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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?

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    The correct version is that we lose $\frac 12$ of regularity. However, it is much easier to show the inequlity when you lose a "unit" of regularity. – 2017-01-12
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    So if we only actually lose $\frac{1}{2}$ regularity how can the inequality above be valid when it involves a loss of an even higher amount of regularity? And what is the definition of the space the extended boundary function lives in, if it isn't in $L^2(\partial U)$ as your comment implies? – 2017-01-12
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    You have a continuous inclusion $L^2\supset H^{1/2}$, which translates into the chain of inequalities $C_{trace\,continuity}\|u\|_{H^1(U)}\ge \|Tu\|_{H^{1/2}(\partial U)} \ge C_{Sobolev\,inclusion} \|Tu\|_{L^2(\partial U)}$ – 2017-01-12

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