Krdytkyu

I need a proof to show that the inequality $m < n$ leads to a contradiction and $P_n$ has $n$ distinct real roots, all of which lie in the open interval $(-1, 1)$.

By Rodrigues formula for Legendre polynomials,
$$displaystyle P_n(x) = frac{1}{2^nn!}frac{d^n}{dx^n} (x^2 - 1)^{n}tag{*1}$$
$P_n(x)$ is the $n^{th}$ derivative of a polynomial with roots at $-1$ and $1$.
Repeat applying Gauss-Lucas theorem $n$ times, we know all roots of $P_n(x)$ lie on the closed line segment
$[-1,1]$ in the complex plane. By direct expansion of $(*1)$, one can check that
$P_n(pm 1) ne 0$, so the roots of $P_n(x)$ are all real and belongs to $(-1,1)$.

Since Legendre polynomials are solutions of the Legendre's differential equation:
$$frac{d}{dx}left[(1-x^2)frac{d}{dx}P_n(x)right] + n(n+1)P_n(x) = 0$$
which is a $2^{nd}$ order ODE, $P_n(x)$ cannot have any double roots. To see this,
let's say $P_n(x)$ has a double root at $alpha in (-1,1)$.
$P_n(x)$ will then be a solution of following initial value problem:

$$frac{d}{dx}left[(1-x^2)frac{d}{dx}y(x)right] + n(n+1)y(x) = 0,quad
begin{cases}y(alpha) = 0,\y'(alpha) = 0end{cases}$$
Apply Picard-Lindelöf theorem to this $2^{nd}$ order ODE, the "uniqueness" part of the theorem tell us $P_n(x)$ vanishes over some neighbor of $alpha$. Since $P_n(x)$ is not identically zero, this is impossible. As a consequence, all roots of $P_n(x)$ are simple.

From the wording of your question I guess you are looking for something like this:

The Legendre polynomial $P_n$ with $n>0$ has $n$ simple roots in $(-1, 1).$

Proof by contradiction: Assume $P_n$ has $m$ with $0 le m lt n;$
pairwise different zeroes $x_1, x_2, dots x_m$ of odd multiplicity in $(-1,1),;$ i.e. $P_n$ changes sign in $x_i$, and consider the polynomial $Z_n(x) = (x-x_1)cdot (x-x_2) dots (x-x_m).;$ Because $P_n$ is orthogonal to $P_0$ the intgeral $int_{-1}^{1} P_n(x) dx$ vanishes and $P_n$ has at least one zero of odd multiplicity in $(-1,1),,$ i.e. $m>0.;$

The polynomial $Z_n(x) P_n(x)$ does not change sign in $(-1,1)$ and therefore $I_n = int_{-1}^{1} Z_n(x) P_n(x) dx ne 0.;$ On the other hand $Z_n$ has degree $m$ and can be written as a linear combination $Z_n = sum_{k=0}^m c_k P_k(x)$. Thus we get the contradiction by means of the orthogonality of the $P_k$:
begin{align}
0 ne I_n &= int_{-1}^{1} Z_n(x) P_n(x) dx\
&= int_{-1}^{1} P_n(x) sum_{k=0}^m c_k P_k(x) dx\
&= sum_{k=0}^m c_k int_{-1}^{1} P_n(x) P_k(x) dx\
&= 0
end{align}

This shows that $P_n$ has $n$ pairwise different zeroes of odd multiplicity, and because $P_n$ has at most $n$ zeroes, the zeroes $x_1,dots, x_n$ are simple.

Note that this kind of proof can be applied to general orthogonal polynomials over intervals $(a,b)$ with weighting function $w(x) ge 0$.

Let Legendre polynomial be $$ f_n(x)=frac{1}{2^n n!}frac{d^{n}}{d x^n}(x^2-1)^n .$$

Note that $ (x^2-1)^n $ has roots $ -1, 1 $ each with multiplicity $ n $, and $$ (x^2-1)Big|frac{d^m}{d x^{m}}(x^2-1)^n, , text{when} m<n .$$

Thus, $ -1, 1 $ are always roots of $ frac{d^{m}}{d x^m}(x^2-1)^n $. And by Rolle's theorem, we know that there are $ n-1 $ distinct zeros $ x_1, x_2, cdots, x_{n-1} $ in the interval $ (-1, 1) $ for the polynomial $ frac{d^{n-1}}{d x^{n-1}}(x^2-1)^n $. Since $ -1, 1 $ are also zeros of $ frac{d ^{n-1}}{d x^{n-1}}(x^2-1)^n $. Use Rolle's theorem once again, we get $ n $ distinct zeros in $ (-1, 1) $ and since $ deg (frac{d^{n}}{d x^{n}}(x^2-1)^n)=n $, we have found all the distinct roots of $ f_n(x) $ in $ (-1, 1) $.