# Lower estimate for a convolution of a specific function and the surface measure on the unit sphere

On p.399 of Stein's book "Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals", one considers the operator

$$f\mapsto \nabla (f * \text{d}\sigma)$$,

where $$\sigma$$ denotes the standard surface measure on the unit sphere $$S^2\subseteq \mathbb{R}^3$$. Now, Stein asserts that, if $$f$$ is compactly supported with $$f(x)=|x|^{-\alpha}$$ for "small" $$x$$, then we have

$$|\nabla (f*\text{d}\sigma)(x)|\gtrsim |1-|x||^{1-\alpha}$$, provided $$1<\alpha<3$$.

However, I am not able to reproduce this estimate. My considerations so far: If $$f$$ has support in $$B_{\varepsilon}$$ (the ball with small radius $$\varepsilon$$ and center in the origin), then it follows immediately that the support of $$f*\text{d}\sigma$$ lies in $$A_{1-\varepsilon, 1+\varepsilon}$$ (the annulus with smaller radius $$1-\varepsilon$$ and larger radius $$1+\varepsilon$$, and center in the origin). If we assume $$f(x)=|x|^{-\alpha} \chi_{B_\varepsilon}(x)$$, then $$\nabla (f*\text{d}\sigma)(x)=-\alpha\int_{S^2\cap \{|x+y|\leq \varepsilon\}} \frac{x+y}{|x+y|^{\alpha+2}} \text{d}\sigma(y)\quad$$ for $$x\in A_{1-\varepsilon, 1+\varepsilon}$$. However, I am not able to estimate the latter integral in a clever way.

Any help would be greatly appreciated!