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What does it mean for a continuous function $ f $ on $ \mathbb{R} $ to be Hölder continuous with exponent $ \alpha $ at a point $ x_0 $ ?

I only now the global definition: A function $ f $ on $ \mathbb{R} $ is (globally) Hölder continuous with exponent $ \alpha $ if

$ \sup_{x \neq y} \frac{| f(x) - f(y) |}{ |x - y|^\alpha} < + \infty $

Thanks for the clarification!

Regards, Si

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    @DavideGiraudo: Hmmm... So if one says that Brownian motion is $P$-a.s. nowhere Hölder continuous with exponent \alpha > 1/2 , is the meaning that $P$-a.s. no point $ x_0 $ has a neighborhood $ V $ such that \sup_{x, y \in V, x \neq y} \frac{|f(x) - f(y)|}{|x - y|^\alpha} < + \infty or that no point $ x_0 $ has a neighborhood $ V $ such that \sup_{x \in V, x \neq x_0} \frac{|f(x) - f(x_0)|}{|x - x_0|^\alpha} < + \infty ? Thanks a lot Davide!2012-05-02

2 Answers 2

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as far as I remember, one calls $f$

  • Hölder continuous of exponent $\alpha$ iff \[ \sup_{x,y\in\mathbb R} \frac{|f(x) - f(y)|}{|x-y|^\alpha} <\infty \]
  • locally Hölder continuous of exponent $\alpha$ iff \[ \sup_{x,y\in K} \frac{|f(x) - f(y)|}{|x-y|^\alpha} <\infty \] for each compact $K \subset \mathbb R$
  • Hölder continuous at $x_0$ of exponent $\alpha$ iff \[ \sup_{x\in U} \frac{|f(x) - f(x_0)|}{|x-x_0|^\alpha} <\infty\] for some neighbourhood $U \ni x_0$.
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    @WillieWong thx.2012-05-02
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As usual, the term "local" (or "locally") means that the definition should be restricted to any neighborhood. In you case, $f$ is locally Hölder continuous if, for every interval $(a,b)$ there exists a constant $C>0$ such that $|f(x)-f(y)| \leq C |x-y|^\alpha$ for every $x$, $y \in [a,b]$. Clearly enough, this amounts to considering $f_{|[a,b]}$ instead of $f$ in the global definition. Notice that the constant $C$ depends on $(a,b)$, so the local definition is, in general, strictly different than the global one.