We know that Legendre's DE
$$ \frac{d}{d\phi}\bigg(sin\phi\frac{dg}{d\phi}\bigg) + \big(\mu - \frac{m^2}{sin\phi}\big)g = 0 $$ can be transformed by letting $x = cos\phi$ into
$$ \frac{d}{dx}\bigg((1-x^2)\frac{dg}{dx}\bigg) + \big(\mu - \frac{m^2}{1-x^2}\big)g = 0 $$
I want to prove, by using Rayleigh's quotient, that $\mu \geq 0$. Can $\mu = 0$ be an eigenvalue? If not, are there extra conditions that asure so?
For a general Sturm-Liouville problem given by $$ \frac{d}{dx}\bigg(p(x)\frac{dg}{dx}\bigg) + \big(\mu\sigma(x) + q(x)\big)g = 0 $$
Rayleigh's Quotient is given by:
$$\mu = \frac{-p(x)g(x)g'(x)|_{-1}^1+\int_{-1}^1[p(x)g_x^2(x)-q(x)g^2(x)]dx}{\int_{-1}^{1}g^2(x)\sigma(x)dx} $$
where $g_x(x) = g'(x)$. In this particular case we have that
$$ p(x) = 1-x^2; \ \sigma(x) = 1; \ q(x) = -\frac{m^2}{1-x^2}$$
Therefore from $p(x)$ it's clear that our expression
$$-p(x)g(x)g'(x)|_{-1}^1 = -p(1)g(1)g'(1) + p(-1)g(-1)g'(-1) = 0$$ and our denominator is positive, so it remains the other integral is positive.
$$ \int_{-1}^1[p(x)g_x^2(x)-q(x)g^2(x)]dx = \int_{-1}^1[(1-x^2)g_x^2(x)+\frac{m^2}{1-x^2}g^2(x)]dx$$
This is where I got stuck, how can I prove this? That said integral is positive.
Also, there's conditions on $g(\pm 1) < \infty$. I was thinking, are there conditions that give $\mu = 0$ as an eigenvalue?
Consider $$ g(x)=\frac{1}{2}\ln\frac{1+x}{1-x}. $$ For this, $$ \frac{dg}{dx}=\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=\frac{1}{1-x^2} \\ \frac{d}{dx}\left((1-x^2)\frac{dg}{dx}\right)=0. $$ So $0$ is an eigenvalue, when $m=0$. The function $g$ is square integrable, and $Lg=0$ where $$ Ly=-\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right). $$ This is in addition to the constant function $1$ also being a solution of $Lg=0$. Both $1$ and $\ln\frac{1+x}{1-x}$ are legitimate solutions of $Lg=0$. The ordinary Legendre problem has $0$ for an eigenvalue, with eigenfunction $g=1$.
You can also show that $$ g_m(x) = (1-x^2)^{m/2}g^{(m)}(x) $$ is a solution of $$ -\frac{d}{dx}\left((1-x)^2\frac{dg_m}{dx}\right)+\frac{m^2}{1-x^2}g_m = 0. $$ This holds for $m=0,1,2,3,\cdots$.