3 The Markov-Bernstein majorant

3.1 What is to solve

Inequalities like $\forall f\in S\; :\; \left\langle f' \vert f' \right\rangle \leq Cte\, \left\langle f \vert f \right\rangle$ are known as "Markov-Bernstein inequality". When dealing with polynomials, the natural choice for the sets $S$ are the $\mathbb{R}_{n}\left[ X\right]$, i.e. the sets of polynomials whose degree is bounded by some integer $n$. Obviously, the bounding factor is an increasing function of $n$, and an asymptotic estimate is of interrest.

Defining $A_{n}$ and $B_{n}$ by $A_{n}\left[ j,\, k\right] =\left\langle F_{j} \vert F_{k} \right\rangle$ and $B_{n}\left[ j,\, k\right] =\left\langle F_{j}' \vert F_{k}' \right\rangle$ where $1\leq j,\, k\leq n$, we obtain to $n\times n$ matrices, the former being positive diagonal and the later being positive symmetric (and half-sparse, due to parity considerations). Defining $\chi _{n}\left( \mu \right)$ as $\det \left( A_{n}\, \mu -B_{n}\right) \div \det \left( A_{n}\right)$, we obtain an unitary real polynomial of degree $n$ whose roots are positive, the greatest of them being the requested Markov-Bernstein factor. In fact, recurrence relations are easier to obtain regarding the $\varphi _{n}$ defined as $\varphi _{n}\left( \lambda \right) =\lambda ^{n}\, \chi _{n}\left( 1/\lambda \right)$, the Markov-Bernstein factor being now the reciprocal of the least root of $\varphi _{n}\left( \lambda \right)$.

Let us take the example $\alpha =0,\, n=4$. One obtains

$$\mu \, A_{4}-B_{4}=\left[ \begin{array}{cccc} \frac{1}{2}\mu -1 &... ...\frac{1}{2}\mu -27 & 0\\ 0 & -16 & 0 & \frac{1}{2}\mu -64 \end{array}\right]$$

leading to $\chi _{4}\left( \mu \right) =\mu ^{4}-200\, \mu ^{3}+9160\, \mu ^{2}-67712\, \mu +73728$, whose roots are $72+8\, \sqrt{65}\approx 136.498$, $72-8\, \sqrt{65}\approx 7.502$, $28+2\, \sqrt{178}=54.683$ and $28-2\, \sqrt{178}=1.317$. The coordinates of the eigenvectors are $\left[ \, 0,\, \frac{-7\pm \sqrt{65}}{4},\, 0,\, 1\, \right]$ and $\left[ \, \frac{-13\pm \sqrt{178}}{3},\, 0,\, 1,\, 0\, \right]$, the corresponding polynomials being $t^{4}+\frac{-23\pm \sqrt{65}}{16}\, t^{2}+\frac{11\mp \sqrt{65}}{32}$ and $t^{3}+\frac{-22\pm \sqrt{178}}{12}\, t$. Therefore the Markov-Bernstein factor (relative to the Chebyschev scalar product) is $28+2\sqrt{178}$ over $\mathbb{R}_{3}\left[ X\right]$ and $72+8\sqrt{65}$ over $\mathbb{R}_{4}\left[ X\right]$.

3.2 Recurence over the $\chi _{n}$

From EQ. 20, it can be seen that $r_{n}\doteq \left\langle F_{m}' \vert F_{n}' \right\rangle \div \left\langle F_{m}' \vert F_{n+2}' \right\rangle$ is, for a given $n$, constant for all $m$ such that $n-m\in 2\mathbb{N}$. Therefore, it is convenient to introduce the triangular sparse matrix $\mathrm{T}_{n}$ defined by $T_{j,\, j}=1$, $T_{j+2,\, j}=-r_{j}=-\frac{\left( j+1\right) \left( j+2\right) }{\left( \alpha +j\right) \left( \alpha +j+1\right) }$ and $T_{j,\, k}=0$ elsewhere. Defining the matrix $\mathrm{M}_{n}\doteq \mathrm{T}_{n}\, \left( \mu \, \mathrm{A}_{n}-\mathrm{B}_{n}\right) \, ^{t}\mathrm{T}_{n}$, we have the obvious relation $\chi _{n}\left( \mu \right) =\det \mathrm{M}_{n}\div \det \mathrm{A}_{n}$, and now $M_{j,\, k}=0$ except when $j-k\in \left\{ -2,\, 0,\, 2\right\}$. Some computations give :

$$\begin{array}{rcl} M_{j,\, j} & = & \left( \frac{\mu \, \lef... ... \, \frac{\alpha !\, j!}{\left( \alpha +j\right) !} \end{array}$$

except from $M_{2,\, 2}$ whose value remains $\mu \frac{2}{\left( \alpha +4\right) \left( \alpha +1\right) }-\frac{4\left( \alpha +2\right) }{\left( \alpha +1\right) ^{2}}$ since no eliminations are occurring.

That special form of matrix $\mathrm{M}_{n}$ induces the recurrence

$$\det \mathrm{M}_{n}=M_{n,\, n}\, \det \mathrm{M}_{n-1}-M_{n,\, n-... ..._{n-3}+\left( M_{n,\, n-2}\, M_{n-1,\, n-3}\right) ^{2}\, \det \mathrm{M}_{n-4}$$

Going back to the $\chi _{n}$, and then to the $\psi _{n}$, we obtain :

$$\chi _{n}=\left\{ \begin{array}{l} \chi _{n-1}\left( \mu \; \frac... ...n+\alpha \right) ^{2}\, \left( \alpha +n-3\right) }\right) \end{array}\right.$$

$$\psi _{n}=\left\{ \begin{array}{l} \psi _{n-1}\left( -\lambda \; ... ...\alpha +n-2\right) ^{2}\, \left( \alpha +n-3\right) }\right) \end{array}\right.$$ (21)

3.3 Factorisation of the recurrence relation

Let us define $\mathbb{E}_{p }\doteq \mathrm{Span}\left( X^{2},\, X^{4},\, \cdots \, X^{2p} \right) =X^{2}\, \mathbb{R}_{p-1}\left[ X^{2}\right]$ (excluding the constants), together with $\mathbb{O}_{p }\doteq \mathrm{Span}\left( X^{1},\, X^{3},\, \cdots \, X^{2p-1} \right) =X\, \mathbb{R}_{p-1}\left[ X^{2}\right]$, and assume $n=2p$. The space $\mathbb{R}_{n}\left[ X\right]$ splits as $\mathbb{R}\oplus \mathbb{O}_{p }\oplus \mathbb{E}_{p }$, while $\mathbb{R}_{n+1}\left[ X\right]$ splits as $\mathbb{R}\oplus \mathbb{O}_{p+1 }\oplus \mathbb{E}_{p }$. Both spaces have the same even subspace in common, and therefore $\psi _{2p}$ and $\psi _{2p+1}$ must have a common factor of degree $p$ (say $\theta _{2p}$). For the same reason, $\psi _{2p-1}$ and $\psi _{2p}$ must have a common factor of degree $p$ (say $\theta _{2p-1}$). Therefore, it can be hoped that each polynomial $\psi _{n}$ splits into $\psi _{n}=\theta _{n}\, \theta _{n-1}$, the $\theta _{n}$ being defined by $\theta _{n}=Cte\, \gcd \left( \psi _{n+1},\, \psi _{n}\right)$ where the $Cte$ is chosen such that $\theta _{n}\left( 0\right) =1$.

This suggest that the $\theta _{n}$ can be related by a recurrence relation of the form $\theta _{n}\left( \lambda \right) =\left( \lambda \, a_{n}+b_{n}\right) \theta _{n-2}\left( \lambda \right) +c_{n}\, \theta _{n-4}\left( \lambda \right)$ i.e. a relation dealing separately with odd and even $\theta _{n}$. Substituting this relation, together with $\psi _{n}=\theta _{n}\, \theta _{n-1}$ into EQ. 21 and identifying, we obtain

$$\theta _{n}=\left\{ \begin{array}{l} \theta _{n-2}\left( -\lambda... ... \, \left( \alpha +n-1\right) \, \left( \alpha +n-2\right) } \end{array}\right.$$ (22)

with respective initial conditions $\theta _{-1}=1,\, \theta _{1}=1-\lambda \left( \alpha +2\right)$ and $\theta _{0}=1,\, \theta _{2}=1-\lambda \left( \alpha +2\right) \frac{2\left( \alpha +4\right) }{\alpha +1}$. A backwards recurrence a la Fermat shows that $\theta _{n}$ and $\theta _{n-2}$ are coprime for all $n$, and therefore $\psi _{n}$ and $\psi _{n-2}$ are also coprime for all $n$, as expected when trying the $\psi _{n}=\theta _{n}\, \theta _{n-1}$ decomposition.

From EQ. 22, the recurrence involving the coefficients of $\theta _{n}$ can easily be obtained. Defining $c_{j}\left( n\right)$ by $\theta _{n}\left( \lambda \right) =\sum c_{j}\left( n\right) \lambda ^{j}$, we have

$$c_{j}\left( n\right) =\left\{ \begin{array}{l} \frac{\left( \alph... ...ft( \alpha +2n\right) }{\alpha +n-1}c_{j-1}\left( n-2\right) \end{array}\right.$$ (23)

As it should be from the very definitions, $c_{0}\left( n\right) =1$ satisfies this relation.

3.4 Even and odd numbers

Let us now examine the even case in more details. Starting with $j=1$, EQ. 23 can be solved by $c_{1}\left( n\right) =polynomial\left( n\right) +u\, f\left( n\right) +v\, g\left( n\right)$ where $f,\, g$ are two independant solutions of the linear equation associated with EQ. 23. Obviously, $f=1$ is one of them, and it can be checked that $g\left( n\right) =\frac{2^{\alpha }\Gamma \left( 2m+3\right) }{8\, \Gamma \left( 2m+\alpha \right) }$ is another ($\alpha =3$ appears to be special). By identification, one obtains $c_{1}\left( n\right) =-\frac{n\left( n+2\right) \left( \alpha +n\right) \left( \alpha +n+2\right) }{4\left( \alpha +1\right) }+u+v\, g\left( n\right)$. Using initial conditions issued from $\theta _{0}$ and $\theta _{2}$, it appears that $u=v=0$, and $c_{1}\left( n\right)$ is a polynomial and $\mathrm{dg}\left( c_{1} \right) =4$. Iterating the process, we obtain that $c_{2}$ is a polynomial, with $\mathrm{dg}\left( c_{2} \right) =8$, etc.

Rewitting EQ. 23 as $c_{j}\left( n\right) +\clubsuit c_{j}\left( n-2\right) +\clubsuit c_{j}\left( n-4\right) =\clubsuit c_{j-1}\left( n-2\right)$ and taking the equivalents $c_{j}\left( n\right) \sim a\, n^{p}$ and $c_{j-1}\left( n\right) \sim b\, n^{q}$, we obtain $n^{p-2}\left( 4\alpha \, p\, a+4p^{2}\, a-12p\, a\right) \sim -4n^{q+2}b$, i.e. $p=q+4$ and $a=-\frac{b}{\left( q+4\right) \, \left( q+\alpha +1\right) }$. By recurrence

$$c_{j}\left( n\right) \sim n^{4j}\: \prod _{k=1}^{j}\frac{-1}{4k\, \left( \alpha +k-3\right) }$$ (24)

Applying the same computations to the odd case, we obtain $\widehat{c_{1}}\left( n\right) =c_{1}\left( n\right) -\frac{\alpha -1}{4}$, where $\widehat{c_{1}}$ is the coefficient when $n$ is odd and $c_{1}\left( n\right)$ is what is obtained when applying the even formula (to odd numbers, i.e. outside it's domain of validity). Since $\widehat{c_{1}}$ and $c_{1}$ are equivalents, the former recurrence shows that EQ. 24 holds for odd numbers too.

3.5 Some results

Let us introduce the new variable $\nu =\frac{n^{4}}{\mu }=n^{4}\, \lambda$. By EQ. 24, the polynomial in $\nu$ defined by $\theta _{n}\left( \nu /n^{4}\right)$ converges towards a series whose coefficients do not depend on $n$, and it is easy to write that series as an hypergeometric function. We obtain

$$\theta _{n}\left( \frac{\nu }{n^{4}}\right) =1-\frac{\nu }{4\left... ...1\right] ,\, \left[ 2,\, \frac{5+\alpha }{4}\right] ,\, -\frac{\nu }{16}\right)$$ (25)

Applying this formula to the Chebyschev's polynomials, the smallest root of $\theta _{n}\left( \frac{\nu }{n^{4}}\right)$ appears to be $\nu \approx 4.5$ (cf FIG. 1) leading to $\mu \approx n^{4}\times .2231266815$.

FIG.  1: Asymptotic behaviour of $\theta _{n}$ in the Chebyschev case.

For other values of $\alpha$, one obtains FIG. 2. It appears that $\alpha =1$ and $\alpha =5$ are special, since the hypergeometric functions can be rewritten in a simpler form : $1-\frac{\nu }{8}\mathrm{hypergeom}\left( \left[ 1\right] ,\, \left[ 3/2,\, 2\right] ,\, -\frac{\nu }{16}\right) =\cos \left( \frac{1}{2}\sqrt{\nu }\right)$ and $1-\frac{\nu }{24}\mathrm{hypergeom}\left( \left[ 1\right] ,\, \left[ 2,\, 5/2\... ...sin \left( \frac{1}{2}\sqrt{\nu }\right) /\left( \frac{1}{2}\sqrt{\nu }\right)$.

FIG.  2: Markov-Bernstein factor for some $\alpha$.