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Let $f: (0,\infty)\times \mathbb {R} \to \mathbb{R}$ be $1$-periodic in the second variable and in $L^\infty((0,\infty)\times \mathbb{R}).$ If it is necessary, we can also assume $f$ to be continuous.

  1. Suppose that $f(t,x) \to a \in \mathbb{R}$ in $L^\infty$ on compact sets as $t \to \infty$. Do we have that $f(t/\epsilon, x/\epsilon^2) \to a$ in $L^\infty$ on compact sets as $\epsilon \to 0$?

  2. Suppose that $f(t/\epsilon, x/\epsilon^2) \to a \in \mathbb{R}$ as $\epsilon \to 0$ in $L^\infty$ on compact sets. Do we have that $f(t,x) \to a$ in $L^\infty$ on compact sets as $t \to \infty$?

  3. If 1. and 2. are not true, is there a reasonable set of assumptions that make the statements true?

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Due to periodicity in $x$ you can write $f$ as a function on $(0,\infty)\times S^1$ so that your 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.

EDIT. Okay, here is a direct argument. $f(t,x)\to a$ (uniformly) on compact sets means that for all compact $K\subseteq \mathbb R$ and all $e>0$ there is $t_0>0$ such that for all $t>0$ and all $x\in K$ one has $|f(t,x)-a|<e$. Use this for $K=[0,1]$ together with the observation that $f(t/\varepsilon,x/\varepsilon^2)=f(t/\varepsilon,\lbrace x/\varepsilon^2\rbrace)$ where $\lbrace y\rbrace$ denotes the fractional part of a real number. This implies 1. since for $t>0$ it is enough to take $\varepsilon$ so small that $t/\varepsilon>t_0$. (If you want to have the convergence in 1 uniformly with respect to $t$, no chance.)

The proof of 2. is similar because every $x\in [0,1]$ can be expressed as $\lbrace y/\varepsilon^2\rbrace$ for some $y\in[0,1]$.

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  • $\begingroup$ To be honest, as it is currently written, your answer is not clear to me. Could you add some more details to your arguments? $\endgroup$
    – user89890
    Commented Jul 12, 2017 at 10:08
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    $\begingroup$ 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.? $\endgroup$
    – user89890
    Commented Jul 12, 2017 at 10:10
  • $\begingroup$ @JochenWengenroth: If we interpret the second argument $x$ as belonging to $S^1$, what meaning do you give to $x/\epsilon$? $\endgroup$
    – Alex M.
    Commented Jul 12, 2017 at 11:03
  • $\begingroup$ 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 $x/\varepsilon$. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 12, 2017 at 11:51
  • $\begingroup$ 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." $\endgroup$
    – user89890
    Commented Jul 12, 2017 at 12:28

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