# Bounding the $L^2$ norm of a polynomial from below

Let $$\sigma >0$$ be fixed. For even $$k \in \mathbb{N} \cup \{0\}$$, we consider the polynomial $$$$\varphi_k(x) = \sum_{j=0}^{k} (-1)^j {k \choose j} b_j \, x^{2j} \quad x \in (-1,1),$$$$ where $$$$b_j = \frac{\Big(k+\sigma+\frac12\Big)_j}{\Big(\frac12 \Big)_j}$$$$ and for $$s \in \mathbb{R}$$, $$(s)_j$$ denotes the Pochhammer symbol $$$$(s)_{j}={\begin{cases}1&j=0\\s(s+1)\cdots (s+j-1)&j>0.\end{cases}}$$$$ In particular, $$\varphi_0(x) =1$$.

My question is the following.

For $$-1 < a < b < 1$$, does there exist $$c>0$$ such that $$$$\int_a^b |\varphi_k(x)|^2 dx \geq c \quad$$$$ for any even $$k \in \mathbb{N} \cup \{0\}$$?

Unless I am mistaken, a straightforward computation yields $$$$\int_a^b |\varphi_k(x)|^2 dx = \sum_{j=0}^k \sum_{\ell=0}^k (-1)^{j+\ell} {k \choose j} {k \choose \ell} \frac{b_j b_{\ell}}{2(j+\ell)+1} \, (b^{2(j+\ell)+1}-a^{2(j+\ell)+1}).$$$$ But I do not see how I may bound this double sum from below.

Remark 1: I dont know if it is of any use, one may notice that $$\varphi_k$$ is a hypergeometric function of the form $${}_{2}F_{1}(-k,k+\sigma+\frac12;\frac12;x^2)$$ (see https://en.wikipedia.org/wiki/Hypergeometric_function).

Remark 2: Using this interpretation as a hypergeometric function (which terminates), it is in fact possible to relate $$\varphi_k$$ to the Jacobi polynomials (https://en.wikipedia.org/wiki/Jacobi_polynomials): $$$$\varphi_k(x) = {}_{2}F_{1}(-k,\sigma +\frac12 +k;\frac12; x^2)={\frac {k!}{(\alpha +1)_{k}}}P_{k}^{(-\frac12 ,\sigma )}(1-2x^2).$$$$ Perhaps this observation may be of use.

Comment: The same question is open for any odd $$k \in \mathbb{N}$$, but this time one considers $$$$\varphi_k(x) = \sum_{j=0}^{k} (-1)^j {k \choose j} c_j \, x^{2j+1},$$$$ with $$c_j = \frac{\Big(k+\sigma+\frac32\Big)_j}{\Big(\frac32 \Big)_j}$$. I suspect that the methodology is similar as for the case where $$k$$ is even.

Notice: I have asked the question on MSE, where it is subject to a bounty: https://math.stackexchange.com/questions/3154394/bounding-a-polynomial-from-below.