[proofplan] The proof is a direct comparison of powers on a bounded set. The case $x=y$ is immediate, and for distinct points we factor $|x-y|^\gamma$ as $|x-y|^\alpha |x-y|^{\gamma-\alpha}$. Boundedness of $X$ gives $|x-y|\le \operatorname{diam}(X)$, and the nonnegative exponent $\gamma-\alpha$ lets us bound the remaining factor uniformly by $\max\{1,\operatorname{diam}(X)^{\gamma-\alpha}\}$. [/proofplan] [step:Verify the estimate when the two points coincide] Let $x,y\in X$ satisfy $x=y$. Then $|f(x)-f(y)|=0$ and $|x-y|^\alpha=0$, so \begin{align*} |f(x)-f(y)|\le C_\gamma\max\{1,\operatorname{diam}(X)^{\gamma-\alpha}\}|x-y|^\alpha. \end{align*} [/step] [step:Factor the higher Hölder power for distinct points] Let $x,y\in X$ satisfy $x\ne y$. Since $0<\alpha\le\gamma$, define \begin{align*} \delta:=\gamma-\alpha. \end{align*} Then $\delta\ge0$, and the positivity of $|x-y|$ gives \begin{align*} |x-y|^\gamma=|x-y|^{\alpha+\delta}=|x-y|^\alpha |x-y|^\delta. \end{align*} [guided] Let $x,y\in X$ be distinct. The reason for separating this case is that $|x-y|>0$, so all powers of $|x-y|$ that appear below are ordinary positive real powers. Define the exponent gap \begin{align*} \delta:=\gamma-\alpha. \end{align*} The hypothesis $0<\alpha\le\gamma$ gives $\delta\ge0$. Since $\gamma=\alpha+\delta$, the usual law of exponents for positive [real numbers](/page/Real%20Numbers) gives \begin{align*} |x-y|^\gamma=|x-y|^{\alpha+\delta}=|x-y|^\alpha |x-y|^\delta. \end{align*} This is the central reduction: the desired $\alpha$-Hölder factor is $|x-y|^\alpha$, and it remains only to control the extra factor $|x-y|^\delta$ by a constant depending on the bounded set $X$. [/guided] [/step] [step:Use boundedness of $X$ to control the extra exponent factor] Let \begin{align*} D:=\operatorname{diam}(X)=\sup\{|a-b|:a,b\in X\}. \end{align*} Because $X$ is bounded and nonempty, $D<\infty$. Since $x,y\in X$, the definition of diameter gives $|x-y|\le D$. If $\delta=0$, then $|x-y|^\delta=1$. If $\delta>0$, monotonicity of the map $t\mapsto t^\delta$ on $[0,\infty)$ gives \begin{align*} |x-y|^\delta\le D^\delta. \end{align*} In both cases, \begin{align*} |x-y|^\delta\le \max\{1,D^\delta\}. \end{align*} [/step] [step:Combine the assumed $\gamma$-Hölder estimate with the diameter bound] Using the assumed estimate for $f:X\to\mathbb R^m$, the factorization from the distinct-point case, and the bound on the extra factor, we obtain \begin{align*} |f(x)-f(y)|\le C_\gamma |x-y|^\gamma. \end{align*} Hence \begin{align*} |f(x)-f(y)|\le C_\gamma |x-y|^\alpha |x-y|^\delta. \end{align*} Therefore \begin{align*} |f(x)-f(y)|\le C_\gamma\max\{1,D^\delta\}|x-y|^\alpha. \end{align*} Substituting $D=\operatorname{diam}(X)$ and $\delta=\gamma-\alpha$ gives \begin{align*} |f(x)-f(y)|\le C_\gamma\max\{1,\operatorname{diam}(X)^{\gamma-\alpha}\}|x-y|^\alpha. \end{align*} Together with the coincident-point case, this proves the asserted estimate for all $x,y\in X$. Thus $f$ is Hölder continuous with exponent $\alpha$ on $X$. [/step]