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\centerline{\Large\bf   An extremal property of Chebyshev Polynomials}


\thispagestyle{empty}

\vskip .5in

\centerline{\sc Carlos Gustavo T. de A. Moreira}

\vskip .8in

\begin{abstract}
Let $B_n=\{p(x)\in\re[x] : |p(x)|\le 1,\,\, \forall\, x\in [-1,1]\}$, and let $f\colon\re_+^{n+1}\to\re$ be a symmetric convex function, increasing in each variable. For $p(x)=\sum\limits_{j=0}^n a_j x^j$ we define $\ov f(p)=f(|a_0|,|a_1|,\dots,|a_n|)$. We prove that $\ov f(p)\le \ov f(T_n)$ for every $p\in B_n$, where $T_n$ is the $n-th$ Chebyshev Polynomial, and the equality holds if and only if $p=\pm T_n$. In particular, we have $\sum\limits_{j=0}^n |a_j| \le \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}2$, when $\sum\limits_{j=0}^n a_j x^j\in B_n$, and the equality holds if and only if $T_n(x)= \sum\limits_{j=0}^n a_j x^j$.

\end{abstract}


\section{Introduction}
It is well known that $\cos(n\theta)=T_n(\cos\theta)$, for a real polynomial $T_n$, the $n-th$ Chebyshev polynomial. Chebyshev polynomials have many extremal properties. For instance, if $B_n=\{p(x)\in\re[x] : |p(x)|\le 1,\,\, \forall\, x\in [-1,1]\} \subset \{p(x)\in\re[x] : |p(\cos \left( \frac{j\pi}n \right)| \le 1,\,\,  0\le j\le n \}=C_n$ then $T_n\in B_n$ and, if $p(x)=\sum\limits_{j=0}^n a_jx^j\in
C_n$, then $|a_n|\le 2^{n-1} $ and the equality holds if and only if $p=\pm T_n$ (see [R], chapter 2). Here we prove the following

\vskip .2in

\noindent
{\bf Proposition 1:} {\it Let  $p(x)=\sum\limits_{j=0}^n a_j x^j\in C_n$ and let $T_n(x)=\sum\limits_{j=0}^n t_j^{(n)} x^j$. If $n$ is even then $|a_0|\le |t_0^{(n)}|=1$ and, for $0\le i\le \frac n2 -1$, $|a_{2i+1}|+|a_{2i+2}|\le |t_{2i+2}^{(n)}|$. Moreover, when $p(x)\in B_n$, $|a_{2i+1}|+|a_{2i+2}|=|t_{2i+2}^{(n)}|$ for some $i$ with $0\le i\le \frac n2 -1$ if and only if $p=\pm T_n$. If $n$ is odd then, for $0\le i\le \frac{n-1}2$, $|a_{2i}|+|a_{2i+1}|\le |t_{2i+1}^{(n)}|$, and when $p(x)\in B_n$, the equality holds for some $i$ with $0\le i\le \frac{n-1}2$ if and only if $p=\pm T_n$\/}.

We use this proposition to prove

\vskip .2in

\noindent
{\bf Theorem 1:} {\it Let $f\colon\re_+^{n+1}\to\re$ be a symmetric convex function, increasing in each variable. For $p(x)=\sum\limits_{j=0}^n a_j x^j$ we define
$\ov f(p)=f(|a_0|,|a_1|,\dots,|a_n|)$. We have $\ov f(p)\le \ov f(T_n)$ for every $p\in C_n \supset B_n$. When $p\in B_n$, the equality holds if and only if $p=\pm T_n$\/}.

\vskip .2in

If $p(x)=\sum\limits_{j=0}^n a_j x^j$, we define $\ell(p)=\sum\limits_{j=0}^n|a_j|$. We conclude from theorem 1 above that if $p\in C_n$ then $\ell(p) \le \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}2=\ell(T_n)$, and, when $p \in B_n$, the equality holds
if and only if $p=\pm T_n$ (Corollary 1). This improves a result from [RZ], where this fact is proved for even or odd polynomials
$p(x)\in B_n$. The equality 
$\ell(T_n) = \frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}2$ follows from $T_{n+2}(x)=2xT_{n+1}(x)-T_n(x)$, which implies $\ell(T_{n+2})=2\ell(T_{n+1})+\ell(T_n)$ (since the non-zero coefficients of $T_n(x)$, which is even for $n$ even and odd for $n$ odd, alternate signs); notice that
$\ell(T_0)=\ell(T_1)=1$. 

\vskip .2in

I would like to thank the participants of the list of discussion of problems of the Brazilian Mathematical Olympiad (see {\tt www.mat.puc-rio.br/$\sim$nicolau/olimp/obm-l.html} and
\linebreak
{\tt www.obm.org.br/lista.htm}), particularly Eduardo Casagrande Stabel, who posed in the list the problem which motivated this note and F\'abio Dias Moreira and Alex Corr\^ea Abreu, who gave valuable suggestions in the preparation of this manuscript. I also would like to thank Professors Ken McLaughlin, Dimitar Dimitrov, Vilmos Totik and Tamás Erdélyi, for the helpful conversations and suggestions about this work.



\section{Proofs} Let us first prove Proposition 1. Let $p\in C_n$. Write $p(x)=g(x)+h(x)$, where $g(x)=\frac{p(x)+p(-x)}2$ is even and $h(x)=\frac{p(x)-p(-x)}2$ is odd. Clearly, $g$ and $h$ also belong to $C_n$. Since $|g(\cos \left( \frac{j\pi}n \right))+h(\cos \left( \frac{j\pi}n \right))|\le 1$ and
$|g(-\cos \left( \frac{j\pi}n \right))+h(-\cos \left( \frac{j\pi}n \right))|=|g(\cos \left( \frac{j\pi}n \right))-h(\cos \left( \frac{j\pi}n \right))|\le 1$, we have $|g(\cos \left( \frac{j\pi}n \right))|+|h(\cos \left( \frac{j\pi}n \right))|\le1$, $0\le j\le n$. We will compute $g(x)$ and $h(x)$ from their values in the points $w_j=\cos \left( \frac{j\pi}n \right)$, $0\le j\le n$. More precisely, we will write $g(x)=\tilde g(x^2)$, $h(x)=x\tilde h(x^2)$, and we will compute $\tilde g(x)$ from its values in $w_j^2$, $0\le j\le \frac n2$ and $\tilde h(x)$ from its values in $w_j^2$, $0\le j<\frac n2$, by Lagrange interpolation (notice that $\text{deg}(\tilde g)\le \frac n2$ and $\text{deg}(\tilde h)<\frac
n2$). We will consider two cases:

\vskip .2in

\noindent
i) $n$ even, say $n=2k$: In this cases, we have
$$
\tilde g(x)=\sum_{j=0}^k  \left(g(w_j) \cdot \prod_{\substack{r=0\\ r\ne j}}^k \left( \frac{x-w_r^2}{w_j^2-w_r^2} \right) \right)\quad \text{ and }\quad 
\tilde h(x)=\sum_{j=0}^{k-1} \left( \frac{h(w_j)}{w_j} \cdot \prod_{\substack{r=0\\ r\ne j}}^{k-1} \left( \frac{x-w_r^2}{w_j^2-w_r^2} \right) \right),
$$
so
$$
 g(x)=\sum_{j=0}^k  \left(g(w_j) \cdot \prod_{\substack{r=0\\ r\ne j}}^k \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right)\quad \text{ and }\quad 
h(x)=x\sum_{j=0}^{k-1} \left( \frac{h(w_j)}{w_j} \cdot \prod_{\substack{r=0\\ r\ne j}}^{k-1} \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right).
$$

\vskip .2in

\noindent
ii) $n$ odd, say $n=2k+1$: In this case, we have
$$
\tilde g(x)=\sum_{j=0}^k  \left(g(w_j) \cdot \prod_{\substack{r=0\\ r\ne j}}^k \left( \frac{x-w_r^2}{w_j^2-w_r^2} \right) \right)\quad \text{ and }\quad 
\tilde h(x)=\sum_{j=0}^{k} \left( \frac{h(w_j)}{w_j} \cdot \prod_{\substack{r=0\\ r\ne j}}^{k} \left( \frac{x-w_r^2}{w_j^2-w_r^2} \right) \right),
$$
so
$$
 g(x)=\sum_{j=0}^k  \left(g(w_j) \cdot \prod_{\substack{r=0\\ r\ne j}}^k \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right)\quad \text{ and }\quad 
h(x)=x\sum_{j=0}^{k} \left( \frac{h(w_j)}{w_j} \cdot \prod_{\substack{r=0\\ r\ne j}}^{k} \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right).
$$

Notice that, since $w_i>w_{i+1}$ for $0\le i<i+1\le\frac n2$, the signs of $\prod\limits_{\substack{r=0\\ r\ne j}}^{k}(w_j^2-w_r^2)$ and of $\prod\limits_{\substack{r=0\\ r\ne j}}^{k-1}(w_j^2-w_r^2)$ are $(-1)^j$. On the other hand, $\prod\limits_{\substack{r=0\\ r\ne j}}^{\ell}(x^2-w_r^2)$ is an even polynomial of degree $2\ell$ with alternating coefficients (the sign of the coefficient of $x^{2\ell-2u}$ is $(-1)^u$, for $0\le u\le \ell$; all of these coefficients are non-zero, with the possible exception of the constant coefficient), where $\ell\in\{k-1,k\}$. For $p(x)=\sum\limits_{j=0}^n a_j x^j$, and $0\le i<n$, we define $\ell_i(p)=|a_i|+|a_{i+1}|$. 

\vskip .2in

In the case i), if $p(x)=T_n(x)$, we have $h(x)\equiv 0$ and $g(w_j)=\cos (j\pi)=(-1)^j$, for $0\le j\le k$, so 
$$
\ell_i(T_n)=\sum\limits_{j=0}^k\ell_i\left( 
\prod_{\substack{r=0\\ r\ne j}}^k \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right),\quad \forall\, i<n
$$
by the  above remarks (since in this case $g(w_j) \cdot \prod_{\substack{r=0\\ r\ne j}}^k ( \frac{1}{w_j^2-w_r^2} )$ is positive for every $j$, with $0\le j\le k$). On the other hand, for an arbitrary polynomial $p\in C_n$, $|g(w_j)|+|h(w_j)|\le 1$,
and, for $0\le j\le k-1$,
$$
\prod_{\substack{r=0\\ r\ne j}}^k \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) = \frac{x^2}{w_j^2} \cdot 
\prod_{\substack{r=0\\ r\ne j}}^{k-1} \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) = \frac x{w_j}
\left( \frac{x}{w_j} \cdot 
\prod_{\substack{r=0\\ r\ne j}}^{k-1}  \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right),
$$
so
$$
\ell_{2i+1} \left(  \prod_{\substack{r=0\\ r\ne j}}^{k}
 \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right)\right) = \frac 1{w_j} \ell_{2i+1} \left( \frac x{w_j} \cdot
\prod_{\substack{r=0\\ r\ne j}}^{k-1}  \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right) \right),\quad \forall\, i<k;
$$
in particular, 
$$
\ell_{2i+1} \left(  \prod_{\substack{r=0\\ r\ne j}}^{k}
 \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right)\right) \ge 
\ell_{2i+1} \left( \frac x{w_j} \cdot
\prod_{\substack{r=0\\ r\ne j}}^{k-1}  \left( \frac{x^2-w_r^2}{w_j^2-w_r^2}
\right) \right),
$$
and the equality holds if and only if $j=0$. This implies that, if $p(x)=\sum\limits_{j=0}^n a_j x^j$ and $T_n(x)=\sum\limits_{j=0}^n t_j^{(n)} x^j$, then $|a_0|\le |t_0^{(n)}|=1$ and, for $0\le i<\frac n2$, $|a_{2i+1}|+|a_{2i+2}|\le |t_{2i+2}^{(n)}|$.
If $|a_{2i+1}|+|a_{2i+2}|=|t_{2i+2}^{(n)}|$ for some $i$ with $0\le i<\frac n2$, we should have $|g(w_j)|=1$ and $h(w_j)=0$ for $1\le j\le k$, and $g(w_{j+1})=-g(w_j)$ for $1\le j<j+1\le k$. We should also have $|g(w_0)|+|h(w_0)|=1$, and $g(w_0)g(w_1)\le 0$. This implies that $p(x)=\pm
\left(T_n(x)+cx(x-1) \prod\limits_{r=1}^{k-1} \left( \frac{x^2-w_r^2}{1-w_r^2} \right) \right)$ or  $p(x)=\pm
\left(T_n(x)+cx(x+1) \prod\limits_{r=1}^{k-1} \left( \frac{x^2-w_r^2}{1-w_r^2} \right) \right)$, for some
$c\in\re$. If $c\ne 0$, we would have $|p(w_1)|=|T_n(w_1)|=1$ and $p'(w_1)\ne 0$, which is not possible when $p\in B_n$.

In the case ii), the analysis is analogous: we use the fact that
$$
\ell_{2i} \left( \frac x{w_j} 
\prod_{\substack{r=0\\ r\ne j}}^{k}
 \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right)\right)=\frac1{w_j}\cdot
\ell_{2i} \left(  \prod_{\substack{r=0\\ r\ne j}}^{k}
 \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right)\right)\ge 
\ell_{2i} \left(  \prod_{\substack{r=0\\ r\ne j}}^{k}
 \left( \frac{x^2-w_r^2}{w_j^2-w_r^2} \right)\right),
$$
for $0\le i\le k$, $0\le j\le k$ (the equalities hold only if $j=0$), and so, if $p(x)=\sum\limits_{j=0}^n a_j x^j$ and $T_n(x)=\sum\limits_{j=0}^n t_j^{(n)} x^j$, then, for $0\le i\le k=\frac{n-1}2$, 
$|a_{2i}|+|a_{2i+1}|\le |t_{2i+1}^{(n)}|$ (the only case in which some of these equalities hold is when $p=\pm T_n$). This concludes the proof of Proposition 1. \qed

\pagebreak

{\bf Proof of Theorem 1:}

We have $\ov f(p)=f(|a_0|,|a_1|,\dots,|a_n|)\le f(1,|a_1|, |t_2^{(n)}|-|a_1|,\dots,|a_{n-1}|,|t_n^{(n)}|-|a_{n-1}|) \le f(1,0,|t_2^{(n)}|,0,|t_4^{(n)}|,\dots,0,|t_n^{(n)}|)=\ov f(T_n)$. Here we used the fact that, if $g(x,y)$ is a convex symmetric function then $g(x,y)\le g(x+y,0)$, for $x,y>0$, since
\begin{align*}
g(x,y) &= g\left( \frac x{x+y} (x+y,0)+\frac y{x+y} (0,x+y) \right) \\
&\le \frac x{x+y} g(x+y,0)+\frac y{x+y} g(0,x+y)=g(x+y,0),
\end{align*}
 since $g(x+y,0)=g(0,x+y)$. Notice that since $f$ is strictly increasing in each variable, the equality only holds when $p=\pm T_n$.

The argument in the case ii) is analogous. \qed


\vskip .2in


\noindent
{\bf Corollary 1:} If $p\in B_n$ then $\ell(p)\le \ell(T_n)=\frac{(1+\sqrt2)^n+(1-\sqrt2)^n}{2}$, and the equality holds if and only if $p=\pm T_n$.

\vskip .2in

\noindent
{\bf Corollary 2:} If $p(x)=\sum\limits_{j=0}^na_jx^j\in B_n$ then $\ell^{(\al)}(p):=\sum\limits_{j=0}^n|a_j|^\al\le \sum\limits_{j=0}^n|t_j^{(n)}|^\al=\ell^{(\al)}(T_n)$, for every $\al\ge 1$. The equality holds if and only if $p=\pm T_n$.

\vskip .2in


\noindent
{\bf Remark:} The conclusion of Corollary 2 is false if $0<\al<1$. Indeed, if we take \linebreak $f_n(x)=\frac{x+1}2 + (-1)^n \cdot \frac{x-1}2$, then for $\ve \in (0,1)$, $(1-\ve) T_n + \ve f_n \in B_n$, and $\ell^{(\al)}((1-\ve) T_n + \ve f_n)= \linebreak =(1-\ve)^\al \ell^{(\al)}(T_n) + \ve ^\al > \ell^{(\al)}(T_n)$, if $\ve > 0$ is small.


\section{General comments and Problems}

\vskip .2in

Professor Tamás Erdélyi remarked that if $0\le a<b$, $B_n(a,b)=\{p(x)\in\re[x] \mid |p(x)|\le 1,\,\, \forall\, x\in [a,b]\}$ and $T_n^{(a,b)}(x)=T\left(\frac{2x-(a+b)}{b-a} \right) = \sum\limits_{j=0}^n t_j^{(n,a,b)} x^j$  is the normalized Chebyshev polynomial then, for any $p(x)=\sum\limits_{j=0}^n a_j x^j \in B_n(a,b)$ we have $|a_j|\le|t_j^{(n,a,b)}|$ for $0\le j\le n$. Prof. Erdélyi's argument is as follows: If $0<a<b$ and $0\le j\le n$ then $\{x^r \mid 0\le r\le n,\,\, r\ne j\}$ is a Chebyshev system with respect to $[a,b]$ (this is easy to check using induction is $n$, by considering the derivative of a polynomial in the space generated by such a set). Now, by Theorem 2.10 of [R], the best approximation on $[a,b]$ to $\la t_j^{(n,a,b)}x^j$ by polynomials of the form $\sum\limits_{i=0}^{j-1} a_ix^i+\sum\limits_{\ell=j+1}^n a_\ell x^\ell$ is $\la\left( t_j^{(n,a,b)} x^j - T_n^{(a,b)}(x) \right)$, for any $\la\ne0$. In particular, if $|\la|>1$ then the maximum norm of this best approximation is larger than 1, and this clearly implies the result. The case $a=0$ follows by continuity. We remark that this result also follows from the technique of the previous section, by using Lagrange interpolation at the points $w_k=w_k^{(a,b)}=\frac{(b-a)\cos\left(\frac{k\pi}n\right)+a+b}2$, $0\le k\le n$ to write $p(x)$, and observing that the norm of each coefficient is maximized when $|p(w_k)|=1$ for $0\le k\le n$ and $p(x_{w+1})=-p(w_k)$ for $0\le k\le n-1$ (here it is not necessary to decompose $p(x)$ in its even and odd part, as we did before). Notice that the same arguments can be used to show the (strongest) results obtained b
y changing $B_n(a,b)$ by $C_n(a,b)=\{p(x)\in\re[x] \mid |p(w_k^{(a,b)})|\le 1,\,\, 0 \le k \le n\}$ in the hypothesis. 

Prof. Erdélyi also observed that this result for $[a,b]=[0,1]$ gives another proof to our estimates in $[-1,1]$: let $f(x)\in B_n=B_n(-1,1)$ (the argument for $f(x)\in C_n=C_n(-1,1)$ is analogous). Write $f(x)=\tilde g(x^2)+x\tilde h(x^2)$ as in Section 2.

Let us first consider the case when the degree of $f(x)$ is even, say $2n:f(x)=\sum\limits_{k=0}^{2n}a_kx^k$. We consider now the polynomials $p_1(x)=\tilde g(x^2)+x^2\tilde h(x^2)$ and $p_2(x)=\tilde g(x^2)-x^2\tilde h(x^2)$. Notice that by the argument in the beggining of Section 2, 
$|p_i(x)|\le|\tilde g(x^2)|+|x^2\tilde h(x^2)|\le|\tilde g(x^2)|+|x\tilde h(x^2)|\le 1$, $\forall \, x\in [-1,1]$, $i=1,2$. We can write $p_i(x)=\tilde p_i(x^2)$, $i=1,2$, where
$$
\tilde p_1(x)=a_0+\sum_{k=1}^n (a_{2k-1}+a_{2k})x^k,\quad \tilde p_2(x)=a_0+\sum_{k=1}^n(a_{2k}-a_{2k-1})x^k.
$$
Now, since $\tilde p_i\in B_n(0,1)$, $i=1,2$, we have $|a_{2k}+a_{2k-1}|\le|t_k^{(n,0,1)}| $ and $|a_{2k}-a_{2k-1}|\le |t_k^{(n,0,1)}|$, and so $|a_{2k}|+|a_{2k-1}|\le |t_k^{(n,0,1)}|$ for $0\le k\le n$. On the other hand, $T_n^{(0,1)}(x)=T_n(2x-1)$, so $T_n^{(0,1)}(x^2)=T_n(2x^2-1)=T_n\circ T_2(x)=T_{2n}(x)$, and so $|t_k^{(n,0,1)}|=|t_{2k}^{(2n)}|$, thus we get $|a_0|\le1$, $|a_{2i+1}|+|a_{2i+2}|\le |t_{2i+2}^{(2n)}|$ for $0\le i< \frac n2$, and we can conclude as in Section 2.

For the case where the degree of $f(x)$ is odd, say $2n+1$: $f(x)=\sum\limits_{k=0}^{2n+1} a_kx^k$, we consider the polynomials $p_1(x)=x\tilde g(x^2)+x\tilde h(x^2)$ and $p_2(x)=x\tilde g(x^2)-x\tilde h(x^2)$. We have $|p_i(x)|\le |x\tilde g(x^2)|+|x\tilde h(x^2)|\le |\tilde g(x^2)|+|x\tilde h(x^2)|\le 1$, $\forall\, x\in [-1,1]$, $i=1,2$. Notice now that $\{x^{2r+1},\, 0\le r\le n,\,\, r\ne j\}$ is a Chebyshev system with respect to $\left[\cos\left(\frac{n\pi}{2n+1} \right), 1\right]$ for any $j$ with $0 \le j\le n$, and so the best approximation on $\left[\cos\left(\frac{n\pi}{2n+1} \right), 1\right]$ to
$t_{2j+1}^{(2n+1)}\cdot x^{2j+1}$ is $\left( t_{2j+1}^{(2n+1)} \cdot x^{2j+1}- T_{2j+1}(x)\right)$.
Since $p_1(x)=\sum\limits_{j=0}^n(a_{2j}+a_{2j+1})x^{2j+1}$ and $p_2(x)=\sum\limits_{j=0}^n(a_{2j}-a_{2j+1})x^{2j+1}$, this implies, as before, $|a_{2j}|+|a_{2j+1}|\le |t_{2j+1}^{(2n+1)}|$ for $0\le j\le n$, and we conclude again as in Section 2.

Now we will consider the corresponding problems for general intervals $[a,b]$. For instance, we want to find polynomials $f(x)=\sum\limits_{j=0}^n a_j x^j$ belonging to $B_n(a,b)=\{p(x)=\sum\limits_{j=0}^n b_j x^j\in\re[x] \mid |p(x)|\le 1,\,\, \forall\, x\in [a,b]\}$ which maximize $\ell(f)=\sum\limits_{j=0}^n |a_j|$.

Let us first notice that if $[a,b]=[-c,c]$, with $0< c\le 1$, and $f\in C_n(a,b) \supset B_n(a,b)$ then $\tilde f(x)=f(cx)$ belongs to $C_n(1,1)=C_n$. If $f(x)=\sum\limits_{j=0}^n a_jx^j$ then $\tilde f(x)=\sum\limits_{j=0}^n a_j \cdot  c^j\cdot x^j$.

If $n$ is even, Proposition 1 gives $|a_{2i+1}\cdot c^{2i+1}|+|a_{2i+2}\cdot c^{2i+2}|\le |t_{2i+2}^{(n)}|$ for $0\le i <\frac n2$, and so $|a_{2i+1}|+|a_{2i+2}|\le \frac{|a_{2i+1}|}{c}+|a_{2i+2}|\le \frac{|t_{2i+2}^{(n)}|}{c^{2i+2}}=|t_{2i+2}^{(n,-c,c)}|$, since $T_n^{(-c,c)}(x)=T_n(\frac xc)$. If $n$ is odd, we have $|a_{2i}\cdot c^{2i}|+|a_{2i+1}\cdot c^{2i+1}|\le |t_{2i+1}^{(n)}|$ for $0\le i< \frac n2$, and so $|a_{2i}|+|a_{2i+1}|\le \frac{|a_{2i}|}c+|a_{2i+1}|\le \frac{|t_{2i+1}^{(n)}|}{c^{2i+1}}=|t_{2i+1}^{(n,-c,c)}|$. This implies that Theorem 1 holds if we replace $B_n and C_n$ by $B_n(-c,c) and C_n(-c,c)$ and $T_n$ by $T_n^{(-c,c)} $ when $0< c\le 1$, and, in particular, $\ell(f)\le\ell(T_n^{(-c,c)})$, for every $f\in C_n(-c,c) \supset B_n(-c,c)$.

However, the situation is quite different for general intervals $[a,b]$, and even for symmetric intervals $[-c,c]$, with $c>1$. For instance, we have the following

\vskip .2in

\noindent
{\bf Proposition 2:} For any $c>1$ and $n\ge1$, there is $f(x)\in B_n(-c,c)$ such that \linebreak $\ell(f)>\ell(T_n^{(-c,c)})$. 

\vskip .2in

\noindent
{\bf Proof:} We will show that if $\delta>0$ is small enough then $\ell(T_n^{(-c,c)}=\ell(T_n(\frac xc))<\ell(T_n(\frac{x+\delta}{c+\delta}))$. Notice that $x\in[-c,c]\Rightarrow \frac{x+\delta}{c+\delta}\in [\frac{\delta-c}{c+\delta}, 1]\subset [-1,1]$, and so $T_n(\frac{x+\delta}{c+\delta})\in B_n(-c,c)$.

In order to do this, notice that $T_n(\frac{x+\delta}{c+\delta})=\sum\limits_{j=0}^n t_n^{(n)} (\frac{x+\delta}{c+\delta})^j = \sum\limits_{j=0}^n \frac{t_j^{(n)}}{c^n} \cdot \frac{(x^j+j\delta x^{j-1}+O(\delta^2))}{(1+\frac{j\delta}c+O(\delta^2))} = 
\sum\limits_{j=0}^n \frac{t_n^{(n)}}{c^n} \cdot (x^j+j\delta x^{j-1}+O(\delta^2))(1-\frac{j\delta}c+O(\delta^2))=\sum\limits_{j=0}^n \frac{t_j^{(n)}}{c^j} (x^j+j\delta(\frac{c-1}c)c^{j-1})+O(\delta^2)$. Therefore, since $T_n$ is an even or odd polynomial, we have $\ell(T_n(\frac{x+\delta}{c+\delta}))=\sum\limits_{j=0}^n \frac{|t_j^{(n)}|}{c^j}+\delta(\frac{c-1}c)\sum\limits_{j=0}^n \frac{j|t_j^{(n)}|}{c^j}+O(\delta^2)>\sum\limits_{j=0}^n \frac{|t_j^{(n)}|}{c^j}=\ell(T_n^{(-c,c)})$ for $\delta$ small enough. \qed

\vskip .2in

\noindent
{\bf Examples:} If $c>1$ and $n=1$ then, $\ell(f)$ for $f\in B_n(-c,c)$ is maximum when $f(x)\equiv 1$ or $f(x)\equiv-1$. Indeed, if $f(x)=ax+b\in B_n(-c,c)$ then $|ax|+|b|\le 1$, $\forall\, x\in [-c,c]$, so $|a|+|b|\le |ac|+|b|\le 1$.

 If $c>1$ and $n=2$ then $\ell(f)$ for $f\in B_n(-c,c)$ is maximum when 
$$f(x)=\pm \frac{2(c+1)^2}{(c^2+2c-1)^2}(\pm x-\frac{c-1}{c+1})^2-1.$$

Indeed, by the Exercise 2.6.11 of [R], we may restrict our attention to polynomials \linebreak $f(x)\in B_n(-c,c)$ such that the total multiplicity with which the values 1 and  $-1$ are assumed by $f(x)$, in $[-c,c]$ is at least 3. By considering $\pm f(-x)$, we may suppose that $f(-c)=1$ and $f(\al)=-1$, $f'(\al)=0$, for some $\al\in[0,c]$, so $f(x)=\frac2{(\al+c)^2} \cdot(x-\al)^2-1$. We have $\ell(f)=\frac2{(\al+c)^2}\cdot(1+2\al)+ |\frac{2\al^2}{(\al+c)^2}-1|=\frac2{(\al+c)^2} \cdot(1+2\al)+1-\frac{2\al^2}{(\al+c)^2}$, which is maximum for $\al=\frac{c-1}{c+1} $. In this case, $\ell(f)=1+\frac4{c^2+2c-1}$. For instance, when $c=2$, $f(x)=(\frac{18}{49})(x-\frac13)^2-1=(\frac{18}{49})x^2-(\frac{12}{49})x-\frac{47}{49}$ and $\ell(f)=\frac{11}7>\frac32=\ell(\frac{x^2}2-1)=\ell(T_2^{(-2,2)})$.

In general, it is an interesting (and apparently difficult) problem to characterize the polynomials $f(x)\in  B_n(a,b)$ which maximize $\ell(f)$ for general intervals $[a,b]$ (and even for symmetric intervals $[-c,c](x)$.


\vskip .5in

\section{References}

\begin{itemize}

\item[{[R]}] Theodore J. Rivlin, Chebyshev Polynomials, Wiley Interscience, 1990.

\item[{[RZ]}] M. Reimer and K. Zeller, Abschätzung der Teilsummen reeler Polynome, Math. Zeitschr. {\bf 99} (1967), 101--104.

\end{itemize}

\centerline{\sl Instituto de Matemática Pura e Aplicada - IMPA}

\centerline{\sl Estrada Dona Castorina, 110, Jardim Botânico}

\centerline{22460-320, Rio de Janeiro - Brasil}

\centerline{\sl e-mail: gugu@impa.br}


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