# Bessel's Inequality Proof

I have a question about a step in my book, "Mathematical Methods" by McQuarrie about Bessels proof.

The solution on Bessel's inequality start with: $$\int^1_{-1} \left[ f (x) - \sum^N_{n=0} a_n P_n(x) \right] ^2 dx \geq 0.$$ Then, \begin{align} \int^1_{-1} {f^2 (x)}dx &- 2\sum^N_{n=0} a_n \int^1_{-1} f(x)P_n(x)dx\\ &+ \sum^N_{n=0}\sum^N_{m=0}a_n a_m \int^1_{-1}P_n(x)P_m(x) dx \geq 0. \end{align} or $$\int^1_{-1}f^2(x)dx \geq \sum^N_{n=0}\frac{2}{2n+1} a^2_n.$$

Where $P_n(x)$ are the Legendre polynomials and, $$a_n=\frac{2n+1}{2}\int_{-1}^1 f(x)P_n(x).$$ I am confused why $$2\sum^N_{n=0} a_n \int^1_{-1} f(x)P_n(x)dx - \sum^N_{n=0}\sum^N_{m=0}a_n a_m \int^1_{-1}P_n(x)P_m(x) dx = \sum^N_{n=0}\frac{2}{2n+1}.$$

• I assume the $P_n(x)$ are the Legendre polynomials. So what are the coefficients $a_n$? Nov 16, 2017 at 0:26
• $a_n=\frac{2n+1}{2}\int_{-1}^1 f(x)P_n(x).$ You are correct to assume $P_n(x)$ is the Legendre polynomials. Nov 16, 2017 at 1:05

Since $$a_n = \frac{2n + 1}{2} \int^1_{-1} f(x) P_n (x) \, dx \quad \Rightarrow \quad \int^1_{-1} f(x) P_n (x) \, dx = \frac{2 a_n}{2n + 1}.$$
Also, from the well-known orthogonality property for the Legendre polynomials, namely \begin{align*} \int^1_{-1} P_m (x) P_n (x) \, dx = \begin{cases} 0, & m \neq n\\[1ex] \displaystyle{\frac{2}{2n + 1}}, & m = n \end{cases} \end{align*} (a proof of this can be found here) all terms appearing in the $m$ sum will be equal to zero except for the $m = n$ term which, from the orthogonality condition, will be equal to $$\int^1_{-1} P_m (x) P_n (x) \, dx = \frac{2}{2n + 1}.$$ Thus \begin{align*} 2\sum^N_{n=0} a_n \int^1_{-1} f(x)P_n(x)dx - \sum^N_{n=0}\sum^N_{m=0}a_n a_m \int^1_{-1}P_n(x)P_m(x) dx &= 2 \sum^N_{n = 0} \frac{2 a^2_n}{2n + 1} - \sum^N_{n = 0} \frac{2 a^2_n}{2n + 1}\\ &= \sum^N_{n=0}\frac{2 a^2_n}{2n+1}, \end{align*} as required.