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If $\Omega$ is an open set in $\mathbb{R}^n$, is the Hölder space $C^{k, \alpha}(\Omega)$ Banach? Or is it only that $C^{k, \alpha}(\overline{\Omega})$ is Banach, like with ordinary continuous functions? If not, why is that??? Norms are $$|f|_{C^{0,\alpha}} = \sup_{x,y \in \Omega ,\ x \neq y} \frac{| f(x) - f(y) |}{|x-y|^\alpha},$$ $$|f|_{C^{k, \alpha}} = \|f\|_{C^k}+\max_{| \beta | = k} | D^\beta f |_{C^{0,\alpha}}$$ where $$|f|_{C^k} = \max_{| \beta | \leq k} \, \sup_{x\in\Omega} |D^\beta f (x)|$$

Thanks for any help

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Can you write the norms you are using? Thanks. –  Davide Giraudo Aug 8 '12 at 18:10
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Hi @DavideGiraudo I just use the Wikipedia norms. –  hopo Aug 8 '12 at 18:20
    
And I guess $C^{k,\alpha}(\Omega)$ is the set of functions $k$ times continuously differentiable, with Hölder continuous derivatives, and such that all the derivative of order $\leq$ are bounded in $\Omega$. –  Davide Giraudo Aug 8 '12 at 18:27
    
If we take a Cauchy sequence in $C^{k,\alpha}(\Omega)$, say $\{f_j\}$, then for each index $\beta$, $\{D^{\beta}f_j\}$ is Cauchy in $C_b(\Omega)$, set of continuous bounded functions on $\Omega$ endowed with the supremum norm. It's a Banach space, hence we can define $g^{\beta}$ as the uniform limit of $D^{\beta}f_j$. Using fundamental theorem of analysis, we can check that $g^{\beta}=D^{\beta}g$. –  Davide Giraudo Aug 8 '12 at 18:35
    
@DavideGiraudo Thanks. Is it correct: if we have an open set, we must constrain the derivatives to be bounded for space to be Banach. If we have a compact set, it doesn't matter since derivatives will be automatically be bounded so we will have banach. Hence it is possible for $C^k(\Omega)$ to be Banach space where $\Omega$ is open. –  hopo Aug 8 '12 at 19:07

2 Answers 2

up vote 2 down vote accepted

Note that all $g\in C^{k,\alpha}(\Omega)$ and $\beta\in\mathbb{N}_0^k$ we have $$ \Vert \partial^\beta g\Vert_{C_b(\Omega)}\leq\Vert g\Vert_{C^{k,\alpha}(\Omega)}\tag{1} $$ $$ |\partial^\beta g|_{C^{0,\alpha}}\leq \Vert g\Vert_{C^{k,\alpha}(\Omega)}\tag{2} $$

Let $\{f_n:n\in\mathbb{N}\}$ be a Cauchy sequence in $C^{k,\alpha}(\Omega)$, then for all $\varepsilon>0$, we have $N\in\mathbb{N}$ such that $m,n>N$ implies $$ \Vert f_n-f_m\Vert_{C^{k,\alpha}(\Omega)}<\varepsilon\tag{3} $$ Fix multi-index $\beta\in\mathbb{N}_0^k$ such that $|\beta|\leq k$. From $(1)$ and $(3)$ it follows that for all $\varepsilon>0$, we have $N\in\mathbb{N}$ such that $m,n>N$ implies $\Vert \partial^\beta f_n-\partial^\beta f_m\Vert_{C_b(\Omega)}<\varepsilon$. This means that $\{\partial^\beta f_n: n\in\mathbb{N}\}$ is a Cauchy sequence in $C_b(\Omega)$. Since $C_b(\Omega)$ is complete, then $\{\partial^\beta f_n: n\in\mathbb{N}\}$ converges to some function in $C_b(\Omega)$. By $\varphi$ we denote limit of $\{ f_n:n\in\mathbb{N}\}$ in $C_b(\Omega)$.

Now we use the following standard result

Let $\{f_n:n\in\mathbb{N}\}\subset C(\Omega)$ be a family of differentiable functions, such that

1) the sequence $\{\partial_{x_i}f_n:n\in\mathbb{N}\}$ converges in $C_b(\Omega)$ to $g\in C_b(\Omega)$.

2) for some point $x\in\Omega$ the sequence $\{f_n(x_0):n\in\mathbb{N}\}$ is convergent

Then $\{f_n:n\in\mathbb{N}\}$ converges in $C_b(\Omega)$ to some differentiable function $f\in C_b(\Omega)$, and moreover $\partial_{x_i} f=g$.

Using this result and induction by multi-indices one can show that for all $\beta\in\mathbb{N}_0^k$ with $|\beta|\leq k$ the sequence $\{\partial^\beta f_n:n\in\mathbb{N}\}$ uniformly converges to $\partial^\beta \varphi$. This means that $$ \lim\limits_{n\to\infty}\Vert f_n-\varphi\Vert_{C^k(\Omega)}=0\tag{4} $$ From $(2)$, $(3)$ and deinition of $|\cdot|_{C^{0,\alpha}(\Omega)}$ it follows that for all $\beta\in\mathbb{N}_0^k$ with $|\beta|= k$ and all $x,y\in\Omega$ such that $x\neq y$ we have $$ \frac{|\partial^\beta f_n(x) - \partial^\beta f_m(y)|}{|x-y|^\alpha}<\varepsilon $$ Let's take $m\to\infty$ in this inequality, then we get $$ \frac{|\partial^\beta f_n(x) - \partial^\beta \varphi(y)|}{|x-y|^\alpha}<\varepsilon $$ Since $\beta\in\mathbb{N}_0^k$ with $|\beta|= k$ and $x,y\in\Omega$ such that $x\neq y$ are arbitrary we can say that $$ \max\limits_{|\beta|=k}|\partial^\beta f_n-\partial^\beta \varphi|_{C^{0,\alpha}(\Omega)}<\varepsilon $$ Thus for all $\varepsilon>0$ we have $N\in\mathbb{N}$ such that $n>N$ implies $$ \max\limits_{|\beta|=k}|\partial^\beta f_n-\partial^\beta \varphi|_{C^{0,\alpha}(\Omega)}<\varepsilon $$ This means that $$ \lim\limits_{n\to\infty}\max\limits_{|\beta|=k}|\partial^\beta f_n-\partial^\beta \varphi|_{C^{0,\alpha}(\Omega)}=0\tag{5} $$ From $(4)$ and $(5)$ it follows that $\{f_n:n\in\mathbb{N}\}$ converges to $\varphi$ in $C^{k,\alpha}(\Omega)$. Since we showed that arbitrary Cauchy sequence in $C^{k,\alpha}(\Omega)$ is convergent, then $C^{k,\alpha}(\Omega)$ is complete.

Proof of the completeness of $C^{k,\alpha}(\overline{\Omega})$ is already discussed in comments.

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Thank you, very much appreciate it. –  hopo Aug 8 '12 at 21:06
    
@hopo, You are welcome! –  userNaN Aug 8 '12 at 21:07

Just so that you will be aware, if both notations $C^{k, \alpha}(\Omega)$ and $C^{k, \alpha}(\bar{\Omega})$ are used in the same book, there is a chance that they are actually different. The latter is the Banach space you have in mind. The former may mean a locally Hölder space, that is $$ C^{k, \alpha}(\Omega) = \{u\in C(\Omega):\textrm{for any compact set }K\subset\Omega, u\in C^{k, \alpha}(K)\}. $$ This is a Fréchet space, and the situation is of course analogous to the difference between $C(\Omega)$ and $C(\bar\Omega)\equiv C_b(\Omega)$.

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Thanks timur! $$ –  hopo Aug 12 '12 at 20:02
    
Please vote up the answer if you think it was helpful. –  timur Aug 12 '12 at 22:04

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