# To what fractional Sobolev spaces does the step function belong? (Sobolev-Slobodeckij norm of step function)

I'm new to fractional Sobolev spaces and I'm curious about the regularity of some simple functions like e.$\,$g. step functions in order to understand these spaces better.

In more detail, for $\Omega = [-1,1]^n \subseteq \mathbb{R^n}$ and $A = [-\frac{1}{2},\frac{1}{2}]^n\subseteq \Omega$ consider the function \begin{align} f \colon \ \Omega & \longrightarrow \mathbb{R} \\ x & \longmapsto \begin{cases} 1 & \text{ for } x \in A \\ 0 & \text{ for } x \notin A\text{.}\end{cases} \end{align} For which $s \in [0,1]$ does $f$ have a finite Sobolev-Slobodeckij norm? The norm that is meant here is defined by $$\Vert f\Vert_{s}^2 := \int_\Omega \int_\Omega \frac{\vert f(x) - f(y)\vert^2}{\Vert x-y\Vert^{2s+n}} \, \mathrm{d}x \, \mathrm{d}y\text{.}$$

Is there a way to determine the value of the integral analytically in dependence of $n$ and $s$? Or can one at least easily determine those $s$ for which this integral would be finite? Can it at least be done for $n=2$?

So far I tried the simple case $n=1$ for which I get that $s\in [0,\frac{1}{2})$ has to be fulfilled. I expect that to be the case for any $n$ but at the moment I'm not quite sure since I did not prove it. For $n=2$ I would try to integrate by hand but with my approach it's about to become a rather long calculation. Is there maybe an elegant way to do it? I don't mind if $A$ is replaced by another set like for example a scaled $n$-Sphere or some simplex.

### Guessing

First I would make a quick guess based on the chart of function space. It groups together Sobolev spaces $W^{s,p}$ with the same value of $\frac n p -s$, because these are related by the embedding theorem. While the inclusion provided by this theorem is strict, the sharpness of the theorem still makes "the spaces with equal $\frac n p -s$ are similar" a useful heuristic.

Your function $f$ fails to be in $W^{1,1}$, because its gradient is not an $L^1$ function but rather a vector-valued measure supported on the boundary of $A$. On the other hand, this measure has finite mass (meaning $f\in BV$), which is pretty close to $W^{1,1}$. So, it seems that $s=1$, $p=1$ is at the edge of spaces to which $f$ belongs. From $$\frac{n}{1}-1 = \frac{n}{2} - s$$ we conclude that for $p=2$ (your question), the relevant $s$ is $s=1-\frac n2$.

This agrees with the situation in 1D, and suggests we won't find anything good when $n\ge 2$. Of course, this is just a guess; it may well be wrong.

### Estimating

Recall the layercake principle: the integral of a nonnegative function $g$ is equal to $\int_0^\infty |\{g>t\}|\,dt$ where $|\cdot|$ stands for the measure of the set. So, let's consider the inequality $$\frac{|f(x)-f(y)|}{\|x-y\|^{2s+n}}>t \tag{1}$$ for large values of $t$ (only they are of interest when the measure space has finite measure).

For (1) to hold, exactly one of $x,y$ must be in $A$; also, both must be within distance $\delta \approx t^{-1/( 2s+n)}$ of the boundary of $A$. This constrains $x$ to a set of measure $\approx \delta$. Also, $y$ must lie in a ball of radius $\delta$ around $x$. So,
$$\left|\left\{ (x,y) : \frac{|f(x)-f(y)}{\|x-y\|^{2s+n}}>t\right\}\right| \approx \delta^2 \approx t^{-(n+1)/(2s+n)}$$ The integral over $t\ge 1$ converges iff $$\frac{ n+1}{2s+n} >1$$ which is equivalent to $s<1/2$.

(In particular, the "guess" wasn't correct after all.)

• Thank you a lot! I like that approach especially since I had not seen that strategy before. But shouldn't it be $\left|\left\{ (x,y) : \frac{|f(x)-f(y)}{\|x-y\|^{2s+n}}>t\right\}\right| \approx t^{-(n+1)/(2s+n)} =: c^{n+1}$? Because, roughly spoken, each $x \in A$ gets a ball of feasible $y$ that has area $c^n$ and the strip of feasible $x\in A$ has an area of order $c$ (because the area of the strip is basically $\vert\partial A\vert \cdot c$). Then we would get $s \in [0,\frac{1}{2})$ for any dimension $n$ and $p = 2$. – Murp Aug 4 '15 at 9:08
• @Murp Good catch; I edited. Left the incorrect guess there since in other situations it actually works. [Now that I think of it more, it works for functions where the singular set is a point.] – user147263 Aug 4 '15 at 15:08

The function belongs to $W^{s,1}$ and BV space as in paper "The fractional Cheeger problem".