When we extend a function to the boundary in a Sobolev space, with $p = 2$, using the Trace Theorem we use a continuous operator $T$ such that
$$||Tu||_{L^2(\partial U)} \le C ||u||_{W^{1,2}(U)}$$
So we have lost a factor of $1$ in regularity by transitioning from $H^1(U) = W^{1,2}(U)$ to $L^2(\partial U)$.
However, I have read in many places that an extension to the boundary costs us $\frac{1}{2}$ regularity. That is we go from $H^1(U)$ to $H^{\frac{1}{2}}(\partial U)$.
So which is correct..is the inconsistency due to different definitions of taking the trace of a function? How can these different values for the loss of regularity be reconciled?