Gradient condition implies Lipschitz condition

Question

Let $K:\mathbb R^n\setminus\{0\}\to \mathbb C$ be a smooth function with the estimate $|\nabla K(x)|\leq C|x|^{-n-1}$ for $x\neq 0$ where $|.|$ is the Euclidean distance function and $\nabla$ gradient. How to prove the following statement? $|K(x-y)-K(x)|\leq C|y|^{\delta}/|x|^{n+\delta}$ for $|x|\geq 2|y|$ and for some choice of $\delta\in(0,1]$ independent of $x$ and $y.$ Using mean value theorem I can show that $|K(x-y)-K(x)|\leq C|y|/|\zeta|^{n+1}$ where $\zeta$ is in the line segment joining $x-y$ and $x.$ I'll be done if $|\zeta|\geq |x|$ whenever $|x|\geq 2|y|$.

Answer 1

Saying $\zeta$ is on the segment from $x-y$ to $x$ says that $$\zeta=x-ty$$for some $t\in[0,1]$. So $$|\zeta|\ge|x|-t|y|\ge|x|-|y|\ge|x|-|x|/2=|x|/2.$$If $\zeta|\ge|x|$ would have been good enough this is presumably god enough as well...