This is an example from Evans book. Let $X=B^n(0,1)$, the n-dimensional unit ball.I know that the function $|x|^{-\alpha}$ is in $W^{1,p}(X)$ if and only if $\alpha<\frac{n-p}{p}$.
Now let $(x_j)_j$ be a countable dense subset of $X$. Define $u(x)=\sum_{j=1}^{\infty}\frac{1}{2^j}|x-x_j|^{-\alpha}$. Then $u\in W^{1,p}(X)$ if and only if $\alpha<\frac{n-p}{p}$ Since \begin{align} \sum_{j=1}^{\infty}||\frac{1}{2^j}|x-x_j|^{-\alpha}||_{1,p}&\le \sum_{j=1}^{\infty}\frac{1}{2^j}|||x-x_j|^{-\alpha}||_{1,p}\\ &=|||x|^{-\alpha}||_{1,p}\sum_{j=1}^{\infty}\frac{1}{2^j}\\ &=|||x|^{-\alpha}||_{1,p} \end{align}
But I don't see why the third equality is true. That is, why is $\sum_{j=1}^{\infty}\frac{1}{2^j}|||x-x_j|^{-\alpha}||_{1,p}=|||x|^{-\alpha}||_{1,p}\sum_{j=1}^{\infty}\frac{1}{2^j}$? Can someone please explain.