Weierstrass Approximation Theorem/Proof 2

Proof
, assume $\Bbb I = \closedint 0 1$

For each $n \in \N$, let:
 * $\ds \map {P_n} x := \sum_{k \mathop = 0}^n \map f {\dfrac k n } \dbinom n k x^k \paren {1 - x}^{n - k}$

We shall show that $\lim_{n \to \infty} \norm { P_n - f}_\infty = 0$.

Let $\epsilon \in \R_{>0}$.

There is a $\delta \in \R_{>0}$ such that:
 * $\forall x,y \in \Bbb I : \size {x - y} \le \delta \implies \size {\map f x - \map f y} \le \epsilon $

Let $p \in \Bbb I$.

Let $\sequence {X_k} _{k \in \N}$ be independent random variables such that:
 * $\forall k \in \N : X_k \sim \Binomial 1 p$

where $\Binomial n p$ denotes the binomial distribution with parameters $n$ and $p$.

Note that such $\sequence {X_k} _{k \in \N}$ exists by Kolmogorov Extension Theorem in a certain probability space.

In the following discussion, we suppress the underlying space.

Let:
 * $\ds Z_n := \dfrac 1 n \sum_{k \mathop = 0}^{n-1} X_k$

Observe that:

By Sum of Independent Binomial Random Variables:
 * $n Z_n \sim \Binomial n p$

Thus:

On the other hand:

Therefore:

Thus for all $n \in \N_{> 2 \delta^2 / \norm f_\infty}$ we have:
 * $\size {\map {P_n} p - \map f p} \le 2 \epsilon$

As the above is true for all $p \in \Bbb I$, we have:
 * $\forall n \in \N_{> 2 \delta^2 / \norm f_\infty} : \norm { P_n - f}_\infty \le 2 \epsilon$