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I wish to prove that the magnitude of extreme values of $U_n(x)$, the Chebyshev polynomial of the second kind, is monotonically increasing on $[-1,1]$. By symmetry it suffices to prove it over $[0,1]$. The following is the graph of $U_8(x)$ on $[0,1]$. enter image description here

Consider three consecutive zeroes of $U_n(x)$ as $x_1=\cos(\frac{(k+1)\pi}{n+1})< x_2=\cos(\frac{k\pi}{n+1})< x_3=\cos(\frac{(k-1)\pi}{n+1})$ and let $y\in(x_1,x_2)$ and $z\in(x_2,x_3)$ be two consecutive extreme points of $U_n(x)$. We want to show that $\lvert U_n(y)\rvert < \lvert U_n(z)\rvert $.

We know that $U'_n(x)=\frac{(n+1)T_{n+1}(x)-xU_n(x)}{x^2-1}$ where $T_{n+1}$ is the Chebyshev polynomial of the first kind, so $\lvert U_n(y)\rvert = \lvert \frac{(n+1)T_{n+1}(y)}{y}\rvert$, hence we need to show $\frac{\lvert T_{n+1}(y)\rvert}{y} < \frac{\lvert T_{n+1}(z)\rvert}{z}$. Since $\frac{T_{n+1}(x_2)}{x_2}=\frac{(-1)^k}{x_2}$, $\lvert T_n(x)\rvert \leq 1$ and $\frac{T_{n+1}(x)}{x}$ is a polynomial, the first idea came to my mind was MVT. Thus there exist $y<c<x_2<d<z $ such that $$ \frac{1}{x_2} - \frac{\lvert T_{n+1}(z)\rvert }{z} = \lvert \frac{1}{x_2} - \frac{\lvert T_{n+1}(z)\rvert }{z} \rvert\leq \lvert f'(d)\rvert (z-x_2)$$ and $$ \lvert \frac{1}{x_2} - \frac{\lvert T_{n+1}(y)\rvert }{y} \rvert\leq \lvert f'(c)\rvert (x_2-y) $$ where $f(x)=\frac{T_{n+1}(x)}{x}$.

Any hint to proceed from this point is really appreciated.

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    $\begingroup$ I believe this can be neatly proved through Turan's inequality. $\endgroup$ – Jack D'Aurizio Feb 25 '18 at 16:51
  • $\begingroup$ @JackD'Aurizio, I tried Turan's inequality for both first and second kinds, but no progress. Could you please help a little more? Thanks! $\endgroup$ – Aleph-null Feb 26 '18 at 23:48
  • $\begingroup$ I found an elementary proof through simple substitutions and an implicit relation, hope it helps. $\endgroup$ – Jack D'Aurizio Feb 27 '18 at 14:28
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Since we are just interested in the magnitude of the extreme values, we may just focus on the values of $f_n(\theta)=\frac{\sin(n\theta)}{\sin\theta}$ at its stationary points in $\left[0,\frac{\pi}{2}\right]$. We obviously have $f_n(0)=n$ and the roots of $f_n'$ occur at the solutions of $n\tan(\theta)=\tan(n\theta)$. We have $$ f_n^2(\theta) = \frac{1-\cos^2(n\theta)}{1-\cos^2\theta}=\frac{1-\frac{1}{1+\tan^2(n\theta)}}{1-\frac{1}{1+\tan^2(\theta)}}$$ hence the claim follows from the fact that $$ g_n(z) = \frac{1-\frac{1}{1+n^2 z^2}}{1-\frac{1}{1+z^2}}=\frac{1+z^2}{\frac{1}{n^2}+z^2} $$ is a decreasing function on $\mathbb{R}^+$ for any $n\geq 1$.

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