Let $K:mathbb R^nsetminus{0}to mathbb C$ be a smooth function with the estimate $|nabla K(x)|leq C|x|^{-n-1}$ for $xneq 0$ 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 $deltain(0,1)$ 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 $|x|geq 2|y|$.