# On the trace theory and restrictions of Sobolev space functions

Consider a connected open bounded subset $D\subset \mathbb{R}^d$ with smooth boundary. It is easier to state for the disc so I will do so. I don't think it would change the argument (except for in our case the torus is one-dim, so the projection on the curve allows us to say it is absolutely continuous? If so it is not my intention, I am interested in boundaries of higher dimension.)

Consider $D:=\{x\in\mathbb{R}^2\mid \|x\|_2<1 \}$ and $\partial D=\mathbb{T}=\{x\in\mathbb{R}^2\mid \|x\|_2=1 \}$.

Let $H^m(D):=W^{m,2}(D)$ the Sobolev space and let $H^{s}(\mathbb{T})$ be the subspace of functions $f\in L^2(\mathbb{T})$ such that

$$\| f \|_{H^s(\mathbb{T})} := \sum_{k\in\mathbb{Z}} (1+k^2)^s|f_k|^2<\infty,$$ where $f_k:=\int_{0}^{2\pi}|f(\theta)e^{ik\theta}|\mathrm{d}\theta$.

Let $f\in H^m(D)$ ($m\in\mathbb{N}$). In general, the trace theorem tells us that $Tf\in H^{m-\frac12}(\mathbb{T})$.

Does the following make sense? Let $f\in C(\overline{D})\cap H^m(D)$. (I do this so that the value at each point make sense.) Let $[g(r)](\cdot):=f(r,\cdot)$, then, for each $r\in[0,1]$.

Q1. Is it true to say $g(r)\in H^{m-\frac12}(\mathbb{T})$ ($r\in[0,1]$)?

Q2. If $f\in C^m(\overline{D})$, then $f(r,\cdot)\in C^m(\mathbb{T})$ $(r\in[0,1])$. But if we have only $f\in H^m(D)\cap C(\overline{D})$, can we not say $f(r,\cdot)\in H^{m}(\mathbb{T})$ even when $r<1$? Can we say $f(r,\cdot)\in H^{m-\frac12}(\mathbb{T})$ ($0\in[0,1]$)?

Q3. This might sound like a stupid question, but: Suppose that firstly we only know $f\in H^{m}(D)$. Suppose that we successfully show that $g(r)\in H^{m}(\mathbb{T})$ ($r\in[0,1]$), when we only knew $f\in H^{m}(D)$ at first. This does not contradict with the trace theorem, as it does not say it is sharp in general? I think it is fine, but the phrase 'we have to lose' from "The reason we have to lose half-derivative when restricting is that we can gain half-derivative when extending." in Intuition behind losing half a derivative via the trace operator confuses me, together with all the equivalence-class-of-functions subtlety.

Short answers. Q1: yes. Q2: no, but yes for almost every $r$. Q3: sharp forms of trace theorem also exists, and the target space in it (a Besov space) has lower regularity.
Q2-3. The reason we may lose regularity when restricting to a surface is that the surface may pass through a singularity of the function, which has greater effect in smaller dimensions. Consider $u(x)=|x|^\alpha$. This is in $H^m$ on $n$-dimensional ball when $2(\alpha-m)>-n$. If the function is restricted to a $k$-dimensional plane passing through $0$, the condition becomes $2(\alpha-m)>-k$, which is more restrictive.
However, one can make a Fubini-type argument that the restriction to almost every plane/sphere must have at least as much regularity as the original function. (To see how, first do this for $L^p$; then apply to derivatives of $u$; fractional-order case is more difficult.)
• Q1: That's what I thought. Q2: Is it because: let $f\in H^m(D)$. Using the chain rule for weak derivatives we have $$\text{(A)}\qquad\qquad \int_{0}^1\!\!\int_{0}^\pi |D_{\theta}^\alpha f(r,\theta)|^2\mathrm{d}r\mathrm{d}\theta<\infty\quad (\alpha\le m).$$ and thus $f(r,\cdot)\in H^m(\mathbb{T})$ for a.e. $r$? So we cannot say "every $r$" because in general we do not know if $D_{\theta}^\alpha f(r,\theta)$ is continuous in $r$. – shall.i.am May 6 '16 at 7:24
• But, in this specific example, we can take a representative of $D_{\theta}^\alpha f(r,\theta)$ ($\alpha\le m-1$) that is absolutely continuous on $(0,1)$ and thus (A) is finite for every $r\in(0,1)$. Oh ok this ($m-1$) is weaker than the trace theorem. Q3: Ok, so one "may" lose the half order on the boundary. I will look for Adams--Fournier for sharp version (if that's not too elementary) Also thank you for the nice example. Regarding the last paragraph starting with "However", could I ask a reference? – shall.i.am May 6 '16 at 7:26