requires uniform continuity to work, doesn't it? – Ivan Neretin Aug 31 '16 at 08:23
  • @Evgeny yes, seems that $|(f(x+T)−f_n(x+T)|$ and $|(f_n(x+T_n)−f(x)|$ is OK, but i don't know how to deal with $| (f_n(x+T)−f_n(x+T_n) |$ – Takanashi Aug 31 '16 at 08:23
  • @IvanNeretin Yes, it does, but I thought that since we are dealing with continuous functions on compact domains, the uniform continuity (for each $n$) holds automatically. – Evgeny Aug 31 '16 at 08:30
  • @Evgeny That's right, but who said we have a compact domain? – Ivan Neretin Aug 31 '16 at 08:32
  • @IvanNeretin Do you have a counterexample? I have a handwaving argument and so might be mistaken. For each $n$ you have a periodic function. You can narrow its domain down to one period, attach its limit points and prove uniform continuity for this segment. Endpoints require extra work, but I think it's manageable. Still might be mistaken. – Evgeny Aug 31 '16 at 08:37
  • @Evgeny I don't have a counterexample, just a hand-waving counterargument. True, if the functions are defined on $\mathbb R$, then periodicity+continuity enforce uniform continuity, and we are done. Still, I don't see how we rule out possible pathological cases (like, domain is neither $\mathbb R$ nor compact). – Ivan Neretin Aug 31 '16 at 08:42
  • @IvanNeretin But aren't these conditions mentioned in the beginning of the question? It is said that all $f_n(x)$ are continuous and periodic with its own periods $T_n$. – Evgeny Aug 31 '16 at 08:44
  • @Evgeny Not quite. $\tan x$ is periodic and continuous within its domain, and the domain is not nice. – Ivan Neretin Aug 31 '16 at 08:52
  • @IvanNeretin Yes, you are right. If domains are tricky it might require extra work. By the way, Zestylemonzi's answer avoids using uniform continuity of $f_n$ on their domains and requires only uniform convergence. – Evgeny Aug 31 '16 at 17:00

2 Answers2

2

Hint:

$$|f(x+T) - f(x)| = |f(x+T) - f(x+T_n) + f(x+T_n) - f_n(x+T_n) + f_n(x+T_n) - f(x)|$$

Since the $f_n$ are continuous and converge uniformly, what do you know about $f$?

Hope this helps.

Zestylemonzi
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There is a catch to the question as we can only assume that $f_n$ converges to $f$ uniformly on compact subsets (for example $\sin (2\pi x/T_n)$ converges uniformly to $\sin(2\pi x/T)$ on compacts but not on ${\Bbb R}$, except if $T_n=T$)

Let $K_N=[-NT, NT]$ (compact interval). Then $f_{|K_{N+2}}=\lim (f_n)_{|K_{N+2}}$ is continuous, being the uniform limit of continuous functions.

Let $\epsilon>0$ and find $0<\delta<T$ so that $x,y\in K_{N+2}, |x-y|< \delta \Rightarrow |f(x)-f(y)| < \epsilon/3$.

For $n$ large enough $\|f-f_n\|_{K_{N+2}}<\epsilon/3$ and $|T-T_n|<\delta<T$. Let $x\in K_N$. Then for any $x\in K_N$ we have $x+T_n, x+T\in K_{N+2}$. So: $$ |f(x)-f(x+T)|\leq \epsilon/3 + |f(x)-f(x+T_n)| \leq \epsilon + |f_n(x)-f(x+T_n)| = \epsilon$$ and $\epsilon>0$ was arbitrary. The above can surely be tidied up somewhat...

H. H. Rugh
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  • Okay, so the problem had two tricks: "usual" trick (when we add and subtract additional terms to form nice pairs which could be easily estimated) and the trick with $K_N$ :) – Evgeny Aug 31 '16 at 09:24