Construct an element of $W^{1,2}(\mathbb{R}^2)$ that is unbounded at the origin Construct an element of $W^{1,2}(\mathbb{R}^2)$ that is unbounded at the origin. I want to use this to show that $W^{1,2}(\mathbb{R}^2)$ is not contained in $C(\mathbb{R}^2)$. (This relates to $W^{1,2}(\mathbb{R})$ being a "borderline" case for Sobolev embedding.)
My attempt: I've been looking at the function
$u(x) = \log\log(1+\frac{1}{|x|})$
because I know that it's in $W^{1,2}(B^0(0,1))$, but I'm having trouble modifying it so that it and its derivative will converge in $L^2$. I considered a piecewise function, constant before and after the zeros of $u$, but then $Du$ will be discontinuous.
Any help appreciated!
 A: When trying to extend a Sobolev function like your $u$ outside some domain you can follow different ways. An abstract one is to use a well known extension operator, that is it exists a continuous linear operator
$$T:W^{1,2}(\Omega)\to W^{1,2}(\mathbb{R}^n)$$
such that $Tf|_\Omega=f$ almost everywhere for all $f\in W^{1,2}(\Omega)$, provided $\Omega\subset\mathbb{R}^n$ is an open set with boundary sufficiently regular like your ball $B_1(0)$ (see for example Brezis - Functional Analysis). Applying this operator to your function $u$ we get a function in $W^{1,2}(\mathbb{R}^n)$ that is discontinuous at the origin.
A more explict way can be multipling by a suitable cut off function. That is, let us take $\eta\in C^\infty_c(B_1(0))$ such that: $\eta|_{B_{1/2}(0)}=1$ and $|\nabla\eta|\le 4$. Then $f=\eta\, u\in W^{1,2}_0(B_1(0))$ and it can be extended to zero outside the ball to get a function in $W^{1,2}(\mathbb{R}^n)$ that is discontinuous at the origin.
Maybe you might be interested in my question here Sobolev spaces on $B_1^2(0)$ don't change if we remove a point from the domain .
