Weierstrass Approximation Theorem

Since $f \equiv 1$, we have $f(k/n) = 1$. Hence $B_n(1)(x) = \sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n-k}$ — just the sum of the basis functions.

By the binomial theorem, $\sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n-k} = (x + (1-x))^n = 1^n = 1$.

So $B_n(1)(x) = 1$ for every $x$ and every $n$ — zero error. The constant is not approximated; it is reproduced.

Setup: fix an integer $n \ge 1$ (for $n = 0$ the values $f(k/n)$ involve division by zero). The target function is $f(t) = t$, so $f(k/n) = k/n$.

For $k \geq 1$, $k\binom{n}{k} = n\binom{n-1}{k-1}$ (the $k = 0$ term vanishes since $\frac{k}{n} = 0$).

Substituting gives $B_n(x)(x) = \sum_{k=1}^{n}\frac{k}{n}\binom{n}{k}x^k(1-x)^{n-k} = x\sum_{j=0}^{n-1}\binom{n-1}{j}x^j(1-x)^{n-1-j}$ (setting $j = k-1$).

By the binomial theorem the inner sum is $(x+(1-x))^{n-1} = 1$, so $B_n(x)(x) = x$ — the identity is also reproduced exactly for every $n \ge 1$.

Let $x_0 \in [a,b]$ be arbitrary and let $\varepsilon > 0$ be given. Since $f_N \to f$ uniformly, pick $N$ with $\lVert f_N - f\rVert_\infty < \frac{\varepsilon}{3}$.

Since $f_N$ is continuous at $x_0$, pick $\delta > 0$ such that $\lvert x - x_0\rvert < \delta$ implies $\lvert f_N(x) - f_N(x_0)\rvert < \frac{\varepsilon}{3}$.

By the triangle inequality, $\lvert f(x) - f(x_0)\rvert \leq \lvert f - f_N\rvert + \lvert f_N(x) - f_N(x_0)\rvert + \lvert f_N - f\rvert < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$ — so $f$ is continuous at $x_0$. Since $x_0 \in [a,b]$ was arbitrary, $f$ is continuous on all of $[a,b]$.

(This is also why the uniform norm $\lVert\cdot\rVert_\infty$ makes $C[0,1]$ a closed space.)