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Question
Show that
i)$$\left|\frac{d^n}{dx^n}(x^2-1)^n\right|\le\sqrt\frac2\pi\cdot\frac{2^nn!}{\sqrt n}\cdot\frac1{\sqrt[4]{1-x^2}},$$ or equivalently $$\left|P_n(x)\right|\le\sqrt\frac2{\pi n}\cdot\frac1{\sqrt[4]{1-x^2}}\text{ (where $P$ denotes Legendre Polynomial)},$$ when $x\in(-1,1)$ and $n\in\mathbb Z^+$.
ii) The constant term $\sqrt\frac2\pi$ in (i) is best.

I have successfully proven statement (ii) by using the asymptotic expansion of $P_n$ given by Wikipedia: $$P_n(\cos \theta)=\sqrt\frac2{\pi n\sin\theta}\cos((n+1/2)\theta-\pi/4)+O(n^{-1}).$$ Taking supremum limit, $$\limsup_{n\to\infty}|P_n(\cos\theta)|\sqrt n\le\sqrt\frac2\pi\sqrt[4]{1-\cos^2\theta}.$$ Substituting $\cos\theta=x$, we can see that (ii) is a weaker result of the limit result above. However, I have no idea how to deduce the explicit bound of $P_n(x)$.

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Remark that it suffices to prove:

\begin{equation*} \forall \theta \in [0, \pi], \lvert P_n(\cos \theta) \rvert \leq \sqrt{\dfrac{2}{n \pi} \dfrac{1}{\sin \theta}} \qquad (1) \end{equation*}

Because then, $\forall x \in [-1, 1], \sin(\arccos x) = \sqrt{1 - x^2}$ will give you $(i)$.

Now, to prove $(1)$, you can use the approach found in Marić, V., & Tomić, M. (1975). A new proof of Bernstein’s inequality for Legendre polynomials. Proceedings of the American Mathematical Society, 48(2), 511-512. for example.

But there are plenty of techniques using or not using complex analysis. Look for Bernstein's inequality on Legendre polynomials.

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