We are set in $\mathbb{R}^3$. Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be a $C^1\_0$ function, i.e. continuously differentiable with compact support. Let $\epsilon > 0$ be small. I need to show that the derivative 
$$ || \partial\_x [\int\_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} f(y)dy] ||\_{\infty}$$ 
of this integral function exists and can be bounded from above by a constant times $||f||\_{\infty}$ for $\epsilon \rightarrow 0$. We can take any $\infty$ norm we want for the derivative, we're differentiating a function $\mathbb{R}^3 \rightarrow \mathbb{R}^3$ so the derivative is a matrix, but we can take for example supremum over its entries, so 
$$ \max\_{i \in \{1,2,3\}} | \partial\_{x\_i} [\int\_{B(x,\epsilon)} \frac{x\_1 - y\_1}{|x-y|^3} f(y)dy] |$$
for example (or something else if it's easier, but I guess it shouldn't differ much). I want to show that this quantity can be bounded from above by $C ||f||\_{\infty}$ with $\epsilon \rightarrow 0$. I can't just push the derivative in and say that $\partial\_x \frac{x-y}{|x-y|^3} \approx |x-y|^{-3}$ because that won't let me conclude - $|x-y|^{-3}$ is not integrable around $0$ in $\mathbb{R}^3$. I was thinking about pushing the derivative inside and computing explicitly using spherical coordiantes but there are two issues: first it seems really cumbersome and I'd rather avoid it and use simpler arguments if possible and second - I'd still have to argue why I can push the derivative inside. Any suggestions are welcome.

edit: when I think about it I imagine an actual difference quotient when I consider two sphere centered at $x$ and $x +h$ and I guess that I could take care of the overlapping part by 'pairing up' the points with factors $f(y)$ and $f(y+h)$, taking supremum of $\nabla f$ and integrating - that would go to $0$ with $\epsilon \rightarrow 0$ and I expect that the parts that I wouldn't get to pair up with anything will have measure suffiently small (so that the integrals are comparable with $|h|$) so that I can use the $||f||\_{\infty}$ bound there but I'm having a hard time writing it down

**Edit**: my more precise although still incomplete try. I show how to bound the derivative assuming it exists:

ok, after all it seems to be not that difficult, which gets me a bit worried so it would be great if someone could point out a mistake I might've made. I guess the only tricky part is showing that the function is indeed differentiable - if we knew that it would suffice to show that 
$$ \frac{1}{|h|} | \int\_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} f(y) dy - \int\_{B(x + h,\epsilon)} \frac{x+h-y}{|x+h-y|^3} f(y) dy| \leq C \epsilon$$
for some constant $C$. a change of variables $y = y-h$ in the second integral gives
$$ \frac{1}{|h|} | \int\_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} f(y) dy - \int\_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} f(y+h) dy|$$
i.e.
$$ | \int\_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} \frac{f(y) - f(y+h)}{|h|} dy | \leq \int\_{B(x,\epsilon)} \frac{1}{|x-y|^2} ||\nabla f||\_{\infty} dy \leq C ||\nabla f||\_{\infty} \epsilon$$ 
because $\int\_{B(x,\epsilon)} \frac{1}{|x-y|^2} = C \epsilon$, hence the whole thing goes to $0$ with $\epsilon$. Nevertheless we still need to know that it is differentiable

however, the paper I'm reading clearly states that we bound it by a constant times $||f||\_{\infty}$ which confuses me since if I knew it was differentiable I could say it tends to $0$ with $\epsilon$ and if I don't know whether it's differentiable then I can't really talk about bounding the derivative

**edit2**: ok, I see where I went wrong - I got the wrong area of integration after my change of variables. I still believe though that if I knew the derivative exists I could replace the second (wrong) integral 
$$
\int\_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} f(y+h) dy| $$
with $$ \int\_{B(x,\epsilon - 2|h|)} \frac{x-y}{|x-y|^3} f(y+h) dy|$$ plus some remainder which would then lead to a bound involving $||f||\_{\infty}$ but we need differentiability to do that anyway.