The Legendre polynomials $P_l(x)$ are complete in that any continuous function on $[-1,1]$ can be expanded as, $$ f(x) = \sum_{l=0}^{\infty} a_l P_l(x) $$ (see here for example). However, what is the corresponding result for the associated Legendre polynomials $P_l^m(x)$? Since $P_l^0(x) = P_l(x)$, this means that any function on $[-1,1]$ can be expanded using only the $P_l^0(x)$, ignoring the other functions with $m \neq 0$. According to the paper Debnath and Harrell 1976, this appears to be true for any $m$. Their equation 3.2 states that an arbitrary function can be expanded as $$ f(x) = (1-x^2)^{m/2} \sum_{l=0}^{\infty} a_l P_l^m(x) $$ for any value of $m$. However I have not seen a proof or discussion of this anywhere (the wiki page does not have any information on this). Does anyone know why this is true? Is there an easy proof to see why this is true?
Debnath, L. and Harrell, C. 1976. The operational calculus of associated Legendre transforms-I. Indian J. Pure and Appl. Math., 7(3): 278–291.
The Associated Legendre operator is $$ L_m f = -\frac{d}{dx}(1-x^2)\frac{df}{dx}+\frac{m^2}{1-x^2}f,\;\;\; m=1,2,3,\cdots. $$ The domain consists of twice absolutely continuous functions $f\in L^2(-1,1)$ for which $L_mf \in L^2(-1,1)$. These operators are self-adjoint for $m=1,2,3,\cdots$, in the strictest sense of self-adjoint. The ordinary Legendre operator $L_0$ is not self-adjoint without endpoint conditions at $\pm 1$. The condition that $f\in\mathcal{D}(L_0)$ remain bounded near $x=\pm 1$ is enough to guarantee that $L_0$ is self-adjoint. Boundedess at the endpoints is the classical condition imposed on the domain of $L_0$ in order to obtain a self-adjoint operator. The operators $L_m$ are self-adjoint for all $m=1,2,3,\cdots$, without any conditions at $\pm 1$.
The reason why $L_0$ is not selfadjoint, is that it is in the limit circle case at $x=\pm 1$, which can be seen by solving $L_0f=0$: $$ ((1-x^2)f')'=0 \\ (1-x^2)f'=A \\ f' = \frac{A}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right) \\ f = \frac{A}{2}\ln\frac{1+x}{1-x}+B $$ All solutions are in $L^2(-1,1)$, which means that $L_0$ is not self-adjoint on $L^2(-1,1)$ without endpoint conditions. This is not often discussed. Notice that imposing boundedness near $\pm 1$ is enough to force $A=0$; that gives a well-posed self-adjoint operator, and it corresponds with the classical condition imposed for $m=0$. But it interesting that other endpoint conditions are possible at $x=\pm 1$, and these will lead to different self-adjoint operators $L_0$.
It's a little tedious, but not difficult to verify, that for $m=1,2,3,\cdots$, $$ L_m(1-x^2)^{m/2}=m(m+1)(1-x^2)^{m/2} \\ L_m(1-x^2)^{-m/2}=m(m-1)(1-x^2)^{-m/2}. $$ The second solution is not in $L^2(-1,1)$ near either endpoint for $m=1,2,3,\cdots$, which is enough to conclude that $L_m$ is in the limit point case at both endpoints of $(-1,1)$ for $m=1,2,3,\cdots$. So $L_m$ is essentially selfadjoint on its natural domain in $L^2(-1,1)$ for $m=1,2,3,\cdots$. However, $L_0$ is not, because there are two independent solutions of $L_0f=0$ that are in $L^2(-1,1)$.
The Associated Legendre Polynomials of order $m$ form an orthogonal basis of eigenfunctions for $m=1,2,3,\cdots$. These are given in Rodrigues form by $$ P_n^m(x) = (1-x^2)^{m/2}\frac{(-1)^m}{2^nn!}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n. $$ For $m=0$, these are the ordinary Legendre polynomials.