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Bernstein's Démonstration du théorème de Weierstrass

 Bernstein's 1912 paper on the Weierstrass Approximation Theorem
Bernstein on the Weierstrass Approximation Theorem (1912) — click image for original.
Démonstration du théorème de Weierstrass fondée sur le calcul des probabilités
(Demonstration of a theorem of Weierstrass based on the calculus of probabilities)
by S. Bernstein
Communications of the Kharkov Mathematical Society, Volume XIII, 1912/13 (p 1-2)

I propose to give a very simple proof of the following theorem of Weierstrass:

If F(x) is any continuous function in the interval [0,1], it is always possible, regardless how small ε, to determine a polynomial En(x)=a0xn+a1xn1++an of degree n high enough such that we have |F(x)En(x)|<ε for every point in the interval under consideration.

To this end, I consider an event A, whose probability is equal to x. Suppose n experiments are conducted and that is agreed to pay a player the sum F(mn), if the event A occurs m times. Under these conditions, the mathematical expectation En for the player will have the value

En=m=nm=0F(mn)Cmnxm(1x)nm.

It follows from the continuity of the function F(x) that it is possible to set a number δ, such that the inequality

|xx0|δ

causes

|F(x)F(x0)|<ε2;

so that, if ¯F(x) is the maximum and F_(x) the minimum of F(x) in the interval (xδ,x+δ), then

¯F(x)F(x)<ε2,F(x)F_(x)<ε2.

Let η be the probability of the inequality |xmn|>δ and L the maximum of |F(x)| in the interval [0,1].

We then have

F_(x)(1η)Lη<En<¯F(x)(1η)+Lη.

But by virtue of a theorem of Bernoulli, we can take n large enough to have

η<ε4L.

Inequality (3) will in turn take the form

F(x)+(F_(x)F(x))η(L+F_(x))<En<F(x)+(¯F(x)F(x))+η(L¯F(x))

and so

F(x)ε22L4Lε<En<F(x)+ε2+2L4Lε;

therefore

|F(x)En|<ε

En is clearly a polynomial of degree n.
The theorem is therefore proved.

I would only add two points.

The approximate polynomials En(x) are especially convenient, it seems to me, when you know exactly or approximately the values of F(x) for x=mn(m=0,1,n).

Formula (1) and inequality (5) show that, for any continuous function F(x):

F(x)=limnm=nm=0F(mn)Cmnxm(1x)nm.

S. Bernstein

Translated by Mike Bertrand (Mar 6, 2015).