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Let $\Omega\subset\mathbb R^n$ be compact and $C^{0,\alpha}(\Omega)$ the space of all $\alpha$-Hölder-continuous functions. Define $||u||_{C^{0,\alpha}(\Omega)}:=||u||_{\sup}+\sup\limits_{{x,y\in \Omega\space\&\space x\ne y}}\frac{|u(x)-u(y)|}{|x-y|^\alpha}$ and consider $(C^{0,\alpha}(\Omega),||u||_{C^{0,\alpha}(\Omega)})$ and $\alpha\in]0,1]$ .

How can you prove that for any sequence in bounded closed set of $(C^{0,\alpha}(\Omega),||u||_{C^{0,\alpha}(\Omega)})$ there exists a convergent subsequence (concerning the uniform norm) and it limes is in $(C^{0,\alpha}(\Omega))$?

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Subsequence of what? – Norbert Jun 15 '12 at 22:10
@Norbert of any function on a bounded closed subset of $(C^{0,\alpha}(\Omega),||u||_{C^{0,\alpha}(\Omega)})$ – sheldoor Jun 15 '12 at 22:16
May be you meant that for any sequence in bounded closed set there exist convergent subsequence such that ... – Norbert Jun 15 '12 at 22:23
Look up Arzelà-Ascola theorem (e.g. on Wikipedia). – D. Thomine Jun 15 '12 at 22:25
@Norbert : I am questioning to myself how could i get equi- continuity ? – Theorem Jun 15 '12 at 22:36
up vote 2 down vote accepted


1) Using estimations of norms prove that if set $F$ is bounded in $(C^{0,\alpha}(\Omega),\Vert\cdot\Vert_{C^{0,\alpha}(\Omega)})$ then it is bounded in $(C^{0}(\Omega),\Vert\cdot\Vert_{C^{0}(\Omega)})$

2) If $F$ is bounded in $(C^{0,\alpha}(\Omega),\Vert\cdot\Vert_{C^{0,\alpha}(\Omega)})$ then $$ \exists C>0\quad\forall u\in F\quad \forall x,y\in\Omega\quad |u(x)-u(y)|\leq C|x-y|^\alpha $$

3) Prove that 2) implies equicontinuity

4) From 1) and 4) you see that $F$ is relatively compact in $(C^{0}(\Omega),\Vert\cdot\Vert_{C^{0}(\Omega)})$.

5) If you get to this paragraph it is remains to prove uniform convergence. Using estimations of norms prove that if sequence $\{u_n:n\in\mathbb{N}\}$ converges in $(C^{0,\alpha}(\Omega),\Vert\cdot\Vert_{C^{0,\alpha}(\Omega)})$, then it converges in $(C^{0}(\Omega),\Vert\cdot\Vert_{C^{0}(\Omega)})$.

6) From statement of paragraph 2) prove that the limit function is $\alpha$-Hölder-continuous.

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I didn't get what $C^0(\Omega)$ means . is it same as taking $\alpha =1$ – Theorem Jun 15 '12 at 23:03
I don't know what is $\gamma$, but $(C^0(\Omega),\Vert\cdot\Vert_{C^0(\Omega)})$ is the space of continuous functions on $\Omega$ with $\sup$ norm. – Norbert Jun 15 '12 at 23:05
sorry , i meant $\alpha$. – Theorem Jun 15 '12 at 23:08
@Norbert isn't it enough to say that all Hölder-continuous functions are uniformly continuous so it follows (1) ? – sheldoor Jun 16 '12 at 20:46
@sheldoor, I don't think so. Generally spaeking the family of uniformly continuous functions is not uniformly bounded - consider constants. – Norbert Jun 16 '12 at 20:49

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