2
$\begingroup$

Let $u$ be a $W^{1,p}$-Sobolev function defined on the half disk $D^2_+ := \{(x,y)\in \mathbb{R}^2| x^2+y^2 < 1, y > 0\}$.

If I define a function $\tilde u$ by $$ \tilde u\colon (x,y) \mapsto \begin{cases} u(x,y) & y > 0 \\ u(x,-y) & y < 0\end{cases}, $$ is this $\tilde u$ a Sobolev $W^{1,p}$-function on the whole unit disk?

It seems obvious to me that if $\tilde u$ does have generalized derivatives, these have to be simply $\partial_x \tilde u (x,y) = \partial_x u (x,|y|)$ and $\partial_y \tilde u(x,y) = \partial_y u(x,y)$ if $y >0$ and $\partial_y \tilde u(x,y) = - \partial_y u(x,-y)$ if $y <0$. These functions are all $L^p$, so the only question that is left is whether these really are the generalized derivatives of $\tilde u$ and how to check this.

Thank you for any help.

$\endgroup$
1
  • $\begingroup$ I just found this problem as an exercise in Giovanni Leoni, "A first course in Sobolev spaces", Graduate Studies in Mathematics, Vol. 105, AMS. See Exercise 10.37.(iii). $\endgroup$ – Klaus Niederkrüger May 9 '18 at 16:38
1
$\begingroup$

Since no one answers, I write some ideas that can be useful.

I prove the analogous problem in $\mathbb{R}$ ( the same idea can be use in your problem ).

I want use that \begin{equation}u\in W^{1,1}((a,b)) \text{ iff } u\in AC((a,b))\\ \text{ (In the sense that there is an absolutely continuous representative function)}\end{equation}

Thm1: Let $u\in W^{1,1}((0,1))$ and $\bar{u}(x):=\begin{cases}u(x)\quad > x\in(0,1)\\ u(-x)\quad x\in(-1,0)\end{cases}$

Then $\bar{u}\in W^{1,1}((-1,1))$.

Proof: \begin{equation}u\in W^{1,1}((0,1)),\ \text{ then }\ u\in AC((0,1))\end{equation}

For how $\bar{u}$ is defined, we have that \begin{equation}\bar{u}\in AC((-1,1))\end{equation} Then $\bar{u}\in W^{1,1}((-1,1))$

Indeed, since $\bar{u}$ is absolutly continuous, we can use integration by parts and we obtain \begin{equation}\bar{u}'(x)=\begin{cases}u'(x)\quad x\in(0,1)\\ -u'(-x)\quad x\in(-1,0).\end{cases}\end{equation}

Now, using Thm1, it's easy prove that

Thm2: Let $u\in W^{1,p}((0,1))$ and $\bar{u}(x):=\begin{cases}u(x)\quad > x\in(0,1)\\ u(-x)\quad x\in(-1,0)\end{cases}$

Then $\bar{u}\in W^{1,p}((-1,1))$.

Check that there are not any mistakes because I'm not used to working with absolutly continuos function.

$\endgroup$
0
$\begingroup$

Thank you Revzora for your answer. I'll just add in a few more details to your solution.

Indeed by a theorem due to Nykodim a function $f$ is Sobolev $W^{1,p}(\Omega)$ if and only if its restriction to almost every line parallel to the coordinate directions is absolutely continuous, and if $f$ and $\nabla f$ are both in $L^p(\Omega)$.

If $u\in W^{1,p}(D_+^2)$ with $D^2_+ := \{(x,y)\in \mathbb{R}^2| x^2+y^2 < 1, y > 0\}$ the half-disk, we define $\bar u$ by $$ \bar u\colon (x,y) \mapsto \begin{cases} u(x,y) & y > 0 \\ u(x,-y) & y < 0\end{cases}. $$

Since $u$ is Sobolev, clearly the restriction of $u$ to almost all lines parallel to the coordinate axis will be absolutely continuous. From this we directly see that the restriction of $\bar u$ is absolutely continuous along almost all lines that are parallel to the $x$-direction, and it only remains to verify the lines in $y$-direction.

The defintion of absolute continuity means for a function $f\colon [a,b] \to \mathbb{R}$ that for every $\epsilon > 0$ there exists a $\delta >0$ with the following property: choose any finite sequence $a \le a_1 < b_1 \le a_2 < b_2 \le a_3 < \dotsb \le a_N < b_N \le b$ with $\sum_{j=1}^N (b_j-a_j) < \delta$, then $$ \sum_{j=1}^N |f(b_j) - f(a_j)| < \epsilon. $$

If $x_0\in (-1,1)$ such that $y\mapsto u(x_0,y)$ is absolutely continuous then $y\mapsto \bar u(x_0,y)$ will inherit this property. In fact, it trivially holds if we only use intervals that do not include the point $0$ at which we have mirrored the half-line, and if one of the intervals $(a_j,b_j)$ does contain that point, there is no danger either, because we can split $(a_j, b_j)$ into $(a_j,0)$ and $(0,b_j)$, and for such intervals the statement holds.

It still remains to see that the restriction to the lines is $L^p$, but this is obvious. To see that the derivatives are $L^p$, we can use that the generalized partial derivatives agree almost everywhere with the classical partial derivatives.

$\endgroup$

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.