ON SOME PROBLEMS OF MAXIMUM STIELTJES

BY
TIBERIU POPOVICIU
(Cluj)

On some theorems of Stieltjes and MI Schur

1.

Let us designate by $V\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)$ the Van Der Monde determinant of numbers (or points) $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ The square of this determinant is the discriminant of the numbers $\alpha_{i}$ , or the discriminant of the degree polynomial $n$ having as zeros the points $\alpha_{l}$ .

In a note Stieljes stated the following theorem ² ):
Theorem 1. When $\alpha>0,\beta>0$ and that the $\mathrm{x}_{1}$ remain within the interval (0,1), the expression

\mathrm{D}=\mathrm{D}\left(x_{1},x_{2},\ldots,x_{n}\right)=\left[\prod_{i=1}^{n}x_{i}^{\alpha}\left(1-x_{i}\right)^{\beta}\right]\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $\mathrm{x}_{1}$ are the zeros of the Jacobi polynomial $\mathrm{P}_{n}(x)=\mathrm{F}(-n,\alpha+\beta+n-1,\alpha,x)=\frac{x^{1-\alpha}(1-x)^{1-\beta}}{\alpha(\alpha+1)\ldots(\alpha+n-1)}\frac{d^{n}}{dx^{n}}x^{\alpha+n-1}(1-x)^{\beta+n-1}$

The value of this maximum is

\mathrm{M}_{n}(\alpha,\beta)=\prod_{i=1}^{n}\frac{i^{i}(\alpha+i-1)^{\alpha+i-1}(\beta+i-1)^{\beta+i-1}}{(\alpha+\beta+n+i-2)^{\alpha+\beta+n+i-2}}

Stieltjes gave the demonstration in another work ³ ), to which we will return later. The calculation of $\mathrm{M}_{n}(x,\beta)$ is not indicated by Stieltjes but it can be done by following the method, developed on certain particular cases, of MI Schur ⁴ ).

1.

I undertook this work following discussions I had with Prof. Th. Angheluta at the Mathematical Institute of the University of Cluj.
² ) Th. J. Stieltjes „On Jacobi Polynomials", CR Acad. Sc., Paris, t. 100 (1885), p. 620.
³ ) Th. J. Stieltjes "On Certain Polynomials Which Verify a Linear Differential Equation of the Second Order and on the Theory of LamÉ Functions". Acta Math., t. 6 (1885), p. 321.
⁴⁾ I. Schur „Uber die Verteilung der Warzeln gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten". Maih. Zeitschrift, t. 1 (1918), p. 377.

2.

Theorem I can be easily proven. Let us first note that $\mathrm{D}\left(x_{1},x_{2},\ldots,x_{n}\right)$ has a maximum for $0\leq x_{i}\leq 1$ which is necessarily achieved and in particular for values $x_{i}$ all distinct and not coinciding with 0 and 1.

It can easily be seen that this maximum can only be achieved for a single point system $\mathrm{x}_{\mathrm{i}}$ . Let us suppose, in fact, that
$\operatorname{max.\penalty 10000\ }\mathrm{D}\left(x_{1},x_{2}\ldots,x_{n}\right)=\mathrm{D}\left(y_{1},y_{2},\ldots,y_{n}\right)=\mathrm{D}\left(y_{1}^{\prime},y_{2}^{\prime},\ldots,y_{n}^{\prime}\right)$
$(0,1)$ and (1)

\begin{gathered}0<y_{1}<y_{2}<\ldots<y_{n}<1;0<y_{1}^{\prime}<y_{2}^{\prime}<\ldots<y_{n}^{\prime}<1\\ \left|y_{1}-y_{1}^{\prime}\right|+\left|y_{2}-y_{2}^{\prime}\right|+\ldots+\left|y_{n}-y_{n}^{\prime}\right|>0\end{gathered}

If we pose

y_{i}^{\prime\prime}=\frac{y_{i}+y_{i}^{\prime}}{2},\quad i=1,2,\ldots,n

We have

\left(y_{i}^{\prime\prime}\right)^{2}\geqslant y_{i}y_{i}^{\prime},\left(1-y_{i}^{\prime\prime}\right)^{2}\geqslant\left(1-y_{i}\right)\left(1-y_{i}^{\prime}\right),\left(y_{i}^{\prime\prime}-y_{j}^{\prime\prime}\right)^{2}\geqslant\left|\left(y_{i}-y_{j}\right)\left(y_{i}^{\prime}-y_{j}^{\prime}\right)\right|

equality, in one of these formulas, can only take place if $y_{i}=y_{i}^{\prime}$ , resp. $y_{i}-y_{j}=y_{i}^{\prime}-y_{j}^{\prime}$ .

It follows that
(2) $\quad\mathrm{D}\left(y_{1}^{\prime\prime},y_{2}^{\prime\prime},\ldots,y_{n}^{\prime\prime}\right)\geq V\overline{\mathrm{D}\left(y_{1},y_{2},\ldots,y_{n}\right)\mathrm{D}\left(y_{1}^{\prime},y_{2}^{\prime},\ldots,y_{n}^{\prime}\right)}$
equality being possible only if the sequences (1) coincide. This is in contradiction with our hypotheses and the reported uniqueness results.

We deduce, in particular, that if $\alpha=\beta$ the points for which the maximum is reached are symmetrically distributed with respect to the middle of the interval $(0,1)$ This results from the fact that if $\alpha=\beta$ , we have
$\mathrm{D}\left(1-x_{1},1-x_{2},\ldots,1-x_{n}\right)=\mathrm{D}\left(x_{1},x_{2},\ldots,x_{n}\right)\quad\left(0\leq x_{i}\leq 1\right)$
3. Stieltjes demonstrated that the determination of the maximum is done using differential calculus. Let us set

\mathrm{P}(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right)

The maximum is given by the system

\frac{\partial\mathrm{D}}{\partial x_{i}}=0,i=1,2,\ldots,n

which admits, as will result later, a single solution under the hypothesis that the $x_{i}$ are distinct from each other. This uniqueness can also be demonstrated directly using inequality (2).

A simple calculation gives us

\frac{1}{\mathrm{D}}\frac{\partial\mathrm{D}}{\partial x_{i}}=\frac{\alpha}{x_{i}}-\frac{\beta}{1-x_{i}}+2\sum_{j=1}^{n}\frac{1}{x_{i}-x_{j}}=\frac{\alpha}{x_{i}}-\frac{\beta}{1-x_{i}}+\frac{\mathrm{P}^{\prime\prime}\left(x_{i}\right)}{\mathrm{P}^{\prime}\left(x_{i}\right)}

$\Sigma^{\prime}$ designating a summation where the value $j=i$ is excluded.

The maximum will therefore be given by the system

\begin{gathered}x_{l}\left(1-x_{i}\right)\mathrm{P}^{\prime\prime}\left(x_{i}\right)+\left[\alpha-(\alpha+\beta)x_{i}\right]\mathrm{P}^{\prime}\left(x_{j}\right)=0\\ i=1,2,\ldots,n.\end{gathered}

We immediately see that the polynomial $\mathrm{P}(x)$ , which is of degree $n$ , must verify the differential equation

x(1-x)y^{\prime\prime}+[\alpha-(\alpha+\beta)x]y^{\prime}+\lambda y=0

$\lambda$ being a constant. We find $\lambda=n(n+\alpha+\beta-1)$ and we see that $\mathrm{P}(x)$ is expressed using the Gauss hypergeometric series, $\mathrm{F}(\alpha,\beta,\gamma,x)$ . We actually find that this polynomial, up to a constant factor, is equal to $\mathrm{F}(-n,\alpha+\beta+n-1,\alpha,x)$ so it is a Jacobi polynomial of degree $n$ .
4. To calculate the value of the maximum $M_{n}(x,\beta)$ We will first make some slightly more general considerations.

Either

\mathrm{Q}_{0}(x),\mathrm{Q}_{1}(x),\ldots,\mathrm{Q}_{n}(x),\ldots

a sequence of polynomials defined by the recurrence relations
(3) $\mathrm{Q}_{0}=1,\mathrm{Q}_{1}=b_{1}x+c_{1},\mathrm{Q}_{n}=\left(b_{n}x+c_{n}\right)\mathrm{Q}_{n-1}+a_{n}\mathrm{Q}_{n-2}$

n=2,3,\ldots

where we assume that the $a_{i}$ and the $b_{t}$ are different from zero.
Let us denote by $\xi_{1}^{(n)},\xi_{2}^{(n)},\ldots,\xi_{n}^{(n)}$ the zeros of the polynomial $Q_{n}(x)$ and by

\Delta_{n}=\prod_{l=1}^{n}\mathrm{Q}_{n}

(i)

the result of the polynomials $Q_{n-1},Q_{n}$ (this resultant being defined in this way).

From (3) we deduce that the coefficient of $x^{n}$ In $\mathrm{Q}_{n}(x)$ is equal to $b_{1}b_{2}\ldots b_{n}$ . Then doing successively $x=\xi_{1}^{(n-1)},\xi_{2}^{(n-1)},\ldots,\xi_{n-1}^{(n-1)}$ , we deduce from this

\prod_{i=1}^{n-1}\mathrm{Q}_{n}\left(\xi_{i}^{(n-1)}\right)=a_{n}^{n-1}\prod_{i=1}^{n-1}\mathrm{Q}_{n-2}\left(\xi_{i}^{(n-1)}\right)\doteq a_{n}^{n-1}\Delta_{n-1}

But

\prod_{l=1}^{n-1}\mathrm{Q}_{n}\left(\xi_{l}^{(n-1)}\right)=\frac{b_{n}^{n-1}}{b_{1}b_{2}\ldots b_{n-1}}\Delta_{n}

\Delta_{n}=\frac{b_{1}b_{2}\ldots b_{n-1}}{b_{n}^{n-1}}a_{n}^{n-1}\Delta_{n-1}

If we now notice that $\Delta_{1}=1$ , we obtain

\Delta_{n}=\frac{a_{2}}{b_{2}}\left(\frac{a_{3}}{b_{3}}\right)^{2}\cdots\left(\frac{a_{n-1}}{b_{n-1}}\right)^{n-2}\left(\frac{a_{n}}{b_{n}}\right)^{n-1}b_{1}^{n-1}b_{2}^{n-2}\ldots b_{n-2}^{2}b_{n-1}

(4)

In particular, polynomials $\mathrm{P}_{n}(x)=\mathrm{F}(-n,\alpha+\beta+n-1,\alpha,x)$ are linked by relations of the form (3). In this case
(5) $\quad b_{n}=-\frac{(\alpha+\beta+2n-2)(\alpha+\beta+2n-3)}{(\alpha+n-1)(\alpha+\beta+n-2)},\left(b_{1}=-\frac{\alpha+\beta}{\alpha}\right)$ and we find

\frac{a_{n}}{b_{n}}=\frac{(n-1)(\beta+n-2)}{(\alpha+\beta+2n-3)(x+\beta+2n-4)}

(6) $\quad\Delta_{n}=(-1)^{\frac{n(n-1)}{2}\prod_{i=1}^{n-1}\frac{i^{i}(\beta+i-1)^{i}}{(\alpha+i-1)^{n-i}(\alpha+\beta+n+i-2)^{i}}\text{. }}$
5. Now let's move on to the calculation of $M_{n}(\alpha,\beta)$ . If $\xi_{1},\xi_{2},\ldots,\xi_{n}$ are the zeros of the polynomial $\mathrm{P}_{n}(x)$ , We have

M_{n}(\alpha,\beta)=\left[\prod_{i=1}^{n}\xi_{i}^{\alpha}\left(1-\xi_{i}\right)^{\beta}\right]V^{2}\left(\xi_{1},\xi_{2},\ldots,\xi_{n}\right).

Relationships

\begin{gathered}P_{n}(0)=1,\quad P_{n}(1)=(-1)^{n}\prod_{i=1}^{n}\frac{\beta+i-1}{\alpha+i-1}\\ P_{n}(x)=(-1)^{n}\left[\prod_{i=1}^{n}\frac{\alpha+\beta+n+i-2}{\alpha+i-1}\right]\left(x-\xi_{1}\right)\left(x-\xi_{2}\right)\ldots\left(x-\xi_{n}\right)\end{gathered}

give us

\left\{\begin{array}[]{c}\xi_{1}\xi_{2}\ldots\xi_{n}=\prod_{i=1}^{n}\frac{\alpha+i-1}{\alpha+\beta+n+i-2}\\ \left(1-\xi_{1}\right)\left(1-\xi_{2}\right)\ldots\left(1-\xi_{n}\right)=\prod_{i=1}^{n}\frac{\beta+i-1}{\alpha+\beta+n+i-2}\end{array}\right.

We now have

\mathrm{V}^{2}\left(\xi_{1},\xi_{2},\ldots,\xi_{n}\right)=(-1)^{\frac{n(n-1)}{2}}\prod_{\frac{i=1}{\left(b_{1}b_{2}\ldots b_{n}\right)^{n}}\mathrm{P}_{n}^{\prime}\left(\xi_{i}\right)}

(8)

But the polymones $\mathrm{P}_{n},\mathrm{P}_{n-1}$ and the derivative of the first still verify the following relation

x(1-x)\mathrm{P}_{n}^{\prime}(x)=\left(d_{n}x+e_{n}\right)\mathrm{P}_{n}(x)-\frac{n(\beta+n-1)}{\alpha+\beta+2n-2}\mathrm{P}_{n-1}(x)

$d_{n},e_{n}$ being two constants of no importance to us.
If we successively do $x=\xi_{1},\xi_{2},\ldots,\xi_{n}$ we get

\prod_{l=1}^{n}\mathrm{P}_{n}^{\prime}\left(\xi_{l}\right)=(-1)^{n}\frac{n^{n}(\beta+n-1)^{n}}{(\alpha+\beta+2n-2)^{n}}\frac{\Delta_{n}}{\prod_{l=1}^{n}\xi_{l}\left(1-\xi_{l}\right)}

(9)

Finally, taking into account (5), (6), (7), (8), (9) we obtain

\mathrm{M}_{n}(\alpha,\beta)=\prod_{i=1}^{n}\frac{i^{i}(\alpha+i-1)^{\alpha+i-1}(\beta+i-1)^{\beta+i-1}}{(\alpha+9+n+i-2)^{\alpha+\beta+n+i-2}}

which is precisely the requested formula.
6. A simple linear transformation allows us to state the following theorem:

Theorem II. When $\alpha>0,\beta>0$ and that the $\mathrm{x}_{\mathrm{i}}$ remain within the interval $(\mathrm{a},\mathrm{b}),\mathrm{a}<\mathrm{b}$ , the expression

\left[\prod_{i=1}^{n}\left(x_{i}-a\right)^{\alpha}\left(b-x_{i}\right)^{\beta}\right]\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $\mathrm{x}_{\mathrm{i}}$ are the zeros of the polynomial

(x-a)^{1-\alpha}(b-x)^{1-\beta}\frac{d^{n}}{dx^{n}}(x-a)^{\alpha+n-1}(b-x)^{\beta+n-1}

The value of this maximum is

(b-a)^{n(\alpha+f+n-1)}\mathrm{M}_{n}(\alpha,\beta)

Especially if $a=-1,b=1,\alpha=\beta=1$ we obtain another theorem of Stieltjes ⁵ ):

Theorem III. When the $\mathbf{x}_{\mathbf{i}}$ remain in the meantime $(-1,1)$ , the expression

\left(1-x_{1}^{2}\right)\left(1-x_{2}^{2}\right)\ldots\left(1-x_{n}^{2}\right)V^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $\mathrm{x}_{\mathrm{i}}$ are the zeros of the polynomial $\mathrm{X}_{\mathrm{n}}$ of legendre of degree n. The value of this maximum is

\frac{2^{4}3^{6}\ldots n^{2n}}{3^{3}5^{5}\ldots(2n-1)^{2n-1}}

7.

Stieltjes further stated the following theorem ⁶ ):

Theorem IV. When the $\mathrm{x}_{\mathrm{i}}$ are in the interval $(-1,1)$ the expression

\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $\mathrm{x}_{\mathrm{i}}$ are the zeros of the polynomial $\mathrm{V}_{n}$ which comes from development

\sqrt{1-2xz+z^{2}}=\sum_{n=0}^{\infty}V_{n}(x)z^{n}

The value of this maximum, which can be designated by $2^{\mathrm{n}(\mathrm{n}-1)}\mathrm{M}_{\mathrm{n}}(0,0)$ , East

\frac{2^{2}3^{3}\ldots(n-2)^{n-2}2^{2}3^{3}\ldots n^{n}}{3^{3}5^{5}\ldots(2n-3)^{2n-3}}

⁰⁰footnotetext: ⁵⁾ Th. J. Stieltjes „On some theorems of algebra", CR Acad. Sc., Paris, t. 100 (1885) p. 439.
⁶ ) Loc cit. ⁵ ).

This theorem corresponds to the case where $\alpha=\beta=0$ . It was demonstrated by MI Schur ⁷ ). The previous results are not applicable without precautions, but we can easily see that this property results from Theorem II. We see, in fact, that this maximum can only be reached if the $x_{i}$ are all distinct and if one coincides with 0 and another with 1. We therefore deduce Theorem IV from Theorem II by $y$ doing $a=-1$ , $b=1,\alpha=\beta=2$ and taking $n-2$ instead $n$ .

Note that

\begin{gathered}\frac{1}{2^{n-1}(n-1)!}\frac{d^{n-2}}{dx^{n-2}}\left(x^{2}-1\right)^{n-1}=-\frac{n\left(x^{2}-1\right)}{2^{n-1}(n-2)!}\frac{d^{n}}{dx^{n}}\left(x^{2}-1\right)^{n-1}=\\ =-\int_{-1}^{x}X_{n-1}(x)dx=V_{n}(x)\end{gathered}

We can also, in the same way, deal with cases where $\alpha=0$ Or $\beta=0$ only. For example the case $\alpha=0,\beta\neq 0$ results from Theorem II in $y$ doing $x=2$ and taking $n-1$ instead of $n$ . We define the numbers in this way $\mathrm{M}_{n}(0,\mathrm{p}),\mathrm{M}_{n}(\alpha,0)$ .

We actually find ⁸ )

M_{n}(\alpha,0)=M_{n}(0,\alpha)=M_{n-1}(\alpha,2)=\prod_{i=1}^{n-1}\frac{i^{i}(i+1)^{i+1}(\alpha+i-1)^{\alpha+i-1}}{(\alpha+n+i-1)^{\alpha+n+i-1}}

8.

On numbers $M_{n}(\alpha,\beta)$ We can make a few remarks. Let us first delimit the number $M_{n}(\alpha,\beta)$ with the help of $M_{n}(0,0)$ .

We have
(10) $M_{n}(x,\beta)<\max_{(0,1)}V^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)=M_{n}(0,0)\quad(\alpha+\beta>0)(0,1)$
Note that

\mathrm{V}^{2}\left(\eta_{1},\eta_{2},\ldots,\eta_{n}\right)=\frac{\mathrm{V}^{2}\left(\eta_{0},\eta_{1},\ldots,\eta_{n},\eta_{n+1}\right)}{\left(\eta_{0}-\eta_{n+1}\right)^{2}\prod_{i=1}^{n}\left(\eta_{0}-\eta_{l}\right)^{2}\left(\eta_{n+1}-\eta_{l}\right)^{2}}

If $\eta_{0}=-1,\eta_{1},\ldots,\eta_{n},\eta_{n+1}=1$ are the zeros of the polynomial $V_{n+2}(x)$ , We have

\begin{gathered}2^{n(\alpha+\beta+n-1)}M_{n}(\alpha,\beta)\geqslant\left[\prod_{i=1}^{n}\left(1+\eta_{i}\right)^{\alpha}\left(1-\eta_{i}\right)^{\beta}\right]V^{2}\left(\eta_{1},\eta_{2},\ldots,\eta_{n}\right)=\\ =2^{(n+2)}(n+1)-2M_{n+2}(0,0)\left[\prod_{i=1}^{n}\left(1+\eta_{i}\right)^{\alpha-2}\left(1-\eta_{i}\right)^{\beta-2}\right]\end{gathered}

⁷ ) Loc cit.⁴ ).

		$\displaystyle\qquad\begin{aligned} \mathrm{M}_{n}(\alpha,0)=\lim_{\beta\rightarrow 0}\mathrm{M}_{n}(\alpha,\beta)&;\mathrm{M}_{n}(0,0)=\lim_{\alpha\rightarrow 0}\mathrm{M}_{n}(\alpha,\beta)\\ &\beta\rightarrow 0\end{aligned}$
		$\displaystyle\text{ compte tenant de }\alpha^{\alpha}\rightarrow 1\text{ lorsque }\alpha\rightarrow 0.$

We have

\mathrm{V}_{n+1}(x)=-\frac{(2n+1)!}{2^{n}n!(n+2)!}x^{n+2}+\cdots

	$\displaystyle\prod_{i=1}^{n}\left(1-\eta_{i}\right)=-\frac{2^{n}n!(n+2)!}{(2n+1)!}\lim$	$\displaystyle\frac{\mathrm{V}_{n+2}(x)}{x^{2}-1}=-\frac{2^{n-1}n!(n+2)!}{(2n+1)!}\mathrm{V}_{n+2}^{\prime}(1)$
	$\displaystyle x\rightarrow 1$
	$\displaystyle\prod_{i=1}^{n}\left(1+\eta_{i}\right)=(-1)^{n+1}\frac{2^{n}n!(n+2)!}{(2n+1)!}$	$\displaystyle\lim\cdot\frac{\mathrm{V}_{n+2}(x)}{x^{2}-1}$
		$\displaystyle=-1$
		$\displaystyle=(-1)^{n+2}\frac{2^{n+1}n!(n+2)!}{(2n+1)!}\mathrm{V}_{n+2}^{\prime}(-1)$

But

\mathrm{V}_{n+2}^{\prime}(1)=-1,\mathrm{\penalty 10000\ V}_{n+2}^{\prime}(-1)=(-1)^{n+2}

\prod_{i=1}^{n}\left(1+\eta_{i}\right)=\prod_{i=1}^{n}\left(1-\eta_{i}\right)=\frac{2^{n-1}n!(n+2)!}{(2n+1)!}

Taking into account the formula
$2^{(n+2)(n+1)}\mathrm{M}_{n+2}(0,0)=\frac{2^{n(n-1)}(n-1)^{n-1}n^{n}(n+1)^{n+1}(n+2)^{n+2}}{(2n-1)^{2n-1}(2n+1)^{2n+1}}\mathrm{M}_{n}(0,0)$ we deduce from this

	$\displaystyle\mathrm{M}_{n}(\alpha,\beta)$	$\displaystyle=\frac{(n-1)^{n-1}n^{n}(n+1)^{n+1}(n+2)^{n+2}}{2^{\alpha+\beta+2(2n+1)}(2n-1)^{2n-1}(2n+1)^{2n+1}}$		(11)
		$\displaystyle\because\left[\frac{n!(n+2)!}{(2n+1)!}\right]^{\alpha+\beta-4}\mathrm{M}_{n}(0,0)$

and this formula is obviously valid for $\alpha\geqslant 0,\beta\geqslant 0$ .
Taking into account Stirling's formula, we see that the root $n(n-1)^{\text{éme }}$ of the coefficient of $\mathrm{M}_{n}(0,0)$ in (11) tends to 1 for $n\longrightarrow\infty$ . On the other hand, MM Fekete ⁹ ) has already noticed that

\begin{gathered}\sqrt[n(n-1)]{\mathrm{M}_{n}(0,0)}\longrightarrow\frac{1}{4}\\ n\longrightarrow\infty\end{gathered}

this number being the transfinite diameter of the segment $0-1$ . Formulas (10), (11) therefore show us that:

Theorem V. $\alpha,\beta$ being two non-negative numbers, the expression

\sqrt[n(n-1)]{\mathrm{M}_{n}(\alpha,\beta)}

⁹ ) M. Fekete “Über die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten” Math. Zeitschrift,t. 17 (1923), p. 228.
tends, for $\mathrm{n}\longrightarrow\infty$ , towards the transfinite diameter $\frac{1}{4}$ of the segment $0-1^{10}$ ).

It is easily established that $M_{n}(\alpha,\beta)$ is a decreasing function of $\alpha$ and of $\beta$ . We immediately check that the function

\begin{gathered}(\alpha+i-1)\log(\alpha+i-1)+(\beta+i-1)\log(\beta+i-1)-\\ -(\alpha+\beta+n+i-2)\log(\alpha+\beta+n+i-2)\end{gathered}

is convex with respect to $\alpha$ And $\beta$ when $i\leq n,\alpha>0,\beta>0$ .
The function $\log M_{n}(\alpha,\beta)$ is a sum of convex functions and we can therefore state the following property:

Theorem VI. The function $M_{n}(\alpha,\beta)$ is decreasing and its logarithm is a convex function with respect to the two variables a and $\beta$ .

It can easily be seen that the property holds for $\alpha\geq 0,\beta\geq 0$ .
The corresponding properties are easily found when, instead of the interval ( 0,1 ), we take any finite interval ( $a,b$ ).
9. Let us take $a=0,b=\beta$ in Theorem II; we deduce that for $0\leq x_{l}\leq\beta$ the expression

\left[\prod_{l=1}^{n}x_{i}^{\alpha}\left(1-\frac{x_{l}}{\beta}\right)^{\beta}\right]\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $x_{i}$ are the zeros of the polynomial (written in a suitable form)

\frac{x^{1-\alpha}\left(1-\frac{x}{\beta}\right)^{1-\beta}}{n!}\frac{d^{n}}{dx^{n}}x^{\alpha+n-1}\left(1-\frac{x}{\beta}\right)^{\beta+n-1}.

If we do now $\beta\longrightarrow\infty$ , we obtain
Theorem VII. When $x>0$ and that the $x_{1}$ remain non-negative, expression
(12) $\quad\left(x_{1}x_{2}\ldots x_{n}\right)^{\alpha}e^{-\left(x_{1}+x_{2}+\ldots+x_{n}\right)}\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)$
¹⁰ ) Mr. Fekete established the existence of the limit $n(n-1)$ that the continuation

\left\{\sqrt[n]{\mathrm{M}_{n}(0,0)}\right\}\quad n=2,3,\ldots

is decreasing. $\alpha,\beta$ given, the more general sequence

\left\{V\overline{M_{n}(\alpha,\beta)}\right\}\quad n=k,k+1,\ldots

is also decreasing if we take the number $k$ large enough.
becomes maximum when the $\mathrm{x}_{\mathrm{i}}$ are the zeros of the generalized Laguerre polynomial

L_{n}^{(\alpha)}(x)=\frac{e^{x}x^{1-\alpha}}{n!}\frac{d^{n}}{dx^{n}}e^{-x}x^{\alpha+n-1}

The value of this maximum is

\mathrm{N}_{n}(\alpha)=e^{-n(\alpha+n-1)}\prod_{i=1}^{n}i^{i}(\alpha+i-1)^{\alpha+i-1}.

The passage to the limit is justified by the fact - which is demonstrated as in No. 2 - that expression (12) admits a maximum reached for a single system of points $\mathrm{x}_{\mathrm{i}}\geq 0$ .

If $\alpha=1$ the corresponding problem is solved by the Laguerre polynomial itself.

The case $\alpha=0$ can also be related to the case $\alpha=2$ . Indeed if $\alpha=0$ for (12) to be maximum one of the points must $x_{i}$ coincides with 0 . So making $\alpha=2$ and taking $n-1$ instead of $n$ we deduce the

Theorem VIII. When the $\mathrm{x}_{1}$ remain non-negative, the expression

e^{-\left(x_{1}+x_{2}+\ldots+x_{n}\right)}\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $\mathrm{x}_{1}$ are the zeros of the polynomial

-\frac{e^{x}x}{n!}\frac{d^{n}}{dx^{n}}e^{-x}x^{n-1}=\frac{e^{x}}{n!}\frac{d^{n-1}}{dx^{n-1}}e^{-x}x^{n}=\int_{0}^{x}\mathrm{\penalty 10000\ L}_{n-1}(x)dx=\mathrm{G}_{n}(x)

The value of this maximum is

\mathrm{N}_{n}(0)=e^{-n(n-1)}2^{2}3^{3}\ldots(n-1)^{n-1}2^{2}3^{3}\ldots n^{n}

$\mathrm{L}_{n-1}(x)$ is the Laquerre polynomial

\mathrm{L}_{n-1}(x)=\frac{e^{x}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}e^{-x}x^{n-1}

of degree $n-1$ .
10. Let us make some remarks about numbers. $\mathrm{N}_{n}(\alpha)$ .

If $\xi_{1},\xi_{2},\ldots,\xi_{n}$ are the zeros of $L_{n}^{(\alpha)}(x)$ , We have

\left.\mathrm{N}_{n}(\alpha)\leq\left(\xi_{1}\xi_{2}\ldots\xi_{n}\right)^{\alpha}\mathrm{N}_{n}(0)^{11}\right)

If $\eta_{0}=0,\eta_{1},\eta_{2},\ldots,\eta_{n}$ are the zeros of $G_{n+1}(x)$

\begin{gathered}\mathrm{N}_{n}(\alpha)\geqslant\left(\eta_{1}\eta_{2}\ldots\eta_{n}\right)^{\alpha}e^{-\left(\eta_{1}+\eta_{2}+\ldots+\eta_{n}\right)}\mathrm{V}^{2}\left(\eta_{1},\eta_{2},\ldots,\eta_{n}\right)=\\ =\left(\eta_{1}\eta_{2}\ldots\eta_{n}\right)^{\alpha-2}e^{-\left(\eta_{0}+\eta_{1}+\ldots+\eta_{n}\right)}\mathrm{V}^{2}\left(\eta_{0},\eta_{1},\ldots,\eta_{n}\right)=\\ =\left(\eta_{1}\eta_{2}\ldots\eta_{n}\right)^{\alpha-2}\mathrm{\penalty 10000\ N}_{n+1}(0)\end{gathered}

But, a simple calculation gives us
$\xi_{1}\xi_{2}\ldots\xi_{n}=\alpha(x+1)\ldots(x+n-1),\eta_{1}\eta_{2}\ldots\eta_{n}=(n+1)!$
so
(13) $[(n+1)!]^{\alpha-2}\mathrm{\penalty 10000\ N}_{n+1}(0)\leq\mathrm{N}_{n}(x)[x(\alpha+1)\ldots(\alpha+n-1)]^{\alpha}\mathrm{N}_{n}(0)$ .
¹¹ ) Equality is only possible if $\alpha=0$ .

However, the order of magnitude of the product $2^{2}3^{3}$ . $n^{n}$ East

n^{\frac{n^{2}}{2}}+\frac{n}{2}+\frac{1}{12}e^{-\frac{n^{2}}{4}}

which means that

\begin{gathered}\frac{\sqrt[n(n-1)]{V2^{2}3^{3}\cdots n^{n}}}{\sqrt{n}}\rightarrow\frac{1}{\sqrt[4]{e}}\\ n\rightarrow\infty.\end{gathered}

We deduce that

	$\displaystyle\frac{\sqrt[n(n-1)]{V}\overline{\mathrm{\penalty 10000\ N}_{n}(0)}}{n}$	$\displaystyle\rightarrow\frac{1}{eV\bar{e}}$
	$\displaystyle n\rightarrow\infty$

Let us also note that $\mathrm{N}_{n}(\alpha)$ is not a monotonic function but its logarithm is convex. Also taking into account (13) and $\mathrm{N}_{n+1}(0)=e^{-2n}n^{n}(n+1)^{n+1}\mathrm{\penalty 10000\ N}_{n}(0)$ we have the

Theorem IX. The logarithm of the function $\mathrm{N}_{n}(\alpha)$ is convex for $\alpha\geqslant 0$ . given that we have

\frac{\sqrt[n(n-1)]{\mathrm{N}_{n}(\alpha)}}{n}\rightarrow\frac{1}{e\sqrt{e}}

11.

Let us now take $a=-\sqrt{2\beta,}b=\sqrt{2\beta,}\alpha=\beta$ in Theorem II. We deduce that for $-\sqrt{2\beta}\leq x_{i}\leq\sqrt{2\beta}$ the expression

\left[\prod_{i=1}^{n}\left(1-\frac{x_{i}^{2}}{2\beta}\right)^{\beta}\right]\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $x_{i}$ are the zeros of the polynomial (written in a suitable form)

(-1)^{n}\left(1-\frac{x^{2}}{2\beta}\right)^{1-\beta}\frac{d^{n}}{dx^{n}}\left(1-\frac{x^{2}}{2\beta}\right)^{\beta+n-1}

Doing $\beta\longrightarrow\infty$ we obtain a third theorem stated by Stieltjes ¹² ).
¹² ) Loc. cit. ⁵ ).

Theorem X. The expression

e^{-\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)}\mathrm{V}^{2}\left(x_{1},x_{2},\ldots x_{n}\right)\quad\left(x_{i}\text{ réels }\right)

becomes maximum when the $\mathrm{x}_{\mathrm{i}}$ are the zeros of the Hermite polynomial

\mathrm{H}_{n}(x)=(-1)^{n}e^{\frac{x^{2}}{2}}\frac{d^{n}}{dx^{n}}e^{-\frac{x^{2}}{2}}

The value of this maximum is

\mathrm{N}_{n}^{*}=e^{-\frac{n(n-1)}{2}}2^{2}3^{3}\ldots n^{n}

The passage to the limit, as is easily demonstrated, is perfectly justified.

We have

\frac{\sqrt[n(n-1)]{\mathrm{N}_{n}^{*}}}{\sqrt{\bar{n}}}\rightarrow\frac{1}{\sqrt[4]{\bar{e}^{3}}}

12.

Note that we have

(-1)^{n}n!\mathrm{L}_{n}^{(\alpha)}(x)=x^{n}-n(\alpha+n-1)x^{n-1}+\cdots

We see that the sum of the zeros of the polynomial $L_{n}^{(\alpha)}(x)$ . is equal to $n(\alpha+n-1)$ . Theorem VII then shows us that if $\alpha>0,x_{i}$ are non-negative and

x_{1}+x_{2}+\ldots+x_{n}\leq n(\alpha+n-1)

(14)

the expression

\left(x_{1}x_{2},\ldots x_{n}\right)^{\alpha}\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

reaches its maximum, equal to $\prod_{i=1}^{n}i^{l}(x+i-1)^{x+i-1}$ , when the $x_{i}$ are the zeros of the polynomial $L_{n}^{(\alpha)}(x)$ . We were able to remove the factor $e^{-\left(x_{1}+x_{2}+\ldots+x_{n}\right)}$ since the function $e^{x}$ grows when $x$ increases from the value zero. We see that we must take the equality in (14) to have the maximum.

The substitution $x\mid n(x+n-1),x$ allows us to state the following theorem:

Theorem XI. When $x>0$ , THE $x_{i}$ are non-negative and
the expression

\begin{gathered}x_{1}+x_{2}+\ldots+x_{n}\leq 1\\ \left(x_{1}x_{2}\ldots x_{n}\right)^{\alpha}\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)\end{gathered}

becomes maximum when the $\mathrm{n}(\alpha+\mathrm{n}-1)\mathrm{x}_{\mathrm{i}}$ are the zeros of the generalized Laguerre polynomial $\mathrm{L}_{\mathrm{n}}^{(\alpha)}(\mathrm{x})$ . The value of this maximum is

M_{n}^{*}(\alpha)=\frac{\prod_{i=1}^{n}i^{i}(\alpha+n-1)^{\alpha+n-1}}{[n(\alpha+n-1)]^{n(\alpha+n-1)}}\ldots

We have, in particular,
Theorem XII. When the $\mathrm{x}_{1}$ are non-negative and

x_{1}+x_{2}+\ldots+x_{n}\leq 1

the discriminant

\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $\mathrm{n}(\mathrm{n}-1)\mathrm{x}_{\mathrm{i}}$ are the zeros of the polynomial $\mathrm{G}_{n}(\mathrm{x})$ . The value of this maximum is

M_{n}^{*}(0)=\frac{2^{2}3^{3}\ldots n^{n}2^{2}3^{3}\ldots(n-1)^{n-1}}{[n(n-1)]^{n(n-1)}}

Similarly, noticing that

\mathrm{H}_{n}(x)=x^{n}-\frac{n(n-1)}{2}x^{n-2}+\ldots

we see that the sum of the squares of the zeros of the polynomial $\mathrm{H}_{n}(x)$ is equal to $n(n-1)$ . We then obtain, as above, the

Theorem XIII. When real numbers $\mathrm{x}_{\mathrm{i}}$ verify the inequality

x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\leq 1

the discriminant

\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

becomes maximum when the $V\overline{\mathrm{n}(\mathrm{n}-1)}\mathrm{x}_{1}$ are the zeros of the Hermite polynomial $\mathrm{H}_{\mathrm{n}}(\mathrm{x})$ . The value of this maximum is

\mathrm{M}_{n}^{*}=\frac{2^{2}3^{3}\cdots n^{n}}{[n(n-1)]^{\frac{n(n-1)}{2}}}

Theorems XII, XIII are due to MI Schur who demonstrated them by another route ¹³ ).
13. Exactly as above we demonstrate that:

Theorem XIV. The function $\mathrm{M}_{n}^{*}(x)$ is decreasing and its logarithm is convex for $\alpha\geq 0$ We have the limit

\begin{gathered}n{}_{n\rightarrow\infty}\overline{M_{n}^{*}(\alpha)}\\ n\rightarrow\frac{1}{\sqrt{e}}\end{gathered}

for any given value of $x$ .

⁰⁰footnotetext: ¹³ ) Loc. cit. ⁴ ).

We also have

\begin{gathered}\sqrt{n}^{\frac{n(n-1)}{V}}\overline{\mathrm{M}}_{n}^{*}\longrightarrow\frac{1}{V^{\frac{4}{e}}}\\ n\longrightarrow\infty\end{gathered}

II. Generalization of the Stieltjes problem

14.

Now consider the expression

\mathrm{E}_{n}=\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};f\right)=f\left(x_{1}\right)f\left(x_{2}\right)\ldots f\left(x_{n}\right)\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

which we will seek to study as much as possible when the $x_{i}$ remain in a finite interval. We can always refer to the interval $(0,1)$ , but it goes without saying that we move on to analogous problems for any interval by simple transformations.

Of course, to have interesting properties we will limit ourselves to the case where the function $f(x)$ has a certain particular form. Let us first clarify the assumptions we make about the function $f(x)$ .

We will say that the function $f(x)$ , real, defined and uniform in the closed interval ( 0,1 ) verifies the conditions ( C ) or that it is a function (C), if:
$1^{0}$ . it is positive in the open interval $(0,1)$ .
$2^{0}$ . it is continuous at points 0 and 1.
$3^{0}.f(0)=f(1)=0$ .
$4^{0}$ . it is exponentially concave in the interval $(0,1)$ .
We will also say that $f(x)$ checks the conditions ( $\mathrm{C}^{\prime}$ ) or that it is a function ( $\mathrm{C}^{\prime}$ ) if, apart from the properties $1^{0},2^{0},3^{0},4^{0}$ , it also verifies the following property:
$5^{0}.\mathrm{f}(\mathrm{x})$ has a derivative at every point in the open interval $(0,1)$ .
The property $4^{0}$ is understood in the restrictive sense (and not in Jensen's sense), therefore

f\left(x_{2}\right)>f\left(x_{1}\right)^{\frac{x_{3}-x_{2}}{x_{3}-x_{1}}}f\left(x_{3}\right)^{\frac{x_{2}-x_{1}}{x_{3}-x_{1}}}

(15)

whatever the points $x_{1}<x_{2}<x_{3}$ of the interval $(0,1)$ .
In particular we have

f\left(\frac{x_{1}+x_{2}}{2}\right)>V\overline{f\left(x_{1}\right)f\left(x_{2}\right)}

(16)

for two points $x_{1},x_{2}$ of $(0,1)$ . The function $\log f(x)$ is concave in any interval completely interior to $(0,1)$ .

Well-known properties result that a function (C) is upper bounded in $(0,1)$ and is continuous not only internally but, as a result of the property $2^{0}$ , everywhere in $(0,1)$ .

The problem we propose to examine is the following:
The function $\mathrm{f}(\mathrm{x})$ verifying the conditions (C), determine and study the maximum of the expression $\mathrm{E}_{\mathrm{n}}$ when the $\mathrm{x}_{\mathrm{i}}$ remain in the meantime $(0,1)$ .

We could study this problem also when the function $f(x)$ does not verify conditions (C), but our hypotheses appear to be the most interesting. For example, the function $f(x)$ equal to 1 for $0<x<1$ And $f(0)=f(1)=0$ . This function checks the properties $1^{0},3^{0},4^{0}$ . In this case we have

\max_{(0,1)}\mathrm{E}\left(x_{1}x_{2},\right)=1

but it is clear that this maximum is not reached.
Let us now move on to the study of our problem.
15. The expression $\mathrm{E}_{n}$ is a symmetric function and is continuous in the domain $0\leq x_{1}\leq x_{2}\leq\ldots\leq x_{n}\leq 1$ . It is zero on the boundary of this domain, therefore

Theorem XV. If $\mathrm{f}(\mathrm{x})$ is a function (C) the expression $\mathrm{E}_{\mathrm{n}}$ admits in $(0,1)$ a maximum reached for at least one point system $\mathrm{x}_{\mathrm{i}}$ distinct and located within the interval $(0,1)$ .

Suppose the maximum is reached for two different point systems $x_{i}$

	$\displaystyle 0<y_{1}<y_{2}<\ldots<y_{n}<1,\quad 0<y_{1}^{\prime}<y_{2}^{\prime}<\ldots<y_{n}^{\prime}<1$		(17)
	$\displaystyle\left\|y_{1}-y_{2}^{\prime}\right\|+\left\|y_{2}-y_{2}^{\prime}\right\|+\ldots+\left\|y_{n}-y_{n}^{\prime}\right\|>0.$

Then taking the points

y_{i}^{\prime\prime}=\frac{y_{i}+y_{i}^{\prime}}{2},\quad i=1,2,\ldots,n

and taking into account (16) we deduce, exactly as in No. 2, that

\mathrm{E}\left(y_{1}^{\prime\prime},y_{n}^{\prime\prime},\ldots,y_{n}^{\prime\prime};f\right)\geq\sqrt{\mathrm{E}\left(y_{1},y_{2},\ldots,y_{n};f\right)\mathrm{E}\left(y_{1}^{\prime},y_{2}^{\prime},\ldots,y_{n}^{\prime};f\right)}

equality being possible only if the two sequences (17) coincide. This is in contradiction with our hypotheses. We therefore have the

Theorem XVI. If $\mathrm{f}(\mathrm{x})$ is a function ( C ) the expression $\mathrm{E}_{\mathrm{n}}$ admits in $(0,1)$ a maximum reached for a single points system $\mathrm{x}_{\mathrm{i}}$ distinct and located within the interval $(0,1)$

We will designate by $0<\xi_{n,1}<\xi_{n,2}<\ldots<\xi_{n,n}<1$ the points for which this maximum is reached. We call see system the maximizing system of $\mathrm{E}_{\mathrm{n}}$ . The polynomial

\mathrm{P}_{n}(x)=\left(x-\xi_{n,1}\right)\left(x-\xi_{n,2}\right)\ldots\left(x-\xi_{n,n}\right)

will be called the maximizing polynomial of $\mathrm{E}_{\mathrm{n}}$ .

Finally we will designate by

\mathrm{M}_{n}(f)=\max_{(0,1)}\mathrm{E}_{n}=\mathrm{E}\left(\xi_{n,1},\xi_{n,2},\ldots,\xi_{n,n},f\right)

which is obviously a positive number.
16. We will now demonstrate that the correspondence between a function $(\mathrm{C})$ and its maximizing system or maximizing polynomial is continuous.

We need the following property:
Lemma I. We can match any positive number $\varepsilon$ another positive number $\eta$ such that if
we also have

\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};f\right)-\mathrm{M}_{n}(f)\mid<\eta

\begin{gathered}\left|x_{i}-\xi_{n,i}\right|<\varepsilon\left(x_{1}<x_{2}<\ldots<x_{n}\right)\\ i=1,2,\ldots,n\end{gathered}

This property is a consequence of uniqueness. Let us suppose, in fact, that the property is not true. By classical reasoning we see that there would then exist a positive number $\varepsilon$ and a series of points systems

x_{1}^{(m)}\leq x_{2}^{(m)}\leq\ldots\leq x_{n}^{(m)}\quad,\quad m=1,2,\ldots

such that we have

\left|\mathrm{E}\left(x_{1}^{(m)},x_{2}^{(m)},\ldots,x_{n}^{(m)};f\right)-\mathrm{M}_{n}(f)\right|<\frac{1}{m}\quad,\quad m=1,2,\ldots

and that at least one of the inequalities

\left|x_{i}^{(m)}-\xi_{n,i}\right|\geq\varepsilon,\quad i=1,2,\ldots,n

be checked for everything $m$ . We deduce the existence of a system

\left.x_{1}\leq x_{2}\leq\ldots\leq x_{n}{}^{14}\right)

such as

\begin{gathered}\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};f\right)=\mathrm{M}_{n}(f)\\ \left|x_{1}-\xi_{n,1}\right|+\left|x_{2}-\xi_{n,2}\right|+\ldots+\left|x_{n}-\xi_{n,n}\right|\geq\varepsilon\end{gathered}

which is in contradiction with Theorem XVI.
Now consider two functions (C), $f(x),g(x)$ .
Let $\xi_{n,1}<\xi_{n,2}<\ldots<\xi_{n,n}$ And $\xi_{n,1}^{\prime}<\xi_{n,2}^{\prime}<\ldots<\xi_{n,n}^{\prime}$ the
corresponding maximizing systems. Let A be the upper bound of the function $f(x)$ and B the upper bound of $|f(x)-g(x)|$ ,

\mathrm{A}=\max.f(x),\quad\mathrm{B}=\max.|f(x)-g(x)|

$(0,1)$
$(0,1)$

⁰⁰footnotetext: ¹⁴⁾ We also know that everywhere we have the sign <.

So we have

f(x)\leq g(x)+\mathrm{B},\quad g(x)\leq f(x)+\mathrm{B}

\begin{gathered}f\left(x_{1}\right)f\left(x_{2}\right)\ldots f\left(x_{n}\right)\leq g\left(x_{1}\right)g\left(x_{2}\right)\ldots g\left(x_{n}\right)+\mathrm{C}\\ g\left(x_{1}\right)g\left(x_{2}\right)\ldots g\left(x_{n}\right)\leq f\left(x_{1}\right)f\left(x_{2}\right)\ldots\left(x_{n}\right)+\mathrm{C}\end{gathered}

Or $\mathrm{C}=(\mathrm{A}+\mathrm{B})^{n}-\mathrm{A}^{n}$ .
We deduce from this
$\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};f\right)\leq\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};g\right)+\mathrm{CV}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)$

\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};g\right)\leq\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};f\right)+\mathrm{CV}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)

from where

		$\displaystyle\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};f\right)\leq\mathrm{M}_{n}(g)+\mathrm{CM}_{n}(0)$
		$\displaystyle\mathrm{E}\left(x_{1},x_{2},\ldots,x_{n};g\right)\leq\mathrm{M}_{n}(f)+\mathrm{CM}_{n}(0)$

We have, in particular,

\begin{gathered}M_{n}(f)\leqslant M_{n}(g)+\mathrm{CM}_{n}(0,0),\quad M_{n}(g)\leqslant M_{n}(f)+\mathrm{CM}_{n}(0,0)\\ \therefore\quad\left|M_{n}(f)-M_{n}(g)\right|\leqslant\mathrm{CM}_{n}(0,0)\end{gathered}

Here $M_{n}(0,0)$ is the number defined in $\S$ previous.
We also have

		$\displaystyle\mathrm{E}\left(\xi_{n,1}^{\prime},\xi_{n,2}^{\prime},\ldots,\xi_{n,1}^{\prime};f\right)\leq\mathrm{M}_{n}(g)+\mathrm{CM}_{n}(0)$
		$\displaystyle\mathrm{M}_{n}(g)\leq\mathrm{E}\left(\xi_{n,1}^{\prime},\xi_{n,2}^{\prime},\ldots\xi_{n,n}^{\prime};f\right)+\mathrm{CM}_{n}(0)$
		$\displaystyle\left\|\mathrm{E}\left(\xi_{n,1}^{\prime},\xi_{n,2}^{\prime},\ldots,\xi_{n,n}^{\prime};f\right)-\mathrm{M}_{n}(g)\right\|\leq\mathrm{CM}_{n}(0).$

We deduce that

\left|E\left(\xi_{n,1}^{\prime},\xi_{n,2}^{\prime},\ldots,\xi_{n,n}^{\prime};f\right)-M_{n}(f)\right|\leq 2\mathrm{CM}_{n}(0,0).

Now, C tends to zero with B. Taking into account Lemma I, it results in the

Theorem XVII. We can match any positive number to another positive number $\eta$ such that, for the function (C) $\mathrm{f}(\mathrm{x})$ and for any other function (C) $\mathrm{g}(\mathrm{x})$ which verifies the inequality
we also have

|f(x)-g(x)|<\eta\text{ dans }(0,1)

\left|\xi_{n,i}-\xi_{n,i}^{\prime}\right|<\varepsilon,i=1,2,\ldots,n

for the corresponding maximizing systems.
This is the property we had in mind.
We deduce from it, in particular,
Theorem XVIII. If a sequence of functions (C)

f_{1}(x),f_{2}(x),\ldots,f_{n}(x),\ldots

converges uniformly to a function (C), $\mathrm{f}(\mathrm{x})$ , the maximizing system $\xi_{\mathrm{n},1}^{(\mathrm{m})}<\xi_{\mathrm{n},2}^{(\mathrm{m})}<\ldots<\xi_{\mathrm{n},\mathrm{n}}^{(\mathrm{m})}$ corresponding to $\mathrm{f}_{\mathrm{m}}(\mathrm{x})$ tends, for
$\mathrm{m}\longrightarrow\infty$ , towards the maximizing system $\xi_{\mathrm{n},1}<\xi_{\mathrm{n},2}\ldots<\xi_{\mathrm{n},\mathrm{n}}$ corresponding to $\mathrm{f}(\mathrm{x})^{15}$ ).

\begin{gathered}\xi_{n,i}^{(m)}\longrightarrow\xi_{n,i}\quad i=1,2,\ldots,n\\ m\longrightarrow\infty\end{gathered}

This property will be fundamental for us in what follows.
17. Let us now suppose that $f(x)$ be a function $\left(\mathrm{C}^{\prime}\right)$ . To determine the maximum then we can apply differential calculus. The maximum is therefore determined by the system

\frac{1}{\mathrm{E}_{n}}\frac{a\mathrm{E}_{n}}{dx_{i}}=\frac{\mathrm{P}^{\prime\prime}\left(x_{i}\right)}{\mathrm{P}^{\prime}\left(x_{i}\right)}+\frac{f^{\prime}\left(x_{i}\right)}{f\left(x_{i}\right)},i=1,2,\ldots,n

(18)

P(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right).

(Or)

It is easy to see that, as a result of the assumptions made and especially because of the property $4^{0}$ (concavity), we have the

Lemma II. If $\mathrm{f}(\mathrm{x})$ is a function ( $\mathrm{C}^{\prime}$ ) the system (18) admits one solution and only one such that the $\mathrm{x}_{1}$ are in the interval $(0,1)$ .

Given Theorem XV, we have
Theorem XIX. If $\mathrm{f}(\mathrm{x})$ is a function $\left(\mathrm{C}^{\prime}\right)$ , We have

\begin{gathered}f\left(\xi_{n,i}\right)\mathrm{P}_{n}^{\prime\prime}\left(\xi_{n,i}\right)+\left(\xi_{n,i}\right)\mathrm{P}_{n}^{\prime}\left(\xi_{n,i}\right)=0\\ i=1,2,\ldots,n\end{gathered}

18.

Before going further, we must show how the demonstrations of the properties stated for the functions (C) and which will follow are deduced from the demonstrations of the corresponding properties for the functions ( $\mathrm{C}^{\prime}$ ).

Let us first prove
Lemma III. If

0<x_{1}<x_{2}<\ldots<x_{m}<1

is a sequence of points located inside the interval $(0,1)$ and if

c_{1}>c_{2}>\ldots>c_{m}

is a sequence of non-increasing numbers, there exists a function ' $\left(\mathrm{C}^{\prime}\right)$ whose logarithmic derivative takes the values $\mathrm{c}_{\mathrm{i}}$ at the corresponding points $\mathbf{x}_{1}$ .

Either $\varphi(x)$ the continuous function defined in the interval ( $x_{1},x_{m}$ ) such as

\varphi\left(x_{l}\right)=c_{i},\quad i=1,2,\ldots,m

$\varphi(x)=$ a linear function in $\left(x_{i},x_{i+1}\right),i=1,2,\ldots,m-1$ .
¹⁵⁾ That the limit is a function (C) is not essential, but we will apply this theorem in this form in the following.

The function

\varphi_{1}(x)=\int_{x_{1}}^{x}\varphi(x)dx

is concave in the interval ( $x_{1},x_{m}$ ).
We will now suitably extend this function in the interval $(0,1)$ . Let us take the function $\varphi_{2}(x)$ defined as follows:

\begin{gathered}\varphi_{2}(x)=\left(\left|c_{1}\right|+1-c_{1}\right)\left(x_{1}-x\right)+x_{1}\left(\left|c_{1}\right|+1\right)-\frac{x_{1}^{2}\left(\left|c_{1}\right|+1\right)}{x},0<x\leq x_{1}\\ \varphi_{2}(x)=\varphi_{1}(x),\quad\leq x\leq x_{m}\\ \varphi_{2}(x)=\left(c_{m}-1-\left|c_{m}\right|\right)\left(x-x_{m}\right)+\varphi_{1}\left(x_{m}\right)+\left(1-x_{m}\right)\left(\left|c_{m}\right|+1\right)-\\ -\frac{\left(1-x_{m}\right)^{2}\left(\left|c_{m}\right|+1\right)}{1-x},\quad x_{m}\leq x<1.\end{gathered}

Finally the function $f(x)$ defined as follows

		$\displaystyle f(0)=f(1)=0$
		$\displaystyle f(x)=e^{\Phi_{2}(x)}\quad\text{ pour }0<x<1$

is a function ( $\mathrm{C}^{\prime}$ ) and verifies the conditions of Lemma III.
The extension of the function $\varphi_{1}(x)$ is done in short by connection using arcs of suitably chosen hyperbolas. If $c_{1}$ is non-negative the function $\varphi_{2}(x)$ is increasing in the interval $\left(0,x_{1}\right)$ and if $c_{m}$ is non-positive $\rho_{2}(x)$ is decreasing in the interval ( $x_{m},1$ ). This remark will be useful to us for the following property.

Lemma III has an important consequence:
If $\mathrm{P}_{\mathrm{n}_{1}}(\mathrm{x}),\mathrm{P}_{\mathrm{n}_{2}}(\mathrm{x}),\ldots,\mathrm{P}_{\mathrm{n}_{\mathrm{k}}}(\mathrm{x})$ are the maximizing polynomials corresponding to a function ( C ) and to k values of n , these polynomials are also the maximizing polynomials corresponding to a function ( $\mathrm{C}^{\prime}$ ) and at the same values of n.

The demonstration presents no difficulty and follows immediately from Lemma II and Theorem XIX.

In the following we will apply this property for polynomials $\mathrm{P}_{n}(x),\mathrm{P}_{n-1}(x)$ .

We still have
Lemma IV. Every function (C) is the limit of a sequence of functions ( $\mathrm{C}^{\prime}$ ) converging uniformly in the interval ( 0,1 ).

First there is a positive number $\rho$ such that in the interval $(0,\rho)$ the function $f(x)$ is non-decreasing and in the interval ( $1-p,1$ ) non-increasing. (We take of course $\rho\leq\frac{1}{2}$ ). Now let's $\varepsilon$ an arbitrary positive number. There exists a positive number $\delta<\rho$ such as

f(x)<\varepsilon

For $0\leq x\leq\delta$ , and for $1-\delta\leq x\leq 1$ .

In the closed interval ( $\delta,1-\delta$ ) the function $\mathrm{F}(x)=\log f(x)$ is continuous and concave. There exists a number A such that in this interval $|\mathrm{F}(x)|\leq\mathrm{A}$ . Now consider the sequence of polynomials ¹⁶ )

\mathrm{Q}_{1}(x),\mathrm{Q}_{2}(x),\ldots,\mathrm{Q}_{m}(x),\ldots

(19)

$\mathrm{Q}_{m}(x)=\frac{1}{(1-2\delta)^{m}}\sum_{i=1}^{m}\mathrm{\penalty 10000\ F}\left(\delta-i\frac{1-2\delta}{m}\right)\binom{m}{i}(x-\delta)^{i}(1-\delta-x)^{m-i}$
which verify the following properties:
$1^{0}.\quad\mathrm{Q}_{m}(\delta)=\mathrm{F}(\delta),\mathrm{Q}_{m}(1-\delta)=\mathrm{F}(1-\delta)$
$2^{0}.\left|\mathrm{Q}_{m}(x)\right|\leq\mathrm{A}$ , In $(\delta,1-\delta)$
$3^{0}$ . $\quad\mathrm{Q}_{m}(x)$ is concave in ( $\delta,1-\delta$ )
4. $\quad\mathbf{Q}_{m}^{\prime}(\delta)=\frac{m}{1-2\delta}\left[\mathrm{\penalty 10000\ F}\left(\delta+\frac{1-2\delta}{m}\right)-\mathrm{F}(\delta)\right]$

\mathrm{Q}_{m}^{\prime}(1-\delta)=\frac{m}{1-2\delta}\left[\mathrm{\penalty 10000\ F}(1-\delta)-\mathrm{F}\left(1-\delta-\frac{1-2\delta}{m}\right)\right]

$5^{0}$ . The sequence (19) converges uniformly to $\mathrm{F}(x)$ throughout the interval ( $\delta,1-\delta$ ).

We extend the polynomial $\mathrm{Q}_{m}(x)$ exactly as we did with $\Phi_{1}(x)$ and we designate by $Q_{m}^{*}(x)$ the function thus defined in the entire open interval (0,1). Finally, let us define the function $f_{m}(x)$ by

		$\displaystyle f_{m}(0)=f_{m}(1)=0$
		$\displaystyle f_{m}(x)=e^{Q^{*}m(x)}\quad,\quad 0<x<1$

The functions $f_{m}(x)$ check the conditions ( $\mathrm{C}^{\prime}$ ).
For $m>\frac{\rho-\delta}{1-2\delta}$ We have $Q_{m}^{\prime}(\delta)\geq 0,Q_{m}^{\prime}(1-\delta)\leq 0$ , SO $f_{m}(x)$ is respectively increasing and decreasing in the intervals $(0,\delta)$ , ( $1-\delta,1$ ), therefore a fortiori

f_{m}(x)<\varepsilon\text{ pour }0\leq x\leq\delta,1-\delta\leq x\leq 1,m>\frac{p-\delta}{1-2\delta}

On the other hand, we can determine a number $m^{\prime}$ such as for $m>m^{\prime}$ we have

\left|\mathrm{Q}_{m}(x)-\mathrm{F}(x)\right|<\log\left(\varepsilon+e^{\mathrm{A}}\right)-\mathrm{A},\quad\delta\leq x\leq 1-\delta

¹⁰ ) Polynomials $\mathrm{Q}_{m}(x)$ are the MS Bernstein polynomials. For the proof of the properties of the text see: Tiberiu Popoviciu „On the approximation of convex functions of higher order" Mathematica t. X (1935), p. 49.

We deduce from this

\left|f_{m}(x)-f(x)\right|<\varepsilon,\quad\delta\leq x\leq 1-\delta,\quad m>m^{\prime}

So finally
$\left|f_{m}(x)-f(x)\right|<\varepsilon$ For $0\leq x\leq 1,m>\max.\left(m^{\prime},\frac{\rho-\delta}{1-2o}\right)$
which proves Lemma IV.
19. We will say that two sequences

x_{1}<x_{2}<\ldots<x_{n}\text{ et }x_{1}^{\prime}<x_{2}^{\prime}<\ldots<x_{n-1}^{\prime}

(20)

of $n$ and of $n-1$ points respectively, separate in the strict sense if we have

x_{1}<x_{1}^{\prime}<x_{2}<x_{2}^{\prime}<\ldots<x_{n-1}<x_{n-1}^{\prime}<x_{n}

We will also say that the two sequences (20) separate in the broad sense if we have

x_{1}\leq x_{1}^{\prime}\leq x_{2}\leq x_{2}^{\prime}\leq\ldots\leq x_{n-1}\leq x_{n-1}^{\prime}\leq x_{n}

and if the sequences have at least one point in common.
We can also say that the zeros of the polynomials

\mathrm{P}(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\ldots\left(x-x_{n}\right),\mathrm{P}_{1}(x)=\left(x-x_{1}^{\prime}\right)\left(x-x_{2}^{\prime}\right)\ldots\left(x-x_{n-1}^{\prime}\right)

separate respectively in the strict and broad senses.
The polynomial $\mathrm{P}_{1}(x)$ can always be written in the form

\mathrm{P}_{1}(x)=\mathrm{P}(x)\sum_{i=1}^{n}\frac{s_{i}}{x-x_{i}}\quad\left(\sum_{i=1}^{n}s_{i}=1\right)

which is none other than the Lagrange interpolation formula. The constants $s_{i}$ are completely determined.

We have
Lemma V. The necessary and sufficient condition for the zeros of the polynomials $\mathrm{P}(\mathrm{x})$ And $\mathrm{P}_{1}(\mathrm{x})$ separate in the strict sense is that the coefficients si are all positive. The necessary and sufficient condition for the zeros of these polynomials to separate in the broad sense is that the si are all non-negative, at least one being zero.

These properties are well known, the demonstration being immediate. Let us now return to the maximizing polynomials $\mathrm{P}_{n}(x),\mathrm{P}_{n-1}(x)$ corresponding to a function ( $\mathrm{C}^{\prime}$ ). We have the

Theorem XX. The Zeros of Maximizing Polynomials $\mathrm{P}_{\mathrm{n}}(\mathrm{x})$ , $\mathrm{P}_{\mathrm{n}-1}(\mathrm{x})$ corresponding to a function ( $\mathrm{C}^{\prime}$ ) cannot separate in the broad sense.

The proof is easy. Let's assume the opposite. We can then write

\mathrm{P}_{n-1}(x)=\mathrm{P}_{n}(x)\sum_{i=1}^{n}\frac{s_{i}}{x-\xi_{n,i}}

Or $s_{i}\geq 0,\sum_{i=1}^{n}s_{i}=1$ . To fix the ideas, let us suppose that $s_{1}=0$ . We then have $\xi_{n,1}=\xi_{n-1,1}$ . Drifting twice in a row we have

\mathrm{P}_{n-1}^{\prime}(x)=\mathrm{P}_{n}^{\prime}(x)\sum_{i=1}^{n}\frac{s_{i}}{x-\xi_{n,i}}-\mathrm{P}_{n}(x)\sum_{i=1}^{n}\frac{s_{i}}{\left(x-\xi_{n,i}\right)^{2}}

	$\displaystyle\mathrm{P}_{n-1}^{\prime\prime}(x)=\mathrm{P}_{n}^{\prime\prime}(x)\sum_{i=1}^{n}\frac{s_{i}}{x-\xi_{n,i}}-2\mathrm{P}_{n}^{\prime}(x)\sum_{i=1}^{n}$	$\displaystyle\frac{s_{i}}{\left(x-\xi_{n,i}\right)^{2}}+$
		$\displaystyle+2\mathrm{P}_{n}(x)\sum_{i=1}^{n}\frac{s_{i}}{\left(x-\xi_{n,i}\right)^{3}}$

from where

\begin{gathered}\mathrm{P}_{n-1}^{\prime}\left(\xi_{n,1}\right)=\mathrm{P}_{n}^{\prime}\left(\xi_{n,1}\right)\sum_{i=2}^{n}\frac{s_{i}}{\xi_{n,1}-\xi_{n,i}}\\ \mathrm{P}_{n-1}^{\prime\prime}\left(\xi_{n,1}\right)=\mathrm{P}_{n}^{\prime\prime}\left(\xi_{n,1}\right)\sum_{i=2}^{n}\frac{s_{i}}{\xi_{n,1}-\xi_{n,i}}-2\mathrm{P}_{n}^{\prime}\left(\xi_{n,1}\right)\sum_{i=2}^{n}\frac{s_{i}}{\left(\xi_{n,1}-\xi_{n,i}\right)^{2}}\end{gathered}

Taking into account $\xi_{n,1}=\xi_{n,1}$ and from Theorem XIX we deduce

f\left(\xi_{n,1}\right)\mathrm{P}_{n}^{\prime}\left(\xi_{n,1}\right)\sum_{i=2}^{n}\frac{s_{i}}{\left(\xi_{n,1}-\xi_{n,l}\right)^{2}}=0

But, $f\left(\xi_{n,1}\right)\mathrm{P}_{n}^{\prime}\left(\xi_{n,1}\right)\neq 0$ , it therefore follows that

s_{2}=s_{3}=\ldots=s_{n}=0

This is in contradiction with the hypothesis $\sum_{i=1}^{n}s_{i}=1$ and Theorem XX is therefore demonstrated.
20. We now propose to demonstrate the property that we had in mind and which is expressed by the

Theorem XXI. The zeros of two consecutive maximizing polynomials $\mathrm{P}_{\mathrm{n}}(\mathrm{x}),\mathrm{P}_{\mathrm{n}-1}(\mathrm{x})$ corresponding to a function $(\mathrm{C})$ separate in the strict sense.

Let us first show that it is sufficient to demonstrate the property for functions $\left(\mathrm{C}^{\prime}\right)$ . Indeed, if the property is true for functions $\left(\mathrm{C}^{\prime}\right)$ , Theorem XVIII and Lemma IV show us that for a function (C) the zeros separate either in the strict sense or in the broad sense. But, by virtue of the consequence of Lemma III, it is indeed the first case that occurs.

It remains to demonstrate ownership in the case where $f(x)$ is a function ( $\mathrm{C}^{\prime}$ ). Let us consider for this the function

\begin{gathered}f_{m}(x)=[x(1-x)]^{1-m}[f(x)]^{m}\\ 0\leq m\leq 1,f_{0}(x)=x(1-x),f_{1}(x)=f(x)\end{gathered}

$f_{m}(x)$ is a function $\left(\mathrm{C}^{\prime}\right)$ . When $m$ increases from 0 to 1, the function $f_{0}(x)$ Or $x(1-x)$ deforms continuously and tends uniformly towards $f_{1}(x)$ Or $f(x)$ .

Theorems XVIII and XX then show us that it is sufficient to demonstrate the property for the function $\mathrm{f}_{0}(\mathrm{x})=\mathrm{x}(1-\mathrm{x})$ . Now, in this case

\mathrm{P}_{n}(x)=\mathrm{F}(-n,n+1,1,x),\mathrm{P}_{n-1}(x)=\mathrm{F}(-n+1,n,1,x)

and it is well known that the zeros of these Jacobi polynomials separate in the strict sense. The theorem is therefore completely demonstrated.
21. We now propose to say a few words about the maximum $M_{n}(f)$ .

Let A be the maximum (or the upper bound) of $f(x)$ . We have

\mathrm{M}_{n}(f)\leqslant\mathrm{A}^{n}.\max_{(0,1)}\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)=\mathrm{A}^{n}\mathrm{M}_{n}(0,0)

(0.1)

MM Fekete's theorem therefore shows us that
(21)

\varlimsup_{n\rightarrow\infty}\sqrt[n(n-1)]{\lim_{n}(f)}\leq\frac{1}{4}

Either $\varepsilon<\frac{1}{4}$ an arbitrary positive number. In the interval $(2\varepsilon,1-2\varepsilon)$ the function $f(x)$ has a positive minimum $a$ . Let $\xi_{1},\xi_{2},\ldots,\xi_{n}$ the points for which

		$\displaystyle\quad\max.\mathrm{V}^{2}\left(x_{1},x_{2},\ldots,x_{n}\right)=m_{n}(\varepsilon)$
		$\displaystyle(2\varepsilon 1-2\varepsilon)$

is reached. We have

f\left(\xi_{1}\right)f\left(\xi_{2}\right)\ldots f\left(\xi_{n}\right)m_{n}(\varepsilon)\leq\mathrm{E}\left(\xi_{1},\xi_{2},\ldots,\xi_{n};f\right)\leq M_{n}(f)

from where

a^{n}m_{n}(\varepsilon)\leq M_{n}(f)

But

\begin{gathered}\sqrt[n(n-1)]{a^{n}}\longrightarrow 1,\text{ pour }n\longrightarrow\infty\\ \sqrt[n(n-1)]{m_{n}(\varepsilon)}\longrightarrow\frac{1}{4}-\varepsilon,\text{ pour }n\longrightarrow\infty\end{gathered}

and it follows that

\underset{n\rightarrow\infty}{\lim.}\sqrt[n(n-1)]{\mathrm{M}_{n}(f)}\geq\frac{1}{4}-\varepsilon

(22)

however small $\varepsilon$ . Relations (21) and (22) therefore show us that:

Theorem XXII. If $\mathrm{f}(\mathrm{x})$ is a function (C) the expression

\sqrt[n(n-1)]{\mathrm{M}_{n}(f)}

a, for $\mathrm{n}\longrightarrow\infty$ , a limit which is equal to the transfinite diameter $\frac{1}{4}$ of the segment 0–1.
22. We will now consider a special case.

Either
$f(x)=\left|x-a_{1}\right|^{\sigma_{1}}\left|x-a_{2}\right|^{\sigma_{2}}\ldots\left|x-a_{h}\right|^{\sigma_{h}}\left|x-a_{h+1}\right|^{\sigma_{h+1}}\ldots\left|x-a_{h+k}\right|^{\sigma_{h+k}}$ Or $a_{1}<a_{2}<\ldots<a_{h-1}<a_{h}=0<a_{h+1}=1<a_{h+2}<\ldots<a_{h+k}$ And $\sigma_{1},\sigma_{2},\ldots,\sigma_{h+k}$ are $h+k$ positive numbers. It is a function ( $\mathrm{C}^{\prime}$ ). The maximizing polynomial $\mathrm{P}_{n}(x)$ is then a solution to the homogeneous linear differential equation of order 2

	$\displaystyle y^{\prime\prime}+\left(\frac{\sigma_{1}}{x-a_{1}}+\frac{\sigma_{2}}{x-a_{2}}+\ldots+\frac{\sigma_{h+k}}{x-a_{h+k}}\right)y^{\prime}+\ldots$		(23)
	$\displaystyle\cdots+\frac{\psi(x)}{\left(x-a_{1}\right)\left(x-a_{2}\right)\ldots\left(x-a_{h+k}\right)}y=0$

Or $\psi(x)$ is a polynomial of degree $h+k-2$ suitably chosen. Our results therefore allow us to say that an equation of the form (23) admits a polynomial solution of degree $n$ and only one such that this polynomial has all its real zeros and is included in one of the intervals ( $a_{i},a_{i+1}$ ). Either $\mathrm{P}_{n}(x)$ the degree solution $n$ which has its zeros in the interval ( $a_{i},a_{i+1}$ ) And $\mathrm{P}_{n-1}(x)$ the degree solution $n-1$ which has its zeros in the same interval. We can then say that the zeros of the polynomials $\mathrm{P}_{\mathrm{n}}(\mathrm{x})$ And $\mathrm{P}_{\mathrm{n}-1}(\mathrm{x})$ separate in the strict sense.

It was Stieltjes who first studied the polynomial solutions of equation (23) ¹⁷ ). He showed, taking into account that the equation has at most

\binom{n+h+k-2}{h+k-2}=\frac{(n+1)(n+2)\cdots(n+h+k-2)}{1.2\cdots(h+k-2)}

polynomial solutions of degree $n$ , that this equation has exactly $\binom{n+h+k-2}{h+k-2}$ solutions of this form which have all their zeros real and included in the interval ( $a_{1},a_{h+k}$ ). The zeros of these polynomials are distributed in all possible ways in the intervals ( $a_{i},a_{i+1}$ ), to each distribution corresponding only one solution. Moreover to each solution corresponds a maximum problem. These maxima problems, which constitute a generalization of the problem studied in this work, as well as other problems on the distribution of the zeros of the polynomial solutions of an equation of the form (23) will be examined by us in another work which will appear soon ¹⁸ ).

⁰⁰footnotetext: ¹⁷ ) Loc. cit. ³ )
18) We have stated more general results in our note: "On a problem of maximum Stielties". CR Acad. Sc., Paris, t. 202 (1936) p. 1645.

On some maximum problems of Stieltjes

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ON SOME PROBLEMS OF MAXIMUM STIELTJES

On some theorems of Stieltjes and MI Schur

II.

Generalization of the Stieltjes problem

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