Bernstein's Démonstration du théorème de Weierstrass

Bernstein on the Weierstrass Approximation Theorem (1912) — click image for original.

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 $\varepsilon$, to determine a polynomial $E_n(x) = {a_0 x^n + a_1 x^{n-1} + \cdots + a_n}$ of degree $n$ high enough such that we have $${|F(x) - E_n(x)|} < \varepsilon$$ 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 $F({m \over n})$, if the event $A$ occurs $m$ times. Under these conditions, the mathematical expectation $E_n$ for the player will have the value

$$\begin{equation}{E_n = {\sum_{m=0}^{m=n} F\left({m \over n }\right) \cdot C_n^m \cdot x^m \cdot ({1-x})^{n-m}}.} \tag{1} \end{equation}$$

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

$${|x - x_0|} \leq \delta$$

causes

$${|F(x) - F(x_0)|} < {\varepsilon \over 2};$$

so that, if $\overline{F}(x)$ is the maximum and $\underline{F}(x)$ the minimum of $F(x)$ in the interval $(x - \delta, x + \delta)$, then

$$\begin{equation}{{\overline{F}(x) - F(x)} < {\varepsilon \over 2}, \;\;\; {F(x) - \underline{F}(x)} < {\varepsilon \over 2}.} \tag{2} \end{equation}$$

Let $\eta$ be the probability of the inequality ${|x - {m \over n}|} > \delta$ and $L$ the maximum of $|F(x)|$ in the interval $[0, 1]$.

We then have

$$\begin{equation}{{\underline{F}(x) \cdot (1 - \eta) - L \cdot \eta} < E_n < {\overline{F}(x) \cdot (1 - \eta) + L \cdot \eta}.} \tag{3} \end{equation}$$

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

$$\begin{equation}{\eta < {\varepsilon \over 4L}.} \tag{4} \end{equation}$$

Inequality (3) will in turn take the form

$${F(x) + (\underline{F}(x) - F(x)) - \eta (L + \underline{F}(x))} < E_n < {F(x) + (\overline{F}(x) - F(x)) + \eta (L - \overline{F}(x))}$$

and so

$${F(x) - {\varepsilon \over 2} - {2L \over 4L} \varepsilon} < E_n < {F(x) + {\varepsilon \over 2} + {2L \over 4L} \varepsilon};$$

therefore

$$\begin{equation}{{|F(x) - E_n|} < \varepsilon} \tag{5} \end{equation}$$

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

I would only add two points.

The approximate polynomials $E_n (x)$ are especially convenient, it seems to me, when you know exactly or approximately the values of $F(x)$ for $x = {m \over n} (m = 0, 1, \cdots n).$

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

$$F(x) = {\displaystyle \lim_{n \rightarrow \infty}}{\sum_{m=0}^{m=n} F\left({m \over n }\right) \cdot C_n^m \cdot x^m \cdot ({1-x})^{n-m}}.$$

Translated by Mike Bertrand (Mar 6, 2015).