Relationship between $f(t,x)$ as $t \to \infty$ and $f(t/\epsilon, x/\epsilon^2)$ as $\epsilon \to 0$ (periodic functions) 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. 


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*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 $\epsilon \to 0$?

*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 $t \to \infty$?

*If 1. and 2. are not true, is there a reasonable set of assumptions that make the statements true?
 A: 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 $y\in[0,1]$.
