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 $$ \underline{F}(x) \cdot(1-\eta)-L \cdot \eta<E_{n}<\bar{F}(x) \cdot(1-\eta)+L \cdot \eta . \tag{3}$$

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 $$ F(x)+(\underline{F}(x)-F(x))-\eta(L+\underline{F}(x))<E_{n}<F(x)+(\bar{F}(x)-F(x))+\eta(L-\bar{F}(x)) $$ and so $$ F(x)-\frac{\varepsilon}{2}-\frac{2 L}{4 L} \varepsilon<E_{n}<F(x)+\frac{\varepsilon}{2}+\frac{2 L}{4 L} \varepsilon $$ 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 Answer 1

  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.

  2. I think it's old name of law of large numbers.

  • $\begingroup$ Can you explain the $\eta$ thing in 3? $\endgroup$ Jul 3, 2022 at 12:15

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