# Bernstein's probabilistic proof of Weierstrass's theorem

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

66 If $$F(x)$$ is any continuous function in the interval $$\left[0,1\right]$$ , it is always possible 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 $$\left|F(x)-E_{n}(x)\right|<\varepsilon$$ for every point in the interval under consideration. 99

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 $$F\left(\frac{m}{n}\right)$$, if the event $$A$$ occurs $$m$$ times. Under these conditions, the mathematical expectation $$E_{n}$$ for the player will have the value $$E_{n}=\sum_{m=0}^{m=n} F\left(\frac{m}{n}\right) \cdot C_{n}^{m} \cdot x^{m} \cdot(1-x)^{n-m} \tag{1}$$ It follows from the continuity of the function $$F(x)$$ that it is possible to set a number $$\delta$$, such that the inequality $$\left|x-x_{0}\right| \leq \delta$$ causes $$\left|F(x)-F\left(x_{0}\right)\right|<\frac{\varepsilon}{2}$$ so that, if $$\bar{F}(x)$$ is the maximum and $$\underline{F}(x)$$ the minimum of $$F(x)$$ in the interval $$(x-\delta, x+\delta)$$, then $$\bar{F}(x)-F(x)<\frac{\varepsilon}{2}, \quad F(x)-\underline{F}(x)<\frac{\varepsilon}{2} \tag{2}$$ Let $$\eta$$ be the probability of the inequality $$\left|x-\frac{m}{n}\right|>\delta$$ and $$L$$ the maximum of $$|F(x)|$$ in the interval $$[0,1]$$ We then have $$\underline{F}(x) \cdot(1-\eta)-L \cdot \eta

But by virtue of a theorem of Bernoulli, we can take $$n$$ large enough to have $$\eta<\frac{\varepsilon}{4 L} \tag{4}$$ Inequality (3) will in turn take the form $$F(x)+(\underline{F}(x)-F(x))-\eta(L+\underline{F}(x)) and so $$F(x)-\frac{\varepsilon}{2}-\frac{2 L}{4 L} \varepsilon therefore $$\left|F(x)-E_{n}\right|<\varepsilon \tag{5}$$ $$E_{n}$$ 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 $$x=\frac{m}{n}(m=0,1, \cdots n)$$. Formula (1) and inequality (5) show that, for any continuous function $$F(x)$$ : $$F(x)=\lim _{n \rightarrow \infty} \sum_{m=0}^{m=n} F\left(\frac{m}{n}\right) \cdot C_{n}^{m} \cdot x^{m} \cdot(1-x)^{n-m}$$ S. Bernstein

Communications of the Kharkov Mathematical Society, Volume XIII, 1912/13 (p 1-2) refer

1. What is the idea behind equation $$(3)$$?

2. What theorem of Bernoulli is being used here?

1. To get lower bound on expectation, we can replace our variable $$E(x)$$ with something smaller - specifically, with $$\underline{F}$$ on $$(x - \delta, x + \delta)$$ and with $$-L$$ outside. Similarly for upper bound.
• Can you explain the $\eta$ thing in 3? Jul 3, 2022 at 12:15