[proofplan] Fix $x \in [0,1]$ and split the sum defining $|f(x) - B_n(f;x)|$ into indices $k$ where $|k/n - x|$ is small (less than $\delta$) and indices where it is large (at least $\delta$). On the small set, the modulus of continuity controls the error. On the large set, the Bernstein variance identity provides a Chebyshev-type bound. Optimising $\delta = 1/\sqrt{n}$ yields the stated estimate with constant $3/2$. Uniform convergence follows because $\omega_f(\delta) \to 0$ as $\delta \to 0$ for any $f \in C([0,1])$. [/proofplan]

[step:Write the error using the partition-of-unity property]Fix $x \in [0,1]$. Since $B_n(1; x) = 1$ by the [Bernstein Polynomial Identities](/theorems/1214), we write \begin{align*} f(x) - B_n(f; x) &= f(x) \sum_{k=0}^{n} b_{n,k}(x) - \sum_{k=0}^{n} f\!\left(\frac{k}{n}\right) b_{n,k}(x) \\ &= \sum_{k=0}^{n} \left[f(x) - f\!\left(\frac{k}{n}\right)\right] b_{n,k}(x), \end{align*} where $b_{n,k}(x) = \binom{n}{k} x^k (1-x)^{n-k} \geq 0$. Taking absolute values and using the triangle inequality, \begin{align*} |f(x) - B_n(f; x)| \leq \sum_{k=0}^{n} \left|f(x) - f\!\left(\frac{k}{n}\right)\right| b_{n,k}(x). \end{align*}[/step]

[guided]The Bernstein operator $B_n(f; x) = \sum_{k=0}^n f(k/n) \, b_{n,k}(x)$ is a weighted average of the values $f(k/n)$, with the Bernstein basis polynomials $b_{n,k}(x) = \binom{n}{k} x^k(1-x)^{n-k}$ serving as weights. The identity $B_n(1;x) = 1$ from the [Bernstein Polynomial Identities](/theorems/1214) tells us these weights sum to $1$, so we can subtract $f(x) = f(x) \cdot 1 = f(x) \sum_k b_{n,k}(x)$ and bring $f(x)$ inside the sum: \begin{align*} f(x) - B_n(f; x) = \sum_{k=0}^{n} \left[f(x) - f\!\left(\frac{k}{n}\right)\right] b_{n,k}(x). \end{align*} Since each $b_{n,k}(x) \geq 0$ (it is a product of non-negative factors on $[0,1]$), the triangle inequality gives \begin{align*} |f(x) - B_n(f; x)| \leq \sum_{k=0}^{n} \left|f(x) - f\!\left(\frac{k}{n}\right)\right| b_{n,k}(x). \end{align*} This reformulation reduces the problem to bounding how well $f(k/n)$ approximates $f(x)$ — a question answered by the modulus of continuity.[/guided]

[step:Split the sum according to $|k/n - x| < \delta$ and $|k/n - x| \geq \delta$]Let $\delta > 0$ be a parameter to be chosen. Define the index sets \begin{align*} S_1 &:= \{k \in \{0, \ldots, n\} : |k/n - x| < \delta\}, \\ S_2 &:= \{k \in \{0, \ldots, n\} : |k/n - x| \geq \delta\}. \end{align*} We split the bound from the previous step: \begin{align*} |f(x) - B_n(f; x)| \leq \underbrace{\sum_{k \in S_1} \left|f(x) - f\!\left(\frac{k}{n}\right)\right| b_{n,k}(x)}_{\displaystyle =: I_1} + \underbrace{\sum_{k \in S_2} \left|f(x) - f\!\left(\frac{k}{n}\right)\right| b_{n,k}(x)}_{\displaystyle =: I_2}. \end{align*}[/step]

[guided]The idea is a Chebyshev-type decomposition: terms where $k/n$ is close to $x$ are controlled by the modulus of continuity, while terms where $k/n$ is far from $x$ are rare (the Bernstein variance identity makes their total weight small). We introduce a free parameter $\delta > 0$ that will be optimised at the end.[/guided]

[step:Bound $I_1$ using the modulus of continuity] The modulus of continuity of $f$ is the function $\omega_f: [0, \infty) \to [0, \infty)$ defined by \begin{align*} \omega_f(\delta) := \sup\{|f(s) - f(t)| : s, t \in [0,1], \, |s - t| \leq \delta\}. \end{align*} For $k \in S_1$, we have $|x - k/n| < \delta$, so $|f(x) - f(k/n)| \leq \omega_f(\delta)$. Therefore \begin{align*} I_1 \leq \omega_f(\delta) \sum_{k \in S_1} b_{n,k}(x) \leq \omega_f(\delta) \sum_{k=0}^{n} b_{n,k}(x) = \omega_f(\delta), \end{align*} where we enlarged the summation domain from $S_1$ to $\{0, \ldots, n\}$ (using $b_{n,k} \geq 0$) and applied $\sum_{k=0}^n b_{n,k}(x) = 1$. [/step]

[step:Bound $I_2$ using the variance identity and Chebyshev's trick]Since $f$ is continuous on the compact set $[0,1]$, it is bounded: let $M := \|f\|_\infty = \max_{[0,1]} |f|$, so $|f(x) - f(k/n)| \leq 2M$ for all $k$. For $k \in S_2$, we have $|k/n - x| \geq \delta$, so $1 \leq (k/n - x)^2 / \delta^2$. Combining, \begin{align*} I_2 &\leq 2M \sum_{k \in S_2} b_{n,k}(x) \leq 2M \sum_{k \in S_2} \frac{(k/n - x)^2}{\delta^2} \, b_{n,k}(x) \\ &\leq \frac{2M}{\delta^2} \sum_{k=0}^{n} \left(\frac{k}{n} - x\right)^2 b_{n,k}(x) = \frac{2M}{\delta^2} \cdot \frac{x(1-x)}{n}, \end{align*} where we enlarged the summation domain from $S_2$ to $\{0, \ldots, n\}$ (all terms are non-negative) and applied the variance identity from the [Bernstein Polynomial Identities](/theorems/1214). Since $x(1-x) \leq 1/4$, \begin{align*} I_2 \leq \frac{M}{2n\delta^2}. \end{align*}[/step]

[guided]The estimate on $S_2$ uses a Chebyshev-type trick: when $|k/n - x| \geq \delta$, the inequality $(k/n - x)^2 \geq \delta^2$ allows us to trade the characteristic function of $S_2$ for a quadratic weight. Concretely, $\mathbb{1}_{S_2}(k) \leq (k/n - x)^2 / \delta^2$, so \begin{align*} \sum_{k \in S_2} b_{n,k}(x) \leq \frac{1}{\delta^2} \sum_{k=0}^{n} \left(\frac{k}{n} - x\right)^2 b_{n,k}(x). \end{align*} The variance identity from the [Bernstein Polynomial Identities](/theorems/1214) states that this last sum equals $x(1-x)/n$. Since $|f(x) - f(k/n)| \leq 2M$ (the function is bounded on the compact interval $[0,1]$, with $M = \|f\|_\infty$), we obtain \begin{align*} I_2 \leq 2M \cdot \frac{1}{\delta^2} \cdot \frac{x(1-x)}{n} \leq \frac{2M}{n\delta^2} \cdot \frac{1}{4} = \frac{M}{2n\delta^2}. \end{align*} This bound says that the "bad" terms contribute at most $O(1/(n\delta^2))$. When $\delta$ is large, this contribution is small, but the modulus-of-continuity bound on $I_1$ becomes large. The next step balances these competing effects.[/guided]

[step:Combine and optimise $\delta = 1/\sqrt{n}$ to obtain the $\frac{3}{2}\,\omega_f(1/\sqrt{n})$ bound]So far we have, for every $\delta > 0$, \begin{align*} |f(x) - B_n(f; x)| \leq \omega_f(\delta) + \frac{M}{2n\delta^2}. \end{align*} We now use the standard subadditivity property of the modulus of continuity: for any $\lambda \geq 1$, \begin{align*} \omega_f(\lambda \delta) \leq (\lfloor \lambda \rfloor + 1) \, \omega_f(\delta) \leq (\lambda + 1) \, \omega_f(\delta). \end{align*} This follows by covering an interval of length $\lambda \delta$ with at most $\lfloor \lambda \rfloor + 1$ subintervals of length $\delta$ and applying the triangle inequality. In the bound $|f(x) - f(k/n)| \leq 2M$ used for $I_2$, we can sharpen this by observing that $2M = 2\omega_f(1) \leq 2(1/\delta + 1)\,\omega_f(\delta)$ when $\delta \leq 1$ (applying subadditivity with $\lambda = 1/\delta$). Returning to the full sum without splitting: \begin{align*} \left|f(x) - f\!\left(\frac{k}{n}\right)\right| &\leq \omega_f\!\left(\left|\frac{k}{n} - x\right|\right) \leq \left(1 + \frac{|k/n - x|}{\delta}\right) \omega_f(\delta), \end{align*} where the second inequality uses $\omega_f(s) \leq (1 + s/\delta)\,\omega_f(\delta)$ for $s \geq 0$ (a consequence of subadditivity applied with $\lambda = s/\delta$ when $s \geq \delta$, and the monotonicity $\omega_f(s) \leq \omega_f(\delta)$ when $s \leq \delta$). Summing over $k$ with weights $b_{n,k}(x)$: \begin{align*} |f(x) - B_n(f;x)| &\leq \omega_f(\delta) \sum_{k=0}^{n} \left(1 + \frac{|k/n - x|}{\delta}\right) b_{n,k}(x) \\ &= \omega_f(\delta) \left(1 + \frac{1}{\delta} \sum_{k=0}^{n} \left|\frac{k}{n} - x\right| b_{n,k}(x)\right). \end{align*} By the Cauchy--Schwarz inequality applied to the sum $\sum_k |k/n - x| \, b_{n,k}(x)$ (viewing it as an inner product of $|k/n - x| \sqrt{b_{n,k}(x)}$ and $\sqrt{b_{n,k}(x)}$): \begin{align*} \sum_{k=0}^{n} \left|\frac{k}{n} - x\right| b_{n,k}(x) &\leq \left(\sum_{k=0}^{n} \left(\frac{k}{n} - x\right)^2 b_{n,k}(x)\right)^{1/2} \left(\sum_{k=0}^{n} b_{n,k}(x)\right)^{1/2} \\ &= \left(\frac{x(1-x)}{n}\right)^{1/2} \leq \frac{1}{2\sqrt{n}}, \end{align*} where we used the variance identity from the [Bernstein Polynomial Identities](/theorems/1214) and $x(1-x) \leq 1/4$. Substituting, \begin{align*} |f(x) - B_n(f;x)| \leq \omega_f(\delta) \left(1 + \frac{1}{2\delta\sqrt{n}}\right). \end{align*} Setting $\delta = 1/\sqrt{n}$: \begin{align*} |f(x) - B_n(f;x)| \leq \omega_f\!\left(\frac{1}{\sqrt{n}}\right) \left(1 + \frac{1}{2}\right) = \frac{3}{2}\,\omega_f\!\left(\frac{1}{\sqrt{n}}\right). \end{align*} Since this holds for every $x \in [0,1]$, we have $\|f - B_n(f;\cdot)\|_\infty \leq \frac{3}{2}\,\omega_f(1/\sqrt{n})$.[/step]

[guided]Let us refine the estimate by avoiding the crude splitting into $S_1$ and $S_2$. Instead, we use the inequality \begin{align*} \left|f(x) - f\!\left(\frac{k}{n}\right)\right| \leq \omega_f\!\left(\left|\frac{k}{n} - x\right|\right) \leq \left(1 + \frac{|k/n - x|}{\delta}\right) \omega_f(\delta), \end{align*} valid for all $k$ and all $\delta > 0$. The first inequality is the definition of the modulus of continuity. The second inequality is a consequence of subadditivity: if $|k/n - x| \leq \delta$, then $\omega_f(|k/n - x|) \leq \omega_f(\delta) \leq (1 + |k/n - x|/\delta)\,\omega_f(\delta)$; if $|k/n - x| > \delta$, then with $\lambda = |k/n - x|/\delta > 1$, subadditivity gives $\omega_f(|k/n - x|) \leq (\lfloor \lambda \rfloor + 1)\,\omega_f(\delta) \leq (\lambda + 1)\,\omega_f(\delta) = (1 + |k/n - x|/\delta)\,\omega_f(\delta)$. Summing over $k$ with weights $b_{n,k}(x)$ and using $\sum_k b_{n,k}(x) = 1$: \begin{align*} |f(x) - B_n(f;x)| \leq \omega_f(\delta) \left(1 + \frac{1}{\delta} \sum_{k=0}^{n} \left|\frac{k}{n} - x\right| b_{n,k}(x)\right). \end{align*} To bound the remaining sum, we apply the Cauchy--Schwarz inequality: viewing $\sum_k |k/n - x| \, b_{n,k}(x)$ as $\sum_k (|k/n-x| \sqrt{b_{n,k}}) \cdot \sqrt{b_{n,k}}$, \begin{align*} \left(\sum_{k=0}^{n} \left|\frac{k}{n} - x\right| b_{n,k}(x)\right)^2 \leq \left(\sum_{k=0}^{n} \left(\frac{k}{n} - x\right)^2 b_{n,k}(x)\right) \cdot \left(\sum_{k=0}^{n} b_{n,k}(x)\right). \end{align*} The first factor on the right is $x(1-x)/n$ by the variance identity from the [Bernstein Polynomial Identities](/theorems/1214), and the second factor is $1$. Taking square roots and using $x(1-x) \leq 1/4$: \begin{align*} \sum_{k=0}^{n} \left|\frac{k}{n} - x\right| b_{n,k}(x) \leq \sqrt{\frac{x(1-x)}{n}} \leq \frac{1}{2\sqrt{n}}. \end{align*} Therefore \begin{align*} |f(x) - B_n(f;x)| \leq \omega_f(\delta)\left(1 + \frac{1}{2\delta\sqrt{n}}\right). \end{align*} We now choose $\delta = 1/\sqrt{n}$ to balance the two terms: \begin{align*} |f(x) - B_n(f;x)| \leq \omega_f\!\left(\frac{1}{\sqrt{n}}\right)\left(1 + \frac{\sqrt{n}}{2\sqrt{n}}\right) = \frac{3}{2}\,\omega_f\!\left(\frac{1}{\sqrt{n}}\right). \end{align*} Since this bound is independent of $x$, taking the supremum over $x \in [0,1]$ gives $\|f - B_n(f;\cdot)\|_\infty \leq \frac{3}{2}\,\omega_f(1/\sqrt{n})$.[/guided]

[step:Deduce uniform convergence from continuity of $f$] Since $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $|s - t| < \delta$ implies $|f(s) - f(t)| < \varepsilon$. This means $\omega_f(\delta) \to 0$ as $\delta \to 0^+$. Since $1/\sqrt{n} \to 0$ as $n \to \infty$, \begin{align*} \|f - B_n(f; \cdot)\|_\infty \leq \frac{3}{2}\,\omega_f\!\left(\frac{1}{\sqrt{n}}\right) \to 0 \quad \text{as } n \to \infty. \end{align*} This establishes that $B_n(f; \cdot) \to f$ uniformly on $[0,1]$. [/step]