# How to show that any Legendre polynomial, over $[-1,1]$, is bounded by one in absolute value?

I'm trying to show that the Legendre polynomials are bounded by 1 using:

$$P_n(x) = \frac{1}{\pi} \int_{0}^{\pi} \big[x + i\sqrt{1+x^2}\cos(\theta)\big]^n d\theta$$

i.e.

$$|P_n(x)| \leq 1$$

using $$\left|\int f(z)\,dz\right | \leq \int|f(z)|\,dz$$

What I've got so far:

I've taken the absolute value of the polynomial to get:

$$|P_n(x)| \leq \frac{1}{\pi} \int_{0}^{\pi} [x^2 + (1+x^2)\cos^2(\theta)]^{n/2} d\theta$$

I've scaled $$\cos(\theta)$$ by $$\sqrt{\frac{(x^2-1)}{x}}$$ which gives me:

$$|P_n(x)| \leq \frac{1}{\pi} \sqrt{\frac{x}{x^2-1}}x^n \int_{a}^{b}\sin^n(\theta)d\theta$$

where $$a/b = \cos^{-1}(\pm \sqrt{\frac{x^2-1}{x}})$$

This gives me a recursion formula which isn't pretty once you've put in the limits.

Can anyone see a better way, or see where I've gone wrong?

• ap.smu.ca/~kbradler/sec3.pdf gives a proof of $|P_n(x)|\leq 1$ over $(-1,1)$ that is based on Cauchy's integral formula and a suitable deformation of the integration path. I bet it is exactly what you need. Sep 9, 2015 at 19:04
• Hi, this looks good, thanks very much. I think I was perhaps over thinking the problem. Sep 9, 2015 at 21:24

Once you've applied the absolute value of $$P_n(x)$$ on both sides of the integral equation and taken that inside the integral sign and got the inequality, consider the integrand:
$$|x^2 + (1-x^2)\cos^2(\theta)|^{n/2}$$
$$-1 \leq x \leq 1$$
If $$y = x^2 + (1-x^2)\cos^2(\theta) = x^2(1-\cos^2(\theta)) + \cos^2(\theta)$$, then the minimum value of $$y$$ is $$\cos^2(\theta) = 0$$ as it is a quadratic centered on $$x = 0$$. The maximum values of $$y$$ always occur when $$x = \pm 1$$ (as it is a quadratic) and this will always be with a value of $$y = 1$$ as the $$\cos^2(\theta)$$ terms cancel.
So $$0 \leq |y|\leq 1$$ therefore $$0 \leq |y|^{n/2} \leq 1$$ so the integral between $$0$$ and $$\pi$$ can be at most $$\pi$$ (assuming a value of $$|y|^{n/2} = 1$$) and so $$|P_{n}(x)| \leq 1$$ .