\epsilon^2)$ as $\epsilon \to 0$ (periodic functions)

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Jul 20, 2017 at 12:53	history	edited	Jochen Wengenroth	CC BY-SA 3.0	added 761 characters in body
Jul 19, 2017 at 22:00	comment	added	user89890		The bounty on the question is about to expire, but your answer is still not clear to me. Would you mind editing it a bit to clarify the doubts I rised in the previous comments?
Jul 12, 2017 at 12:28	comment	added	user89890		Thanks. But I'm still not clear about the sentence "assumption in 1. becomes just uniform convergence on $S^1$ and this implies $f(t/\varepsilon,\cdot)\to a$ uniformly and hence $f(t/\varepsilon,x/\epsilon)\to a$. The same argument gives the second statement."
Jul 12, 2017 at 11:51	comment	added	Jochen Wengenroth		Writing $f$ as a function on $(0,\infty)\times S^1$ means that for $\varphi(x)=\exp(2\pi i x)$ the function $\tilde f:(0,\infty)\times S^1\to\mathbb R$, $(t,\varphi(x)) \mapsto f(t,x)$ is well defined (and continuous if so is $f$). This should also clarify the meaning of $x/\varepsilon$.
Jul 12, 2017 at 11:03	comment	added	Alex M.		@JochenWengenroth: If we interpret the second argument $x$ as belonging to $S^1$, what meaning do you give to $x/\epsilon$?
Jul 12, 2017 at 10:10	comment	added	user89890		That is, why does "assumption in 1. become just uniform convergence on $S^1$ and this implies $f(t/\varepsilon,\cdot)\to a$ uniformly and hence $f(t/\varepsilon,x/\epsilon)\to a$"? And how does the same argument apply to 2.?
Jul 12, 2017 at 10:08	comment	added	user89890		To be honest, as it is currently written, your answer is not clear to me. Could you add some more details to your arguments?
Jul 12, 2017 at 7:18	history	answered	Jochen Wengenroth	CC BY-SA 3.0