How to prove that Legendre polynomials satisfy $P_n(x)\leq 1$ How to prove that Legendre polynomials satisfy:
$$
|P_n(x)|\leq1
$$
for $-1\leq x\leq 1$.
Thanks 
My idea: 
Consider the equation $S_n(x)\equiv P_n(x)-1=0$. Then (I don't know if this result is true in the form I preset there) if $S_n(1)S_n(-1)> 0$ we do not have any root of $S_n$ inside the interval $[-1,1]$. If $S_n(1)S_n(-1)=0$  we have a root in $1$ or in $-1$ or two roots. In particular for even $n$ if i prove that the sum of the coefficients of the Legendre polynomials is positive I've done maybe. For odd $n$ maybe I should try to compute $S_n(0)*S_n(1)$ and then $S_n(0)S_n(-1)$....
 A: There's a proof I've seen in the web you might like (I did): let's define an auxiliary function,
$$h_n(x)=n(n+1)\,P_n(x)^2+(1-x^2)\,P'_n(x)^2.$$ Obviously, $h_n(\pm1)=n(n+1)$.
Now, we calculate
$$h'_n(x)=2n(n+1)\,P'_n(x)P_n(x)-2\,x\,P'_n(x)^2+2\,(1-x^2)\,P''_n(x)\,P'_n(x).$$
But Legendre polynomials satisfy the differential equation
$$(1-x^2)\,P''_n(x)=2\,x\,P'_n(x)-n(n+1)\,P_n(x),$$ so
$$h'_n(x)=2\,n(n+1)\,P'_n(x)\,P_n(x)-2\,x\,P'_n(x)^2+2\,(2\,x\,P'_n(x)-n(n+1)\,P_n(x))\,P'_n(x).$$ This simplifies to
$$h'_n(x)=2\,x\,P'_n(x)^2,$$ so $h'_n(x)\ge0$ for $x\in[0,1]$, and $h'_n(x)\le0$ for $x\in[-1,0]$.
Thus $$h_n(x)\le h_n(1)=n(n+1)$$ for $x\in[0,1]$, and $$h_n(x)\le h_n(-1)=n(n+1)$$ for $x\in[-1,0]$. Combining both, we get
$$n(n+1)\,P_n(x)^2+(1-x^2)\,P'_n(x)^2\le n(n+1)$$ for $x\in[-1,1]$.  Since $(1-x^2)\,P'_n(x)^2\ge0$, this implies $|P_n(x)|\le1$ for $x\in[-1,1]$.
A: Here's a proof that starts with Laplace's integral formula for the Legendre polynomials of
$$P_n (x) = \frac{1}{\pi} \int_0^\pi \left (x + \sqrt{x^2 - 1} \cos \theta \right )^n \, d\theta, \quad n = 0,1,2,\ldots$$
As the integrand will not be real when $|x| < 1$ we rewrite it as
$$P_n(x) = \frac{1}{\pi} \left (x + i \sqrt{1 - x^2} \cos \theta \right )^n \, d\theta.$$
Now
\begin{align*}
|P_n(x)| &= \left |\frac{1}{\pi} \int_0^\pi \left (x + i \sqrt{1 - x^2} \cos \theta \right )^n \, d\theta \right |\\
&\leqslant \frac{1}{\pi} \int_0^\pi \left |x + i \sqrt{1 - x^2} \cos \theta \right |^n \, d\theta\\
&= \frac{1}{\pi} \int_0^\pi (\cos^2 \theta + x^2 \sin^2 \theta)^{n/2} \, d\theta\\
&\leqslant \frac{1}{\pi} \int_0^\pi (\cos^2 \theta + \sin^2 \theta)^{n/2} \, d\theta, \quad \text{since}\,\, |x| \leqslant 1\\
&= \frac{1}{\pi} \int_0^\pi d\theta = 1.
\end{align*}
Thus $|P_n (x)| \leqslant 1$, as required to show.
