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I want to show that the fuction $\frac{1}{1+x^2}$ is locally Hölder continuous, I used the mean value theorem, but does not work. Tools that I need to resolve?.

Also does definitión is:

if $f$ is a function $(f:\Omega\longrightarrow\mathbb{R})$, then $f$ locally Hölder continuous iff there are $M$, $\alpha$ such that for each $k\subseteq \Omega $, $k$ compact $$|f(x)-f(y)|\leq M|x-y|^\alpha\;\;\;\;\;\;\forall x,y\in k?$$

Thanks for your help, I do not speak good English and I'm new.

José Carlos Santos
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user317892
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1 Answers1

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The function is locally Hölder continuous because it is $C^1$. In fact it is locally Lipschitz continuous (that is, you can take $\alpha=1$ in each compact set).

John B
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  • Thank you. To show it, I proceed as follows: if $f:(0,+\infty)\longrightarrow \mathbb{R}$ such that $f(x)=\frac{1}{1+x^2}$, then $f$ is continuos in $a,b$ and differentiable in $(a,b)$, so mean value theorem is valid and existe $c\in(a,b)$ such that $$|f(b)-f(a)|\leq f'(c)|b-a|$$, $if ;;;;;;M=f'(c);;;;;;;;|f(x)-f(y)|\leq M|x-y|;;;;;\forall x,y\in[a,b]$$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$Is this right? – user317892 Feb 26 '16 at 03:47
  • Yes, like that is fine. – John B Feb 26 '16 at 09:13
  • Is there another way? – user317892 Feb 26 '16 at 11:31
  • I don't understand what you mean, this is pretty immediate. – John B Feb 26 '16 at 12:17
  • Like this is OK, thanks. – user317892 Feb 26 '16 at 12:38