[proofplan] We prove the estimate directly from the definitions of the supremum norm and the Hölder seminorm. For two points $x,y\in X$, we split the product difference into one term measuring the oscillation of $g$ and one term measuring the oscillation of $f$. The supremum norms control the non-oscillating factors, and taking the supremum over distinct pairs gives the desired seminorm bound. Boundedness of the pointwise product follows from boundedness of $f$ and $g$, so the product belongs to the same Hölder space. [/proofplan]

[step:Define the product and verify boundedness] Define the pointwise product map $h:X\to\mathbb R$ by \begin{align*} h(x)=f(x)g(x). \end{align*} Since $f,g\in C^{0,\gamma}(X;\mathbb R)$, the functions $f$ and $g$ are bounded on $X$. Thus their supremum norms \begin{align*} \|f\|_\infty=\sup_{x\in X}|f(x)|,\qquad \|g\|_\infty=\sup_{x\in X}|g(x)| \end{align*} are finite. For every $x\in X$, the absolute value on $\mathbb R$ is multiplicative, so \begin{align*} |h(x)|=|f(x)g(x)|=|f(x)|\,|g(x)|\le \|f\|_\infty\|g\|_\infty. \end{align*} Taking the supremum over $x\in X$ gives \begin{align*} \|h\|_\infty\le \|f\|_\infty\|g\|_\infty<\infty. \end{align*} Hence $h$ is bounded on $X$. [/step]

[step:Split the product difference into two Hölder-controlled terms]Let $x,y\in X$ with $x\ne y$. Using addition and subtraction of the intermediate scalar $f(x)g(y)$, we obtain \begin{align*} h(x)-h(y)=f(x)g(x)-f(y)g(y)=f(x)(g(x)-g(y))+g(y)(f(x)-f(y)). \end{align*} Taking absolute values and applying the triangle inequality in $\mathbb R$ gives \begin{align*} |h(x)-h(y)|\le |f(x)|\,|g(x)-g(y)|+|g(y)|\,|f(x)-f(y)|. \end{align*} By the definition of the supremum norm and the Hölder seminorm, \begin{align*} |f(x)|\le \|f\|_\infty,\qquad |g(y)|\le \|g\|_\infty, \end{align*} and \begin{align*} |g(x)-g(y)|\le [g]_{C^{0,\gamma}(X)}|x-y|^\gamma,\qquad |f(x)-f(y)|\le [f]_{C^{0,\gamma}(X)}|x-y|^\gamma. \end{align*} Substituting these four bounds into the previous inequality yields \begin{align*} |h(x)-h(y)|\le \left(\|f\|_\infty [g]_{C^{0,\gamma}(X)}+\|g\|_\infty [f]_{C^{0,\gamma}(X)}\right)|x-y|^\gamma. \end{align*}[/step]

[guided]We want to estimate the Hölder seminorm of the product $h=fg$, so we must control the quotient \begin{align*} \frac{|h(x)-h(y)|}{|x-y|^\gamma} \end{align*} for arbitrary distinct points $x,y\in X$. The basic difficulty is that $h(x)-h(y)$ contains two functions varying at once. We separate these variations by adding and subtracting $f(x)g(y)$: \begin{align*} h(x)-h(y)=f(x)g(x)-f(y)g(y)=f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y). \end{align*} Factoring each pair gives \begin{align*} h(x)-h(y)=f(x)(g(x)-g(y))+g(y)(f(x)-f(y)). \end{align*} This identity isolates the oscillation of $g$ in the first term and the oscillation of $f$ in the second term. Applying the triangle inequality for [real numbers](/page/Real%20Numbers) gives \begin{align*} |h(x)-h(y)|\le |f(x)|\,|g(x)-g(y)|+|g(y)|\,|f(x)-f(y)|. \end{align*} Now the non-oscillating factors are controlled by the supremum norms: \begin{align*} |f(x)|\le \|f\|_\infty,\qquad |g(y)|\le \|g\|_\infty. \end{align*} The oscillating factors are controlled by the Hölder seminorms, because $f,g\in C^{0,\gamma}(X;\mathbb R)$: \begin{align*} |g(x)-g(y)|\le [g]_{C^{0,\gamma}(X)}|x-y|^\gamma,\qquad |f(x)-f(y)|\le [f]_{C^{0,\gamma}(X)}|x-y|^\gamma. \end{align*} Combining these estimates gives \begin{align*} |h(x)-h(y)|\le \|f\|_\infty [g]_{C^{0,\gamma}(X)}|x-y|^\gamma+\|g\|_\infty [f]_{C^{0,\gamma}(X)}|x-y|^\gamma. \end{align*} Factoring out $|x-y|^\gamma$ yields \begin{align*} |h(x)-h(y)|\le \left(\|f\|_\infty [g]_{C^{0,\gamma}(X)}+\|g\|_\infty [f]_{C^{0,\gamma}(X)}\right)|x-y|^\gamma. \end{align*} This is exactly the pointwise Hölder estimate needed before taking the supremum over all distinct pairs.[/guided]

[step:Take the supremum over distinct pairs] For every $x,y\in X$ with $x\ne y$, the Euclidean distance satisfies $|x-y|>0$. Dividing the estimate from the previous step by $|x-y|^\gamma$ gives \begin{align*} \frac{|h(x)-h(y)|}{|x-y|^\gamma}\le \|f\|_\infty [g]_{C^{0,\gamma}(X)}+\|g\|_\infty [f]_{C^{0,\gamma}(X)}. \end{align*} Taking the supremum over all distinct $x,y\in X$ gives \begin{align*} [h]_{C^{0,\gamma}(X)}\le \|f\|_\infty [g]_{C^{0,\gamma}(X)}+\|g\|_\infty [f]_{C^{0,\gamma}(X)}. \end{align*} The right-hand side is finite because $f,g\in C^{0,\gamma}(X;\mathbb R)$. Together with the boundedness of $h$ proved above, this shows $h\in C^{0,\gamma}(X;\mathbb R)$. Since $h=fg$, the claimed estimate follows. [/step]