I wish to find a mollifier $\psi\in C_0^{d+1}(-1,1)$ such that $$ \int_{-1}^1 x^k \psi(x)dx = \begin{cases} 1, & k=0;\\ 0, & k=1,\dots,d. \end{cases} $$

This paper (https://home.cscamm.umd.edu/publications/Gibbs_phenomenon_Tadmor_Acta07_final_CS-07-07.pdf) considers this problem in Section 10.1. It uses Gegenbauer polynomials $C_k^{(\alpha)}(x)$, which are orthogonal with respect to the weight function $w(x) = (1-x^2)^{\alpha-1/2}$, to construct $$ \psi_p(x) = c_{\alpha,d}(1-x^2)^{\alpha-\frac{1}{2}} C_d^{(\alpha)}(x),\quad -1<x<1. $$

It does not seem entirely correct to me because $C_d^{(\alpha)}$ is also orthogonal to $C_0^{(\alpha)}$, so it cannot satisfy the moment condition for $k=0$.

Question: Is there a neat form of $\psi$ that satisfies the $(d+1)$ moment conditions above?

  • $\begingroup$ The set of polynomials $x^k$, for $k \in \{0, \ldots, d\}$, is a $d+1$ dimensional space, and it can also be spanned by $C^{(\alpha)}_{k}$ for $k \in \{0, \ldots, d\}$, since each of the $C^{(\alpha)}_k$ is a polynomial of degree $k$. So if you expand $x^k$ in terms of the Gegenbaur polynomials bases, you are now down to solving a finite dimensional linear equation. So while the formula listed explicitly is not quite right, you can certainly write $\psi_p$ as a linear combination of the Gegenbaur polynomials. $\endgroup$ – Willie Wong Oct 15 '19 at 15:46
  • $\begingroup$ @WillieWong Yes, of course, I was hoping that something can be found in the literature.. I am sure that the coefficients will be very messy... Not sure if they are tractable following this way. $\endgroup$ – user58955 Oct 15 '19 at 15:58

This is too long for a comment, but not a complete answer.

If $P_n(x) = P_n^{d,\alpha}$ is a non-zero polynomial of degree $n$ orthogonal to $x^k$, $k = 0, 1, \ldots, n - 1$, with respect to the measure $$\mu_d(x) = |x|^{-2d-2\alpha-1} (x^2 - 1)^{\alpha - 1/2} \mathbb{1}_{\mathbb{R} \setminus (-1,1)}(x)dx,$$ then $$ \int_{-1}^1 (x^d P_d(1/x)) x^{d - k} (x^2 - 1)^{\alpha - 1/2} dx = \int_{\mathbb{R} \setminus (-1,1)} y^{-d} P_d(y) y^{k - d} y^{1 - 2 \alpha} (1 - y^2)^{\alpha - 1/2} y^{-2} dy = 0 $$ for $k = 0, 1, 2, \ldots, d - 1$. Therefore, $c_{d,\alpha} (1 - x^2)^{\alpha - 1/2} x^d P_d(1/x)$ seems to be the function with all the desired properties.

I do not know if the (finite!) sequence $P_n^{d,\alpha}$ has any name or has it been studied before.

  • $\begingroup$ Thanks! It seems that generalized Gegenbauer polynomial is orthogonal with respect to the weight function $(1-x^2)^{\lambda-1/2}|x|^{2\mu}$, so if I pick $\mu=1$ that seems to solve the problem? $\endgroup$ – user58955 Oct 16 '19 at 0:35
  • $\begingroup$ So it seems to me that $(1-x^2)^{d+1}P_{d/2}^{(d+1,\frac{1}{2})}(2x^2-1)$ (assuming that $d$ is even) satisfies the moment constraints, where $P_k^{(a,b)}$ is the Jacobi polynomial. $\endgroup$ – user58955 Oct 16 '19 at 1:10
  • $\begingroup$ I believe one needs here polynomials orthogonal in the complement of the interval $(-1,1)$ rather than in $(-1,1)$. I will try to have a look at it later today. $\endgroup$ – Mateusz Kwaśnicki Oct 16 '19 at 5:40
  • $\begingroup$ I think the Jacobi polynomial thing works. All terms in the polynomial are of even order, so it satisfies the moment condition of odd orders; and $x^{2k}$ is a linear combination of even order polynomials so I think the moment condition also follows from the Jacobi orthogonality. $\endgroup$ – user58955 Oct 16 '19 at 9:13
  • $\begingroup$ @user58955: But your candidate seems to have integral zero (that is, be orthogonal to constants), right? $\endgroup$ – Mateusz Kwaśnicki Oct 16 '19 at 11:07

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