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Let $B\subset \mathbb{R}^2$ be the unit "open" ball with centre in origin. Define a function $u:\overline{B}\to \mathbb{R}$ in the folllowing way: let $(x,y)=(r\cos\theta,r\sin\theta)$ with $r\in [0,1]$ and $\theta\in [0,2\pi]$.

For $r\in [0,1]$ and $\theta\in [\pi,2\pi]$, $u(r,\theta)=0$.

Now consider the segment $\ell=\{te_2+(1-t)e_1, \ t\in [0,1]\}$. For each $\theta\in [0,\pi/2]$, let $r_\theta\in [0,1]$ be the number for which $r_\theta\theta\in \ell$.

For $\theta\in [0,\pi/2]$, define $u(r,\theta)=2r\theta/\pi $, if $r\in [0,r_\theta]$.

For $\theta\in [0,\pi/2]$, define $u(r,\theta)=2r_\theta\theta/\pi-2r_\theta\theta/((1-r_\theta)\pi)(r-r_\theta)$, if $r\in [r_\theta,1]$.

We define $u(r,\theta)$ for $\theta\in [\pi/2,\pi]$ and $r\in [0,1]$ in an analogous way, by considering the segment $\ell'=\{te_2-(1-t)e_1,\ \forall\ t\in [0,1]\}$.

My question is: does $u\in W_0^{1,p}(\Omega)$ for some $p\in [1,\infty)$. If not, does anyone know of a function $u$ satisfying $$u\in (W_0^{1,p}(B)\cap C(B))\setminus C(\overline{B})$$

Remark: Note that my function $u$ satisfies $u\in C(B)\setminus C(\overline{B})$.

Remark 1: The graph of $u$ is piecewise linear.

Tomás
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2 Answers2

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Near $e_2$, you have the function approaching $1$ along the segment $\ell$; yet it's zero on the boundary of the disk. This will make $|\nabla u|$ blow up like $|x-e_2|^{-1}$ in between, guaranteeing $u\notin W^{1,p}$ for $p\ge 2$. I think $u\in W^{1,p}$ for $1\le p<2$. A more transparent description of (some version of) your function could be:

Let $\Omega$ be an open disk, and let $Q$ be a square inscribed in $\Omega$. Define $$ u(x) = \frac{\operatorname{dist}(x,\Omega^c)}{\operatorname{dist}(x,\Omega^c)+\operatorname{dist}(x,Q)},\quad x\in \Omega \tag{1} $$ Note that $u\equiv 1$ on $Q$, and there are discontinuities at the points where the square touched the boundary of the disk. Since the distance function is $1$-Lipschitz, one can see from (1) that $u$ is locally Lipschitz in $\Omega$ and $$|\nabla u|\lesssim (\operatorname{dist}(x,\Omega^c)+\operatorname{dist}(x,Q))^{-1}$$ The measure of the set where $\operatorname{dist}(x,\Omega^c)+\operatorname{dist}(x,Q)<\epsilon$ is $O(\epsilon^{-2})$, which means $|\nabla u|\in L^{2,\infty}$, hence $|\nabla u|\in L^p$ for $1\le p<2$.

The function is in $W^{1,p}_0$ because its zero extension to $\mathbb R^2$ has the ACL property and the derivatives are integrable, as stated above. I omit the details because the goal is easier to achieve with a different construction which I describe below.

Let $$ u(x) = \sum_{n=1}^\infty \left(1-M_n|x-x_n| \right)^+ $$ where $x_n\to e_1$ and $M_n\to \infty$. Every partial sum is a Lipschitz function with compact support on $\Omega$. The $n$th term of the series has $W^{1,p}$ norm $O(M_n^{1-2/p})$; hence, the series can be made convergent in $W^{1,p}$ for every $p<2$. And since $u(x_n) = 1$ for all $n$, the sum is not continuous at $e_1$.

  • That's the first time I see someone proving that $f\in L^p$, by showing first that it is in some weak Lebesgue space. For this particular case, is there some justification behind this idea? – Tomás Sep 10 '14 at 12:23
  • @Tomás If a function is geometrically described, it may be easier to determine the geometry of level sets than to actually integrate it. And that naturally leads to weak $L^p$ estimates. BTW I added a remark about $W^{1,p}_0$: this is the case, because the zero extension is ACL. –  Sep 10 '14 at 13:00
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Such a function could also be constructed the lines of Boundedness of functions in $W_0^{1,p}(\Omega)$ if one chooses the functions $f_i$ to be continuous which is clearly possible. For any $x \in B$, there is a neighborhood which intersects the support of finitely many of the $f_i$, hence the limit $f$ is continuous at $x$. And since $f$ is not bounded on $B$, it can not be extended continuously to $\bar B$.

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