Let $u\in L^\infty(\mathbb{R}, \mathbb{R}^n)$ with $n\in\mathbb{N}$. During class, it has been said that it implies that the set of accumulation points of $u(x)$ as $x\to +\infty$ (or equivalently as $x\to -\infty$) is non empty.
Anyone could please explain me why that is true?
Thank you in advance!
$\bf{EDIT:}$ I suspect that something was stated wrong. I think that it was referred to "set of limits points as $u(x)$ as $x\to \infty$". In that case, does $u\in L^\infty$ imply that it is not empty?
You haven’t stated the problem in a way that’s clear to me. So this answer is based on a guess of the intended meaning.
For example, with $u=\sin$ this interpretation suggests the accumulation points at infinity are precisely $[-1,1]$.
In this case, an essentially bounded $u$ shall always have at least one such point. There is a subset $A$ of $\Bbb R$ with null complement and a bound $M>0$ with $u(A)\subseteq[-M,M]^n$. The right hand side is sequentially (and topologically) compact. In fact, I believe the conclusion is true for any integrable function $u$ too. If $A:=u^{-1}[-1,1]^n$ has $\mu(\Bbb R\setminus A)=+\infty$ then $u$ would obviously fail to be integrable. So $\Bbb R\setminus A$ has finite measure.
In both cases ($u$ integrable or $u$ essentially bounded):
$A$ must contain a sequence $(y_n)_{n\in\Bbb N}$ that tends to positive infinity. By sequential compactness, we have - potentially passing to a subsequence - that $(u(y_n))_{n\in\Bbb N}$ is convergent.
$A$ must then be nonempty, so there is some $y_1\in A$. If there existed no $y_2\in A$ with $y_2>1+y_1$, that would imply $\Bbb R\setminus A\supseteq(1+y_1,\infty)$ which contradicts the finiteness of measure. Thus there is such a $y_2\in A$. The same argument inductively provides a $y_3\in A$ with $y_3>1+y_2$, a $y_4\in A$ with $y_4>1+y_3$, etc., producing a strictly increasing and divergent sequence $(y_n)_{n\in\Bbb N}\subseteq A$.