My question is, that is the subspace $$ \left\{ u \in H^1(U) : \int_U u(x) dx=0, \int_{\partial U} \frac{\partial u(x)}{\partial n}dx =0 \right\} $$ bigger than merely the space {0}? Boundary $\partial U$ is assumed to be $C^1$. Boundary values are in the trace sense $u|_{\partial U} = Tu$.
I came up with this queston when applying Lax-Milgram to Helmholtz equation.
NOTE. This is a follow-up of gerw's comment to the OP.
The linear form $$ u\mapsto \int_{\partial \Omega} \nabla u \cdot n\, dS, $$ where $n$ denotes the outward unit normal vector, is not well defined on $H^1(\Omega)$. If it were then it would be continuous, meaning that a constant $C>0$ would exist such that $$\tag{!!} \left| \int_{\partial \Omega} \nabla u \cdot n\, dS\right|\le C \|u\|_{H^1(\Omega)}.$$ But no constant exists such that (!!) holds. Indeed, if $u$ is smooth and bounded, then $$ \int_{\partial \Omega} \nabla u \cdot n\, dS = \int_{\Omega} \Delta u\, dV, $$ and clearly there is no way of controlling that right hand side with the $H^1(\Omega)$ norm alone.
P.S.: Another way of seeing this, as we discussed in a recent question, is contained in the informal phrase "taking the trace loses half derivative". So if you only have one derivative in $\Omega$, as you assume, your trace won't have one derivative on the boundary but "only half". That's why you cannot control $\nabla u \cdot n$ on $\partial \Omega$.