# Bounding 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 $$c = c(a, b, \sigma)>0$$ 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.

• Is c allowed to depend on k,a,$\sigma$ and b? – ChocolateRain Mar 20 at 21:18
• @ChocolateRain Not on $k$, as stated. – bgsk Mar 20 at 21:38
• @ChocolateRain Thanks for pointing that out, you are right. I will edit my post. – bgsk Mar 20 at 22:48
• Also posted to MO, mathoverflow.net/questions/326000/… – Gerry Myerson Mar 21 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 at 15:12