Let $$u(x)\in L^1_s(\mathbb{R}^n) = \left\{ u\in L^1_{\text{loc}}(\mathbb{R}^n; s\in(0,1); \int_{\mathbb{R}^n} \frac{|u(x)|}{1 + |x|^{n+2s}}\ \text{d}x < +\infty \right\} $$
And let $$A_r(y) = \begin{cases} C \dfrac{r^{2s}}{(|y|^2- r^2)^2|y|^n} & y\in\mathbb{R}^n\smallsetminus \bar{B}_r \\\\ 0 & y\in \bar{B}_r \end{cases} $$
where $C$ is a constant. Also it holds $\int_{\mathbb{R}^n} A_r = 1$.
So now my problem: $u(x)$ is such that $u(x) = A_r * u(x)$ (convolution). This property is called $s$-mean.
I cannot procede forward in understanding why
$$u(x) - \int_{\mathbb{R}^n / B_r} A_r(y) u(x-y)\ \text{d}y = C r^{2s} \int_{\mathbb{R}^n / B_r} \frac{u(x) - u(x-y)}{(|y|^2- r^2)^2|y|^n}\ \text{d}y$$
Which more clearly: why can we just take $u(x)$ inside the fraction, in the integral, at the numerator in such way?
Also they say that since $r > 0$ (which is also arbitrary small) we have
$$\int_{\mathbb{R}^n / B_r} \frac{u(x) - u(x-y)}{(|y|^2- r^2)^2|y|^n}\ \text{d}y = 0$$
Can someone please provide me some clarification/motivation?
Thank you!
First of all, you have a small typo: $$ A_r(y) = \frac{Cr^{2s}}{(\vert y \vert^2-r^2)^s \vert y \vert^n} \qquad \text{in } \mathbb R^n \setminus B_r.$$ (Note the power of $s$ in the denominator instead of 2). As you have noted $$\int_{\mathbb R^n}A_r(y)\,dy=1 $$ so $$u(x) =u(x)\int_{\mathbb R^n}A_r(y)\,dy = \int_{\mathbb R^n \setminus B_r} \frac{Cr^{2s}u(x)}{(\vert y \vert^2-r^2)^s \vert y \vert^n} \, dy.$$ Thus, \begin{align*}u(x)-(A_r\ast u)(x)&=u(x)-\int_{\mathbb R^n \setminus B_r} A_r(y)u(x-y)\,dy \\ &=\int_{\mathbb R^n \setminus B_r} \frac{Cr^{2s}u(x)}{(\vert y \vert^2-r^2)^s \vert y \vert^n} \, dy-\int_{\mathbb R^n \setminus B_r} A_r(y)u(x-y)\,dy \\ &=Cr^{2s}\int_{\mathbb R^n \setminus B_r}\frac{u(x)-u(x-y)}{(\vert y \vert^2-r^2)^s \vert y \vert^n} \, dy \end{align*} as required.
For your second question, what you have stated is not true in general. I think what you are trying to say is $$\lim_{r\to0^+}Cr^{2s}\int_{\mathbb R^n \setminus B_r}\frac{u(x)-u(x-y)}{(\vert y \vert^2-r^2)^s \vert y \vert^n} \, dy=0,$$that is, $ u(x) = \lim_{r\to0^+} (A_r \ast u)(x). $ In fact a more general statement holds: \begin{align*} \frac{u(x)-(A_r\ast u)(x)}{r^{2s}} &= C\int_{\mathbb R^n \setminus B_r}\frac{u(x)-u(x-y)}{(\vert y \vert^2-r^2)^s \vert y \vert^n} \, dy \to (-\Delta)^su(x) \end{align*} as $r\to0^+$ where $(-\Delta)^su(x)$ is the fractional Laplacian (of course assuming $u$ has enough regularity for this to exist). To prove this use dominated convergence and the definition of the fractional Laplacian.
As an extra note, it is well-known that if what you said is true $(u(x)=(A_r \ast u)(x)$ for all $r>0$ sufficiently small) then this is equivalent to $(-\Delta)^su(x)=0$. This is the s-harmonic analogue of the mean value formula for harmonic functions. For more details you should check out Chapter 15 of Fractional Thoughts.