Continuous Markov derivative bound on an interval

By [F1], $p$ and each polynomial $r$ below are finite real polynomials; [F2] and [F9] make $p,p^{\mathrm{\prime }}$ continuous and attain their extrema on $[-1,1]$. For any such $r$, $S_r(\theta ):=\sin \theta r(\cos \theta )$ is continuous by [F6], [F9], and [F10], and is $2\pi$-periodic by [F15]; hence [F9] gives a maximum of $\mathrm{\mid }{S}_r\mathrm{\mid }$ on $[0,2\pi ]$. If $d=0$ or $p=0$, then $p^{\mathrm{\prime }}=0$ and $h=0$; assume $d\ge 1$. For $j=1,\dots ,d$, put ${\theta }_j=(2j-1)\pi \mathrm{/}(2d)$ and $x_j=\cos {\theta }_j$. By [F4], $T_d(x_j)=\cos (d{\theta }_j)=0$. Monotonicity and reflection in [F5] show $x_1>\cdots >x_d$ in $(-1,1)$ and $x_d=-x_1$. Also [F5] gives $\sin {\theta }_j>0$. Since $d{\theta }_j=(j-1)\pi +\pi \mathrm{/}2$, repeated use of the shift-by-$\pi$ identity in [F5], with finite induction [F13], gives $\sin (d{\theta }_j)=(-1{)}^{j-1}$. The chain rule [F6] applied to [F4] now gives $T_d^{\mathrm{\prime }}(x_j)=d\sin (d{\theta }_j)\mathrm{/}\sin {\theta }_j=d(-1{)}^{j-1}\mathrm{/}\sin {\theta }_j\mathrm{\ne }0$.

Fix any polynomial $r$ supported in degrees at most $d-1$. For $x\in [-x_1,x_1]$, choose $\theta \in [0,\pi ]$ with $x=\cos \theta$. Monotonicity and reflection in [F5] imply $\pi \mathrm{/}(2d)\le \theta \le \pi -\pi \mathrm{/}(2d)$, so $\sin \theta \ge \sin (\pi \mathrm{/}(2d))$. For every positive integer $n$, $\mathrm{\mid }\sin (n\varphi )\mathrm{\mid }\le n\mathrm{\mid }\sin \varphi \mathrm{\mid }$: equality holds for $n=1$; the sine addition formula [F8] and $\mathrm{\mid }\sin \mathrm{\mid },\mathrm{\mid }\cos \mathrm{\mid }\le 1$ from [F5] give $\mathrm{\mid }\sin ((n+1)\varphi )\mathrm{\mid }\le \mathrm{\mid }\sin (n\varphi )\mathrm{\mid }+\mathrm{\mid }\sin \varphi \mathrm{\mid }$, so induction [F13] proves the claim. At $\varphi =\pi \mathrm{/}(2d)$ this yields $1=\mathrm{\mid }\sin (d\varphi )\mathrm{\mid }\le d\sin \varphi$, hence $\sin (\pi \mathrm{/}(2d))\ge 1\mathrm{/}d$. Therefore $\mathrm{\mid }r(x)\mathrm{\mid }\le d\mathrm{\mid }\sin \theta r(\cos \theta )\mathrm{\mid }\le d{\max }_{\eta \in [0,2\pi ]}\mathrm{\mid }{S}_r(\eta )\mathrm{\mid }$ throughout this central interval.

By the factor theorem [F7] and $T_d(x_j)=0$ from step 1.1, each quotient $T_d(X)\mathrm{/}(X-x_j)$ is a polynomial of degree $d-1$, so the Lagrange polynomials $L_j(X):=T_d(X)\mathrm{/}((X-x_j)T_d^{\mathrm{\prime }}(x_j))$ have degree at most $d-1$. At $x_i$ with $i\mathrm{\ne }j$, $L_j(x_i)=0$; at $x_j$, factorization $T_d(X)=(X-x_j)s_j(X)$ and product rule [F11] give $s_j(x_j)=T_d^{\mathrm{\prime }}(x_j)$, so $L_j(x_j)=1$. Thus every $r$ as in step 2.1 satisfies $r(X)={\sum }_{j=1}^{d}r(x_j)L_j(X)$, since the difference has degree at most $d-1$ and $d$ distinct roots, hence vanishes by [F7]. Repeated factorization by [F7] gives $T_d(X)=c{\prod }_j(X-x_j)$ with $c\mathrm{\ne }0$ (degree $d$ by [F3]); repeated product rule gives $T_d^{\mathrm{\prime }}(X)=c{\sum }_j{\prod }_{k\mathrm{\ne }j}(X-x_k)$. Therefore the polynomial ${\sum }_j{T}_d(X)\mathrm{/}(X-x_j)-T_d^{\mathrm{\prime }}(X)$ vanishes for every $X$ away from the nodes, an infinite set, so it is zero by the root bound [F7]. Substituting $1\mathrm{/}{T}_d^{\mathrm{\prime }}(x_j)=(-1{)}^{j-1}\sin {\theta }_j\mathrm{/}d$ from step 1.1 gives $r(X)=\frac{1}{d}{\sum }_{j=1}^d(-1{)}^{j-1}\sin {\theta }_{j}r(x_j)\frac{T_d(X)}{X-x_j}$.

Let $W:={\max }_{\eta \in [0,2\pi ]}\mathrm{\mid }{S}_r(\eta )\mathrm{\mid }$. At each node, $\sin {\theta }_j\mathrm{\mid }r(x_j)\mathrm{\mid }=\mathrm{\mid }{S}_r({\theta }_j)\mathrm{\mid }\le W$. For $x\in (x_1,1]$, all $x-x_j>0$, and the interpolation formula of step 3.1 gives $\mathrm{\mid }r(x)\mathrm{\mid }\le (W\mathrm{/}d)\mathrm{\mid }{T}_d(x)\mathrm{\mid }{\sum }_{j}1\mathrm{/}(x-x_j)=(W\mathrm{/}d)\mathrm{\mid }{T}_d^{\mathrm{\prime }}(x)\mathrm{\mid }$. For $x\in [-1,-x_1)$ all denominators are negative, so the summands $T_d(x)\mathrm{/}(x-x_j)$ have a common sign and their absolute values sum to $\mathrm{\mid }{T}_d^{\mathrm{\prime }}(x)\mathrm{\mid }$; the same interpolation and node bounds give $\mathrm{\mid }r(x)\mathrm{\mid }\le (W\mathrm{/}d)\mathrm{\mid }{T}_d^{\mathrm{\prime }}(x)\mathrm{\mid }$. For interior $x=\cos \theta$, $0<\theta <\pi$, differentiating [F4] by [F6] gives $\mathrm{\mid }{T}_d^{\mathrm{\prime }}(x)\mathrm{\mid }=d\mathrm{\mid }\sin (d\theta )\mathrm{\mid }\mathrm{/}\sin \theta \le d^2$ by the sine estimate of step 2.1. Continuity extends this bound to the outer endpoints; the central boundary nodes are covered by step 2.1. Hence $\mathrm{\mid }r(x)\mathrm{\mid }\le dW$ on the outer intervals. Combining this with step 2.1 proves $\mathrm{\parallel }r{\mathrm{\parallel }}_{[-1,1]}\le d{\max }_{\eta \in [0,2\pi ]}\mathrm{\mid }{S}_r(\eta )\mathrm{\mid }$ on all of $[-1,1]$.

For continuous $g$ on $[-1,1]$ write $\mathrm{\parallel }g\mathrm{\parallel }:={\max }_{[-1,1]}\mathrm{\mid }g\mathrm{\mid }$; for continuous $2\pi$-periodic $S$ write $\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}:={\max }_{[0,2\pi ]}\mathrm{\mid }S\mathrm{\mid }$. Let $S$ be a real trigonometric polynomial of degree at most $d$, meaning a finite real linear combination of $1,\cos (k\theta ),\sin (k\theta )$ for $1\le k\le d$. Its sine and cosine terms and their derivatives are continuous by [F6] and [F10], and are $2\pi$-periodic by [F15], so all maxima just defined, including that of $\mathrm{\mid }{S}^{\mathrm{\prime }}\mathrm{\mid }$, exist by [F9]. For fixed $\alpha$, $F_{\alpha }(\theta ):=(S(\alpha +\theta )-S(\alpha -\theta ))\mathrm{/}2$ is a linear combination of $\sin (k\theta )$: the constant term cancels, each cosine term contributes $-\sin (k\alpha )\sin (k\theta )$, and each sine term contributes $\cos (k\alpha )\sin (k\theta )$ by the addition formulas [F8] and parity [F14]. By [F4], $\sin (k\theta )=\sin \theta U_{k-1}(\cos \theta )$, so $F_{\alpha }(\theta )=\sin \theta r(\cos \theta )$ for a polynomial $r$ supported in degrees at most $d-1$ by [F3]. By the chain rule [F6], differentiating the centered difference at $0$ gives $F_{\alpha }^{\mathrm{\prime }}(0)=(S^{\mathrm{\prime }}(\alpha )+S^{\mathrm{\prime }}(\alpha ))\mathrm{/}2=S^{\mathrm{\prime }}(\alpha )$. Differentiating the factorization using polynomial differentiability [F2], [F6], and [F11] gives $F_{\alpha }^{\mathrm{\prime }}(0)=r(1)$, because $\sin 0=0$ and $\cos 0=1$ by [F6]. Thus $S^{\mathrm{\prime }}(\alpha )=r(1)$. By steps 2.1 and 4.1, $\mathrm{\mid }r(1)\mathrm{\mid }\le \mathrm{\parallel }r\mathrm{\parallel }\le d\mathrm{\parallel }{F}_{\alpha }{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}$. Also $\mathrm{\parallel }{F}_{\alpha }{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}$ by [F12], since $F_{\alpha }$ is the half-difference of two values of $S$. As this holds for every $\alpha$, $\mathrm{\parallel }{S}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}\le d\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}$.

Put $S(\theta ):=p(\cos \theta )$. Applying [F8] at $(u,v)$ and $(u,-v)$ and using [F14] gives $\cos u\cos v=(\cos (u+v)+\cos (u-v))\mathrm{/}2$. Induction now gives that each power ${\cos }^k\theta$ is a linear combination of $\cos (j\theta )$ for $0\le j\le k$: multiplying a term $\cos (j\theta )$ by $\cos \theta$ gives frequencies $j+1$ and $j-1$ when $j\ge 1$, while the constant term gives $\cos \theta$. Thus $S$ is a real trigonometric polynomial of degree at most $d$. Cosine decreases from $1$ to $-1$ on $[0,\pi ]$ by [F5] and its endpoint values [F6], [F5], so its range on $[0,2\pi ]$ is $[-1,1]$; with [F15] this gives $\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}=\mathrm{\parallel }p\mathrm{\parallel }$. The chain rule [F6] and polynomial differentiability [F2] give $S^{\mathrm{\prime }}(\theta )=-\sin \theta p^{\mathrm{\prime }}(\cos \theta )$, so step 5.1 gives ${\max }_{\theta \in [0,2\pi ]}\mathrm{\mid }\sin \theta p^{\mathrm{\prime }}(\cos \theta )\mathrm{\mid }\le d\mathrm{\parallel }p\mathrm{\parallel }$. Applying the inequality proved in steps 2.1 and 4.1 to $r=p^{\mathrm{\prime }}$, whose degree is at most $d-1$ by [F2], yields $\mathrm{\parallel }{p}^{\mathrm{\prime }}\mathrm{\parallel }\le d{\max }_{\theta \in [0,2\pi ]}\mathrm{\mid }\sin \theta p^{\mathrm{\prime }}(\cos \theta )\mathrm{\mid }\le d^2\mathrm{\parallel }p\mathrm{\parallel }$. Finally, let $M=\max p$ and $m=\min p$, attained by [F9], and apply this derivative bound to $p-(M+m)\mathrm{/}2$. Its range height is $h$ and its supremum norm is $h\mathrm{/}2$, while its derivative is $p^{\mathrm{\prime }}$. Thus $\max \mathrm{\mid }{p}^{\mathrm{\prime }}\mathrm{\mid }\le d^{2}h\mathrm{/}2$, as claimed.