TITLE: Proof that Legendre Polynomials are Complete
QUESTION [8 upvotes]: Can somebody either point me to, or show me a proof, that the Legendre polynomials, or any set of eigenfunctions, are complete?

REPLY [8 votes]: We follow J. Weidmanns Linear Operators in Hilbert Spaces. We find in section $3.2$ Orthonormal systems and orthonormal bases:

Example 6: In $L_2(-1,1)$ the set $F=\{f_n:n\in \mathbb{N}_0\}$ with $f_n(x)=x^n$ is a linearly independent system.
The application of the Gram-Schmidt orthogonalization process 
  provides an orthonormal system (ONS) $M=\{p_n:n\in\mathbb{N}_0\}$, where $p_n(x)=\sum_{j=0}^na_{n,j}x^j$ holds with $a_{n,n}>0$; i.e., $p_n$ is a polynomial of degree $n$ with a positive leading coefficient.
These polynomials are called the Legendre polynomials. As $F$ is total, the Legendre polynomials constitute an orthonormal basis (ONB) in $L_2(-1,1)$.
The polynomials $p_n$ can be given explicitly:
  \begin{align*}
p_n(x)=\frac{1}{2^nn!}\left(\frac{2n+1}{2}\right)^{\frac{1}{2}}\frac{d^n}{dx^n}\left(x^2-1\right)^n,\qquad\qquad n\in\mathbb{N}_0\tag{1}
\end{align*}

$L_2(M)$ denotes a Lebesgue space for a Lebesgue measurable subset $M$ of $\mathbb{R}^m$. If $B$ is a subset of a Hilbert space $H$, then it is said to be total if the linear hull $L(B)$ (the set of finite linear combinations of elements of $B$) is dense in $H$, i.e. ${\overline{L(B)}}=H$.
In order to prove formula (1) it is sufficient to show that the expression given for $p_n$ is a polynomial of degree $n$ whose leading coefficient is positive and that $\langle p_n,p_m\rangle=\delta_{n,m}$.

The first assertion is obvious. For $j<m$ we obtain by a $(j+1)$-fold integration by parts (the integrated terms vanish) that
  \begin{align*}
\int_{-1}^1x^jp_m(x)\,dx&=C_m\int_{-1}^1x^j\frac{d^m}{dx^m}\left(x^2-1\right)^m\,dx\\
&=\ldots\\
&=C_m(-1)^{j+1}\int_{-1}^1\frac{d^{j+1}}{dx^{j+1}}(x^j)\frac{d^{m-j-1}}{dx^{m-j-1}}\left(x^2-1\right)^m\,dx\\
&=0
\end{align*}
  This implies that $\langle p_n,p_m\rangle=0$ for $n\ne m$. It remains to prove that $\|p_n\|=1$. By integration by parts we obtain that
  \begin{align*}
\int_{-1}^1&\left[\frac{d^n}{dx^n}\left(x^2-1\right)^n\right]^2\,dx\\
&=(-1)^n\int_{-1}^1\left(x^2-1\right)^n\frac{d^{2n}}{dx^{2n}}\left(x^2-1\right)^n\,dx\\
&=(2n)!\int_{-1}^1(1-x)^n(1+x)^n\,dx\\
&=(2n)!\frac{n}{n+1}\int_{-1}^1(1-x)^{n-1}(1+x)^{n+1}\,dx\\
&=(2n)!\frac{n(n-1)}{(n+1)(n+2)}\int_{-1}^1(1-x)^{n-2}(1+x)^{n+2}\,dx\\
&=\ldots\\
&=(2n)!\frac{n(n-1)\cdots 1}{(n+1)(n+2)\cdots 2n}\int_{-1}^1(1+x)^{2n}\,dx\\
&=(2n)!\frac{1}{2n+1}\left[(1+x)^{2n+1}\right]_{-1}^1\\
&=(n!)^2(2n+1)^{-1}2^{2n+1}
\end{align*}
From this it follows that $\|p_n\|=1$ and the Legendre polynomials constitute an ONB in $L_2(-1,1)$.