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.

• I'm having trouble seeing why the integral exists in general. It seems it would be infinity for every $x$. What am I missing? – Alex Schiff Jun 18 '14 at 2:42
• well, $\frac{1}{|x|^2}$ is integrable around $0$ in $\mathbb{R}^3$ so the function is well defined. it should also be differentiable and the way I see it is explained in the last paragraph, although it still isn't a formal argument – mm-aops Jun 18 '14 at 12:46
• The only thing I can think of is differentiating under the integral. Not sure if this applies in this situation... – Alex Schiff Jun 18 '14 at 17:10
• not really, the function under the integral is pretty singular at $0$ so one would need much general results – mm-aops Jun 18 '14 at 17:33
• Sorry, clearly I need to refresh my memory on integration in $\mathbb{R}^n$ for $n\geq 2$. – Alex Schiff Jun 18 '14 at 17:35

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

1. There will be boundary terms.
2. 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.

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?

• I allowed myself to improve the formatting of your post. One problem that I have with your approach is that even if $\int_{B(x,\varepsilon)} \frac{x-y}{|x-y|^3} f(y) \, dy \approx \frac{1}{|B(x,\varepsilon)|} \int_{B(x,\varepsilon)} (x-y)d(y) \, dy$ were to hold, this would not per se imply that the same holds after taking derivatives. – PhoemueX Jun 19 '14 at 7:38
• Thank you for editing. $lim_{\epsilon \rightarrow \infty}1/\epsilon...$ is the definition of derivative – claire Jun 19 '14 at 7:43
• Yes, but for example $1 \approx 1 + \frac{1}{1000} \cdot \cos(\frac{1}{x})$, but the derivative on the left hand side vanishes, whereas the derivative on the right hand side gets quite large for $x \rightarrow 0$. – PhoemueX Jun 19 '14 at 8:30
• I am talking about the FIRST "equality", i.e. the $\approx$, not(!) about the estimate in the second line. – PhoemueX Jun 19 '14 at 8:36
• I don't really get this answer. First of all it seems to me you also need to assume that the derivative exists to proceed with taking the limit but I also agree with @PhoemueX as to his counterexample and doubts. Could you elaborate more on it? – mm-aops Jun 19 '14 at 22:17