# Holder condition?

• Nov 3rd 2010, 07:17 AM
zebra2147
Holder condition?
My professor did not discuss this topic in class but it came up in his notes. Any help would be appreciated.

He states that a function $f$ satisfies a Holder condition of order $\alpha$, if there are constants $\alpha >0$ and $M>0$, such that $|f(x)-f(y)|\leq M|x-y|^\alpha$ for all $x$ and $y$.

Now, if $f$ is Holder, prove that $f$ is uniformly continuous.
• Nov 3rd 2010, 08:09 AM
Drexel28
Quote:

Originally Posted by zebra2147
My professor did not discuss this topic in class but it came up in his notes. Any help would be appreciated.

He states that a function $f$ satisfies a Holder condition of order $\alpha$, if there are constants $\alpha >0$ and $M>0$, such that $|f(x)-f(y)|\leq M|x-y|^\alpha$ for all $x$ and $y$.

Now, if $f$ is Holder, prove that $f$ is uniformly continuous.

What leads to you have? This is (at least my solution is) a messy problem. I broke it into two cases dependent upon $\alpha>1$ and $\alpha\geqslant 1$. Can you possibly do either of those and we'll help with the other?
• Nov 3rd 2010, 09:16 AM
Jose27
Quote:

Originally Posted by Drexel28
What leads to you have? This is (at least my solution is) a messy problem. I broke it into two cases dependent upon $\alpha>1$ and $\alpha\geqslant 1$. Can you possibly do either of those and we'll help with the other?

Why messy? Just use the same proof as in the case of Lipschitz functions (with obvious adjustments).

Spoiler:
Actually for $\alpha >1$ your function is differentiable and this last vanishes identically
• Nov 3rd 2010, 09:28 AM
Drexel28
Quote:

Originally Posted by Jose27
Why messy? Just use the same proof as in the case of Lipschitz functions (with obvious adjustments).

Spoiler:
Actually for $\alpha >1$ your function is differentiable and this last vanishes identically

Perhaps you're right. Now that I look at it again, I see I took an extra unnecessary step.
• Nov 3rd 2010, 03:00 PM
zebra2147
Well I have never heard of Lipschitz functions and we never talked about them in class but I looked it up and it seems to be exactly the same type of problem that I have here. Could you point me in the direction on hoe to use it? I found a proof similar to this...Does it work?

$|f(x)-f(c)|\leq M|x-c|^\alpha$
$limx_{x\rightarrow c}|f(x)-f(c)|\leq Mlim_{x\rightarrow c}|x-c|^\alpha=0$
This implies
$lim_{x\rightarrow c}f(x) = f(c)$.
• Nov 3rd 2010, 03:21 PM
Jose27
Quote:

Originally Posted by zebra2147
Well I have never heard of Lipschitz functions and we never talked about them in class but I looked it up and it seems to be exactly the same type of problem that I have here. Could you point me in the direction on hoe to use it? I found a proof similar to this...Does it work?

$|f(x)-f(c)|\leq M|x-c|^\alpha$
$limx_{x\rightarrow c}|f(x)-f(c)|\leq Mlim_{x\rightarrow c}|x-c|^\alpha=0$
This implies
$lim_{x\rightarrow c}f(x) = f(c)$.

That proves that the function is continous, to show it's uniformly cont. you need to show a $\delta$ that works for all c. To see this pick an arbitrary c and in the definition of continuity pick $\delta < \left( \frac{\varepsilon }{M} \right) ^{\frac{1}{\alpha }}$ (notice this does not depend on c) and finish the argument.