[proofplan]
We prove [uniform continuity](/page/Uniform%20Continuity) directly from the global Hölder estimate. The only point requiring separation is the case $C=0$, where all image distances vanish and any positive $\delta$ works. When $C>0$, for a prescribed error $\varepsilon>0$ we choose $\delta=(\varepsilon/C)^{1/\gamma}$, so the Hölder bound turns the condition $d_X(x,y)<\delta$ into $d_Y(f(x),f(y))<\varepsilon$.
[/proofplan]
[step:Unpack the global Hölder estimate]
By hypothesis, there is a constant $C\ge 0$ such that for every pair $x,y\in X$,
\begin{align*}
d_Y(f(x),f(y))\le C\,d_X(x,y)^\gamma.
\end{align*}
We will show that for every $\varepsilon>0$ there exists $\delta>0$ such that, for all $x,y\in X$,
\begin{align*}
d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon.
\end{align*}
This is exactly the definition of uniform continuity of $f:X\to Y$ on $X$.
[/step]
[step:Handle the zero Hölder constant case]
Assume first that $C=0$. Let $\varepsilon>0$ be arbitrary, and define $\delta:=1$. For any $x,y\in X$ satisfying $d_X(x,y)<\delta$, the Hölder estimate gives
\begin{align*}
d_Y(f(x),f(y))\le 0\cdot d_X(x,y)^\gamma=0.
\end{align*}
Since a metric is nonnegative, $d_Y(f(x),f(y))=0$, and hence
\begin{align*}
d_Y(f(x),f(y))<\varepsilon.
\end{align*}
Thus the uniform continuity condition holds when $C=0$.
[/step]
[step:Choose the uniform continuity scale from the Hölder constant]
Assume now that $C>0$. Let $\varepsilon>0$ be arbitrary, and define the positive number
\begin{align*}
\delta:=\left(\frac{\varepsilon}{C}\right)^{1/\gamma}.
\end{align*}
This is well-defined and positive because $\varepsilon>0$, $C>0$, and $\gamma>0$.
Let $x,y\in X$ satisfy $d_X(x,y)<\delta$. Since the function $t\mapsto t^\gamma$ is increasing on $[0,\infty)$ for $\gamma>0$, we have
\begin{align*}
d_X(x,y)^\gamma<\delta^\gamma.
\end{align*}
By the definition of $\delta$,
\begin{align*}
\delta^\gamma=\frac{\varepsilon}{C}.
\end{align*}
Applying the Hölder estimate to the pair $x,y$ gives
\begin{align*}
d_Y(f(x),f(y))\le C\,d_X(x,y)^\gamma<C\,\delta^\gamma=\varepsilon.
\end{align*}
Therefore the same $\delta$ works for all $x,y\in X$.
[guided]
We now prove the main case in the uniform continuity definition. Assume $C>0$, and fix an arbitrary tolerance $\varepsilon>0$. Uniform continuity requires a single number $\delta>0$ that works simultaneously for all pairs $x,y\in X$, so $\delta$ may depend on $\varepsilon$, $C$, and $\gamma$, but it must not depend on $x$ or $y$.
The Hölder estimate tells us that image distances are controlled by $C\,d_X(x,y)^\gamma$. To force this quantity to be smaller than $\varepsilon$, we choose $\delta$ so that $C\delta^\gamma=\varepsilon$. Thus define
\begin{align*}
\delta:=\left(\frac{\varepsilon}{C}\right)^{1/\gamma}.
\end{align*}
This number is positive because $\varepsilon>0$, $C>0$, and $\gamma>0$.
Now let $x,y\in X$ be any points satisfying $d_X(x,y)<\delta$. Distances in a [metric space](/page/Metric%20Space) are nonnegative, so $d_X(x,y)\in[0,\infty)$. Since $\gamma>0$, the power function $t\mapsto t^\gamma$ is increasing on $[0,\infty)$, and therefore
\begin{align*}
d_X(x,y)^\gamma<\delta^\gamma.
\end{align*}
By the definition of $\delta$,
\begin{align*}
\delta^\gamma=\frac{\varepsilon}{C}.
\end{align*}
Using the Hölder estimate for this pair $x,y$, we obtain
\begin{align*}
d_Y(f(x),f(y))\le C\,d_X(x,y)^\gamma<C\,\delta^\gamma=\varepsilon.
\end{align*}
This proves exactly what uniform continuity asks for: for the given $\varepsilon>0$, the single positive number $\delta=(\varepsilon/C)^{1/\gamma}$ works for every pair of points in $X$.
[/guided]
[/step]
[step:Conclude uniform continuity on the whole space]
In the case $C=0$, the preceding zero-constant argument provides a valid $\delta>0$ for every $\varepsilon>0$. In the case $C>0$, the scale
\begin{align*}
\delta=\left(\frac{\varepsilon}{C}\right)^{1/\gamma}
\end{align*}
provides a valid $\delta>0$ for every $\varepsilon>0$. Hence in all cases, for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in X$,
\begin{align*}
d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon.
\end{align*}
Therefore $f:X\to Y$ is uniformly continuous on $X$.
[/step]
