Let $[-\alpha, \alpha] \subset \mathbb{R}$, and let \begin{equation} \text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}dt. \end{equation}
given projections of $\text{erf}(x)$ onto the first $k$ Chebyshev polynomials, forming some linear combination of Chebyshev polynomials I'll call $P_k(x)$, I would like to know what can be said about \begin{equation} \int_{-1}^1 |\text{erf}(\alpha x) - P_k(x)|^2 dx \end{equation} as a function of $\alpha$ and $k$.
I know that theorem 8.2 from ATAP promises that for an analytic function on $[-1,1]$ that can be analytically continuable to the open Bernstein ellipse $E_\rho$, where it satisfies $|f(x)|\leq M$ for some $M$, then its Chebyshev coefficients satisfy $|a_0| \leq M$ and \begin{equation} |a_k| \leq 2 M \rho^{-k}. \end{equation}
I know that for $\text{erf}(x)$ on $\mathbb{R}$ it's upper bounded by $1$, which should be the $M$ referenced in that theorem. What I don't know is how to compute the Bernstein ellipse radius $\rho$.
For complex $x$, the error function is unbounded. I don't know if there is a "nice" bound over the ellipse $E_\rho$, but constructing a bound over a closed disc shouldn't be too hard. You can then find, based on the radius of said disc, the largest ellipse that fits inside it. Use that to get a proxy for $M$. As $\rho\rightarrow\infty$, the ellipse (hence also, the containing disc) will cover the plane, so while the resulting $M$ will be a bit conservative, it should behave well asymptotically for large $\alpha$.
I suspect that the contribution to $M$ will be greatest along the imaginary axis, but that's just a very naive guess.