[proofplan] We prove the Hölder estimate directly from the two defining inequalities. For arbitrary points $x,x'\in X$, first apply the Hölder estimate for $g$ to the pair $f(x),f(x')\in Y$. Then apply the Hölder estimate for $f$ and use monotonicity of the map $t\mapsto t^\beta$ on $[0,\infty)$. Combining the constants and powers gives the desired exponent $\alpha\beta$ and constant $C_gC_f^\beta$. [/proofplan]

[step:Apply the Hölder estimate for $g$ to the images of two points in $X$] Let $x,x'\in X$ be arbitrary. Since $f:X\to Y$, both $f(x)$ and $f(x')$ belong to $Y$. Applying the Hölder estimate for $g:Y\to Z$ with $y=f(x)$ and $y'=f(x')$ gives \begin{align*} d_Z(g(f(x)),g(f(x')))\le C_g d_Y(f(x),f(x'))^\beta. \end{align*} Equivalently, \begin{align*} d_Z((g\circ f)(x),(g\circ f)(x'))\le C_g d_Y(f(x),f(x'))^\beta. \end{align*} [/step]

[step:Raise the Hölder estimate for $f$ to the power $\beta$]The Hölder estimate for $f:X\to Y$ gives \begin{align*} d_Y(f(x),f(x'))\le C_f d_X(x,x')^\alpha. \end{align*} Both sides are nonnegative because metrics take values in $[0,\infty)$ and $C_f\ge 0$. Since $0<\beta\le 1$, the function $\varphi:[0,\infty)\to[0,\infty)$ given by $\varphi(t)=t^\beta$ is increasing. Therefore raising the preceding inequality to the power $\beta$ preserves the inequality: \begin{align*} d_Y(f(x),f(x'))^\beta\le \left(C_f d_X(x,x')^\alpha\right)^\beta. \end{align*} Using the laws of exponents for nonnegative [real numbers](/page/Real%20Numbers), this becomes \begin{align*} d_Y(f(x),f(x'))^\beta\le C_f^\beta d_X(x,x')^{\alpha\beta}. \end{align*}[/step]

[guided]We now convert the Hölder control of $f$ into the exact power needed for the estimate of $g\circ f$. The defining estimate for $f$ says that, for the already chosen points $x,x'\in X$, \begin{align*} d_Y(f(x),f(x'))\le C_f d_X(x,x')^\alpha. \end{align*} The estimate for $g$ contains the quantity $d_Y(f(x),f(x'))^\beta$, so we need to raise this inequality to the power $\beta$. This is legitimate because both sides are nonnegative: $d_Y(f(x),f(x'))\ge 0$ and $d_X(x,x')^\alpha\ge 0$ by the metric axioms and by $0<\alpha\le 1$, while $C_f\ge 0$ by hypothesis. Also, since $0<\beta\le 1$, the map $\varphi:[0,\infty)\to[0,\infty)$ given by $\varphi(t)=t^\beta$ is increasing on $[0,\infty)$. Hence applying $\varphi$ to both sides preserves the inequality: \begin{align*} d_Y(f(x),f(x'))^\beta\le \left(C_f d_X(x,x')^\alpha\right)^\beta. \end{align*} Finally, because $C_f$ and $d_X(x,x')^\alpha$ are nonnegative real numbers, the exponent laws apply without sign ambiguity: \begin{align*} \left(C_f d_X(x,x')^\alpha\right)^\beta=C_f^\beta d_X(x,x')^{\alpha\beta}. \end{align*} Thus \begin{align*} d_Y(f(x),f(x'))^\beta\le C_f^\beta d_X(x,x')^{\alpha\beta}. \end{align*}[/guided]

[step:Combine the two estimates and conclude Hölder continuity of the composition] Substituting the bound from the previous step into the estimate obtained from the [Hölder continuity](/page/H%C3%B6lder%20Continuity) of $g$, and using that $C_g\ge 0$ so multiplication preserves the inequality, we get \begin{align*} d_Z((g\circ f)(x),(g\circ f)(x'))\le C_g C_f^\beta d_X(x,x')^{\alpha\beta}. \end{align*} The points $x,x'\in X$ were arbitrary. Therefore $g\circ f:X\to Z$ is Hölder continuous with exponent $\alpha\beta$ and constant $C_gC_f^\beta$. [/step]