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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 $\int k(u)du=1$.

Oliver
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    Of course not: there is no opportunity for cancellation. – kimchi lover Dec 27 '19 at 13:37
  • I am not sure what you mean by "cancellation"? – Oliver Dec 27 '19 at 15:16
  • One way an integral can vanish is if the positive values of the integrand cancel the negative values, as with $\int_0^{2\pi} \sin x dx=0$. But if the integrand is non-negative this phenomenon cannot occur. How else can the integral be $0$? If the integrand vanishes over the range of integration. – kimchi lover Dec 27 '19 at 15:21
  • You are right, I mis-spoke initially and have now corrected the question so that the integrand has both positive and negative values. This updated version is the question I indeed intended to pose initially, and I thank you for having pointed out the initial mis-statement. – Oliver Dec 27 '19 at 15:40
  • What's the fourier transform of your $f$? – kimchi lover Dec 27 '19 at 15:43
  • I am not completely sure. There is a formula for the Fourier transform of the semi-circular law here: http://mathworld.wolfram.com/FourierSeriesSemicircle.html but it involves a Bessel function of the first kind, and I do not know if it can be modified to pertain to the first incomplete moment. – Oliver Dec 27 '19 at 15:49

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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.

kimchi lover
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  • I think you mean $\hat{f}(t)=\int_{-1}^{+1}x\sqrt{1-x^2}dx$. But nonetheless your demonstration is correct: there is no integrable kernel that would work. I just have a feeling that there is a non-integrable one, i.e., one that decays of order $1/|x|$ as $|x|\to\infty$. But that was not the question I asked, so I mark your answer as accepted . – Oliver Dec 28 '19 at 03:56
  • You're right. Corrected, and thanks for the catch! – kimchi lover Dec 28 '19 at 15:02