Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to prove that Hölder space is a Banach space under the "Hölder Norm" ie. $\|\cdot\|_{C^{k,\alpha}}$. Any hints would be appreciable .

share|improve this question
3  
Have you tried to? What have you tried to do? –  Siminore May 7 '12 at 16:53
    
There is a proof of this in Evans's "Partial Differential Equations", at chapter five. He discuss there lots of properties of space of functions. –  matgaio May 7 '12 at 16:53
1  
thats the exact book that i am following , and its left as exercise . I am stuck with what idea to start or on which line i should be thinking . –  Theorem May 7 '12 at 17:02
2  
Sorry, I thought it was there as a theorem. The average way of proving completeness of space of functions is by taking a Cauchy sequence and electing a candidate to be the limit function. In this case, $\|u\|_{C^{k,\alpha}}=\sum_{|\alpha|\leqslant k}\|D^\alpha u\|_\infty+\sum_{|\alpha|=k}[D^\alpha u]_{C^{0,\alpha}}$. Then, if you take a Cauchy sequence $u_n$, it will be uniformly convergent to a function $u$ up to derivatives of order lesser than $k$, because of the first term of the norm. You need now to work on proving that this limit is in the Hölder space. –  matgaio May 7 '12 at 17:14

1 Answer 1

up vote 7 down vote accepted

HINT:

If I assume you already have proved or known that

Denote $C^k(\Omega)$ as the space for functions bounded and continuous up to $k$-th derivative, $C^k(\Omega)$ equipped with the $\sum\limits_{|\alpha|\leq k} \sup\limits_{x\in \Omega} |\partial_{\alpha}(\cdot)|$ is a Banach space.

Then you need to check if the following facts still hold after the introduction of that extra $C^{0,\gamma}(\Omega)$-seminorm:

  • Check if $[\cdot]_{\gamma}$ is a seminorm on $C^{0,\gamma}(\Omega)$, ie triangle inequality and linearity, this would imply that $\| u\|_{C^k(\Omega)} + \sum\limits_{|\alpha|= k}[\partial_{\alpha} u]_{\gamma}$ is a norm.

  • Check for any Cauchy sequence $\{u_n\}$ in $C^{k,\gamma}(\Omega)$, it will converge to a limit also lying in $C^{k,\gamma}(\Omega)$. This could be done as matgaio suggested in his comment, Being Cauchy in $C^{k,\gamma}(\Omega)$-norm implies being Cauchy in $C^{k}(\Omega)$-norm, knowing the completeness of $C^{k}(\Omega)$, we know there exists a limit $u$, we would like to show this $u$ lies in $C^{k,\gamma}(\Omega)$ too, ie for any $|\alpha| = k$, $[\partial_{\alpha} u]_{\gamma} < \infty$.

  • Last but not least, because the previous argument only deals with the $C^k(\Omega)$-limit $u$ of a $C^{k,\gamma}(\Omega)$ Cauchy sequence lies in $C^{k,\gamma}(\Omega)$, we now need to check the $C^{k,\gamma}(\Omega)$-limit of $u_n$ is still $u$, since the first $k$-derivative's convergence is already guaranteed, it suffices to show that for any $|\alpha|=k$, $[\partial_{\alpha} u_n - \partial_{\alpha} u]_{\gamma} \to 0$.

And these three together with $C^{k}(\Omega)$ being Banach would imply $C^{k,\gamma}(\Omega)$ is Banach. If you have any question about some specific proof, I could edit my answer with some details about the proof.


EDIT: How do we prove

$C^k(\Omega)$ equipped with the $\sum\limits_{|\alpha|\leq k} \sup\limits_{x\in \Omega} |\partial_{\alpha}(\cdot)|$ is a Banach space.

For this claim we first need to prove that $(C(\Omega), \sup\limits_{x\in \Omega} |\cdot|)$ is Banach, normally we say here $C(\Omega)$ denotes the bounded continuous function, if $\Omega\subset \mathbb{R}^d$ is closed and bounded already, we could remove the "bounded" part. To prove this, we need to check:

  • $\sup\limits_{x\in \Omega} |\cdot|$ is a norm on $C(\Omega)$, ie the validities of following relations are left for you to check $$ \sup\limits_{x\in \Omega} |u+v| \leq \sup\limits_{x\in \Omega} |u| + \sup\limits_{x\in \Omega} |v| $$

$$ \sup\limits_{x\in \Omega} |u| = 0 \text{ if and only if } u = 0 $$

$$ \sup\limits_{x\in \Omega} |\lambda u| = |\lambda|\,\sup\limits_{x\in \Omega} | u| $$

  • You need to verify that under this supreme norm, $C(\Omega)$ is complete, ie, choose any Cauchy sequence $\{u_n\}\subset C(\Omega)$ under $\sup\limits_{x\in \Omega} |\cdot|$-norm, the limit is still a bounded continuous function. To check this, define $$ u(x) = \lim_{n\to\infty} u_n(x) $$ as the pointwise limit, and we need to show that $u(x)$ is also a bounded continuous function, ie $u\in C(\Omega)$, you might wanna recall the technique in proving a uniformly convergent sequence of continuous functions converge to a continuous function.

If above are checked, then we could say $(C(\Omega), \sup\limits_{x\in \Omega} |\cdot|)$ is Banach, and for $\left(C^k(\Omega),\sum\limits_{|\alpha|\leq k} \sup\limits_{x\in \Omega} |\partial_{\alpha}(\cdot)|\right)$, the sum of the supreme of every derivative's absolute value being a norm is not hard to check. For the completeness part, the sequence $\{u_n\}$ being Cauchy implies that $\{\partial_{\alpha} u_n\}_{|\alpha| = i}$ for any $i\leq k$ is Cauchy, use above argument for $C(\Omega)$, we know that $\{u_n\}$ and every $\{\partial_{\alpha} u_n\}_{|\alpha| = i}$ would converge to a bounded continuous function, the rest is to check the limits coincide, ie: $$ \text{If } u = \lim_{n\to \infty} u_n, v = \lim_{n\to \infty} \partial_{\alpha}u_n. \;\text{Then } v = \partial_{\alpha} u $$

Once you have checked all of these, you have shown $$\left(C^k(\Omega),\sum\limits_{|\alpha|\leq k} \sup\limits_{x\in \Omega} |\partial_{\alpha}(\cdot)|\right)\text{ is a Banach space.}$$ Then refer to the first part about $C^{k,\gamma}(\Omega)$.

share|improve this answer
    
it would be nice if you could explain the first line , ie what u have assumed i already know. –  Theorem May 7 '12 at 19:12
    
@Vedananda I have edited more instructions and hints into my answer, please let me know if you have trouble proving anything specific. –  Shuhao Cao May 8 '12 at 1:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.