Convolution kernel for semi-circular law

Question

For the first incomplete moment of the semi-circular law $$f(t)=t\times\sqrt{\max(1-t^2,0)}~,$$ is there a kernel function $k(\cdot)$ that integrates it to zero? $$\int k(x-t)\; f(t) dt=0\quad \forall x\in[-1,1]~.$$ It would have to be a proper kernel in the sense that $k(u)\geq0$ and $\int k(u)du=1$.

Answer 1

Assuming your convolution is over the reals.

The Fourier transform $\hat f(t) = \int_{-\infty}^\infty f(x)\exp(-itx)dx=\int_{-1}^1 {\sqrt{1-x^2}} \exp(-itx)\,dx$ is real-analytic and hence has isolated roots, and hence is non-vanishing on a dense set of reals. Since $k$ is integrable, its Fourier transform $\hat k$ is continuous. For the convolution $f*k$ to vanish you would need the pointwise product $\hat f(t) \hat k(t)$ to vanish for all $t$. This implies $\hat k$ is identically $0$ which implies $k$ must be indentically $0$. So there is no non-trivial integrable function $k$ for which $f*k$ vanishes everywhere. The restriction that $k$ be non-negative is not needed.

The real-analyticity of $\hat f$ follows from the DCT or by direct calculations with formulas related to Hankel transforms applied to the indicator function of the unit disk in the plane.