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My analysis is really rusty, so apologies if this is a stupid question.

If $f\in C^1$ in a compact set $\Omega$, does this mean $f$ is Holder continuous for any $\alpha$ in $\Omega$?

I have tried googling but I couldn't find this result,

I have tried to do this myself. take an arbitrary $x$ then we have, for any $\epsilon >0$, we can choose a $\delta\leq 1$

$f'(x) -\epsilon < \dfrac{f(x)-f(y)}{x-y}<f'(x)+\epsilon$

so we have $|f(x)-f(y)|\leq K|x-y|\leq K|x-y|^\alpha$

but this only shows $f$ is Holder at every point inside $K$? Can we show $f$ is Holder on K?

Also are there weaker assumptions than $C^1$, I could have assumed here?

I assume $f$ is $\alpha$ Holder means it is $f$ is $\beta$ Holder for all $\beta<\alpha$?

Lost1
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1 Answers1

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This is a good question.

In particular, you can conclude that $f$ is Lipschitz continuous (Holder continuous with $\alpha = 1$).

Because the derivative is continuous on a compact set, it is uniformly bounded, i.e. $|f'(x)| < K$ for all $x\in\Omega$.

Thus, by the mean value theorem, for any $x,y$, $\exists c$ such that $$ |f(x) - f(y)| \leq |f'(c)||x-y| < K |x-y|. $$

BaronVT
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  • Let me try to piece this together to show Holder continuity, 1 second... – Lost1 Nov 13 '13 at 17:24
  • I think what I really want to ask is why Lipschitz imply Holder... – Lost1 Nov 13 '13 at 17:29
  • Intuitively, the points $x$ and $y$ should be close to each other, in which case $|x-y|<|x-y|^\alpha$ follows directly. In the general case, we just have to throw another constant in there to fix things. In particular $|x-y| = |x-y|^{1-\alpha}|x-y|^\alpha$, so just take the constant to be $C = \textrm{diam}(\Omega)^{1-\alpha} > |x-y|^{1-\alpha}$. – BaronVT Nov 13 '13 at 17:44
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    yeah ^ I figured that after I typed it, but thank you. The points further away are not the problem on a bounded domain – Lost1 Nov 13 '13 at 17:59