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Let $\Omega\subset \Bbb R^d$ be a $C^1$ 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

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

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 $\nabla [|x|^\alpha]= \alpha x|x|^{\alpha-2}$,

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

Therefore my initial question could be resumed to the computation 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 $x$. I may wrong expectation.

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.

Let $\Omega\subset \Bbb R^d$ be a $C^1$ 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

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

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 $\nabla [|x|^\alpha]= \alpha x|x|^{\alpha-2}$,

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

Therefore my initial question could be resumed to the computation 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 $x$. I may wrong expectation.

Let $\Omega\subset \Bbb R^d$ be a $C^1$ 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

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

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 $\nabla [|x|^\alpha]= \alpha x|x|^{\alpha-2}$,

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

Therefore my initial question could be resumed to the computation 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 $x$. I may wrong expectation.

Let $\Omega\subset \Bbb R^d$ be a $C^1$ 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

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

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 $\nabla [|x|^\alpha]= \alpha x|x|^{\alpha-2}$,

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

Therefore my initial question could be resumed to the computation of

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

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 $x$. I may wrong expectation.

Let $\Omega\subset \Bbb R^d$ be a $C^1$ 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

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

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 $\nabla [|x|^\alpha]= \alpha x|x|^{\alpha-2}$,

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

Therefore my initial question could be resumed to the computation 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 $x$. I may wrong expectation.