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I am trying to prove theorem 17 (h) of chapter 7 from Kaplan's Advanced Calculus book. Being $P_{n}$ a legendre polynomial and $P^{\prime}_{n}$ its derivative, I need to prove:

$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le 1 \hspace{1cm} (n \ge 1 , |x| \le 1)$$

I already proved that (see scholia):

$\tag{1}\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} = \frac{1-x^2}{n^2}P^{\prime 2}_{n-1}+P^{2}_{n-1}$

So I tried to do a "recursive prove":

$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le \frac{1-x^2}{(n-1)^2}P^{\prime 2}_{n-1}+P^{2}_{n-1} \le \frac{1-x^2}{(n-2)^2}P^{\prime 2}_{n-2}+P^{2}_{n-2} \le ... \le (1-x^2)P^{\prime 2}_{1}+P^{2}_{1}$$

And as:

$$(1-x^2)P^{\prime 2}_{1}+P^{2}_{1} = (1-x^2) +x^2 = 1$$

It seems to me that:

$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le 1 \hspace{1cm} (n \ge 1)$$

Should be valid for any value of $x$. Not only for $|x| \le 1$. I am afraid I am missing some part of the whole history here. Can anyone help me?

Scholia:

As some people doubted (1) I will show the prove for it.

As cited in Kaplan's book the Legendre polynomials have these two properties:

$\tag{2}P^{\prime}_{n}(x) = xP^{\prime}_{n-1}(x)+nP_{n-1}(x)$

And

$\tag{3}P_{n}(x) = \frac{(x^{2}-1)}{n}P^{\prime}_{n-1}(x)+xP_{n-1}(x)$

So we can substitute (2) and (3) on the LHS of (1) and get:

$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} = \frac{(1-x^2)}{n^2}(x^{2}P^{\prime 2}_{n-1} + 2xnP^{\prime}_{n-1}P_{n-1} + n^{2}P^{2}_{n-1}) + \frac{(x^{2}-1)^2}{n^2}P^{\prime 2}_{n-1}+ 2x\frac{(x^{2}-1)}{n}P^{\prime}_{n-1}P_{n-1}+x^{2}P^{2}_{n-1}$$

$$\frac{P^{\prime 2}_{n-1}}{n^2}(-x^{4}+x^{2}+x^{4}-2x^{2}+1)+P^{2}_{n-1}(1-x^{2}+x^{2}) = \frac{1-x^2}{n^2}P^{\prime 2}_{n-1}+P^{2}_{n-1}$$

And we get (1). The properties (2) and (3) I used in order to demonstrate (1) can be demonstrated using Rodrigues formula and Leibnitz product rule for $n$th order derivatives. I can provide the step-by-step if needed.

user2288551
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    Do you mean $n \ge 1$? Also your second equation has $n^2$ in the denominator on both sides of the equal sign. Finally, assuming that those are typos, all the inequalities could then be made equalities, so there is indeed something wrong. – NickD Mar 22 '18 at 14:29
  • The $n\le1$ was a typo, but the second equation is right, so the inequalities still hold. – user2288551 Mar 22 '18 at 14:33
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    OK - so the reason for $|x| < 1$ is to keep $1-x^2$ positive. – NickD Mar 22 '18 at 16:19
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    Indeed, based on (1), the step $$\frac{1-x^2}{n^2}P^{\prime 2}{n}+P^{2}{n} \le \frac{1-x^2}{(n-1)^2}P^{\prime 2}{n-1}+P^{2}{n-1}$$ requires that $$\frac{1-x^2}{n^2} \le \frac{1-x^2}{(n-1)^2}$$ which requires that $$1-x^2\ge0$$ – Did Mar 22 '18 at 17:29
  • That answers my question. Could you post it as an answer so that I can mark it as answered? – user2288551 Mar 22 '18 at 18:57

1 Answers1

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Given the identity $$\begin{eqnarray*} \frac{1-x^2}{n^2}P_n'^2+P_n^2 &=& \frac{1-x^2}{n^2}P_{n-1}'^2+P_{n-1}^2\\&=&\left(1-\frac{1}{n}\right)^2\left(\frac{1-x^2}{(n-1)^2}P_{n-1}'^2+P_{n-1}^2\right)+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\\\\&\stackrel{\text{Ind.Hyp.}}{\leq}&\left(1-\frac{1}{n}\right)^2+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\end{eqnarray*}$$ ($\text{Ind.Hyp}$ stands for Inductive Hypothesis) the claim is straightforward, since $P_{n-1}(x)^2$ attains its maximum value over $[-1,1]$ at the endpoints of such interval, and such maximum value is $1$.

The inequality can be refined for $x$-points belonging to a neighbourhood of the origin, since as shown here we have $$ P_n(x)^2\leq \frac{4}{\pi(2n+1)}\cdot\frac{1}{\sqrt{1-x^2}}.$$

Jack D'Aurizio
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