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$W^{1,p}_{per}(Y)$ is the closure of $C^\infty_{per}(Y)$ (functions $C^\infty$ Y-periodicm where Y is a rectangular block) in the $W^{1,p}$ norm. Of course, the trace of of a function in this space in opposite sides of the rectangular block Y are the same.

Lemma: Let $g\in L^p(Y,\mathbb{R}^n)$ such that $\displaystyle\int_Y g\nabla\varphi=0, \forall\varphi\in W^{1,p}_{per}(Y)$. Then $g$ can be extended by periodicity to an element of $L^P_{loc}(\mathbb{R}^n,\mathbb{R}^n)$, still denoted by $g$, such that -div $g=0$ in $\mathcal{D}'(\mathbb{R}^n)$.

How to prove this lemma? I have no idea. Can anybody help me?

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    By Green we have $0=\int_Y g \nabla \varphi=-\int_Y \varphi \ \text{div} g + \int_{\partial Y} g \varphi$. Now, if we could get the boundary integral to $0$ (e.g. by periodicity) we could argue by the fundamental lemma of calculus to conclude that $\text{div} g=0$ in distributional sense. Notice that this has to be made explicit. Do we know how $g$ behaves on the boundary of $Y$? And is this $p$ of $L^p$ and $W^{1,p}$ the same? Since the integral has to exist, I'd estimate it by something like $||g||_{L^p(Y)} ||\varphi||_{W^{1,q}(Y)}$ where $1/p+1/q=1$. – 2017-02-17
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    No, we do not have any conditions about g on the boundary of Y. Sorry, my bad. It's p'. – 2017-02-18

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