Hölder continuous functions are uniformly continuous

Question

I want to show that Hölder continuous functions are uniformly continuous using $\epsilon-\delta$. Is it sufficient to find a $\delta>0$ that does not depend on $"x"$ and depends only on $"\epsilon"$? Furthermore is my method correct? Here is the context:

Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces.

Let $\quad f:X\rightarrow Y \quad$ be a Holder continuous function.

Let $\,\epsilon>0,\quad\alpha>0,\quad C>0, \quad \delta =\frac\epsilon{2C}, \quad \gamma>0,\quad x_1,x_2\in X$

Let $\gamma=\delta^{\frac1 \alpha}<1$

As $f$ is continuous we have: $\,\,d_x(x_1,x_2)<\gamma\implies d_y(f(x_1),f(x_2))<\epsilon$

So in particular: $\,\,d_x(x_1,x_2)^{\alpha}<\delta\implies d_y(f(x_1),f(x_2))<\epsilon$

As $f$ is Hölder continuous, we have:

$\,\,d_y(f(x_1),f(x_2)\le C*d_x(x_1,x_2)^{\alpha}<C*\delta=C*\frac\epsilon{2C}<\epsilon$

As $\delta$ does not depend on $"x"$, and this is true for all $"\epsilon>0"$, then $f$ is uniformly continuous. $CQFD.$

Answer 1

As $f$ is continuous we have: $d_x(x_1,x_2)<\gamma\implies d_y(f(x_1),f(x_2))<\epsilon$

There are two problems here. First, why is this true for this specific $\gamma$ you chose? Second, if it's true for some choice of $\gamma$ then you already have uniform continuity, so why continue the proof?

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Correct approach. The input is: $\alpha, C$ such that $d_y(f(x_1),f(x_2)\le C\,d_x(x_1,x_2)^{\alpha}$, and an arbitrary $\epsilon>0$.

From this we must come up with $\delta$. I would use $\delta = (\epsilon/C)^{1/\alpha}$; then the verification of $$d_x(x_1,x_2)<\delta\implies d_y(f(x_1),f(x_2))<\epsilon$$ should be easy.