Show that a nonlocal operator with symmetric kernel is well defined

Question

Assume that $J:\mathbb{R}^N\backslash \{0\} \to (0, \infty)$ is a symmetric function; that is, $J(x)=J(-x)$ for any $x \in\mathbb{R}^N$. Moreover, we assume that there is a constant $J_0 > 0$ and $0 < s < 1$ such that \begin{equation}\label{CondJ1} J_0 \leqslant J(x)|x|^{N + 2s}, \end{equation} almost everywere in $\mathbb{R}^N$. We also assume that \begin{equation}\label{CondJ2} Jm \in L^1(\mathbb{R}^N), \end{equation} where $m(x) = \min\{1, |x|^2\}$. Now, consider the nonlocal operador $$\mathcal{L}_Ju(x)= 2\int_{\mathbb{R}^N} J(x - y)(u(x)-u(y))dy.$$I need to show that $\mathcal{L}_J u \in L^2(\mathbb{R}^N)$ if $u \in C^{\infty}_0(\mathbb{R}^N)$.This indeed should happen, as mentioned on page 4 of the paper: https://arxiv.org/pdf/1612.05696.pdf.

Attempt: If $u \in C^{\infty}_0(\mathbb{R}^N)$, then $|u(x) - u(y)| \leqslant C|x - y|$, for some $C > 0$ and $|u(x) - u(y)| \leqslant 2\|u\|_{L^{\infty}(\Omega)}$. Thus $$|u(x) - u(y)| \leqslant K\sqrt{m(x - y)},$$ for some $K > 0$. Consequently

\begin{split} \int_{\mathbb{R}^N}\Big|2\int_{\mathbb{R}^N}J(x - y)(u(x) - u(y))dy\Big|^2dx &\leqslant 4K^2\int_{\mathbb{R}^N}\Big|\int_{\mathbb{R}^N}J(x - y)\sqrt{m(x - y)}dy\Big|^2dx.\\ \end{split} But I only know that $Jm \in L^1(\mathbb{R}^N)$.

Answer 1

I gave a thought on your problem. I hope that this may help you. I am pretty sure that it can be simplified, but I don't have time right now.

Set (I dropped the $2$) $$ \mathcal{L}_n u(x)=\int_{\mathbb{R}^N\setminus B(0,1/n)} J(y)(u(x)-u(x-y))dy. $$ Let us show that $\mathcal{L}_n u$ is in $L^{2}(\mathbb{R}^N)$. First note that $J\in L^{1}(\mathbb{R}^N\setminus B(0,1/n))$ since that $J(x)\leq n^{2} m(x)J(x)$ for all $x$ such that $|x|\geq\frac{1}{n}$. Hence, $$ x\to u(x)\int_{\mathbb{R}^N\setminus B(0,1/n)} J(y)dy $$ is in $L^{2}(\mathbb{R}^N)$ because of the compact support of $u$. The fact that $$ x\to \int_{\mathbb{R}^N}\mathbf{1}_{ B(0,1/n)}(y) J(y)u(x-y)dy $$ is in $L^2(\mathbb{R}^N)$ can be deduced from Young convolution inequality.

So at this point, we know that $\mathcal{L}_n u$ is in $L^{2}(\mathbb{R}^N)$. Using the same idea as before, we also have that, for any constant $c\in\mathbb{R}^N$, $(x\ \cdot c)J(x)$ is in $L^{1}(\mathbb{R}^N\setminus B(0,1/n))$ and, because of the symetry of $J$, we have that $$ \int_{\mathbb{R}^N\setminus B(0,1/n)} J(y)(y\cdot c)dy=0. $$ So, it follows that $$ \mathcal{L}_n u(x)=\int_{\mathbb{R}^N\setminus B(0,1/n)} J(y)\left(u(x)-u(x-y)-y\ \cdot \nabla u(x)\right)dy. $$ Now, we need to remark that $$ I_n(x)=\int_{B(0,1/n)} J(y)\left(u(x)-u(x-y)-y\ \cdot \nabla u(x)\right)dy. $$ is a well-defined element of $L^2(\mathbb{R}^N)$. Indeed, $$ \left|u(x)-u(x-y)-y\ \cdot \nabla u(x)\right|\leq |y|^2 \sup_{t\in B(x,1/n) }\|\mathcal{H}(u)(t)\| $$ where $\|\mathcal{H}(u)(t)\|$ is the operator norm of the Hessian of $u$ at $t$. Hence, $$ \left|\int_{B(0,1/n)} J(y)\left(u(x)-u(x-y)-y\ \cdot \nabla u(x)\right)dy\right|\leq \left(\int_{B(0,1/n)} |y|^{2}J(y)dy\right)\sup_{t\in B(x,1) }\|\mathcal{H}(u)(t)\| $$ but $$ x\to\sup_{t\in B(x,1) }\|\mathcal{H}(u)(t)\| $$ is compactly supported, hence the r.h.s. of the above inequality is $L^2(\mathbb{R}^N)$ (w.r.t. $x$). Moreover, $$ \int_{B(0,1/n)} |y|^{2}J(y)dy\xrightarrow[n\to \infty]{}0, $$ which implies that $I_n$ converges to 0 in $L^2(\mathbb{R}^N)$. It foolows from all that that $I_n+\mathcal{L}_n u$ is a well defined element of $L^{2}$ which in fact does not depends on $n$, and $I_n$ goes to $0$ in $L^2$.

Answer 2

Since $|u(x)-u(y)|\le C|x-y|$ and $|u(x)-u(y)|\le C$, you have that $|u(x)-u(y)|\le C\min\{|x-y|,1\}$. You will need to break the integral inside into the integral over $B(x,1)$ and the integral over the complement of the ball.

Edit If $\mathcal{L}_J(u)$ were in $L^2$, then $\mathcal{L}_J(u)(x)$ would be finite for a.e. $x$, which means that the Lebesgue integral $\mathcal{L}_J(u)(x)$ would be well-defined. But this means that $$\int_{\mathbb{R}^N}|u(x)-u(y)|J(x-y)|\,dy<\infty$$ for a.e. $x$.

Take $N=1$, $J(x)=1/|x|^{1+2s}$ for $1/2<s<1$ and $u(x)=x$ for $x\in (0,1)$. Then $$\int_{x-1/4}^{x+1/4}\frac{|x-y|}{|x-y|^{1+2s}}\,dy\le \int_{\mathbb{R}}|u(x)-u(y)|J(x-y)|\,dy<\infty$$ for $x\in (1/4,3/4)$, which is a contradiction.

For the fractional Laplacian, when $s\in (1/2,1)$, the trick is to use the fact that the operator is even and Taylor's formula to show that $$\text{pv} \mathcal{L}_J(u)(x):=\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^N\setminus B(x,\varepsilon)}(u(x)-u(y))J(x-y)\,dy$$ exists.