Trying to compute limit of singular integrals : $L= \lim_{s\to 1}(1-s)\int_{\Omega}\frac{(u(x)-u(y))}{|x-y|^{d+2s}} d y.$

Question

Let $\Omega\subset \Bbb R^d$ be a bounded $C^1$ domain. Let $u:\Bbb R^d\to \Bbb R$ be a function in $C^2_b(\Bbb R^d)$. I would like to compute the following limit: for $x\in \partial \Omega$

$$L= \lim_{s\to 1}(1-s)\int_{\Omega}\frac{(u(x)-u(y))}{|x-y|^{d+2s}} d y. $$

Here is what I did so far:

Let $r>0$ be arbitrarily small enough. Then as $u$ is bounded, we have

$$\begin{align}&\lim_{s\to 1}(1-s)\int_{\Omega\cap \{|x-y|\geq r\}}\frac{|u(x)-u(y)|}{|x-y|^{d+2s}}dy\\&\leq C \lim_{s\to 1}(1-s)\int_{|x-y|\geq r}\frac{dy}{|x-y|^{d+2s}} \\&= Cc_d\lim_{s\to 1}(1-s) \int_r^\infty t^{-2s-1} dt= 0. \end{align}$$

so that

$$L= \lim_{s\to 1}(1-s)\int_{\Omega \cap B_r(x)}\frac{(u(x)-u(y))}{|x-y|^{d+2s}} d y. $$

Using the fundamental theorem of calculus and the relation $\nabla [|x|^\alpha]= \alpha x|x|^{\alpha-2}$,

$$\begin{align}&(1-s)\int_{\Omega \cap B_r(x)}\frac{(u(x)-u(y))}{|x-y|^{d+2s}} d y\\ & = -\int_0^1 dt (1-s)\int_{\Omega \cap B_r(x)}\frac{\nabla u(x+ t(y-x))\cdot (y-x)}{|x-y|^{d+2s}}dy\\&= \int_0^1 dt\frac{ (1-s)}{d-2(1-s))}\int_{\Omega \cap B_r(x)} \nabla u(x+ t(y-x))\cdot \nabla_y [|x-y|^{-d+2(1-s)}]dy \end{align}$$

Therefore my initial question could be resumed to computaion of

$$\lim_{s\to 1} (1-s)\int_{\Omega \cap B_r(x)} \nabla u(x+ t(y-x))\cdot \nabla_y [|x-y|^{-d+2(1-s)}]dy.$$

Any idea on to move further?

My feeling is the should be a multiple factor of $\frac{\partial u}{\partial n}(x)= \nabla u(x).n(x)$ where $n(x)$ is the normal derivative on $\partial \Omega$ at the point $x$. Don't be mind corrupted, I may have wrong expectation.

Answer 1

Here's a sketch. I have not finished the last step yet, but maybe this will be useful anyhow... I will replace $d$ in your notation by $n$ to avoid confusion with derivatives and differentials. Also, the exponent of the denominator should be $n+s$ for this question to make sense, and the limit must be one sided from below.

The key idea is that since the quantity of interest $L$ is local we can transfer the problem to a space which is locally the same, but easier to understand (a subset of the half space). Since the domain is $C^1$, there is a $C^1$ chart $\phi$ mapping a neighborhood $U\subset H_n$ of the origin diffeomorphically to a neighborhood of $x$, where $H_n$ is an n dimensional half space, and such that $\phi(0)=x$.

By your first observation we can take the integral to be over a small ball $B_r$ of radius $r$ near $x$, small enough that it lies in the image $\phi(U)$. Then we can do a change of variables to rewrite $L$ as $$L=\lim_{s\rightarrow 1} (1-s)\int_{\phi^{-1}(B_r)} \frac{u(\phi(0))-u(\phi(v))}{|\phi(0)-\phi(v)|^{n+s}} J(v) dv$$ where $J(v)=|D\phi(v)|$ is the Jacobian of $\phi$. We can restrict the domain of this new integral in the same way as before, because under $\phi$ the exterior of any ball around the origin is mapped to the exterior of a ball about $x$ in the original space (and the integral over such a region vanishes in the limit). So replace the domain of integration by a new ball $B_R\cap H_n\subset \phi^{-1}(B_r)$.

We approximate $$u(\phi(0))-u(\phi(v)) = -D(u\circ \phi)(0) V + \mathcal{o}(v) = -Du(x)D\phi(0)v + \mathcal{o}(v)$$ $$J(v) = J(0) + \mathcal{o}(v)$$ and lastly $$\phi(0)-\phi(v) = -D\phi(0)v + \eta(v)$$ with $\eta(v) = \mathcal{o}(v)$. We give a name to this last error term, because it will require some extra consideration.

Writing $v=t w$ with $t=|v|$ and $w\in S^{n-1}\cap H_n$, we transform the integral to spherical coordinates:

$$ L= \lim_{s\rightarrow 1}(1-s) \int_{S^{n-1}\cap H_n} dw \int_0^R dt \left[ \frac{-Du(x) D\phi(0)w t^n J(0) +\mathcal{o}(t) t^{n-1} }{|D\phi(0) w t + \eta(wt)|^{n+s}} \right]$$

We want to show that the $\mathcal{o}(v)$ terms can be ignored, and evaluate what remains. We estimate the integral in several steps:

1. The trickiest part to estimate is the denominator. Note that $D\phi(0)$ is an isomorphism, since $\phi$ is a diffeomorphism. So $D\phi(0)w$ obtains a non-zero minimum on $S^{n-1}$, which I'll call $m$. Then for any $\epsilon$, we can choose sufficiently small $R$ so that $\eta(wt)<\epsilon m t \leq \epsilon |D\phi(0) w t|$ for all $t\leq R$. Hence $ (1+\epsilon) |D\phi(0)wt| > |D\phi(0) wt + \eta(wt)|> (1-\epsilon) |D\phi(0)wt|$.
2. For the $\mathcal{o}(t)$ in the numerator, for any $\epsilon$ we choose $R$ so that this term $\mathcal{o}(t)<\epsilon t$ for all $t\leq R$. Then together with the previous point we have $$\int_0^R \frac{\mathcal{o}(t)t^{n-1} dt}{|D\phi(0) w t + \eta(wt)|^{n+s}} < \int_0^R \frac{\epsilon t^n dt}{(1-\epsilon)^{n+s} |D\phi(0) wt|^{n+s}} \leq \int_0^R \frac{\epsilon t^n dt}{(1-\epsilon)^{n+s} (mt)^{n+s}} = \frac{\epsilon R^{1-s}}{(1-\epsilon) (1-s) m^{n+s}}$$ which is uniformly small for small enough $R$. Hence this term vanishes in the limit as $s\rightarrow 1$.
3. The term left is $\int \left[ \frac{-Du(x) D\phi(0)w t^n J(0)}{|D\phi(0) w t + \eta(wt)|^{n+s}} \right]$. We want to use our estimate for the denominator from part 1 above to show that the $\eta$ term vanishes, but the problem is that we don't know a priori whether to use an upper or lower bound on the denominator in approximating it. To get around this, note that the sign of the integrand depends only on $w$. Hence we can break the integral over $S^{n-1}\cap H_n$ into a positive and negative part. Each such part can be approximated arbitrarily well by replacing the denominator by $|D\phi(0)wt|^{n+s}$ and taking $R$ sufficiently small, and thus the total integral can also be approximated well with this denominator and small $R$. (This is a bit sketchy, but it's late and I can't flesh it out any more right now. If you bug me, I would probably make it more rigorous...)
4. Altogether, factoring out a constant $Du(x)$ and cancelling $t^n$, we have reduced the integral to the following: $$\begin{eqnarray} L & = & -Du(x) \lim_{s\rightarrow 1} (1-s)\int_{S^{n-1}\cap H_n} dw \int_0^R dt \frac{D\phi(0) w t^{-s} J(0)}{|D\phi(0) w|^{n+s}} \\ & = & Du(x) \lim_{s\rightarrow 1} R^{1-s}\int_{S^{n-1}\cap H_n} dw \frac{D\phi(0) w J(0)}{|D\phi(0) w|^{n+s}} \\ & = & Du(x) \int_{S^{n-1}\cap H_n} dw \frac{D\phi(0) w J(0)}{|D\phi(0) w|^{n+1}} \\ \end{eqnarray}$$