Prove that $|k(x)|\le C|x|^{-n}$ under suitable hypothesis on $k\in\mathcal{C}^1(\Bbb R^n\setminus\{0\})$ 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
 A: Assume the gradient condition. For any function $f : \mathbb{R}^n \to \mathbb{R}$, define $f_\varepsilon$ by $f_\varepsilon(x) = \varepsilon^{-n} f(x/\varepsilon)$. Note
$$
\int_{S^{n-1}} k_\varepsilon(x) d\sigma_{n-1}(x) = \int_{[\lvert x \rvert = \varepsilon^{-1}]} k(x) d\sigma_{n-1}(x) = 0 \quad \forall \varepsilon > 0
$$
and
$$
\lvert (\nabla k_\varepsilon)(x) \rvert = \lvert \varepsilon^{-n-1} (\nabla k)(x/\varepsilon) \rvert \leq 1 \quad \forall x \in S^{n-1}.
$$
Now for any $x,y \in S^{n-1}$, we can find a path along a great circle of $S^{n-1}$ connecting $x$ and $y$ of length $\leq \pi$. So by the fundamental theorem of calculus,
$$
\lvert k_\varepsilon(x) - k_\varepsilon(y) \rvert \leq \pi \sup_{z \in S^{n-1}} \lvert (\nabla k_\varepsilon)(z) \rvert \leq \pi \quad \forall x,y \in S^{n-1}.
$$
But if $\lvert k_\varepsilon(x) \rvert > 2 \pi$ for any $x \in S^{n-1}$, then $k_\varepsilon$ is positive on $S^{n-1}$, and so cannot have mean value $0$. Thus $\lvert k_\varepsilon(x) \rvert \leq 2 \pi$ for all $x \in S^{n-1}$. We have
$$
\lvert k(x) \rvert = \lvert x \rvert^{-n} \lvert k_{\lvert x \rvert^{-1}}(x/\lvert x \rvert)\rvert \leq 2\pi \lvert x \rvert^{-n}
$$
as desired.
Now assume the $\alpha$ condition. Observe how the $\alpha$ bound reacts to $k_\varepsilon$:
$$
\lvert k_\varepsilon(x + h) - k_\varepsilon(x) \rvert \leq {\lvert h \rvert^\alpha} \quad \forall x \in S^{n-1}, \forall \lvert h \rvert \leq 2^{-1}.
$$
Now for any two points $x,y \in S^{n-1}$, connect them by $k$ additions of some $h_1, \ldots, h_k$ with $\lvert h_j \rvert < 1/2$ with $x + \sum_{i=1}^j h_i \in S^{n-1}$ and $x + \sum_{i=1}^k h_i = y$ for all $1 \leq j \leq k$, where $k$ does not depend on $x$ and $y$. Splitting up with the triangle inequality shows that $\lvert k_\varepsilon(x) - k_\varepsilon(y) \rvert$ is bounded independently of $x, y \in S^{n-1}$. The rest of the proof is the same as in the gradient case.
A: 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
$$|k(x)| \le r^{-n-1} \int_0^1 |\gamma'(t)|\,dt \le r^{-n-1} \pi r = \pi r^{-n}.$$
