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

My question is the following.

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

Unless I am mistaken, a straightforward computation yields \begin{equation} \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}). \end{equation} 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): \begin{equation} \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). \end{equation} 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 \begin{equation} \varphi_k(x) = \sum_{j=0}^{k} (-1)^j {k \choose j} c_j \, x^{2j+1}, \end{equation} 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.

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  • Is c allowed to depend on k,a,$\sigma$ and b? – ChocolateRain Mar 20 '19 at 21:18
  • @ChocolateRain Not on $k$, as stated. – bgsk Mar 20 '19 at 21:38
  • @ChocolateRain Thanks for pointing that out, you are right. I will edit my post. – bgsk Mar 20 '19 at 22:48
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    Also posted to MO, https://mathoverflow.net/questions/326000/lower-bound-for-the-l2-norm-of-a-polynomial – Gerry Myerson Mar 21 '19 at 21:16
  • Why do you write $|\varphi_k(x)|^2$ instead of $\varphi_k(x)^2$ since all numbers are real? Also, for $k=0$ we know $\varphi_0(x)=1$ and thus the integral is $b-a$ whose only lower bound is $c=0$ unless $c$ is **allowed** to depend on $a,b$. – Somos Mar 27 '19 at 15:12
  • @Somos The lower bound $c$ is allowed (and in fact **expected**) to depend on $a, b, \sigma$. I only want it to be uniform in $k$. This is insinuated in the statement of my question. As for the absolute value, you are right. I use this notation for consistency, as sometimes I work in higher dimensions. I have edited the question to be more precise. – bgsk Mar 27 '19 at 15:31
  • I guess that [asymptotics of Jacobi polynomials](https://en.wikipedia.org/wiki/Jacobi_polynomials#Asymptotics_of_Jacobi_polynomials) can be helpful. Also computer calculations of the integrals for different $a$, $b$, $\sigma$ and $k$ tending to infinity still can provide some suggestions. – Alex Ravsky Mar 28 '19 at 01:00

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