Integral inequality involving polynomial

Question

I saw the following inequality on another website without any solution: for any $n\ge 0$, show that $$\min_{a_1,a_2,\cdots,a_n\in\mathbb{R}}\int_0^{+\infty}(1+a_1x+a_2x^2+\cdots+a_nx^n)^2e^{-x}dx=\frac{1}{n+1}.$$ At the minimum $\{\bar a_j\}_{j=1}^n$, we have for $k=1\dots n$: $$\int_0^{+\infty}(1+\bar a_1x+\bar a_2x^2+\cdots+\bar a_nx^n)x^ke^{-x}dx=k!+\sum_{j = 1}^n \bar a_j (j + k)! =0.$$ The conditions above together with integration by parts on the original integral lead to $$\int_0^{+\infty}(1+\bar a_1x+\bar a_2x^2+\cdots+\bar a_nx^n)^2e^{-x}dx=\frac{1}{1-2\bar a_1}.$$ However, I don't see an easy way to find all the $\bar a_j$'s. Is there a way to prove the inequality without finding the minimum point?

Answer 1

Using orthogonal polynomials instead of the usual basis ($1, x, x^2, \dots, x^n$) will make the computation much simpler.

Let $\mathcal{P}_n$ be the space of all polynomials of degree $\leq n$, and define inner product $\langle f, g \rangle = \int_0^\infty f(x) g(x) e^{-x} dx$. We want to show that $$\inf_{f \in \mathcal{P}_n, f(0) = 1} \|f\|_2^2 = \frac{1}{n + 1}.$$

The orthonormal basis for $\mathcal{P}_n$ under this inner product is the Laguerre polynomials: \begin{align*} L_0(x) & = 1\\ L_1(x) & = -x + 1\\ L_2(x) & = \frac{1}{2}(x^2 - 4x + 2)\\ & \vdots\\ L_n(x) & = \frac{e^x}{n!}\frac{d^n}{dx^n}(e^{-x}x^n). \end{align*} Its properties are well-studied in applied mathematics. In particular, we have $L_j(0) = 1$, $\|L_j\| = 1$ for each $j$.

Each $f \in\mathcal{P}_n$ with $f(0) = 1$ can be written as $f = \beta_0 L_0 + \beta_1 L_1 + \dotsb + \beta_n L_n$, with $f(0) = \beta_0 + \beta_1 + \dotsb + \beta_n = 1$. By orthogonality \begin{align*} \|f\|^2 & = \beta_0^2 \|L_0\|^2 + \beta_1^2 \|L_1\|^2 + \dotsb + \beta_n^2 \|L_n\|^2\\ & = \beta_0^2 + \beta_1^2 + \dotsb + \beta_n^2. \end{align*}

So we only need to solve the optimization problem \begin{align*} \min & \quad \beta_0^2 + \beta_1^2 + \dotsb + \beta_n^2\\ \text{subject to} & \quad \beta_0 + \beta_1 + \dotsb + \beta_n = 1 \end{align*}

Optimality is achieved when $\beta_0 = \beta_1 = \dotsb = \beta_n = \frac{1}{n + 1}$, and the optimal value is $\Big(\frac{1}{n + 1}\Big)^2 + \Big(\frac{1}{n + 1}\Big)^2 + \dotsb + \Big(\frac{1}{n + 1}\Big)^2 = \frac{1}{n + 1}$.