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If $u\in W^{1,p}(\mathbb{R}^N)$ is a solution of a pde problem , what it means $$\lim_{|x|\rightarrow+\infty}u(x)=0$$

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    It usually means that the sequence $\sup\{|u(x)|\mid \|x\|>n\}$ converges to zero. – 2017-01-06
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    @Vrouvrou: Just for curiosity, is there any relation given between $p$ and $N$? – 2017-01-06
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    @s.harp: The trouble is that $u$ is not a function (with pointwise values), but an element of $L^p$, so it doesn't make sens to speak of $u(x)$. That's why I use the *essential* supremum in my answer, and not just the (usual) supremum. – 2017-01-06
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    @AlexM. $p – 2017-01-06

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One posible interpretation is the following: for every $\epsilon > 0$, there exist a compact $K_\epsilon \subset \Bbb R ^N$ such that $\text{ess sup} \left| u \big|_{\Bbb R^N \setminus K_\epsilon} \right| < \epsilon$, where $\text{ess sup}$ is the essential supremum.

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    and what we can deduce for this solution ? – 2017-01-06
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    @Vrouvrou: Sorry, your question is quite vague, can you be more precise please? – 2017-01-06
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    We win what when we have this property? What is its importance? – 2017-01-06
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    @Vrouvrou: Usually, when you search for solutions in a space of functions that "vanish at the boundary" (and in this case the boundary is $\infty$), *sometimes* you get uniqueness, which is a very nice property - but this depends on the equation under consideration. – 2017-01-06