What kind of functions lie in $L^2(\Omega )-H^{1/2}(\Omega )$?

Question

It's well known that a discontinuous function does not lie in $H^{1/2}(\Omega )$ if $\Omega\subset\mathbb{R}$ is one dimension.

My question is,

For $\Omega\subset \mathbb{R}^3$, are there any simple examples of $L^2(\Omega )- H^{1/2}(\Omega )$?

Answer 1

There are examples that are simple to state (in $n$ dimensions): $$f(x) = |x|^{\alpha-n}\chi_{|x|\le 1},\qquad \frac{n}{2}<\alpha\le \frac{n+1}{2}$$ But proving that $f\notin H^{1/2}$ is a bit awkward; it's best done by using the divided-difference-integral characterization of $H^{1/2}$ or a related, simplified formulation in Exercise 33 here.

Then there are examples for which the proof is simple: $$f(x) = G_\alpha(x),\qquad \frac{n}{2}<\alpha\le \frac{n+1}{2}$$ where $G_\alpha$ is the Bessel kernel. Indeed, the defining feature of $G_\alpha$ is $$\widehat{G_\alpha}(\xi) = \frac{1}{(1+|\xi|^2)^{\alpha/2}}$$ (up to normalizing constants; see Stein's Singular Integrals, for example). The above choice of $\alpha$ is made so that $\widehat{G_\alpha}\in L^2$ yet $$ \int |\xi|\ |\widehat{G_\alpha}(\xi)|^2\,d\xi = \infty $$ signifying $G_\alpha\notin H^{1/2}$.

The two examples are related in that their asymptotic behavior as $x\to 0$ is the same.