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?