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

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


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

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?

  • 1
    $\begingroup$ 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. $\endgroup$ Sep 9, 2015 at 19:04
  • $\begingroup$ Hi, this looks good, thanks very much. I think I was perhaps over thinking the problem. $\endgroup$ Sep 9, 2015 at 21:24

1 Answer 1


Ok, thanks to @Jack D'Aurizio for the source, here's my answer.

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 $|P_{n}(x)| \leq 1$ .


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