0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

Explanation: Q1. You can apply trace theorem on a smaller disk.

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.)

$\endgroup$
  • $\begingroup$ 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 $r$. $\endgroup$ – shall.i.am May 6 '16 at 7:24
  • $\begingroup$ 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? $\endgroup$ – shall.i.am May 6 '16 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.