everyone
If I set values of $u \in L^2(\partial \Omega)$, $\Omega \subset R^3$, at several points on $\partial \Omega$, I will not not define $u$, because it is a set of zero measure.
Am I correct that it holds for $H^{1/2}(\partial \Omega)$ ? That is, does prescribing $H^{1/2}(\partial \Omega)$ on a set of zero measure define $u$?
I would appreciate a reference to a textbook. Thank you.
I do not understand what it means to set values of $u\in L^2(\partial \Omega)$, or $H^{s}(\partial\Omega)$. Furthermore you only do so "at several points"? How? Measure zero w.r.t which measure, Lebesgue on $\mathbb R^3$? Or Hausdorff on $\partial\Omega$? I suppose you mean you want the trace of $u$ on $\partial\Omega$ to be in $L^2(\partial\Omega)$?
In any case, the values on the boundary never uniquely define an arbitrary function in any Sobolev space. This non-uniqueness can be directly seen by the existence of smooth, compactly supported functions. Namely, both the zero function and any bump function supported inside $\Omega$ have zero boundary values.
Perhaps you are thinking of solving Poisson's equation, or more generally a second order elliptic PDE. Here's Poisson:
\begin{align} -\Delta u &= f \\ u|_{\partial \Omega} &= g \end{align} Weak solutions to this (which are unique) are in $H^1$ if say $f\in H^{-1}$ and the boundary is $C^1$, so here we need $g$ to be the trace of a $H^1$ function. Such functions are indeed in $H^{1/2}(\partial \Omega)$.
To solve this, you first convert this into the homogeneous boundary equation by subtracting off a function $G\in H^1$ whose boundary values is $g$, say $G|_{\partial \Omega} = g$, $v := u - G$: \begin{align} -\Delta v &= f + \Delta G\\ v|_{\partial \Omega} &= 0\end{align} and then you can apply the theory developed in Chapter 6 of Evans "Partial Differential Equations" to get a weak solution $v\in H^1$ and therefore $u\in H^1$.