TITLE: Prove that $|k(x)|\le C|x|^{-n}$ under suitable hypothesis on $k\in\mathcal{C}^1(\Bbb R^n\setminus\{0\})$
QUESTION [12 upvotes]: DON'T BE AFRAID FROM THE +500 BOUNTY: it doesn't matter that I KNOW this problem is really hard, I put it only because I need to solve the problem really URGENTLY!
Let $n\ge2$; given a kernel $k\in\mathcal{C}^1(\Bbb R^n\setminus\{0\})$ such that
$$|\nabla k(x)|\le|x|^{-n-1}\;\;\; \mbox{for} \;\;x\neq0  \tag{1}$$
  $$\int_{|x|=r}k(x)\,d\sigma_n(x)=0\;\;\;\;\;\forall r>0  \tag{2}$$
then I must prove that
$$
|k(x)|\le C|x|^{-n}\;\;\;\;\mbox{for}\;\;x\neq 0.
$$
Is this still true if the gradient condition $(1)$ is replaced with
$$
|k(x+h)-k(x)|\le\frac{|h|^{\alpha}}{|x|^{n+\alpha}}\;\;\;\;\mbox{for}\;\;|h|<\frac12|x| \tag{3}
$$
where $0<\alpha\le 1$?
I tried to work with the convolution operator $A_Kf:=K*f$ and taking its Fourier transform to get some information, but I'm totally lost, I don't really know where to start

REPLY [5 votes]: I'm only writing a small addition to Eric's nice proof as an answer because it's too big for a comment. I think the following may be a little simpler: Suppose $|x|=r.$ Because $rS$ is connected and the integral of $k$ over $rS$ is $0,$ there exists $x_0\in rS$ such that $k(x_0) = 0.$ Now any two points on $rS$ can be connected by a smooth path in $rS$ of length $\le \pi r.$ So let $\gamma :[0,1]\to rS$ be such a path connecting $x_0$ with $x,$ with $\gamma (0) = x_0, \gamma (1)=x.$ Define $g(t) = k(\gamma(t)).$ Then
$$k(x) = g(1)-g(0) = \int_0^1 g'(t) \, dt = \int_0^1 \nabla k(\gamma(t)) \cdot \gamma'(t) \, dt.$$
Take absolute values to see
$$|k(x)| \le r^{-n-1} \int_0^1 |\gamma'(t)|\,dt \le r^{-n-1} \pi r = \pi r^{-n}.$$