Note: (a) is a repeat.
Let \(f: [0, 1]\to \RR\) be continuous, and prove the Weierstrass approximation theorem: for any \(\eps> 0\) there exists a polynomial \(P\) such that \(\norm{f - P}_{\infty} < \eps\).
Solution
For \(n \geq 1\) let \(p_{n,k}(x) = \binom{n}{k}x^k(1-x)^{n-k}\) and let \(B_n(x) = \sum_{k=0}^n f(k/n)\,p_{n,k}(x)\), the Bernstein polynomial of \(f\), of degree at most \(n\).
1. \(\sum_{k=0}^n p_{n,k}(x) = 1\) and \(\sum_{k=0}^n (k - nx)^2 p_{n,k}(x) = nx(1-x)\) for \(x \in [0,1]\).
The first is the binomial theorem for \((x + (1-x))^n\). The second is the variance of the binomial distribution with parameters \(n\) and \(x\); it follows by differentiating \(\sum_k \binom nk s^k t^{n-k} = (s+t)^n\) once and twice in \(s\) and setting \(s = x\), \(t = 1-x\).
2. \(|B_n(x) - f(x)| \le \sum_{k=0}^n |f(k/n) - f(x)|\,p_{n,k}(x)\).
By step 1, \(B_n(x) - f(x) = \sum_k (f(k/n) - f(x))\,p_{n,k}(x)\), and \(p_{n,k}(x) \geq 0\).
3. Given \(\eps > 0\), there is \(\delta > 0\) such that the terms of step 2 with \(|k/n - x| < \delta\) sum to less than \(\eps/2\).
\(f\) is uniformly continuous on the compact interval \([0,1]\), so there is \(\delta > 0\) with \(|f(u) - f(v)| < \eps/2\) for \(|u - v| < \delta\). By step 1 those terms sum to less than \(\frac\eps2\sum_k p_{n,k}(x) = \frac\eps2\).
4. The terms of step 2 with \(|k/n - x| \ge \delta\) sum to at most \(\frac{\|f\|_\infty}{2n\delta^2}\).
Each such term has \(|f(k/n) - f(x)| \le 2\|f\|_\infty\) and \(1 \le \frac{(k - nx)^2}{n^2\delta^2}\), so by step 1 they sum to at most \(2\|f\|_\infty\cdot\frac{nx(1-x)}{n^2\delta^2} \le \frac{2\|f\|_\infty}{4n\delta^2}\).
5. Q.E.D.
For \(n > \|f\|_\infty/(\eps\delta^2)\), steps <1>2–<1>4 give \(|B_n(x) - f(x)| < \eps\) for every \(x \in [0,1]\), so \(P = B_n\) works.