Equality cannot hold unless $x \in \{-1,1\}$ and/or Wronskian is not zero [closed]

Question

By playing around with assoc. Legendre polynomials, I arrived at

$$((l+1)+m) (P_l^m(x))^2+((l+1)-m)(P_{l+1}^m(x))^2 = 2(l+1)x P_l^m(x)P_{l+1}^m(x).$$

Now, I want to show that we don't have equality for $x \in (-1,1).$

I undertook quite some computations in order to be sure that this is really the case, but I currently don't see why this is true.

There are some simple ways to start with:

For $x=0$ the inequality is obvious and both sides are even functions.

The remarkable remark: A remarkable, but maybe not obvious fact is that this equation is equivalent to the fact that the Wronskian of the functions $P_l^m$ and $P_{l+1}^m$ does not vanish inside $(-1,1)$, so if it is possible to show this, then you showed the inequality too. Maybe this has be shown somewhere, I don't know, I could not find it.

So equivalently, I want to understand why $W(P_l^m,P_{l+1}^m) \neq 0$ in $(-1,1)$.

Answer 1

Let $y_1< z_1< y_2< z_2< \dots< y_{l-m}\leq z_{l-m}< y_{l-m+1}$ be roots of polynomials $f=(1-x^2)^{-m/2} P_l^m$ and $g=(1-x^2)^{-m/2} P_{l+1}^m$ ($y$'s are roots of $g$, $z$'s are roots of $f$). They are real, belong to $(-1,1)$ and alternate as written because $f$, $g$ are orthogonal polynomial in weighted $L^2_w(-1,1)$ with weight $w=(1-x^2)^m$.

Wronskian vanishes (in a point different from $\pm 1$) iff $(P_{l+1}^m/P_l^m)'=0$, i.e. iff $(g/f)'=0$, i.e. iff $f'/f=g'/g$ i.e. iff $$ \sum \frac1{x-y_i}=\sum \frac1{x-z_i}. $$ But it is impossible for real $x$. If, say, $y_k<x<z_k$ for some $k$, then $1/(x-y_i)>1/(x-z_{i-1})$ for $i=2,\dots,{l-m+1}$ and $1/(x-y_1)>0$, thus LHS exceed RHS.