Holder continuity of $\frac{x}{|x|^3} \ast f$ with $f \in C^1_0$ in $\mathbb{R}^3$

Question

Ok, so I need to show that for $f \in C^1_0(\mathbb{R}^3)$ the convolution with $k(x) := \frac{x}{|x|^3}$ is Holder continuous. The exponent doesn't matter much as long as I can bound it using quantities like $||f||_{\infty}$ or $||f||_{L^1}$ - same goes for the Holder norm.

I've managed to show that there is a constant $C$ such that $$ |E(x) - E(y)| \leq |x-y| \ln (\frac{1}{|x-y|})$$ as long as $|x-y| \leq e^{-1}$, but I don't really know if that helps with anything. We could use it to show the Holder 'continuity' for$x$ and $y$'s close to eachother, but this doesn't help at all for those that are further away.

Answer 1

As pointed out in my comment, you can use the easy bound

$$ |E(x) - E(y)| \leq \frac{2}{C^\alpha} \Vert E \Vert_\infty \cdot |x - y|^\alpha $$

for $|x-y| \geq C$, so that only smal values are important (at least if you can show that $E$ is bounded).