# Norm on a Hölder's space

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 .

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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
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
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

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 $C^{k,\gamma}(\Omega)$.

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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