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. $CQFD.$


1 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?

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.

  • $\begingroup$ Thank you for your reply! So to be sure, you are saying my $\gamma$ could still be dependent on $"x"$ and $"\epsilon"$ which means my $"\delta"$ could still be dependent on $"x"$ and $"\epsilon"$ when we want it to depend ONLY on $"\epsilon"$. @bro $\endgroup$
    – Kam
    Apr 23, 2018 at 21:37
  • 1
    $\begingroup$ The way you defined $\gamma$, it does not depend on $x$, but you haven't explained how you know $d_x(x_1,x_2)<\gamma\implies d_y(f(x_1),f(x_2))<\epsilon$. If the reason is "because $f$ is continuous" (which is given in that sentence) then $\gamma$ would have to be chosen based on $x_1$ or $x_2$. $\endgroup$
    – user357151
    Apr 23, 2018 at 23:34
  • $\begingroup$ I see exactly where I went wrong!! Instead of supposing I had $d_x(x_1,x_2)<\gamma$ and tried to prove $d_y(f(x_1),f(x_2))<\epsilon$ to be true, I assumed the implication to be true already! (I hope I am correct in this realization haha) Thanks a lot anyway for the help!! $\endgroup$
    – Kam
    Apr 24, 2018 at 10:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .