[proofplan] We approximate $f$ at dyadic degrees and write $f$ as a uniformly convergent telescoping series of dyadic polynomial increments. Each increment has rapidly decaying uniform norm, and its derivative is controlled by Markov's inequality. For two points $x,y$, we split the dyadic scales at the scale comparable to $h^{1/2}$, where $h = |x-y|$; the high-scale tail is controlled directly by uniform norms, while the low-scale part is controlled by derivative estimates. The two resulting geometric sums are both bounded by a constant multiple of $h^{\alpha/2}$. [/proofplan]

[step:Choose dyadic near-best polynomial approximants and form the telescoping series]For each integer $n \geq 0$, let $\mathcal{P}_n$ denote the [vector space](/page/Vector%20Space) of real algebraic polynomials on $[-1,1]$ of degree at most $n$, and define the best uniform approximation error $E_n(f)$ by \begin{align*} E_n(f) := \inf_{p \in \mathcal{P}_n} \|f-p\|_{C([-1,1])}. \end{align*} For $0<\beta\leq 1$, we use $f \in \operatorname{Lip}(\beta)$ to mean membership in the [Hölder space](/page/Holder%20Space) of exponent $\beta$: there is a constant $L_\beta>0$ such that \begin{align*} |f(s)-f(t)| \leq L_\beta |s-t|^\beta \end{align*} for all $s,t \in [-1,1]$. For each integer $m \geq 0$, choose a polynomial $p_m \in \mathcal{P}_{2^m}$ satisfying \begin{align*} \|f - p_m\|_{C([-1,1])} \leq 2C 2^{-m\alpha}. \end{align*} This is possible by the definition of $E_{2^m}(f)$ and the hypothesis $E_{2^m}(f) \leq C2^{-m\alpha}$. Define $u_0: [-1,1] \to \mathbb{R}$ by $u_0 = p_0$, and for $m \geq 1$ define the map $u_m: [-1,1] \to \mathbb{R}$ by \begin{align*} u_m(t) := p_m(t) - p_{m-1}(t). \end{align*} Then $u_m \in \mathcal{P}_{2^m}$ for every $m \geq 0$. For $m \geq 1$, the triangle inequality gives \begin{align*} \|u_m\|_{C([-1,1])} \leq \|p_m - f\|_{C([-1,1])} + \|f - p_{m-1}\|_{C([-1,1])}. \end{align*} Hence \begin{align*} \|u_m\|_{C([-1,1])} \leq 2C2^{-m\alpha} + 2C2^{-(m-1)\alpha}. \end{align*} Define \begin{align*} B := 2C(1 + 2^\alpha). \end{align*} Then, for every $m \geq 1$, \begin{align*} \|u_m\|_{C([-1,1])} \leq B2^{-m\alpha}. \end{align*} Moreover, since $\|f-p_m\|_{C([-1,1])} \to 0$, the telescoping identity \begin{align*} p_M = u_0 + \sum_{m=1}^{M} u_m \end{align*} implies that \begin{align*} f = u_0 + \sum_{m=1}^{\infty} u_m \end{align*} uniformly on $[-1,1]$.[/step]

[guided]The dyadic decomposition is the central device of the proof. Instead of comparing $f$ directly with one polynomial, we compare successive approximants at degrees $1,2,4,8,\dots$. This produces polynomial increments whose sizes decay geometrically. For each integer $n \geq 0$, $E_n(f)$ denotes the best uniform approximation error \begin{align*} E_n(f) := \inf_{p \in \mathcal{P}_n} \|f-p\|_{C([-1,1])}, \end{align*} where $\mathcal{P}_n$ is the space of real algebraic polynomials on $[-1,1]$ of degree at most $n$. Also, $f \in \operatorname{Lip}(\beta)$ means membership in the [Hölder space](/page/Holder%20Space) of exponent $\beta$: $|f(s)-f(t)| \leq L_\beta |s-t|^\beta$ for all $s,t \in [-1,1]$ and some constant $L_\beta>0$. For each integer $m \geq 0$, the hypothesis gives \begin{align*} E_{2^m}(f) \leq C2^{-m\alpha}. \end{align*} Because $E_{2^m}(f)$ is an infimum over $\mathcal{P}_{2^m}$, we may choose $p_m \in \mathcal{P}_{2^m}$ with \begin{align*} \|f - p_m\|_{C([-1,1])} \leq 2C2^{-m\alpha}. \end{align*} The factor $2$ has no significance; it only avoids discussing whether the infimum is attained. Now define the dyadic increments. Let $u_0: [-1,1] \to \mathbb{R}$ be $u_0 = p_0$, and for each $m \geq 1$ define the map $u_m: [-1,1] \to \mathbb{R}$ by \begin{align*} u_m(t) := p_m(t) - p_{m-1}(t). \end{align*} Since $p_m \in \mathcal{P}_{2^m}$ and $p_{m-1} \in \mathcal{P}_{2^{m-1}} \subset \mathcal{P}_{2^m}$, we have $u_m \in \mathcal{P}_{2^m}$. We next estimate the size of $u_m$. For $m \geq 1$, the triangle inequality gives \begin{align*} \|u_m\|_{C([-1,1])} \leq \|p_m - f\|_{C([-1,1])} + \|f - p_{m-1}\|_{C([-1,1])}. \end{align*} Using the defining bounds for $p_m$ and $p_{m-1}$, we obtain \begin{align*} \|u_m\|_{C([-1,1])} \leq 2C2^{-m\alpha} + 2C2^{-(m-1)\alpha}. \end{align*} Since $2^{-(m-1)\alpha} = 2^\alpha 2^{-m\alpha}$, this becomes \begin{align*} \|u_m\|_{C([-1,1])} \leq 2C(1+2^\alpha)2^{-m\alpha}. \end{align*} Thus, with \begin{align*} B := 2C(1+2^\alpha), \end{align*} we have \begin{align*} \|u_m\|_{C([-1,1])} \leq B2^{-m\alpha} \end{align*} for every $m \geq 1$. Finally, the finite telescoping identity is \begin{align*} u_0 + \sum_{m=1}^{M} u_m = p_0 + \sum_{m=1}^{M}(p_m - p_{m-1}) = p_M. \end{align*} Since $\|f-p_M\|_{C([-1,1])} \leq 2C2^{-M\alpha} \to 0$, the partial sums converge uniformly to $f$. Therefore \begin{align*} f = u_0 + \sum_{m=1}^{\infty} u_m \end{align*} uniformly on $[-1,1]$.[/guided]

[step:Record the polynomial derivative bounds used at each scale]Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. We use two classical polynomial inequalities on $[-1,1]$ as external inputs: Bernstein's inequality for algebraic polynomials and the [Markov inequality for polynomials](/theorems/514). We state the exact forms used here. If $q \in \mathcal{P}_N$, then \begin{align*} \sqrt{1-t^2}\, |q'(t)| \leq N \|q\|_{C([-1,1])} \end{align*} for every $t \in (-1,1)$, and \begin{align*} \|q'\|_{C([-1,1])} \leq N^2 \|q\|_{C([-1,1])}. \end{align*} For later use, fix $m \geq 1$. Since $u_m \in \mathcal{P}_{2^m}$ and $\|u_m\|_{C([-1,1])} \leq B2^{-m\alpha}$, Bernstein's inequality gives \begin{align*} |u_m'(t)| \leq B2^{m(1-\alpha)}(1-t^2)^{-1/2} \end{align*} for every $t \in (-1,1)$. Markov's inequality gives \begin{align*} \|u_m'\|_{C([-1,1])} \leq B2^{m(2-\alpha)}. \end{align*}[/step]

[guided]This step records exactly which outside polynomial estimates enter the proof. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. We use Bernstein's inequality for algebraic polynomials and the [Markov inequality for polynomials](/theorems/514) on $[-1,1]$ as named external inputs, and we state the precise forms needed. If $q \in \mathcal{P}_N$, Bernstein's inequality says that \begin{align*} \sqrt{1-t^2}\, |q'(t)| \leq N \|q\|_{C([-1,1])} \end{align*} for every $t \in (-1,1)$. Markov's inequality says that \begin{align*} \|q'\|_{C([-1,1])} \leq N^2 \|q\|_{C([-1,1])}. \end{align*} Now fix $m \geq 1$. From the previous step, $u_m \in \mathcal{P}_{2^m}$ and \begin{align*} \|u_m\|_{C([-1,1])} \leq B2^{-m\alpha}. \end{align*} Applying Bernstein's inequality with $q=u_m$ and $N=2^m$ gives, for each $t \in (-1,1)$, \begin{align*} \sqrt{1-t^2}\, |u_m'(t)| \leq 2^m B2^{-m\alpha}. \end{align*} Since $t \in (-1,1)$ implies $1-t^2>0$, division by $\sqrt{1-t^2}$ gives \begin{align*} |u_m'(t)| \leq B2^{m(1-\alpha)}(1-t^2)^{-1/2}. \end{align*} Applying Markov's inequality with the same $q$ and $N$ gives \begin{align*} \|u_m'\|_{C([-1,1])} \leq (2^m)^2 B2^{-m\alpha}=B2^{m(2-\alpha)}. \end{align*}[/guided]

[step:Estimate one dyadic increment between two points]Let $x,y \in [-1,1]$ with $x \neq y$, and set \begin{align*} h := |x-y|. \end{align*} Assume $x < y$. For $m \geq 1$, we estimate $|u_m(y)-u_m(x)|$. Define the endpoint distance function $\rho: [-1,1] \to [0,\infty)$ by \begin{align*} \rho(t) := \min\{t+1, 1-t\}. \end{align*} Since $1-t^2 = (1-t)(1+t)$, we have \begin{align*} 1-t^2 \geq \rho(t) \end{align*} for every $t \in [-1,1]$. Hence Bernstein's inequality gives \begin{align*} |u_m'(t)| \leq B2^{m(1-\alpha)} \rho(t)^{-1/2} \end{align*} for every $t \in (-1,1)$. Split the interval $[x,y]$ into the endpoint layer \begin{align*} A_m := \{t \in [x,y] : \rho(t) < 2^{-2m}\} \end{align*} and its complement \begin{align*} G_m := \{t \in [x,y] : \rho(t) \geq 2^{-2m}\}. \end{align*} On $G_m$, Bernstein's inequality gives \begin{align*} |u_m'(t)| \leq B2^{m(1-\alpha)}2^m = B2^{m(2-\alpha)}. \end{align*} On $A_m$, Markov's inequality gives the same bound: \begin{align*} |u_m'(t)| \leq B2^{m(2-\alpha)}. \end{align*} Therefore \begin{align*} |u_m(y)-u_m(x)| \leq \int_x^y |u_m'(t)|\, d\mathcal{L}^1(t) \leq B2^{m(2-\alpha)}h. \end{align*} This estimate is sufficient for very low dyadic scales. For the sharper interior contribution, Bernstein's inequality also gives \begin{align*} \int_{G_m} |u_m'(t)|\, d\mathcal{L}^1(t) \leq B2^{m(1-\alpha)} \int_x^y \rho(t)^{-1/2}\, d\mathcal{L}^1(t). \end{align*} Since $\rho(t)^{-1/2}$ has integrable endpoint singularities and is monotone toward each endpoint, the elementary one-dimensional estimate \begin{align*} \int_x^y \rho(t)^{-1/2}\, d\mathcal{L}^1(t) \leq 4h^{1/2} \end{align*} holds for all $-1 \leq x < y \leq 1$. Hence \begin{align*} \int_{G_m} |u_m'(t)|\, d\mathcal{L}^1(t) \leq 4B2^{m(1-\alpha)}h^{1/2}. \end{align*} On the endpoint layer $A_m$, Markov's inequality and $\mathcal{L}^1(A_m) \leq 2^{1-2m}$ give \begin{align*} \int_{A_m} |u_m'(t)|\, d\mathcal{L}^1(t) \leq B2^{m(2-\alpha)}2^{1-2m} = 2B2^{-m\alpha}. \end{align*} Thus, for every $m \geq 1$, \begin{align*} |u_m(y)-u_m(x)| \leq 4B2^{m(1-\alpha)}h^{1/2} + 2B2^{-m\alpha}. \end{align*}[/step]

[guided]Fix $x,y \in [-1,1]$ with $x<y$ and define \begin{align*} h := |x-y|. \end{align*} We want an estimate for the single increment $u_m(y)-u_m(x)$, where $m\geq1$. Since $u_m$ is a polynomial, it is continuously differentiable on $[-1,1]$, so the [fundamental theorem of calculus](/theorems/632) on the interval $[x,y]$ gives \begin{align*} |u_m(y)-u_m(x)| \leq \int_x^y |u_m'(t)|\, d\mathcal{L}^1(t). \end{align*} The difficulty is that Bernstein's inequality has an endpoint singularity. Define the endpoint distance function $\rho: [-1,1]\to[0,\infty)$ by \begin{align*} \rho(t) := \min\{t+1,1-t\}. \end{align*} Because $1-t^2=(1-t)(1+t)$ and both factors are non-negative on $[-1,1]$, the product is at least the smaller factor: \begin{align*} 1-t^2 \geq \rho(t). \end{align*} Thus Bernstein's inequality for $u_m\in\mathcal{P}_{2^m}$ and the bound $\|u_m\|_{C([-1,1])}\leq B2^{-m\alpha}$ imply \begin{align*} |u_m'(t)| \leq B2^{m(1-\alpha)}\rho(t)^{-1/2} \end{align*} for every $t\in(-1,1)$. Now split $[x,y]$ into the endpoint layer \begin{align*} A_m := \{t\in[x,y] : \rho(t)<2^{-2m}\} \end{align*} and the complementary good set \begin{align*} G_m := \{t\in[x,y] : \rho(t)\geq2^{-2m}\}. \end{align*} On $G_m$, the lower bound on $\rho(t)$ removes Bernstein's singular factor, giving \begin{align*} |u_m'(t)| \leq B2^{m(1-\alpha)}2^m = B2^{m(2-\alpha)}. \end{align*} On $A_m$, we instead use the global [Markov inequality for polynomials](/theorems/514), which gives \begin{align*} |u_m'(t)| \leq \|u_m'\|_{C([-1,1])} \leq B2^{m(2-\alpha)}. \end{align*} Combining these two bounds over all of $[x,y]$ yields the coarse estimate \begin{align*} |u_m(y)-u_m(x)| \leq B2^{m(2-\alpha)}h. \end{align*} For the sharper estimate, keep Bernstein's singular weight on $G_m$. We have \begin{align*} \int_{G_m}|u_m'(t)|\, d\mathcal{L}^1(t) \leq B2^{m(1-\alpha)}\int_x^y \rho(t)^{-1/2}\, d\mathcal{L}^1(t). \end{align*} The function $\rho(t)^{-1/2}$ has integrable singularities at $-1$ and $1$, and the elementary endpoint estimate gives \begin{align*} \int_x^y \rho(t)^{-1/2}\, d\mathcal{L}^1(t) \leq 4h^{1/2}. \end{align*} Therefore \begin{align*} \int_{G_m}|u_m'(t)|\, d\mathcal{L}^1(t) \leq 4B2^{m(1-\alpha)}h^{1/2}. \end{align*} On $A_m$, Markov's inequality and the measure bound $\mathcal{L}^1(A_m)\leq2^{1-2m}$ give \begin{align*} \int_{A_m}|u_m'(t)|\, d\mathcal{L}^1(t) \leq B2^{m(2-\alpha)}2^{1-2m}=2B2^{-m\alpha}. \end{align*} Adding the contributions from $G_m$ and $A_m$ gives \begin{align*} |u_m(y)-u_m(x)| \leq 4B2^{m(1-\alpha)}h^{1/2}+2B2^{-m\alpha}. \end{align*}[/guided]

[step:Split the dyadic series at the scale comparable to $h$]Choose the integer $M \geq 0$ such that \begin{align*} 2^{-(M+1)} < h \leq 2^{-M} \end{align*} if $0<h\leq 1$. If $1<h\leq 2$, the desired estimate follows from boundedness: \begin{align*} |f(x)-f(y)| \leq 2\|f\|_{C([-1,1])} \leq 2\|f\|_{C([-1,1])}h^{\alpha/2}. \end{align*} Thus assume $0<h\leq 1$. Using the uniformly convergent expansion of $f$, we have \begin{align*} |f(x)-f(y)| \leq |u_0(x)-u_0(y)| + \sum_{m=1}^{M} |u_m(x)-u_m(y)| + \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)|. \end{align*} Since $u_0$ is a fixed polynomial, define \begin{align*} D_0 := \|u_0'\|_{C([-1,1])}. \end{align*} Then \begin{align*} |u_0(x)-u_0(y)| \leq D_0 h \leq D_0 h^\alpha. \end{align*} For the high-scale tail, use the uniform norm estimate: \begin{align*} \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)| \leq 2\sum_{m=M+1}^{\infty} \|u_m\|_{C([-1,1])}. \end{align*} Therefore \begin{align*} \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)| \leq 2B\sum_{m=M+1}^{\infty} 2^{-m\alpha}. \end{align*} Evaluating the geometric series gives \begin{align*} \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)| \leq \frac{2B}{1-2^{-\alpha}}2^{-(M+1)\alpha}. \end{align*} Since $2^{-(M+1)} < h$, this implies \begin{align*} \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)| \leq \frac{2B}{1-2^{-\alpha}}h^\alpha. \end{align*}[/step]

[guided]The point of the dyadic split is to separate two different mechanisms. At degrees much larger than $1/h$, the polynomial increments oscillate at scales smaller than the distance between $x$ and $y$, so derivative estimates are inefficient. For those terms, the uniform size estimate is stronger. At degrees up to roughly $1/h$, derivative estimates are useful. Assume first that $0<h\leq 1$, where \begin{align*} h := |x-y|. \end{align*} Choose $M \geq 0$ so that \begin{align*} 2^{-(M+1)} < h \leq 2^{-M}. \end{align*} This says that $2^M$ is comparable to $1/h$. The [uniform convergence](/page/Uniform%20Convergence) of \begin{align*} f = u_0 + \sum_{m=1}^{\infty} u_m \end{align*} allows us to subtract the two series term by term: \begin{align*} |f(x)-f(y)| \leq |u_0(x)-u_0(y)| + \sum_{m=1}^{M} |u_m(x)-u_m(y)| + \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)|. \end{align*} The term $u_0$ is harmless because it is a fixed polynomial. Define \begin{align*} D_0 := \|u_0'\|_{C([-1,1])}. \end{align*} By the mean value estimate for continuously differentiable functions on an interval, \begin{align*} |u_0(x)-u_0(y)| \leq D_0 h. \end{align*} Since $0<h\leq 1$ and $0<\alpha<1$, we have $h \leq h^\alpha$, so \begin{align*} |u_0(x)-u_0(y)| \leq D_0 h^\alpha. \end{align*} Now consider the high-scale tail, where $m \geq M+1$. Here we do not differentiate. We use only \begin{align*} |u_m(x)-u_m(y)| \leq |u_m(x)| + |u_m(y)| \leq 2\|u_m\|_{C([-1,1])}. \end{align*} The dyadic norm estimate gives \begin{align*} \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)| \leq 2B\sum_{m=M+1}^{\infty}2^{-m\alpha}. \end{align*} This is a geometric series with ratio $2^{-\alpha}$, so \begin{align*} \sum_{m=M+1}^{\infty} |u_m(x)-u_m(y)| \leq \frac{2B}{1-2^{-\alpha}}2^{-(M+1)\alpha}. \end{align*} Finally, the choice of $M$ gives $2^{-(M+1)} < h$, hence \begin{align*} 2^{-(M+1)\alpha} < h^\alpha. \end{align*} Thus the high-scale tail is bounded by \begin{align*} \frac{2B}{1-2^{-\alpha}}h^\alpha. \end{align*}[/guided]

[step:Bound the low-scale contribution by summing Markov estimates]Let $K \geq 0$ be the integer such that \begin{align*} 2^{-(K+1)} < h^{1/2} \leq 2^{-K}. \end{align*} The earlier choice of $M$ gives $2^{-(M+1)} < h \leq 2^{-M}$. Since $0<h\leq 1$ implies $h\leq h^{1/2}$, these two dyadic choices force $K\leq M$: if $K\geq M+1$, then $h^{1/2}\leq 2^{-K}\leq 2^{-(M+1)}<h$, contradicting $h\leq h^{1/2}$. For $1 \leq m \leq K$, the global derivative estimate and the one-dimensional mean value estimate give \begin{align*} |u_m(y)-u_m(x)| \leq B2^{m(2-\alpha)}h. \end{align*} Therefore \begin{align*} \sum_{m=1}^{K}|u_m(y)-u_m(x)| \leq Bh\sum_{m=1}^{K}2^{m(2-\alpha)}. \end{align*} Since $2^{2-\alpha}>1$, evaluating the finite geometric series gives \begin{align*} \sum_{m=1}^{K}|u_m(y)-u_m(x)| \leq \frac{B2^{2-\alpha}}{2^{2-\alpha}-1}h2^{K(2-\alpha)}. \end{align*} The defining inequality $2^{-(K+1)} < h^{1/2}$ implies $2^K < 2h^{-1/2}$, hence \begin{align*} h2^{K(2-\alpha)} \leq 2^{2-\alpha}h^{\alpha/2}. \end{align*} Consequently \begin{align*} \sum_{m=1}^{K}|u_m(y)-u_m(x)| \leq \frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1}h^{\alpha/2}. \end{align*} For $K < m \leq M$, use the uniform norm estimate instead of differentiating: \begin{align*} |u_m(y)-u_m(x)| \leq |u_m(y)| + |u_m(x)| \leq 2B2^{-m\alpha}. \end{align*} Thus \begin{align*} \sum_{m=K+1}^{M}|u_m(y)-u_m(x)| \leq 2B\sum_{m=K+1}^{\infty}2^{-m\alpha}. \end{align*} Evaluating the geometric series gives \begin{align*} \sum_{m=K+1}^{M}|u_m(y)-u_m(x)| \leq \frac{2B}{1-2^{-\alpha}}2^{-(K+1)\alpha}. \end{align*} Since $2^{-(K+1)} < h^{1/2}$, we obtain \begin{align*} \sum_{m=K+1}^{M}|u_m(y)-u_m(x)| \leq \frac{2B}{1-2^{-\alpha}}h^{\alpha/2}. \end{align*} Combining the two ranges, \begin{align*} \sum_{m=1}^{M}|u_m(y)-u_m(x)| \leq \left(\frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1} + \frac{2B}{1-2^{-\alpha}}\right)h^{\alpha/2}. \end{align*}[/step]

[guided]The endpoint behaviour of algebraic polynomials prevents a global derivative estimate of order $2^m\|u_m\|_{C([-1,1])}$. Markov's inequality only gives order $2^{2m}\|u_m\|_{C([-1,1])}$, and that loss of one power is exactly why this argument proves the exponent $\alpha/2$. Choose $K \geq 0$ so that \begin{align*} 2^{-(K+1)} < h^{1/2} \leq 2^{-K}. \end{align*} This new scale is no larger than the previous scale $M$. Indeed, $2^{-(M+1)}<h\leq2^{-M}$ and $0<h\leq1$ imply $h\leq h^{1/2}$. If $K\geq M+1$, then $h^{1/2}\leq2^{-K}\leq2^{-(M+1)}<h$, which contradicts $h\leq h^{1/2}$. Therefore $K\leq M$. For the lower dyadic levels $1 \leq m \leq K$, we differentiate. The polynomial $u_m$ belongs to $\mathcal{P}_{2^m}$ and satisfies $\|u_m\|_{C([-1,1])} \leq B2^{-m\alpha}$. Markov's inequality therefore gives \begin{align*} \|u_m'\|_{C([-1,1])} \leq B2^{m(2-\alpha)}. \end{align*} Applying the mean value estimate on the interval with endpoints $x$ and $y$ gives \begin{align*} |u_m(y)-u_m(x)| \leq B2^{m(2-\alpha)}h. \end{align*} Summing this estimate for $1 \leq m \leq K$ yields \begin{align*} \sum_{m=1}^{K}|u_m(y)-u_m(x)| \leq Bh\sum_{m=1}^{K}2^{m(2-\alpha)}. \end{align*} The ratio of this geometric series is $2^{2-\alpha}>1$, so \begin{align*} \sum_{m=1}^{K}|u_m(y)-u_m(x)| \leq \frac{B2^{2-\alpha}}{2^{2-\alpha}-1}h2^{K(2-\alpha)}. \end{align*} The choice of $K$ gives $2^K < 2h^{-1/2}$, and hence \begin{align*} h2^{K(2-\alpha)} \leq 2^{2-\alpha}h^{\alpha/2}. \end{align*} Thus the differentiated part is bounded by \begin{align*} \frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1}h^{\alpha/2}. \end{align*} For the remaining levels $K<m\leq M$, differentiating would be too expensive. Instead, use the size of the increments: \begin{align*} |u_m(y)-u_m(x)| \leq |u_m(y)| + |u_m(x)| \leq 2B2^{-m\alpha}. \end{align*} Therefore \begin{align*} \sum_{m=K+1}^{M}|u_m(y)-u_m(x)| \leq 2B\sum_{m=K+1}^{\infty}2^{-m\alpha}. \end{align*} This geometric series has ratio $2^{-\alpha}$, so \begin{align*} \sum_{m=K+1}^{M}|u_m(y)-u_m(x)| \leq \frac{2B}{1-2^{-\alpha}}2^{-(K+1)\alpha}. \end{align*} Since $2^{-(K+1)} < h^{1/2}$, this becomes \begin{align*} \sum_{m=K+1}^{M}|u_m(y)-u_m(x)| \leq \frac{2B}{1-2^{-\alpha}}h^{\alpha/2}. \end{align*} Adding the two estimates gives \begin{align*} \sum_{m=1}^{M}|u_m(y)-u_m(x)| \leq \left(\frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1} + \frac{2B}{1-2^{-\alpha}}\right)h^{\alpha/2}. \end{align*}[/guided]

[step:Combine the dyadic estimates to obtain the Hölder bound]Combining the estimates for $u_0$, the low-scale sum, and the high-scale tail, we obtain \begin{align*} |f(x)-f(y)| \leq Lh^{\alpha/2} \end{align*} for every $x,y \in [-1,1]$ with $0<h=|x-y|\leq 1$, where \begin{align*} L := D_0 + \frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1} + \frac{4B}{1-2^{-\alpha}}. \end{align*} Here we used $h^\alpha \leq h^{\alpha/2}$ for $0<h\leq 1$ to absorb the high-scale tail estimate. For $1<h\leq 2$, increasing $L$ if necessary by $2\|f\|_{C([-1,1])}$ gives the same inequality. Therefore there exists $L>0$ such that \begin{align*} |f(x)-f(y)| \leq L|x-y|^{\alpha/2} \end{align*} for all $x,y \in [-1,1]$. Hence $f \in \operatorname{Lip}(\alpha/2)$, as required.[/step]

[guided]We now assemble the three estimates. Assume first that $0<h=|x-y|\leq 1$. The fixed initial polynomial contributed \begin{align*} |u_0(x)-u_0(y)| \leq D_0h^\alpha. \end{align*} Since $0<h\leq 1$ and $0<\alpha<1$, we have $h^\alpha \leq h^{\alpha/2}$, so this is also bounded by $D_0h^{\alpha/2}$. The low-scale estimate gives \begin{align*} \sum_{m=1}^{M}|u_m(y)-u_m(x)| \leq \left(\frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1} + \frac{2B}{1-2^{-\alpha}}\right)h^{\alpha/2}. \end{align*} The high-scale tail estimate gives \begin{align*} \sum_{m=M+1}^{\infty}|u_m(y)-u_m(x)| \leq \frac{2B}{1-2^{-\alpha}}h^\alpha. \end{align*} Again $h^\alpha \leq h^{\alpha/2}$, so the high-scale tail is bounded by \begin{align*} \frac{2B}{1-2^{-\alpha}}h^{\alpha/2}. \end{align*} Adding the three bounds gives \begin{align*} |f(x)-f(y)| \leq Lh^{\alpha/2}, \end{align*} where \begin{align*} L := D_0 + \frac{B2^{2(2-\alpha)}}{2^{2-\alpha}-1} + \frac{4B}{1-2^{-\alpha}}. \end{align*} It remains only to cover pairs with $1<h\leq 2$. For such pairs, \begin{align*} |f(x)-f(y)| \leq 2\|f\|_{C([-1,1])} \leq 2\|f\|_{C([-1,1])}h^{\alpha/2}, \end{align*} because $h^{\alpha/2}\geq 1$. Increasing $L$ by $2\|f\|_{C([-1,1])}$ if necessary gives \begin{align*} |f(x)-f(y)| \leq L|x-y|^{\alpha/2} \end{align*} for all $x,y \in [-1,1]$. By the definition of $\operatorname{Lip}(\alpha/2)$ stated at the beginning of the proof, this proves $f \in \operatorname{Lip}(\alpha/2)$.[/guided]