Does the expression
$$ d^m_xP_l(x) d^{m+1}_xP_{l+1}(x)- d^m_xP_{l+1}(x) d^{m+1}_xP_{l}(x)$$ always have a fixed sign ( so is it always positive or negative) on the interval(-1,1)?. $P_l$ is the l-th Legendre polynomial and $m \in \{0,...,l\}$. $d_x^m$ is the m-th derivative with respect to the variable x.
Notice, that this derivative is closely related to the definition of the associated Legendre function mathworld.
I was unable to test this even for simple expressions with wolframalpha, cause he did not get my entries. But apparently if we take $m=0$ and $l=0$, this expression is one, so in case that this is somehow true, then the expression should always be positive, I think.
Edit: Okay, now I checked quite a few cases and it seems as if this is correct.
If anything is unclear, please let me know.
A partial answer for now.
Let: $$B_{m,l}(x)\triangleq\frac{d^m}{dx^m}P_l\frac{d^{m+1}}{dx^{m+1}}P_{l+1}-\frac{d^m}{dx^m}P_{l+1}\frac{d^{m+1}}{dx^{m+1}}P_{l}.$$ We have: $$B_{0,l}(x) = P_l P_{l+1}'-P_{l+1}P_l'$$ so: $$(1-x^2) B_{0,l}(x) = P_l ((1-x^2)P_{l+1}') - P_{l+1}((1-x^2)P_{l}') $$ and since the Legendre differential equation gives: $$\frac{d}{dx}\left((1-x^2)P_l'\right)=l(l+1)P_l\tag{1}$$ we have: $$\frac{d}{dx}(1-x^2)B_{0,l}(x) = (l+1)(l+2)P_l P_{l+1} - l(l+1) P_{l}P_{l+1} =2(1+l)P_l P_{l+1}\tag{2} $$ that is an odd function. Moreover, since $P_l'(0)=-(l+1)P_{l+1}(0)$, we have: $$ B_{0,l}(0)= -(l+2) P_l P_{l+2}(0)+(l+1)P_{l+1}(0)^2\approx\frac{2}{\pi} \tag{3}$$ where $\frac{1}{\sqrt{1+t^2}}=\sum_{l\geq 0}P_l(0)t^n$, giving $P_{2l}(0)=\frac{(-1)^l}{4^l}\binom{2l}{l}$ and $P_{2l+1}(0)=0$.
The key step is now to state $(3)$ as a lower bound for $B_{0,l}(0)$ and to prove that: $$ \left|\int_{0}^{u}2(l+1)P_l P_{l+1}\,dx\right|\leq\frac{2}{\pi}\tag{4} $$ through the Cauchy-Schwarz inequality. The two inequalities together give $(1-x^2)B_{0,l}(x)>0$ for $|x|\leq 1$. At last, we have to find a generalization of this approach for other values of $m$. I am wondering if it is possible to use some kind of induction. Anyway, all the main ingredients should be already here.
EDIT: Ok, got it. By using the generalized Rodrigues formula, it is easy to see that our inequality is just a generalization of the well-known Turan's inequality for Legendre polynomials: $$ \forall x\in(-1,1),\qquad P_n(x)^2 > P_{n-1}(x)P_{n+1}(x).$$ By looking at the fourth proof of such inequality given by Szego, we can see that it just depends on the identity: $$ \sum_{n=0}^{+\infty}\frac{P_n(x)}{n!} z^n = e^{xz}\,J_0\left(z\sqrt{1-x^2}\right)\tag{5}$$ and from the fact that the Bessel function $J_0(z)$ has only real roots. The interesting fact is that we can follow the same steps of Polya and Szego once we differentiate $(5)$ with respect to $x$ $m$ times. That gives a proof of the non-vanishing of $B_{m,l}(x)$ over $(-1,1)$.