Posts by Tiberiu Popoviciu

Tiberiu Popoviciu

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T. Popoviciu, Remarque sur les polynômes de meilleure approximation, Mathematica, 4 (1930), pp. 73-80 (in French).

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Note: this paper is a republishing of:  T. Popoviciu, Remarque sur les polynômes de meilleure approximation, Buletinul Soc. de Ştiinţe din Cluj, 5 (1930), pp.279-286 (in French)

[JFM 56.0933.01]

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1930 b -Popoviciu- Mathematica - Remark on best approximation polynomials
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NOTE ON BEST APPROXIMATION POLYNOMIALS

byTiberiu PopoviciuStudent at the Ecole Normale Supérieure, Paris.

Received on February 21, 1930.
  1. We consider the class (b) of functions defined and continuous in the finite and closed interval ( has , b ) ; ( has < b ) ( has , b ) ; ( has < b ) (a,b);(a < b)(a, b); (a<b)(has,b);(has<b). Properties of - these functions, let us remember that they can vanish at a finite or infinite number of points, but the set of these points always forms a finite number of intervals which can also be reduced to a single point.
Let us now consider a system of n + 1 n + 1 n+1n+1n+1functions of class (b)
(1)
f 1 , f 2 , … , f n f 1 , f 2 , … , f n f_(1),f_(2),dots,f_(n)f_{1}, f_{2}, \ldots, f_{n}f1,f2,…,fn
linearly independent, that is to say, an equality such that
(2) P = c 0 f 0 + c 1 f 1 + ⋯ + c n f n = 0 (2) P = c 0 f 0 + c 1 f 1 + ⋯ + c n f n = 0 {:(2)P=c_(0)f_(0)+c_(1)f_(1)+cdots+c_(n)f_(n)=0:}\begin{equation*} \mathrm{P}=c_{0} f_{0}+c_{1} f_{1}+\cdots+c_{n} f_{n}=0 \tag{2} \end{equation*}(2)P=c0f0+c1f1+⋯+cnfn=0
c 0 , c 1 , … , c n c 0 , c 1 , … , c n c_(0),c_(1),dots,c_(n)c_{0}, c_{1}, \ldots, c_{n}c0,c1,…,cnbeing constants, cannot take place everywhere in ( has , b ) ( has , b ) (a,b)(a, b)(has,b)that if
c 0 = c 1 = … = c n = 0 . c 0 = c 1 = … = c n = 0 . c_(0)=c_(1)=dots=c_(n)=0.c_{0}=c_{1}=\ldots=c_{n}=0 .c0=c1=…=cn=0.
Note that system (1) may not be linearly independent in an interval ( α , β ) ( α , β ) (alpha, beta)(\alpha, \beta)(α,β)interior to ( has , b ) ( has , b ) (a,b)(a, b)(has,b)[such as | has − α | + | β − b | ≠ 0 ] | has − α | + | β − b | ≠ 0 ] |a-alpha|+|beta-b|!=0]|a-\alpha|+|\beta-b| \neq 0]|has−α|+|β−b|≠0].
Any homogeneous linear expression such as (2) will be called a polynomial of the system (1). To any polynomial P we can associate a point M M MMMcoordinates c 0 , c 1 , … , c n c 0 , c 1 , … , c n c_(0),c_(1),dots,c_(n)c_{0}, c_{1}, \ldots, c_{n}c0,c1,…,cnin ordinary space n + 1 n + 1 n+1n+1n+1dimensions. We call the point M the image of the corresponding polynomial P.
The fact that the functions (1) are linearly independent can be expressed as follows:
He y y yyya one-to-one correspondence between polynomials P P PPPand their images M M MMM.
For the points where P P PPPIf this is cancelled, we make the following agreements: 1 1 ^(1){ }^{\mathbf{1}}1)
1 0 1 0 1^(0)1^{0}10We call a simple zero a point inside the interval ( has , b has , b a,ba, bhas,b), where P vanishes by changing sign, or else an endpoint ( has has hashashasor b) where P vanishes.
20. We call a double zero, and we count it as two single zeros, an interior point where P P PPPcancels out without changing sign.
A system (1) is a Chebyshev system or system (T). if any polynomial P P PPPcannot have more n n nnnzeros. It follows that for a system not to be a system (T), it is necessary and sufficient that there be at least one polynomial having at least n + 1 n + 1 n+1n+1n+1zeros.
For example, the system
1 , x , x 2 , … , x n 1 , x , x 2 , … , x n 1,x,x^(2),dots,x^(n)1, x, x^{2}, \ldots, x^{n}1,x,x2,…,xn
is a system (T) in any finite interval. Conversely, the system
1 , si ⁡ x , cos ⁡ x , … , si ⁡ n x , cos ⁡ n x 1 , si ⁡ x , cos ⁡ x , … , si ⁡ n x , cos ⁡ n x 1,sin x,cos x,dots,sin nx,cos nx1, \sin x, \cos x, \ldots, \sin nx, \cos nx1,si⁡x,cos⁡x,…,si⁡nx,cos⁡nx
is not a system (T) in every interval ( 0 , 2 π 0 , 2 π 0.2ft0.2 \pi0,2π), but there is one in an interval of shorter length than 2 π 2 π 2ft2 π2π2.
We will demonstrate the following auxiliary property:
If (1) is not a system (T) we can find n + 1 n + 1 n+1n+1n+1distinct points x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, ..., x_{n+1}x1,x2,…,xn+1completely inside ( has , b has , b a,ba, bhas,b) such as the determinant:
(3) | f 0 ( x 1 ) f 1 ( x 1 ) … f n ( x 1 ) f 0 ( x 2 ) f 1 ( x 2 ) … f n ( x 2 ) ⋅ ⋅ ⋅ ⋅ f 0 ( x n + 1 ) f 1 ( x n + 1 ) … f n ( x n + 1 ) | (3) f 0 x 1 f 1 x 1 … f n x 1 f 0 x 2 f 1 x 2 … f n x 2 ⋅ ⋅ ⋅ ⋅ f 0 x n + 1 f 1 x n + 1 … f n x n + 1 {:(3)|[f_(0)(x_(1)),f_(1)(x_(1)),dots,f_(n)(x_(1))],[f_(0)(x_(2)),f_(1)(x_(2)),do ts,f_(n)(x_(2))],[*,*,*,*],[f_(0)(x_(n+1)),f_(1)(x_(n+1)),dots,f_(n)(x_(n+1))]|:}\left|\begin{array}{cccc} f_{0}\left(x_{1}\right) & f_{1}\left(x_{1}\right) & \ldots & f_{n}\left(x_{1}\right) \tag{3}\\ f_{0}\left(x_{2}\right) & f_{1}\left(x_{2}\right) & \ldots & f_{n}\left(x_{2}\right) \\ \cdot & \cdot & \cdot & \cdot \\ f_{0}\left(x_{n+1}\right) & f_{1}\left(x_{n+1}\right) & \ldots & f_{n}\left(x_{n+1}\right) \end{array}\right|(3)|f0(x1)f1(x1)…fn(x1)f0(x2)f1(x2)…fn(x2)⋅⋅⋅⋅f0(xn+1)f1(xn+1)…fn(xn+1)|
is zero.
Indeed, there exists a polynomial having at least n + 1 n + 1 n+1n+1n+1zeros. If such a polynomial vanishes at at least n + 1 n + 1 n+1n+1n+1distinct points, none of which coincides with an endpoint, the property is obvious.
The points where P is zero can be placed into 3 categories.
1 0 1 0 1^(0)1^{0}10Or P becomes zero by changing sign; let
(4) x 1 , x 2 , … , x m x 1 , x 2 , … , x m quadx_(1),x_(2),dots,x_(m)\quad x_{1}, x_{2}, \ldots, x_{m}x1,x2,…,xm
these points.
2 0 2 0 2^(0)2^{0}20Or P vanishes without changing sign; let
(5)
x 1 ′ , x 2 ′ , … , x k ′ x 1 ′ , x 2 ′ , … , x k ′ x_(1)^('),x_(2)^('),dots,x_(k)^(')x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{k}^{\prime}x1′,x2′,…,xk′
(1) S. Bernstein "Lectures on Extremal Properties ... etc." p. 1; J.W. Young "General Theory of Approximation by Functions Involving a Given Number of Arbitrary Parameters" Transactions of the Amer. Math. Soc. 8 (1907), p. 331.
These points. Finally:
3 0 3 0 3^(0)3^{0}30The extremes where P vanishes
x 1 ′ ′ , x 2 ′ ′ , … , x t ′ ′ ( i = 0 , 1 , 2 ) x 1 ′ ′ , x 2 ′ ′ , … , x t ′ ′ ( i = 0 , 1 , 2 ) x_(1)^(''),x_(2)^(''),dots,x_(t)^('')quad(i=0,1,2)x_{1}^{\prime \prime}, x_{2}^{\prime \prime}, \ldots, x_{t}^{\prime \prime} \quad(i=0,1,2)x1′′,x2′′,…,xt′′(i=0,1,2)
We have by hypothesis:
Suppose that:
therefore a fortiori
+ 2 k + i ≥ n + + 2 k + i ≥ n + +2k+i >= n++2 k+i \geq n++2k+i≥n+
m + k < n + 1 m + k < n + 1 m+k < n+1m+k<n+1m+k<n+1
k < n + 1 k < n + 1 k < n+1k<n+1k<n+1
Let us first assume that
and consider the table:
(7) | f 0 ( x ′ 1 ) f 1 ( x ′ 1 ) … f n ( x ′ 1 ) f 0 ( x ′ 2 ) f 1 ( x ′ 2 ) … f n ( x ′ 2 ) ⋮ f 0 ( x ′ k ) f 1 ( x ′ k ) … f n ( x ′ k ) f 0 ( x ′ ′ 1 f 1 ( x ′ ′ 1 ) … f n ( x ′ ′ 1 ) ⋅ ⋅ ⋅ ⋅ f 0 ( x ′ ′ i ) f 1 ( x ′ ′ i ) … f n ( x ′ ′ i ) | (7) f 0 x ′ 1 f 1 x ′ 1 … f n x ′ 1 f 0 x ′ 2 f 1 x ′ 2 … f n x ′ 2 ⋮ f 0 x ′ k f 1 x ′ k … f n x ′ k f 0 x ′ ′ 1 f 1 x ′ ′ 1 … f n x ′ ′ 1 ⋅ ⋅ ⋅ ⋅ f 0 x ′ ′ i f 1 x ′ ′ i … f n x ′ ′ i {:(7)|[f_(0)(x^(')_(1)),f_(1)(x^(')_(1)),dots,f_(n)(x^(')_(1))],[f_(0)(x^(')_(2)) ,f_(1)(x^(')_(2)),dots,f_(n)(x^(')_(2))],[vdots,,,],[f_(0)(x^(')_(k)),f_(1)(x^(')_ (k)),dots,f_(n)(x^(')_(k))],[f_(0)(x^('')_(1):},f_(1)(x^('')_(1)),dots,f_(n)(x^(' ')_(1))],[*,*,*,*],[f_(0)(x^('')_(i)),f_(1)(x^('')_(i)),dots,f_(n)(x^('')_(i))]|:}\left|\begin{array}{cccc} f_{0}\left(x^{\prime}{ }_{1}\right) & f_{1}\left(x^{\prime}{ }_{1}\right) & \ldots & f_{n}\left(x^{\prime}{ }_{1}\right) \tag{7}\\ f_{0}\left(x^{\prime}{ }_{2}\right) & f_{1}\left(x^{\prime}{ }_{2}\right) & \ldots & f_{n}\left(x^{\prime}{ }_{2}\right) \\ \vdots & & & \\ f_{0}\left(x^{\prime}{ }_{k}\right) & f_{1}\left(x^{\prime}{ }_{k}\right) & \ldots & f_{n}\left(x^{\prime}{ }_{k}\right) \\ f_{0}\left(x^{\prime \prime}{ }_{1}\right. & f_{1}\left(x^{\prime \prime}{ }_{1}\right) & \ldots & f_{n}\left(x^{\prime \prime}{ }_{1}\right) \\ \cdot & \cdot & \cdot & \cdot \\ f_{0}\left(x^{\prime \prime}{ }_{i}\right) & f_{1}\left(x^{\prime \prime}{ }_{i}\right) & \ldots & f_{n}\left(x^{\prime \prime}{ }_{i}\right) \end{array}\right|(7)|f0(x′1)f1(x′1)…fn(x′1)f0(x′2)f1(x′2)…fn(x′2)⋮f0(x′k)f1(x′k)…fn(x′k)f0(x′′1f1(x′′1)…fn(x′′1)⋅⋅⋅⋅f0(x′′i)f1(x′′i)…fn(x′′i)|
which contains at least as many columns as rows.
If in table (7) all the determinants formed by k + i k + i k+ik+ik+iIf any columns are zero, the desired property follows. But suppose there is a non-zero determinant, for example:
(8) | f 0 ( x 1 ′ ) f 1 ( x 1 ′ ) … f k + i ( x 1 ′ ) f 0 ( x 2 ′ ) f 1 ( x 2 ′ ) … f k + i ( x 2 ′ ) ⋅ ⋅ ⋅ ⋅ f 0 ( x i ′ ′ ) f 1 ( x i ′ ′ ) … f k + i ( x i ′ ′ ) | ≠ 0 (8) f 0 x 1 ′ f 1 x 1 ′ … f k + i x 1 ′ f 0 x 2 ′ f 1 x 2 ′ … f k + i x 2 ′ ⋅ ⋅ ⋅ ⋅ f 0 x i ′ ′ f 1 x i ′ ′ … f k + i x i ′ ′ ≠ 0 {:(8)|[f_(0)(x_(1)^(')),f_(1)(x_(1)^(')),dots,f_(k+i)(x_(1)^('))],[f_(0)(x_(2)^(')),f_(1)(x_(2)^(')),dots,f_(k+i)(x_(2)^('))],[*,*,*,*],[f_(0)(x_(i)^('')),f_(1)(x_(i)^('')),dots,f_(k+i)(x_(i)^(''))]|!=0:}\left|\begin{array}{cccc} f_{0}\left(x_{1}^{\prime}\right) & f_{1}\left(x_{1}^{\prime}\right) & \ldots & f_{k+i}\left(x_{1}^{\prime}\right) \tag{8}\\ f_{0}\left(x_{2}^{\prime}\right) & f_{1}\left(x_{2}^{\prime}\right) & \ldots & f_{k+i}\left(x_{2}^{\prime}\right) \\ \cdot & \cdot & \cdot & \cdot \\ f_{0}\left(x_{i}^{\prime \prime}\right) & f_{1}\left(x_{i}^{\prime \prime}\right) & \ldots & f_{k+i}\left(x_{i}^{\prime \prime}\right) \end{array}\right| \neq 0(8)|f0(x1′)f1(x1′)…fk+i(x1′)f0(x2′)f1(x2′)…fk+i(x2′)⋅⋅⋅⋅f0(xi′′)f1(xi′′)…fk+i(xi′′)|≠0
At each point x ′ ; x ′ ; x^(');x^{\prime} ;x′;And x ′ ′ ; x ′ ′ ; x^('');x^{\prime \prime} ;x′′;let's assign a non-zero number a ′ ; a ′ ; a^(');a^{\prime} ;has′;resp. a ′ ′ a ′ ′ a^('')a^{\prime \prime}has′′; such that its sign is that of the polynomial P in the neighborhood of x ′ j x ′ j x^(')_(j)x^{\prime}{ }_{j}x′jresp. x ′ ′ x ′ ′ x^('')x^{\prime \prime}x′′[The neighborhood of an endpoint is counted only towards the interior of the interval]. We can then determine a system of non-zero numbers
such that
λ 0 , λ 1 , … , λ k + 1 λ 0 , λ 1 , … , λ k + 1 lambda_(0),lambda_(1),dots,lambda_(k+1)\lambda_{0}, \lambda_{1}, \ldots, \lambda_{k+1}λ0,λ1,…,λk+1
λ 0 f 0 ( x 1 ′ ) + λ 1 f 1 ( x 1 ′ ) + ⋯ + λ k + i f k + i ( x 1 ′ ) = a 1 ′ λ 0 f 0 ( x 2 ′ ) + λ 1 f 1 ( x 2 ′ ) + ⋯ + λ k + i f k + i ( x 2 ′ ) = a 2 ′ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ λ 0 f 0 ( x i ′ ′ ) + λ 1 f 1 ( x i ′ ′ ) + ⋯ + λ k + i f k + i ( x i ′ ′ ) = a i ′ ′ . λ 0 f 0 x 1 ′ + λ 1 f 1 x 1 ′ + ⋯ + λ k + i f k + i x 1 ′ = a 1 ′ λ 0 f 0 x 2 ′ + λ 1 f 1 x 2 ′ + ⋯ + λ k + i f k + i x 2 ′ = a 2 ′ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ λ 0 f 0 x i ′ ′ + λ 1 f 1 x i ′ ′ + ⋯ + λ k + i f k + i x i ′ ′ = a i ′ ′ . {:[lambda_(0)f_(0)(x_(1)^('))+lambda_(1)f_(1)(x_(1)^('))+cdots+lambda_(k+i)f_(k+i)(x_(1)^('))=a_(1)^(')],[lambda_(0)f_(0)(x_(2)^('))+lambda_(1)f_(1)(x_(2)^('))+cdots+lambda_(k+i)f_(k+i)(x_(2)^('))=a_(2)^(')],[*****************************************],[lambda_(0)f_(0)(x_(i)^(''))+lambda_(1)f_(1)(x_(i)^(''))+cdots+lambda_(k+i)f_(k+i)(x_(i)^(''))=a_(i)^('').]:}\begin{aligned} & \lambda_{0} f_{0}\left(x_{1}^{\prime}\right)+\lambda_{1} f_{1}\left(x_{1}^{\prime}\right)+\cdots+\lambda_{k+i} f_{k+i}\left(x_{1}^{\prime}\right)=a_{1}^{\prime} \\ & \lambda_{0} f_{0}\left(x_{2}^{\prime}\right)+\lambda_{1} f_{1}\left(x_{2}^{\prime}\right)+\cdots+\lambda_{k+i} f_{k+i}\left(x_{2}^{\prime}\right)=a_{2}^{\prime} \\ & \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\ & \lambda_{0} f_{0}\left(x_{i}^{\prime \prime}\right)+\lambda_{1} f_{1}\left(x_{i}^{\prime \prime}\right)+\cdots+\lambda_{k+i} f_{k+i}\left(x_{i}^{\prime \prime}\right)=a_{i}^{\prime \prime} . \end{aligned}λ0f0(x1′)+λ1f1(x1′)+⋯+λk+ifk+i(x1′)=has1′λ0f0(x2′)+λ1f1(x2′)+⋯+λk+ifk+i(x2′)=has2′⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅λ0f0(xi′′)+λ1f1(xi′′)+⋯+λk+ifk+i(xi′′)=hasi′′.
Either ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0and consider the polynomial
P 1 = P − ε ( λ 0 f 0 + λ 1 f 1 + ⋯ + λ k + i f k + i ) P 1 = P − ε λ 0 f 0 + λ 1 f 1 + ⋯ + λ k + i f k + i P_(1)=P-epsi(lambda_(0)f_(0)+lambda_(1)f_(1)+cdots+lambda_(k+i)f_(k+i))\mathrm{P}_{1}=\mathrm{P}-\varepsilon\left(\lambda_{0} f_{0}+\lambda_{1} f_{1}+\cdots+\lambda_{k+i} f_{k+i}\right)P1=P−ε(λ0f0+λ1f1+⋯+λk+ifk+i)
If ε = 0 ε = 0 epsi=0\varepsilon=0ε=0We have P 1 ≡ P P 1 ≡ P P_(1)-=P\mathrm{P}_{1} \equiv \mathrm{P}P1≡P. If ε ε epsi\varepsilonεis small but not zero. P 1 P 1 P_(1)\mathrm{P}_{1}P1is close to P. We are sure that if we take e sufficiently small, at each zero x j x j x_(j)x_{j}xjP will correspond to a zero x j x j x_(j)x_{j}xjof P 1 P 1 P_(1)\mathrm{P}_{1}P1neighbor of x j x j x_(j)x_{j}xjWe can also see this because of the choice of numbers. a j ′ , a j ′ ′ a j ′ , a j ′ ′ a_(j)^('),a_(j)^('')a_{j}^{\prime}, a_{j}^{\prime \prime}hasj′,hasj′′that at each zero x j ′ x j ′ x_(j)^(')x_{j}^{\prime}xj′correspond to at least two simple zeros for P 1 P 1 P_(1)\mathrm{P}_{1}P1and to zero x ′ ′ j x ′ ′ j x^('')_(j)x^{\prime \prime}{ }_{j}x′′jcorresponds to at least one zero inside ( a , b a , b a,ba, bhas,b).
So, finally, ε ε epsi\varepsilonεbeing sufficiently small, the polynomial P I P I P_(I)P_{\mathbf{I}}PIcancels out at least m + 2 k + i m + 2 k + i m+2k+im+2 k+im+2k+itimes; the property results immediately.
In the case k = n k = n k=nk=nk=nit suffices to introduce at most one point x i ′ ′ x i ′ ′ x_(i)^('')x_{i}{ }^{\prime \prime}xi′′3.
Either f ( x ) f ( x ) f(x)f(x)f(x)a function of class (b) and consider the expression:
I ( c 0 , c 1 , … , c n ) = Max ( a , b ) | f − c 0 f 0 − c 1 f 1 − ⋯ − c n f n | I c 0 , c 1 , … , c n = Max ( a , b ) f − c 0 f 0 − c 1 f 1 − ⋯ − c n f n I(c_(0),c_(1),dots,c_(n))=Max_((a,b))|f-c_(0)f_(0)-c_(1)f_(1)-cdots-c_(n)f_(n)|\mathrm{I}\left(c_{0}, c_{1}, \ldots, c_{n}\right)=\operatorname{Max}_{(a, b)}\left|f-c_{0} f_{0}-c_{1} f_{1}-\cdots-c_{n} f_{n}\right|I(c0,c1,…,cn)=Max(has,b)|f−c0f0−c1f1−⋯−cnfn|
To simplify the writing we can write
I ( c 0 , c 1 , … , c n ) = Max ( a , b ) | f − P | = I ( M ) I c 0 , c 1 , … , c n = Max ( a , b ) | f − P | = I ( M ) I(c_(0),c_(1),dots,c_(n))=Max_((a,b))|f-P|=I(M)\mathrm{I}\left(c_{0}, c_{1}, \ldots, c_{n}\right)=\operatorname{Max}_{(a, b)}|f-\mathrm{P}|=\mathrm{I}(\mathrm{M})I(c0,c1,…,cn)=Max(has,b)|f−P|=I(M)
Let M be the image of the polynomial P. Then, the function I(M) is continuous with respect to the coefficients. c 0 , c 1 , … , c n c 0 , c 1 , … , c n c_(0),c_(1),dots,c_(n)c_{0}, c_{1}, \ldots, c_{n}c0,c1,…,cnfor everything a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq bhas≤x≤b.
The expression I(M) has a lower limit, and we know that this limit is attained by at least one polynomial P P PPP, therefore by at least one system of finite values γ 0 , γ 1 , … , γ n γ 0 , γ 1 , … , γ n gamma_(0),gamma_(1),dots,gamma_(n)\gamma_{0}, \gamma_{1}, \ldots, \gamma_{n}γ0,γ1,…,γnquantities c 0 , c 1 , … c n c 0 , c 1 , … c n c_(0),c_(1),dotsc_(n)c_{0}, c_{1}, \ldots c_{n}c0,c1,…cn:
lim I ( M ) = I ( M 1 ) = I ( γ 0 , γ 1 , … , γ n ) lim I ( M ) = I M 1 = I γ 0 , γ 1 , … , γ n limI(M)=I(M_(1))=I(gamma_(0),gamma_(1),dots,gamma_(n))\lim \mathrm{I}(\mathrm{M})=\mathrm{I}\left(\mathrm{M}_{1}\right)=\mathrm{I}\left(\gamma_{0}, \gamma_{1}, \ldots, \gamma_{n}\right)limitI(M)=I(M1)=I(γ0,γ1,…,γn)
M 1 M 1 M_(1)\mathrm{M}_{1}M1is then a minimizing point, and the corresponding polynomial is a minimizing polynomial. In general, it is difficult to decide whether the minimizing polynomial is unique or not, but:
If there are two distinct minimizing polynomials, it y y yyyhas an infinite number of them.
Indeed M 1 , M 2 M 1 , M 2 M_(1),M_(2)M_{1}, M_{2}M1,M2two distinct minimizing points and consider the point
M 3 = M 1 + λ M 2 1 + λ , λ ⩾ 0 M 3 = M 1 + λ M 2 1 + λ , λ ⩾ 0 M_(3)=(M_(1)+lambdaM_(2))/(1+lambda),quad lambda >= 0M_{3}=\frac{M_{1}+\lambda M_{2}}{1+\lambda}, \quad \lambda \geqslant 0M3=M1+λM21+λ,λ⩾0
on the segment M 1 M 2 M 1 M 2 M_(1)M_(2)\mathrm{M}_{1} \mathrm{M}_{2}M1M2the symbolic representation above being clear.
It is easy to see that:
I ( M 3 ) ≤ I ( M 1 ) + λ I ( M 2 ) 1 + λ = I ( M 1 ) I M 3 ≤ I M 1 + λ I M 2 1 + λ = I M 1 I(M_(3)) <= (I(M_(1))+lambdaI(M_(2)))/(1+lambda)=I(M_(1))\mathrm{I}\left(\mathrm{M}_{3}\right) \leq \frac{\mathrm{I}\left(\mathrm{M}_{1}\right)+\lambda \mathrm{I}\left(\mathrm{M}_{2}\right)}{1+\lambda}=\mathrm{I}\left(\mathrm{M}_{1}\right)I(M3)≤I(M1)+λI(M2)1+λ=I(M1)
but by hypothesis,
therefore
I ( M 3 ) ⩾ I ( M 1 ) I ( M 3 ) = I ( M 1 ) I M 3 ⩾ I M 1 I M 3 = I M 1 {:[I(M_(3)) >= I(M_(1))],[I(M_(3))=I(M_(1))]:}\begin{aligned} & \mathrm{I}\left(\mathrm{M}_{3}\right) \geqslant \mathrm{I}\left(\mathrm{M}_{1}\right) \\ & \mathrm{I}\left(\mathrm{M}_{3}\right)=\mathrm{I}\left(\mathrm{M}_{1}\right) \end{aligned}I(M3)⩾I(M1)I(M3)=I(M1)
It follows that any point of the segment M 1 M 2 M 1 M 2 M_(1)M_(2)\mathrm{M}_{1} \mathrm{M}_{2}M1M2is minimizing. On the other hand, any minimizing point is at a finite distance, because for a point with unbounded coordinates I ( M ) I ( M ) I(M)I(M)I(M)becomes infinite and at the same time:
lim I ( M ) ≤ Max ( a , b ) | f | = quantité bornée. lim I ( M ) ≤ Max ( a , b ) | f | =  quantité bornée.  limI(M) <= Max_((a,b))|f|=" quantité bornée. "\lim \mathrm{I}(\mathrm{M}) \leq \operatorname{Max}_{(a, b)}|f|=\text { quantité bornée. }limitI(M)≤Max(has,b)|f|= bounded quantity. 
We can therefore state the property:
Minimizing points form a closed and bounded convex domain.
4. We propose to prove the following property:
The necessary and sufficient condition for there to be only one minimizing polynomial, regardless of the function f f fffof class (b), is that system (1) is a system ( T T TTT).
We know that this condition is sufficient (1). We will show that if the system is not a Chebyshev system, we can find a function f f fffadmitting an infinite number of minimizing polynomials.
We know that there is a polynomial
P = c 0 f 0 + c 1 f 1 + ⋯ + c n f n P = c 0 f 0 + c 1 f 1 + ⋯ + c n f n P=c_(0)f_(0)+c_(1)f_(1)+cdots+c_(n)f_(n)\mathrm{P}=c_{0} f_{0}+c_{1} f_{1}+\cdots+c_{n} f_{n}P=c0f0+c1f1+⋯+cnfn
canceling out in at least n + 1 n + 1 n+1n+1n+1distinct points within the interval ( a , b a , b a,ba, bhas,b). If x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,…,xn+1are n + 1 n + 1 n+1n+1n+1points satisfying this condition, the determinant (3) is zero because we assume, of course, that P ≡∣≡ 0 P ≡∣≡ 0 P≡∣≡0\mathrm{P} \equiv \mid \equiv 0P≡∣≡0.
We can then find n + 1 n + 1 n+1n+1n+1Numbers a 1 , a 2 , … , a n + 1 a 1 , a 2 , … , a n + 1 a_(1),a_(2),dots,a_(n+1)a_{1}, a_{2}, \ldots, a_{n+1}has1,has2,…,hasn+1such as the system
E 1 = c ¯ 0 f 0 ( x 1 ) + c ¯ 1 f 1 ( x 1 ) + ⋯ + c ¯ n f n ( x 1 ) − a 1 = 0 E 2 = c ¯ 0 f 0 ( x 2 ) + c ¯ 1 f 1 ( x 2 ) + ⋯ + c ¯ n f n ( x 2 ) − a 2 = 0 ⋯ ⋯ E n + 1 = c ¯ 0 f 0 ( x n + 1 ) + c ¯ 1 f 1 ( n + 1 ) + ⋯ + c ¯ n f n ( x n + 1 ) − a n + 1 = 0 E 1 = c ¯ 0 f 0 x 1 + c ¯ 1 f 1 x 1 + ⋯      + c ¯ n f n x 1 − a 1 = 0 E 2 = c ¯ 0 f 0 x 2 + c ¯ 1 f 1 x 2 + ⋯      + c ¯ n f n x 2 − a 2 = 0 ⋯      ⋯ E n + 1 = c ¯ 0 f 0 x n + 1 + c ¯ 1 f 1 ( n + 1 ) + ⋯ + c ¯ n f n x n + 1 − a n + 1 = 0 {:[E_(1)= bar(c)_(0)f_(0)(x_(1))+ bar(c)_(1)f_(1)(x_(1))+cdots,+ bar(c)_(n)f_(n)(x_(1))-a_(1)=0],[E_(2)= bar(c)_(0)f_(0)(x_(2))+ bar(c)_(1)f_(1)(x_(2))+cdots,+ bar(c)_(n)f_(n)(x_(2))-a_(2)=0],[cdots,cdots],[E_(n+1)= bar(c)_(0)f_(0)(x_(n+1))+ bar(c)_(1)f_(1)(n+1)+cdots+ bar(c)_(n)f_(n)(x_(n+1))-a_(n+1)=0]:}\begin{array}{ll} \mathrm{E}_{1}=\bar{c}_{0} f_{0}\left(x_{1}\right)+\bar{c}_{1} f_{1}\left(x_{1}\right)+\cdots & +\bar{c}_{n} f_{n}\left(x_{1}\right)-a_{1}=0 \\ \mathrm{E}_{2}=\bar{c}_{0} f_{0}\left(x_{2}\right)+\bar{c}_{1} f_{1}\left(x_{2}\right)+\cdots & +\bar{c}_{n} f_{n}\left(x_{2}\right)-a_{2}=0 \\ \cdots & \cdots \\ \mathrm{E}_{n+1}=\bar{c}_{0} f_{0}\left(x_{n+1}\right)+\bar{c}_{1} f_{1}(n+1)+\cdots+\bar{c}_{n} f_{n}\left(x_{n+1}\right)-a_{n+1}=0 \end{array}E1=c¯0f0(x1)+c¯1f1(x1)+⋯+c¯nfn(x1)−has1=0E2=c¯0f0(x2)+c¯1f1(x2)+⋯+c¯nfn(x2)−has2=0⋯⋯En+1=c¯0f0(xn+1)+c¯1f1(n+1)+⋯+c¯nfn(xn+1)−hasn+1=0
is incompatible. In this case, the expression
Max ( | E 1 | , | E 2 | , … , | E n + 1 | ) ( 2 ) Max E 1 , E 2 , … , E n + 1 2 Max(|E_(1)|,|E_(2)|,dots,|E_(n+1)|)(^(2))\operatorname{Max}\left(\left|E_{1}\right|,\left|E_{2}\right|, \ldots,\left|E_{n+1}\right|\right)\left({ }^{2}\right)Max(|E1|,|E2|,…,|En+1|)(2)
which is a continuous function of c ¯ 0 , c ¯ 1 , … , c ¯ n c ¯ 0 , c ¯ 1 , … , c ¯ n bar(c)_(0), bar(c)_(1),dots, bar(c)_(n)\bar{c}_{0}, \bar{c}_{1}, \ldots, \bar{c}_{n}c¯0,c¯1,…,c¯n, admits a non-zero positive minimum. Let m m mmmthe value of this minimum; it is attained for an infinite number of value systems c ¯ 0 , c ¯ 1 , … , c ¯ n c ¯ 0 , c ¯ 1 , … , c ¯ n bar(c)_(0), bar(c)_(1),dots, bar(c)_(n)\bar{c}_{0}, \bar{c}_{1}, \ldots, \bar{c}_{n}c¯0,c¯1,…,c¯nbecause if
c i ― = c ∗ i i = 0 , 1 , … , n c i ¯ = c ∗ i i = 0 , 1 , … , n bar(c_(i))=c^(**)i quad i=0,1,dots,n\overline{c_{i}}=c^{*} i \quad i=0,1, \ldots, nci—=c*ii=0,1,…,n
(1) S. Bernstein loc. cit. p. 3; JW Joung loc. cit.
( 2 2 ^(2){ }^{2}2Max ( λ 1 , λ 2 , … , λ n λ 1 , λ 2 , … , λ n lambda_(1),lambda_(2),dots,lambda_(n)\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}λ1,λ2,…,λn) means the value of the largest quantity λ i λ i lambda_(i)\lambda_{i}λi.
is a minimizing system, the systems
c l = c l ∗ + λ c l i = 0 , 1 , 2 , … , n c l = c l ∗ + λ c l i = 0 , 1 , 2 , … , n c_(l)=c_(l)^(**)+lambdac_(l)quad i=0,1,2,dots,nc_{l}=c_{l}^{*}+\lambda c_{l} \quad i=0,1,2, \ldots, ncL=cL*+λcLi=0,1,2,…,n
will also minimize λ λ lambda\lambdaλbeing arbitrary.
Now if f f fffis a continuous function taking the values a l a l a_(l)a_{l}hasLto the points x i ( i = 1 , 2 , … , n + 1 ) x i ( i = 1 , 2 , … , n + 1 ) x_(i)(i=1,2,dots,n+1)x_{i}(i=1,2, \ldots, n+1)xi(i=1,2,…,n+1), we will have for toat polynomial Q Q QQQ
Max | f − Q | ≥ m [ dans ( a , b ) ] Max | f − Q | ≥ m [ dans ( a , b ) ] Max|f-Q| >= m quad[dans(a,b)]\operatorname{Max}|f-Q| \geq m \quad[\operatorname{dans}(a, b)]Max|f−Q|≥m[In(has,b)]
Let's ask
P ∗ = c 0 ∗ f n + c 1 ∗ f 1 + ⋯ + c n ∗ f n P ∗ = c 0 ∗ f n + c 1 ∗ f 1 + ⋯ + c n ∗ f n P^(**)=c_(0)^(**)f_(n)+c_(1)^(**)f_(1)+cdots+c_(n)^(**)f_(n)\mathrm{P}^{*}=c_{0}^{*} f_{n}+c_{1}^{*} f_{1}+\cdots+c_{n}^{*} f_{n}P*=c0*fn+c1*f1+⋯+cn*fn
[one can also have P ∗ ≡ 0 P ∗ ≡ 0 P^(**)-=0\mathrm{P}^{*} \equiv 0P*≡0].
(9)
To clarify, let's assume that the polynomial P, taken initially, vanishes at a finite number of points; these are first the points x 1 , x 2 , … , x n + 1 x 1 , x 2 , … , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,…,xn+1and then some other points x 1 ′ , x 2 ′ , … , x r ′ x 1 ′ , x 2 ′ , … , x r ′ x_(1)^('),x_(2)^('),dots,x_(r)^(')x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{r}^{\prime}x1′,x2′,…,xr′(which may not exist at all).
Let's take the function f f fffsuch as:
(10) f ( x i ) = P ∗ ( x i ) − a i i = 1 , 2 , … , n + 1 f ( x i ′ ) = 0 i = 1 , 2 , … , r . (10) f x i = P ∗ x i − a i i = 1 , 2 , … , n + 1 f x i ′ = 0 i = 1 , 2 , … , r . {:(10){:[f(x_(i))=P^(**)(x_(i))-a_(i),i=1","2","dots","n+1],[f(x_(i)^('))=0,i=1","2","dots","r.]:}:}\begin{array}{cc} f\left(x_{i}\right)=\mathrm{P}^{*}\left(x_{i}\right)-a_{i} & i=1,2, \ldots, n+1 \\ f\left(x_{i}^{\prime}\right)=0 & i=1,2, \ldots, r . \tag{10} \end{array}(10)f(xi)=P*(xi)−hasii=1,2,…,n+1f(xi′)=0i=1,2,…,r.
The points x i , x i ′ x i , x i ′ x_(i),x_(i)^(')x_{i}, x_{i}^{\prime}xi,xi′share the interval ( a , b a , b a,ba, bhas,b) in a number of subintervals ( n + r , n + r + 1 n + r , n + r + 1 n+r,n+r+1n+r, n+r+1n+r,n+r+1Or n + r + 2 n + r + 2 n+r+2n+r+2n+r+2) such that in each of them the polynomial P P PPPmaintains a constant sign.
Either
Max ( a , b ) | P | = A Max ( a , b ) | P | = A Max_((a,b))|P|=A\operatorname{Max}_{(a, b)}|\mathrm{P}|=\mathrm{A}Max(has,b)|P|=HAS
Let's take
0 > λ > − m A 0 > λ > − m A 0 > lambda > -(m)/(A)0>\lambda>-\frac{m}{A}0>λ>−mHAS
SO
| λ P | < m | λ P | < m |lambdaP| < m|\lambda \mathrm{P}|<m|λP|<m
  • We determine f f fffin each subinterval such that it is continuous and such that (9) and (10) are satisfied. To do this, we will examine the various kinds of subintervals that can arise.
    1 0 1 0 1^(0)1^{0}10. Interval ( a , x ′ 1 a , x ′ 1 a,x^(')_(1)a, x^{\prime}{ }_{1}has,x′1). Within such an interval we will take
f = m 2 A P f = m 2 A P f=(m)/(2(A))Pf=\frac{m}{2 \mathrm{~A}} \mathrm{P}f=m2 HASP
then in ( a , x ′ 1 a , x ′ 1 a,x^(')1a, x^{\prime} 1has,x′1)
  • And
λ < 0 , | f − − λ P | = | f | + | λ P | = { m 2 A + | λ | } | P | λ < 0 , | f − − λ P | = | f | + | λ P | = m 2 A + | λ | | P | lambda < 0,quad|f--lambdaP|=|f|+|lambdaP|={(m)/(2(A))+|lambda|}|P|\lambda<0, \quad|f--\lambda \mathrm{P}|=|f|+|\lambda \mathrm{P}|=\left\{\frac{m}{2 \mathrm{~A}}+|\lambda|\right\}|\mathrm{P}|λ<0,|f−−λP|=|f|+|λP|={m2 HAS+|λ|}|P|
| f − λ P | < m | f − λ P | < m |f-lambdaP| < m|f-\lambda \mathrm{P}|<m|f−λP|<m
provided that
− m 2 A < λ < 0 − m 2 A < λ < 0 -(m)/(2(A)) < lambda < 0-\frac{m}{2 \mathrm{~A}}<\lambda<0−m2 HAS<λ<0
The same construction is used for an interval ( x r ′ , b ) x r ′ , b (x_(r)^('),b)\left(x_{r}^{\prime}, b\right)(xr′,b).
2 0 2 0 2^(0)2^{0}20. Interval ( a , x 1 ) a , x 1 (a,x_(1))\left(a, x_{1}\right)(has,x1)[Or ( x n + 1 , b ) ] x n + 1 , b {:(x_(n+1),b)]\left.\left(x_{n+1}, b\right)\right](xn+1,b)]. We have
| P ∗ ( x 1 ) − a 1 | ≤ m P ∗ x 1 − a 1 ≤ m |P^(**)(x_(1))-a_(1)| <= m\left|\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right| \leq m|P*(x1)−has1|≤m
We will take from ( a , x 1 a , x 1 a,x_(1)a, x_{1}has,x1)
f = P ∗ ( x 1 ) − a 1 − m A P si P [ P ∗ ( x 1 ) − a 1 ] > 0 f = P ∗ ( x 1 ) − a 1 si P [ P ∗ ( x 1 ) − a 1 ] < 0 f = P ∗ x 1 − a 1 − m A P       si       P P ∗ x 1 − a 1 > 0 f = P ∗ x 1 − a 1       si       P P ∗ x 1 − a 1 < 0 {:[f=P^(**)(x_(1))-a_(1)-(m)/((A))P," si ",P[P^(**)(x_(1))-a_(1)] > 0],[f=P^(**)(x_(1))-a_(1)," si ",P[P^(**)(x_(1))-a_(1)] < 0]:}\begin{array}{lll} f=\mathrm{P}^{*}\left(x_{1}\right)-a_{1}-\frac{m}{\mathrm{~A}} \mathrm{P} & \text { si } & \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0 \\ f=\mathrm{P}^{*}\left(x_{1}\right)-a_{1} & \text { si } & \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]<0 \end{array}f=P*(x1)−has1−m HASP if P[P*(x1)−has1]>0f=P*(x1)−has1 if P[P*(x1)−has1]<0
we then see that
(11) Max | f + λ P | ≤ m Max | f + λ P | ≤ m Max|f+lambdaP| <= m\operatorname{Max}|f+\lambda \mathrm{P}| \leq mMax|f+λP|≤m

provided that

(12)
30. Interval ( x 1 , x 1 ′ x 1 , x 1 ′ x_(1),x_(1)^(')x_{1}, x_{1}^{\prime}x1,x1′).
0 > λ > − m A 0 > λ > − m A 0 > lambda > -(m)/(A)0>\lambda>-\frac{m}{A}0>λ>−mHAS
We can take
f = ( x − x 1 ′ ) P ∗ ( x 1 ) − a 1 x 1 − x 1 ′ − m A P si P [ P ∗ ( x 1 ) − a 1 ] > 0 f = ( x − x 1 ′ ) P ∗ ( x 1 ) − a 1 x 1 − x 1 ′ , si P [ P ∗ ( x 1 ) − a 1 ] < 0 f = x − x 1 ′ P ∗ x 1 − a 1 x 1 − x 1 ′ − m A P  si  P P ∗ x 1 − a 1 > 0 f = x − x 1 ′ P ∗ x 1 − a 1 x 1 − x 1 ′ ,  si  P P ∗ x 1 − a 1 < 0 {:[f=(x-x_(1)^('))(P^(**)(x_(1))-a_(1))/(x_(1)-x_(1)^('))-(m)/((A))Pquad" si "quadP[P^(**)(x_(1))-a_(1)] > 0],[f=(x-x_(1)^('))(P^(**)(x_(1))-a_(1))/(x_(1)-x_(1)^('))","quad" si "quadP[P^(**)(x_(1))-a_(1)] < 0]:}\begin{aligned} & f=\left(x-x_{1}^{\prime}\right) \frac{\mathrm{P}^{*}\left(x_{1}\right)-a_{1}}{x_{1}-x_{1}^{\prime}}-\frac{m}{\mathrm{~A}} \mathrm{P} \quad \text { si } \quad \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0 \\ & f=\left(x-x_{1}^{\prime}\right) \frac{\mathrm{P}^{*}\left(x_{1}\right)-a_{1}}{x_{1}-x_{1}^{\prime}}, \quad \text { si } \quad \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]<0 \end{aligned}f=(x−x1′)P*(x1)−has1x1−x1′−m HASP if P[P*(x1)−has1]>0f=(x−x1′)P*(x1)−has1x1−x1′, if P[P*(x1)−has1]<0
and we still have (11) under conditions (12).
4 0 4 0 4^(0)4^{0}40. Interval ( x 1 ′ , x 2 ′ ) x 1 ′ , x 2 ′ (x_(1)^('),x_(2)^('))\left(x_{1}^{\prime}, x_{2}^{\prime}\right)(x1′,x2′)All you have to do is take
f = − m A P f = − m A P f=-(m)/((A))Pf=-\frac{m}{\mathrm{~A}} \mathrm{P}f=−m HASP
and (11) will be satisfied with (12).
5 0 5 0 5^(0)5^{0}50. Interval ( x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2Let's assume
g = x [ P ∗ ( x 1 ) − a 1 − P ∗ ( x 2 ) + a 2 ] + x 2 [ P ∗ ( x 1 ) − a 1 ] − x 1 [ P ∗ ( x 2 ) − a 2 ] x 1 − x 1 g = x P ∗ x 1 − a 1 − P ∗ x 2 + a 2 + x 2 P ∗ x 1 − a 1 − x 1 P ∗ x 2 − a 2 x 1 − x 1 g=(x[P^(**)(x_(1))-a_(1)-P^(**)(x_(2))+a_(2)]+x_(2)[P^(**)(x_(1))-a_(1)]-x_(1)[P^(**)(x_(2))-a_(2)])/(x_(1)-x_(1))g=\frac{x\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}-\mathrm{P}^{*}\left(x_{2}\right)+a_{2}\right]+x_{2}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]-x_{1}\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]}{x_{1}-x_{1}}g=x[P*(x1)−has1−P*(x2)+has2]+x2[P*(x1)−has1]−x1[P*(x2)−has2]x1−x1
We then ask
f = g − m A P si P [ P ∗ ( x 1 ) − a 1 ] > 0 , P [ P ∗ ( x 2 ) − a 2 ] > 0 ou bien si [ P ∗ ( x 1 ) − a 1 ] [ P ∗ ( x 2 ) − a 2 ] < 0 f = g si P [ P ∗ ( x 1 ) − a 1 ] > 0 , P [ P ∗ ( x 2 ) − a 2 ] < 0 . f = g − m A P       si  P P ∗ x 1 − a 1 > 0 , P P ∗ x 2 − a 2 > 0       ou bien si  P ∗ x 1 − a 1 P ∗ x 2 − a 2 < 0 f = g       si  P P ∗ x 1 − a 1 > 0 , P P ∗ x 2 − a 2 < 0 . {:[f=g-(m)/((A))P," si "P[P^(**)(x_(1))-a_(1)] > 0","P[P^(**)(x_(2))-a_(2)] > 0],[," ou bien si "[P^(**)(x_(1))-a_(1)][P^(**)(x_(2))-a_(2)] < 0],[f=g," si "P[P^(**)(x_(1))-a_(1)] > 0","P[P^(**)(x_(2))-a_(2)] < 0.]:}\begin{array}{ll} f=g-\frac{m}{\mathrm{~A}} \mathrm{P} & \text { si } \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0, \mathrm{P}\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]>0 \\ & \text { ou bien si }\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]<0 \\ f=g & \text { si } \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0, \mathrm{P}\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]<0 . \end{array}f=g−m HASP if P[P*(x1)−has1]>0,P[P*(x2)−has2]>0 or if [P*(x1)−has1][P*(x2)−has2]<0f=g if P[P*(x1)−has1]>0,P[P*(x2)−has2]<0.
We can still determine one λ 1 > 0 λ 1 > 0 lambda_(1) > 0\lambda_{1}>0λ1>0such that if
0 > λ > − λ 1 0 > λ > − λ 1 0 > lambda > -lambda_(1)0>\lambda>-\lambda_{1}0>λ>−λ1
we still have
Max | f − λ P | ≤ m Max | f − λ P | ≤ m Max|f-lambdaP| <= m\operatorname{Max}|f-\lambda \mathrm{P}| \leq mMax|f−λP|≤m
in the meantime ( x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2This property results
from the following remark:
Let g ( x ) g ( x ) g(x)g(x)g(x)a continuous function in ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)(we take the internal ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)(to simplify the explanation) positive (negative) in this interval and:
g ( 0 ) = g ( 1 ) = 1 g ( 0 ) = g ( 1 ) = 1 g(0)=g(1)=1g(0)=g(1)=1g(0)=g(1)=1
So, to clarify, a ′ > 0 , b ′ < 0 a ′ > 0 , b ′ < 0 a^(') > 0,b^(') < 0a^{\prime}>0, b^{\prime}<0has′>0,b′<0We can then determine a λ 1 > 0 λ 1 > 0 lambda_(1) > 0\lambda_{1}>0λ1>0such as for 0 < λ < λ 1 0 < λ < λ 1 0 < lambda < lambda_(1)0<\lambda<\lambda_{1}0<λ<λ1we have
a ′ + x ( b ′ − a ′ ) + λ g < a a ′ + x b ′ − a ′ + λ g < a a^(')+x(b^(')-a^('))+lambda g < aa^{\prime}+x\left(b^{\prime}-a^{\prime}\right)+\lambda g<ahas′+x(b′−has′)+λg<has
In ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)The demonstration is immediate.
The function f f fffConstructed in this way, it answers the question because λ λ lambda\lambdaλbeing between
0 , − λ 1 0 , − λ 1 0,-lambda_(1)0,-\lambda_{1}0,−λ1
Or λ 1 λ 1 lambda_(1)\lambda_{1}λ1is a positive number, we have
| f − λ P | ≤ m | f − λ P | ≤ m |f-lambdaP| <= m|f-\lambda \mathrm{P}| \leq m|f−λP|≤m
In ( a , b ) ( a , b ) (a,b)(a, b)(has,b)and at least in one respect x i x i x_(i)x_{i}xiwe have equality
| f − λ P | = m | f − λ P | = m |f-lambdaP|=m|f-\lambda \mathrm{P}|=m|f−λP|=m
On the other hand, any other polynomial gives
| f − Q | ≥ m | f − Q | ≥ m |f-Q| >= m|f-Q| \geq m|f−Q|≥m
  1. If P P PPPwas zero throughout an entire interval ( α , β α , β alpha,beta\alpha, \betaα,βFor example, we could still do the construction very easily. The points x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2, … . x n + 1 … . x n + 1 dots.x_(n+1)\ldots . x_{n+1}….xn+1can be taken inside ( α , β α , β alpha,beta\alpha, \betaα,β). So apart from ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β)We are doing the same construction as before. In ( α , β α , β alpha,beta\alpha, \betaα,β) we still keep conditions (9); and in each interval x i , x i + 1 x i , x i + 1 x_(i),x_(i+1)x_{i}, x_{i+1}xi,xi+1we take the function f f ffflinear; and constant in ( α , x 1 α , x 1 alpha,x_(1)\alpha, x_{1}α,x1) And ( x n + 1 , β x n + 1 , β x_(n+1),betax_{n+1}, \betaxn+1,β), The function thus obtained still answers the question.

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Abstract AuthorsT. Popoviciu Keywords? Paper coordinatesT. Popoviciu, Remarques sur les polynômes binomiaux, Mathematica, 6 (1932), pp. 8-10 (in French). PDFhttps://ictp.acad.ro/wp-content/uploads/2025/10/1932-e-Popoviciu-Mathematica-Remarques-sur-les-polynomes-binomiaux.pdf…

Remarks on the binomial polynomials

Abstract AuthorsT. Popoviciu Keywords? Paper coordinatesT. Popoviciu, Remarques sur les polynômes binomiaux, Buletinul Soc. de Ştiinţe din Cluj, 6, pp.146-148…

On a theorem of Laguerre

Abstract AuthorsT. Popoviciu Keywords? Paper coordinatesT. Popoviciu, Sur un théorème de Laguerre, Buletinul Soc. de Ştiinţe din Cluj, 8 (1934),…