Charcteristic function not in a fractional Sobolev space

Question

I am trying to show that for any Lebesgue measurable set of finite positive measure $E$, the characteristic function $\chi_E$ is not in $H^{\frac{1}{2}}(\mathbb{R}^n)$. I found somewhere that it would be enough to show instead that

$$ \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \frac{\vert \chi_E(x)-\chi_E(y) \vert^2}{\Vert x-y \Vert^{n+1}} dx dy $$

is infinite. I think that the numerator is simply the sum

$$ \chi_{E\times E^c}(x,y)+\chi_{E^c\times E}(x,y) $$ which simplifies the problem to showing that

$$ \int_{E} \int_{E^c} \frac{1}{\Vert x-y \Vert^{n+1}} dx dy + \int_{E^c} \int_{E} \frac{1}{\Vert x-y \Vert^{n+1}} dx dy $$

is infinite, and using Fubini, I think it is enough to show that the first term is infinite. However I am having trouble trying to simplify it further, and think that I should eventually use an integral of the form $\int_1^\infty \frac{1}{r^p}dr$ somehow.

I would appreciate any hints or helpful remarks, including those telling me that this attempt is inherently flawed.

Answer 1

So with your method, the difficulty is the fact that you have no knowledge of the set $E$ at which the singularity occurs. However, the $H^{1/2}$ seminorm (the quantity you are trying to compute, that I will denote $\|\cdot\|_{\dot{H}^{1/2}}$) decreases when one takes a symmetric decreasing rearrangement (see e.g. Lemma 7.17 in the book Analysis by Lieb & Loss). Therefore, taking the ball $B$ centered in $0$ with the same measure as $E$, we have that $$ \|\chi_E\|_{\dot{H}^{1/2}} \geq \|\chi_B\|_{\dot{H}^{1/2}} $$ From there, one way could be to use the Fourier transform definition of $H^{1/2}$ and the exact Fourier transform of $\chi_B$ (see e.g. Fourier transform of the indicator of the unit ball). This gives us $$ \|\chi_B\|_{\dot{H}^{1/2}} = \int_{\mathbb{R}^n} |J_{n/2}(|x|)|^2 \,|x|^{1-n}\,\mathrm{d}x = C_d\int_0^\infty |J_{n/2}(r)|^2\,\mathrm{d}x $$ which is infinite since (see https://en.wikipedia.org/wiki/Bessel_function) $$ J_{n/2}(r) = (\tfrac{2}{πr})^{1/2} \cos(r-\tfrac{(n+1)\pi}{4}) + O_{r\to\infty}(\tfrac{1}{r}) $$ Therefore your integral (i.e. the seminorm) is infinite and thus $\chi_E$ is not in $H^{1/2}$.

---

Remark: If you want to use your computation, from my first equation you can now restrict to a ball and you have $$ \|\chi_B\|_{\dot{H}^{1/2}} = 2\int_{B} \int_{B^c} \frac{1}{|x-y |^{n+1}} \,\mathrm{d}x \,\mathrm{d}y $$ which might be easier to estimate. I suppose one can restrict this integral on a neighborhood of a point of the sphere and then say that the ball is flat near this point to also get estimates on why this integral is infinite?

Answer 2

This nice question led me to some thoughts, which I am going to write down in this answer.

NOTATION. The letter $C$ will always denote an irrelevant positive constant, whose value can change from line to line.

---

We are concerned here with the Gagliardo seminorms, which I will denote by $\dot{H}^s(\mathbb R^d)$ in compatibility with LL3.14’s answer, and which can be defined in two equivalent ways (see also this answer); $$ \lVert f\rVert_{\dot{H}^s}^2:=\int_{\mathbb R^d} \lvert \xi \rvert^{2s}\lvert \hat{f}(\xi)\rvert^2\, d\xi =C\iint_{\mathbb R^d\times \mathbb R^d} \frac{\lvert f(x+y)-f(x) \rvert^2}{\lvert y \rvert^{d+2s}}\, dxdy.$$ Here $\hat{f}$ denotes the Fourier transform. We let $$ \omega_f(y):=\int_{\mathbb R^d} \lvert f(x+y)-f(x)\rvert^2\, dx,\quad\text{ so that }\quad \lVert f \rVert_{\dot{H}^s}^2=C \int_{\mathbb R^d}\frac{\omega_f(y)}{ \lvert y \rvert^{d+2s}}\, dy. $$ The heuristic emerging from these formulations is that, for $f\in L^2$, the seminorm $\lVert f\rVert_{\dot{H}^s}^2$ is finite if and only if $\omega_f$ decays sufficiently fast at $0$, which happens if and only if $\hat{f}$ decays sufficiently fast at infinity.

---

In what follows, we will consider $f=\chi_E$ for some $E\subset \mathbb R^d$. In particular, we let $B$ denote the unit ball. In the nice answer of LL 3.14, the seminorm $\lVert \chi_B\rVert_{\dot{H}^2}$ is studied via the explicit decay rate of $\hat{\chi_B}$. Here we will perform the same analysis, but studying $\omega_{\chi_B}$ instead. This approach is perhaps more adherent to the OP’s initial thoughts.

Expanding the square, we see that $$\tag{1}\omega_{\chi_B}(y)=2\lvert B\rvert -2\lvert B\cap (B-y)\rvert,$$ so we are reduced to study the measure of the intersection $B\cap (B-y)$. As the following crude picture shows,

such intersection is made of two equal spherical caps. Writing the volume of such caps as an integral, we obtain $$ \lvert B\cap (B-y)\rvert = 2\lvert B^{d-1}\rvert \int_{\lvert y \rvert /2}^1 (1-z^2)^{\frac{d-1}{2}}\, dz.$$ Here $\lvert B^{d-1}\rvert $ denotes the volume of the $d-1$ dimensional ball, but it is not relevant for what follows. Indeed, we do not need an exact expression of $\lvert B\cap B-y\rvert$; an approximation to first order at $y\to 0$ will suffice. To compute such approximation, we note that $$\lvert B\cap (B-y)\rvert \Big\rvert_{y=0}=\lvert B\rvert,$$ and it is clear from the integral that $\nabla_y \lvert B\cap (B-y)\rvert $ exists and it is not zero at $y=0$. We conclude that $$\lvert B\cap (B-y)\rvert =\lvert B\rvert -C\lvert y \rvert + O(\lvert y\rvert^2), $$ which, by (1), gives $$\tag{2} \omega_{\chi_B}(y)= C\lvert y \rvert + O(\lvert y \rvert^2).$$ This is all we need, as it immediately implies that $$ \lVert \chi_B\rVert_{\dot{H}^s}^2= C \int_{\mathbb R^d} \frac{\omega_{\chi_B}(y)}{\lvert y\rvert^{d+2s}}\, dy <\infty \quad \iff \quad s<\frac12.$$

---

For an arbitrary $E\subset \mathbb R^d$ of finite measure, the result is that $$\lVert \chi_E\rVert_{\dot{H}^s}^2<\infty \quad \Longrightarrow \quad s<\frac12.$$ This follows from the above by the symmetric rearrangement, which gives $\lVert \chi_E\rVert_{\dot{H}^s}\ge C\lVert \chi_B\rVert_{\dot{H}^s}$, as cleverly shown by LL 3.14.

I have tried to bypass the symmetric rearrangement. The above argument would push through if we could show that $$ \omega_{\chi_E}(y)\ge C\lvert y \rvert + O(\lvert y \rvert^2), $$ but I couldn’t find a way to prove this. I don’t even know if this is true, actually.