Understanding quadrature rule of a function multiplied by another $C^{\infty}$ function

Question

Define a function $f \in C^m[-1,1],m \in \mathbb{N}$ and $g\in C^{\infty}[-1,1]$. Also define a quadrature rule $Q$ for approximating the integral $\int_{-1}^1 h(x)dx$ for some function integrable function $h$ as $Q_n[h] = \sum_{j=1}^{n} w_j h(x_j)$ where $w_j$ are called weights and $x_j$ are nodes of the quadrature scheme. For simplicity one can take this as Fejer or Clenshaw curtis quadrature [1]. I want to show the following (also trying to find a reference if it's true): There exists $N \in \mathbb{N}$ such that $$ \left| \int_{-1}^1 f(x)g(x) dx - Q_n[fg] \right| \le c\left| \int_{-1}^1 f(x) dx -Q_n[f] \right| \quad \forall\, n \ge N \quad(1)$$ where $c$ is some positive constant (dependent on function $g$). The inequality implies that multiplying $f$ by $g$ doesn't worsen it's quadrature convergence because $g$ is smooth in the domain. (1) seems trivial to me, however I am unable to show this for the Fejer or Clenshaw curtis quadrature. For these quadrature schemes, the convergence of $Q_n$ is linked to convergence of Chebyshev coefficients $c_n$ of the integrand. If we define $c_n$ and $c_n'$ as Chebyshev coefficients of $f$ and $fg$ respectively, then I have tried showing that $c_n' = O(c_n)$, but I wasn't able to prove it.

Edit : Some clarifications: i. $n$ is tending to infinity in (1). It's not fixed. ii. $f \notin C^{m+1}[-1,1]$ (implying $f$ can't be a polynomial).

Answer 1

If I understand your question correctly, we could have $g(x_j) = 0$ for all $j = 1, \ldots, n$. Thus, $Q_n[f g] = 0$. Further, if $f$ is a low degree polynomial, the right-hand side of (1) vanishes. However, the left-hand side does not necessarily vanish. This shows that (1) is, in general, not true.