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I am reading the book "Applied Analysis" by Hunter, and it says that:

Informally, the Sobolev space $W^{k,p}_0$ can be viewed as $W^{k,p}$-functions whose derivatives of order less than or equal to $k-1$ vanish at the boundary

But, why does it say orders less than or equal to $k-1$, instead of $k$?

The $C^{\infty}_c$ functions are in $W^{k,p}_0$, and their derivatives of all orders vanish, and since the space $W^{k,p}_0$ is the completion of such smooth functions, so should't it say all orders of less than $k$?

yumiko
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1 Answers1

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I believe this can be made clear via the Trace Theorem. I will consider $p=2$ for simplicity. The Trace Theorem provides existence of the trace operator $T$ and, more importantly, the bound

$$\|Tu\|_{L^2(\partial \Omega)} \leq C \|u\|_{H^1(\Omega)}.$$

Note that we lose a derivative on the boundary: we only control the trace in the space $L^2$. In general it is easy to see that in the space $H^k$ we can expect control of the trace operator only in the space $H^{k-1}$.

Matt
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