Let $\map f t: \Bbb I = \closedint a b \to \R$ be a continuous function. Introduce $\map x t$ with a rescaled domain: :$\map f t \mapsto \map x {a + t \paren {b - a} } : \closedint a b \to \closedint 0 1$. From now on we will work with $x: \closedint 0 1 \to \R$, which is also continuous. Let $n \in \N$. For $t \in \closedint 0 1$ consider the Bernstein polynomial: :$\displaystyle \map {B_n x} t = \sum_{k \mathop = 0}^n \map x {\frac k n} \binom n k t^k \paren {1 - t}^{n - k}$ For $t \in \closedint 0 1$, $0 \le k \le n$, let: :$\displaystyle \map {p_{n, k} } t := \binom n k t^k \paren {1 - t}^{n - k}$ By the binomial theorem: :$\displaystyle \sum_{k \mathop = 0}^n \map {p_{n, k} } t = 1$ === Lemma 1 === {{:Weierstrass Approximation Theorem/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Weierstrass Approximation Theorem/Lemma 2}}{{qed|lemma}} Now we construct the estimates. {{begin-eqn}} {{eqn | l = \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \map {p_{n, k} } t       | o = \le       | r = \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \map {p_{n, k} } t \frac {\paren {k - n t}^2} {\delta^2 n^2}       | c = $\displaystyle \frac {\paren {k - n t}^2} {\delta^2 n^2} \ge 1$ }} {{eqn | o = \le       | r = \frac 1 {n^2 \delta^2} \sum_{k \mathop = 0}^n \paren{k - n t}^2 \map {p_{n, k} } t }} {{eqn | r = \frac {t \paren {1 - t} } {n \delta^2}       | c = Lemma 2 }} {{eqn | o = \le       | r = \frac 1 {4 n \delta^2}       | c = $\forall t \in \closedint 0 1: \frac 1 4 \ge t \paren {1 - t}$ }} {{end-eqn}} Here $\displaystyle \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n$ denotes the summation over those values of $k \in \N$, $k \le n$, which satisfy the inequality $\displaystyle \size {\frac k n - t} \ge \delta$. For some $\delta > 0$ denote: :$\displaystyle \map {\omega_\delta} x := \sup_{\size {t - s} < \delta} \size {\map x s - \map x t}$ Then: {{begin-eqn}} {{eqn | l = \size {\map {B_n x} t - \map x t}       | r = \size {\map {B_n x} t - \map x t \sum_{k \mathop = 0}^n \map {p_{n, k} } t} }} {{eqn | r = \size {\sum_{k \mathop = 0}^n \map x {\frac k n} \map {p_{n, k} } t - \map x t \sum_{k \mathop = 0}^n \map {p_{n, k} } t} }} {{eqn | o = \le       | r = \sum_{k \mathop = 0}^n \size {\map x {\frac k n}  - \map x t} \map {p_{n, k} } t       | c = as $\map {p_{n, k} } t \ge 0$ }} {{eqn | r = \sum_{k \mathop : \size {\frac k n \mathop - t} < \delta}^n \size {\map x {\frac k n}  - \map x t} \map {p_{n, k} } t + \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \size {\map x {\frac k n}  - \map x t} \map {p_{n, k} } t }} {{eqn | o = \le       | r = \map {\omega_\delta} x \sum_{k \mathop : \size {\frac k n \mathop - t} < \delta}^n \map {p_{n, k} } t + 2 \norm x_\infty \frac 1 {4 n \delta^2} }} {{eqn | o = \le        | r = \map {\omega_\delta} x \cdot 1 + \frac {\norm x_\infty} {2 n \delta^2} }} {{end-eqn}} where $\norm {\,\cdot \,}_\infty$ denotes the supremum norm. Let $\epsilon > 0$. By Continuous Function on Closed Interval is Uniformly Continuous, $\map x t$ is uniformly continuous. We choose $\delta > 0$ such that $\displaystyle \map {\omega_\delta} x < \frac \epsilon 2$. Choose $\displaystyle n > \frac {\norm x_\infty} {\epsilon \delta^2}$ Then: :$\displaystyle \norm {\map {B_n x} t - \map x t}_\infty < \epsilon$. {{qed}}