[proofplan] We first reduce the theorem from an arbitrary finite closed interval $[a,b]$ to the interval $[0,1]$ by an affine change of variables, treating the degenerate case $a=b$ separately. On $[0,1]$, we approximate the transported function by its Bernstein polynomials. The Bernstein error is written as a weighted average, split into indices near $t$ and far from $t$, with [uniform continuity](/page/Uniform%20Continuity) controlling the near part and the binomial variance identity controlling the far part uniformly in $t$. [/proofplan] [step:Dispose of the degenerate interval case] If $a=b$, define the polynomial \begin{align*} p: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto f(a). \end{align*} Then for the only point $x \in [a,b]$, namely $x=a$, we have $|f(x)-p(x)|=0 \leq \varepsilon$. Hence the theorem holds when $a=b$. [/step] [step:Transport the problem to the unit interval] Assume now that $a<b$. Define the affine map \begin{align*} A: [0,1] &\to [a,b] \\ t &\mapsto a+(b-a)t, \end{align*} and define \begin{align*} g: [0,1] &\to \mathbb{R} \\ t &\mapsto f(A(t)). \end{align*} Since $A$ is [continuous](/page/Continuity) and $f \in C[a,b]$, the composition $g=f\circ A$ belongs to $C[0,1]$. It is enough to prove that for every $\varepsilon>0$ there is a polynomial $q:\mathbb{R}\to\mathbb{R}$ such that $|g(t)-q(t)|\leq\varepsilon$ for all $t\in[0,1]$, because then \begin{align*} p: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto q\left(\frac{x-a}{b-a}\right) \end{align*} is a polynomial and, for every $x\in[a,b]$, the point $t=(x-a)/(b-a)$ lies in $[0,1]$, so \begin{align*} |f(x)-p(x)|=|g(t)-q(t)|\leq \varepsilon. \end{align*} [/step] [step:Define the Bernstein polynomial and express its error as a weighted average] Fix $\varepsilon>0$. For each $n\in\mathbb{N}$, define the polynomial \begin{align*} B_n: \mathbb{R} &\to \mathbb{R} \\ t &\mapsto \sum_{k=0}^{n} g\left(\frac{k}{n}\right)\binom{n}{k}t^k(1-t)^{n-k}. \end{align*} For $t\in[0,1]$, define the Bernstein weight \begin{align*} w_{n,k}(t):=\binom{n}{k}t^k(1-t)^{n-k} \end{align*} for each integer $k$ with $0\leq k\leq n$. By the [binomial theorem](/theorems/750), \begin{align*} \sum_{k=0}^{n} w_{n,k}(t) = \sum_{k=0}^{n}\binom{n}{k}t^k(1-t)^{n-k} = (t+(1-t))^n = 1. \end{align*} Therefore, for every $t\in[0,1]$, \begin{align*} B_n(t)-g(t) &= \sum_{k=0}^{n}\left[g\left(\frac{k}{n}\right)-g(t)\right]w_{n,k}(t). \end{align*} [/step] [step:Estimate the far weights by the binomial variance identity] For every $n\in\mathbb{N}$ and every $t\in[0,1]$, \begin{align*} \sum_{k=0}^{n}\left(\frac{k}{n}-t\right)^2w_{n,k}(t) = \frac{t(1-t)}{n} \leq \frac{1}{4n}. \end{align*} Indeed, differentiating the identity $(s+1)^n=\sum_{k=0}^{n}\binom{n}{k}s^k$ after multiplying by suitable powers of $t$ and $1-t$ gives \begin{align*} \sum_{k=0}^{n}k\,w_{n,k}(t)&=nt,\\ \sum_{k=0}^{n}k(k-1)\,w_{n,k}(t)&=n(n-1)t^2. \end{align*} Hence \begin{align*} \sum_{k=0}^{n}k^2w_{n,k}(t) &= n(n-1)t^2+nt, \end{align*} and subtracting the square of the first moment gives \begin{align*} \sum_{k=0}^{n}\left(\frac{k}{n}-t\right)^2w_{n,k}(t) &= \frac{1}{n^2}\sum_{k=0}^{n}k^2w_{n,k}(t)-2t\frac{1}{n}\sum_{k=0}^{n}k w_{n,k}(t)+t^2\sum_{k=0}^{n}w_{n,k}(t)\\ &= \frac{t(1-t)}{n}. \end{align*} Since $0\leq t\leq 1$, we have $t(1-t)\leq 1/4$. [guided] The purpose of this step is to quantify how much Bernstein weight can sit far from the point $t$. The weights $w_{n,k}(t)$ behave like the probability weights of a binomial random variable divided by $n$, but we only need the elementary finite-sum identities. For each $n\in\mathbb{N}$, each $t\in[0,1]$, and each integer $k$ with $0\leq k\leq n$, the weight is \begin{align*} w_{n,k}(t)=\binom{n}{k}t^k(1-t)^{n-k}. \end{align*} The binomial theorem gives \begin{align*} \sum_{k=0}^{n}w_{n,k}(t)=1. \end{align*} To compute the second central moment, we use the standard binomial derivative identities. First, \begin{align*} \sum_{k=0}^{n}k\,w_{n,k}(t)=nt. \end{align*} Second, \begin{align*} \sum_{k=0}^{n}k(k-1)\,w_{n,k}(t)=n(n-1)t^2. \end{align*} Adding the first identity to the second gives the second moment: \begin{align*} \sum_{k=0}^{n}k^2w_{n,k}(t) = n(n-1)t^2+nt. \end{align*} Now expand the square in the central moment: \begin{align*} \sum_{k=0}^{n}\left(\frac{k}{n}-t\right)^2w_{n,k}(t) &= \frac{1}{n^2}\sum_{k=0}^{n}k^2w_{n,k}(t)-2t\frac{1}{n}\sum_{k=0}^{n}k w_{n,k}(t)+t^2\sum_{k=0}^{n}w_{n,k}(t)\\ &= \frac{n(n-1)t^2+nt}{n^2}-2t^2+t^2\\ &= \frac{t(1-t)}{n}. \end{align*} Finally, the elementary inequality $t(1-t)\leq 1/4$ for $t\in[0,1]$ gives \begin{align*} \sum_{k=0}^{n}\left(\frac{k}{n}-t\right)^2w_{n,k}(t) \leq \frac{1}{4n}. \end{align*} This is the variance estimate that will force the far part of the Bernstein average to vanish uniformly in $t$. [/guided] [/step] [step:Split the Bernstein error into near and far indices] Since $g\in C[0,1]$ and $[0,1]$ is compact, the [Heine-Cantor Theorem](/theorems/280) implies that $g$ is uniformly continuous. Apply uniform continuity with tolerance $\varepsilon/2$: there exists $\delta>0$ such that for all $s,t\in[0,1]$, \begin{align*} |s-t|<\delta \quad\Longrightarrow\quad |g(s)-g(t)|<\frac{\varepsilon}{2}. \end{align*} Define \begin{align*} M:=\max_{t\in[0,1]}|g(t)|. \end{align*} The maximum exists by the [Extreme Value Theorem for compact spaces](/theorems/304), because $g$ is continuous on the [compact](/page/Compact%20Space) interval $[0,1]$. For fixed $n\in\mathbb{N}$ and $t\in[0,1]$, split the indices into the near set \begin{align*} N_{n,t}:=\left\{k\in\{0,\dots,n\}:\left|\frac{k}{n}-t\right|<\delta\right\} \end{align*} and the far set \begin{align*} F_{n,t}:=\left\{k\in\{0,\dots,n\}:\left|\frac{k}{n}-t\right|\geq\delta\right\}. \end{align*} Using the weighted-average formula for the error, the triangle inequality, the near estimate, and $|g(k/n)-g(t)|\leq 2M$, we obtain \begin{align*} |B_n(t)-g(t)| &\leq \sum_{k=0}^{n}\left|g\left(\frac{k}{n}\right)-g(t)\right|w_{n,k}(t)\\ &= \sum_{k\in N_{n,t}}\left|g\left(\frac{k}{n}\right)-g(t)\right|w_{n,k}(t) + \sum_{k\in F_{n,t}}\left|g\left(\frac{k}{n}\right)-g(t)\right|w_{n,k}(t)\\ &\leq \frac{\varepsilon}{2}\sum_{k\in N_{n,t}}w_{n,k}(t) + 2M\sum_{k\in F_{n,t}}w_{n,k}(t)\\ &\leq \frac{\varepsilon}{2} + 2M\sum_{k\in F_{n,t}}w_{n,k}(t). \end{align*} For $k\in F_{n,t}$, we have $\delta^2\leq (k/n-t)^2$, so \begin{align*} \sum_{k\in F_{n,t}}w_{n,k}(t) &\leq \frac{1}{\delta^2}\sum_{k\in F_{n,t}}\left(\frac{k}{n}-t\right)^2w_{n,k}(t)\\ &\leq \frac{1}{\delta^2}\sum_{k=0}^{n}\left(\frac{k}{n}-t\right)^2w_{n,k}(t)\\ &\leq \frac{1}{4n\delta^2}. \end{align*} Therefore, for every $t\in[0,1]$, \begin{align*} |B_n(t)-g(t)| \leq \frac{\varepsilon}{2}+\frac{M}{2n\delta^2}. \end{align*} [guided] We now turn the variance estimate into an approximation estimate. The obstacle is that uniform continuity controls $|g(k/n)-g(t)|$ only when $k/n$ is close to $t$. Therefore we split the finite sum into indices close to $t$ and indices far from $t$. Because $g\in C[0,1]$ and $[0,1]$ is compact, the Heine-[Cantor Theorem](/theorems/759) gives uniform continuity. Applying it with tolerance $\varepsilon/2$, choose $\delta>0$ such that for all $s,t\in[0,1]$, \begin{align*} |s-t|<\delta \quad\Longrightarrow\quad |g(s)-g(t)|<\frac{\varepsilon}{2}. \end{align*} Also define \begin{align*} M:=\max_{t\in[0,1]}|g(t)|. \end{align*} This maximum exists by the [Extreme Value Theorem for compact spaces](/theorems/304), since $g$ is continuous on the compact interval $[0,1]$. Fix $n\in\mathbb{N}$ and $t\in[0,1]$. Define the near indices by \begin{align*} N_{n,t}:=\left\{k\in\{0,\dots,n\}:\left|\frac{k}{n}-t\right|<\delta\right\}, \end{align*} and define the far indices by \begin{align*} F_{n,t}:=\left\{k\in\{0,\dots,n\}:\left|\frac{k}{n}-t\right|\geq\delta\right\}. \end{align*} The Bernstein error is \begin{align*} B_n(t)-g(t) = \sum_{k=0}^{n}\left[g\left(\frac{k}{n}\right)-g(t)\right]w_{n,k}(t). \end{align*} Taking absolute values and applying the triangle inequality gives \begin{align*} |B_n(t)-g(t)| &\leq \sum_{k=0}^{n}\left|g\left(\frac{k}{n}\right)-g(t)\right|w_{n,k}(t). \end{align*} On the near set $N_{n,t}$, the definition of $\delta$ gives \begin{align*} \left|g\left(\frac{k}{n}\right)-g(t)\right|<\frac{\varepsilon}{2}. \end{align*} On the far set $F_{n,t}$, we use the uniform bound $|g|\leq M$, which gives \begin{align*} \left|g\left(\frac{k}{n}\right)-g(t)\right| \leq \left|g\left(\frac{k}{n}\right)\right|+|g(t)| \leq 2M. \end{align*} Thus \begin{align*} |B_n(t)-g(t)| &\leq \frac{\varepsilon}{2}\sum_{k\in N_{n,t}}w_{n,k}(t) + 2M\sum_{k\in F_{n,t}}w_{n,k}(t)\\ &\leq \frac{\varepsilon}{2} + 2M\sum_{k\in F_{n,t}}w_{n,k}(t), \end{align*} because the weights are non-negative and their total sum is $1$. It remains to bound the far weight. If $k\in F_{n,t}$, then \begin{align*} \delta^2\leq \left(\frac{k}{n}-t\right)^2. \end{align*} Multiplying by the non-negative weight $w_{n,k}(t)$ and summing over $F_{n,t}$ gives \begin{align*} \sum_{k\in F_{n,t}}w_{n,k}(t) &\leq \frac{1}{\delta^2}\sum_{k\in F_{n,t}}\left(\frac{k}{n}-t\right)^2w_{n,k}(t)\\ &\leq \frac{1}{\delta^2}\sum_{k=0}^{n}\left(\frac{k}{n}-t\right)^2w_{n,k}(t). \end{align*} The variance estimate from the previous step gives \begin{align*} \sum_{k\in F_{n,t}}w_{n,k}(t) \leq \frac{1}{4n\delta^2}. \end{align*} Substituting this into the error estimate yields \begin{align*} |B_n(t)-g(t)| \leq \frac{\varepsilon}{2}+\frac{M}{2n\delta^2}. \end{align*} This estimate is uniform in $t\in[0,1]$, which is the essential point needed for uniform approximation. [/guided] [/step] [step:Choose a Bernstein polynomial that achieves the required accuracy] Choose $n\in\mathbb{N}$ large enough that \begin{align*} \frac{M}{2n\delta^2}\leq \frac{\varepsilon}{2}. \end{align*} Then the previous estimate gives, for every $t\in[0,1]$, \begin{align*} |B_n(t)-g(t)|\leq \varepsilon. \end{align*} Thus $q:=B_n$ is a polynomial satisfying $|g(t)-q(t)|\leq\varepsilon$ for all $t\in[0,1]$. By the affine reduction step, the polynomial \begin{align*} p: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto q\left(\frac{x-a}{b-a}\right) \end{align*} satisfies $|f(x)-p(x)|\leq\varepsilon$ for all $x\in[a,b]$. This proves the theorem. [/step]