In which sense is it possible to solve $\Delta u=0$, $\partial\_\nu u=\phi$, for $\int\phi=0$ on a closed domain, say a ball $B^3\subset\mathbb R^3$? 


For example would a $\phi\in L^p(\partial B^3)$, $1< p<2$ make sense? 


In other words, is $W^{1,p}$ really the right trace space, or else, which is? 


Where can I find this kind of results?


Thanks!



Trace space and Neumann boundary condition 
You can solve the problem with even less regularity than in Rekalo's answer. If $u\in W^{1,p}(\Omega)$, it does not have a normal trace in general. But if you assume in addition that $\Delta u\in L^p(\Omega)$, then the normal trace is well-defined and belongs to $W^{-1/p',p}(\partial\Omega)$, where $p'$ is the conjugate exponent. This space is of negative order, thus is not contained in any $L^q$. It is defined as the dual space of $W^{1/p',p'}(\partial\Omega)$. Since the latter contains the function ${\bf 1}$, it makes sense to say that the integral of $\phi$ is zero: just write $\langle\phi,{\bf 1}\rangle=0$. Under this condition, the Neumann boundary value problem admits a solution, unique up to an additive constant.


