Theorem 7.4.2 (Weierstrass Approximation Theorem).label For each $n \in \natz$, let
\[\phi_{n}: C([0, 1]; \real) \to C([0, 1]; \real) \quad \phi_{n}(f)(t) = \sum_{k = 0}^{n}{n \choose k}f\paren{\frac{k}{n}}t^{k}(1-t)^{n-k}\]
then:
- (1)
For every $n \in \natp$, $\phi_{n}$ is a positive map.
- (2)
For every $n \in \natp$,
\[\phi_{n}(1) = 1 \quad \phi_{n}(t) = t \quad \phi_{n}(t^{2}) = t^{2} + \frac{t - t^{2}}{n}\] - (3)
For every $f \in C([0, 1]; \real)$, $\phi_{n}(f) \to f$ uniformly as $n \to \infty$.
- (4)
$\real[t]$ is dense in $C([0, 1]; \real)$.
Proof. (1): For each $n \in \natz$ and $0 \le k \le n$, $t^{k}(1 - t)^{n-k}\ge 0$ on $[0, 1]$.
(2): Let $n \in \natp$, then by the Binomial formula,
\[\phi_{n}(1) = \sum_{k = 0}^{n}{n \choose k}t^{k}(1 - t)^{n-k}= (t + (1 - t))^{n} = 1\]
and
\begin{align*}\phi_{n}(t)&= \sum_{k = 1}^{n}\frac{n!}{k!(n-k)!}\frac{k}{n}t^{k}(1 - t)^{n-k}\\&= \sum_{k = 1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}t^{k}(1 - t)^{n-k}\\&= t\sum_{k = 0}^{n-1}\frac{(n-1)!}{k!(n-k-1)!}t^{k}(1 - t)^{n-k-1}= t\end{align*}
Finally,
\begin{align*}\phi_{n}(t^{2})&= \sum_{k = 1}^{n} \frac{n!}{k!(n-k)!}\frac{k^{2}}{n^{2}}t^{k}(1 - t)^{n-k}\\&= \sum_{k = 1}^{n} \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k}{n}t^{k}(1 - t)^{n-k}\\&= t \sum_{k = 0}^{n-1}\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{k+1}{n}t^{k}(1 - t)^{n-k}\\&= t \phi_{n-1}\paren{\frac{(n-1)t + 1}{n}}= t^{2} + \frac{t - t^{2}}{n}\end{align*}
(3): Since $\seq{\phi_n}$ are positive with $\phi_{n}(1) \to 1$, $\phi_{n}(t) \to t$, and $\phi_{n}(t^{2}) \to t^{2}$ uniformly as $n \to \infty$, $\phi_{n} \to \text{Id}$ pointwise by Korovkin’s Theorem.$\square$
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