Let $\phi(x)$ be defined by $ \phi(x)=1\text{ if }|x|<1, $ and otherwise $\phi(x)=\frac{1}{(1+(|x|^2-1)^4)^\frac{n+ps}{8}}$, where $n\in\mathbb{N}$ and $0<s<1$. Then $\phi$ satisfies the following properties, for some constants $c,d$ depending on $p,s,n$:
(i) For every $x\in \mathbb R^n$,$$ |\phi(x)|\leq\frac{c}{|x|^{n+ps}}\forall\,x\in\mathbb{R}^n, $$ (ii) For every $x\in \mathbb R^n$ with $|x|\ge 1$, $$ |\phi(x)|\geq\frac{d}{|x|^{n+ps}}\forall\,|x|\geq 1. $$ For $|x|\geq 1$, I want the following estimate for $I$, where $$ I=\Big|\int_{\{|y|<\frac{|x|}{2}\}}\frac{\phi(x)-\phi(y)}{|x-y|^{n+ps}}\,dy\Big|\leq_\text{(step 1)}(\frac{2}{|x|})^{n+ps}\int_{\{|y|<\frac{|x|}{2}\}}\phi(y)\,dy\leq_\text{(step 2)}\frac{C(n,p,s)}{|x|^{n+ps}}, $$ where in step 1, we have used that $|y|<\frac{|x|}{2}$ implies $\phi(x)\leq\phi(2y)\leq\phi(y)$ which implies $|\phi(x)-\phi(y)|\leq\phi(y)$ and the fact $|y-x|>\frac{|x|}{2}$. I am facing problem how to proceed to get step 2. I tried to use the property (i) of $\phi$, but still struggling. Can somebody help? Thanks.
The remaining integral to estimate is $L^1(\mathbb R^n)$ uniformly in $x$, $$ \int_{|y|<|x|/2} \phi (y) dy = \int_{|y|<1} dy + \int_{1\le |y|\le |x|/2} \phi(y) dy \le c_{n} + c\int_{|y|>1} \frac{dy}{|y|^{n+ps}} = c_n + c'_n \frac{r^{-ps }}{-ps}\Big|_{r=1}^\infty < \infty$$
By the way, can I ask what this relates to, and where is it from? This is a pointwise in $x$ estimate of a Gagliardo seminorm...?