simple way to show $|| \partial_x \int_{B(x,\epsilon)} \frac{x-y}{|x-y|^3} f(y) dy||_{\infty} = O(||f||_{\infty})$ in $\mathbb{R}^3$ 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.
 A: I can't give you a complete solution, but maybe the following thoughts are helpful to you:
Using the change of variables $z = x-y$, you can rewrite your original function as follows:
$$
F(x) := \int_{B(x,\varepsilon)} \frac{x-y}{|x-y|^3} \cdot f(y) \, dy = \int_{B(0,\varepsilon)} \frac{z}{|z|^3} \cdot f(x-z) \,dz .
$$
Using the fact that $x \mapsto f(x-z)$ is (continuously) differentiable and that $z \mapsto \frac{z}{|z|^3} \chi_{B(0,\varepsilon)}(z) \in L^1$, you can conclude (using standard theorems), that $F$ is indeed differentiable with derivative
$$
\partial_i F(x) = \int_{B(0,\varepsilon)} \frac{z}{|z|^3} (\partial_i f)(x-z) \,dz.
$$
Using this, you can also derive $\Vert \partial_i F \Vert_\infty \lesssim \Vert \partial_i f\Vert_\infty$, which is strictly weaker than your target estimate.
But also note that the above estimate even implies
$$
\Vert \partial_i F(x) \Vert_\infty \leq \int_{B(0,\varepsilon)} \frac{1}{|z|^2} \, dz \cdot \Vert \partial_i f\Vert_\infty \xrightarrow[\varepsilon \rightarrow 0]{} 0.
$$
Thus, if you are only interested in the case $\varepsilon \rightarrow 0$, this should yield what you want :)
One of the things you could try now (although I could not make it work, but I did not try too hard) is to write
$$
\partial_i F(x) = \lim_{\delta \downarrow 0} \int_{B(0,\varepsilon) \setminus B(0,\delta)} \frac{z}{|z|^3} (\partial_i f)(x-z) \, dz
$$
and then use partial integration (i.e. the divergence theorem) to pass on the derivative on $f$ to $\frac{z}{|z|^3}$. Note here, that


*

*There will be boundary terms.

*You can not do the same without the limit, because $\frac{z}{|z|^3}$ is not smooth in the origin and (even worse) the derivative is not integrable in a neighborhood of the origin.


Finally, it could be helpful to know where you found the statement. The context could perhaps indicate the method to use.
A: We have that
$$
\begin{align}
\partial_x\int_{B(x,\epsilon)} \frac{x-y}{|x-y|^3}f(y)dy
& \approx \partial_x\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}(x-y)f(y) \\
 &\leq \lim_{\epsilon->0}\frac{1}{\epsilon}\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}|\epsilon f(y)|dy\\
&=lim _{\epsilon->0}\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}| f(y)|dy
\end{align}
$$
The first equality $\approx$ holds because:
$|B(x,\epsilon)|$ is the volume of the ball centered at $x$ with radius $\epsilon$.
$|B(x,\epsilon)|=4/3\pi \epsilon^3\approx |x-y|^3$ since $y\in B(x,\epsilon)$
definition of derivatives can be found here: 
http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
The ineuqlity $\leq$ holds because by definition of derivative, the derivative with respect to $x$ is  defined to be $\partial_x g(x)=lim_{\epsilon->0}g(x)$ for $g:\mathbb{R^3}->\mathbb{R}$, since $\mathbb{R}$ is one dimension. And also $|x-y|\approx \epsilon$ hence two $\epsilon $ are cancelled with each other.
By strong version of Hardy-Littlewood theorem, http://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function, The result holds.
Is this time better?
