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I am trying to show that the Hölder space $C^{k,\gamma}(\bar{U})$ is a Banach space. To do this, I successfully proved that the mapping $\| \quad \| : C^{k,\gamma}(\bar{U}) \to [0,\infty)$ is a norm, by proving its properties.

But how do I show that the sequence $\{u_k\}_{k=1}^\infty \subset C^{k,\gamma}(\bar{U})$ converges to $u \in C^{k,\gamma}(\bar{U})$, that is, how do I show that $$\lim_{k \to \infty}\|u_k-u\|=0,$$ which would mean the normed linear space is *complete*, and hence a Banach space?

Here are the following taken from PDE Evans, 2nd edition, page 255:

- Cancellation law for Minkowski sums
- Non-closed subspace of a Banach space
- Positive operator is bounded
- Prove that $C^1()$ with the $C^1$- norm is a Banach Space
- Nonnegative linear functionals over $l^\infty$
- Weak net convergence in $\ell_p$, where $1 < p < \infty$.

Definition.The Hölder space $$C^{k,\gamma}(\bar{U})$$ consists of all functions $u \in C^k(\bar{U})$ for which the norm $$\|u\|_{C^{k,y}(\bar{U})}:= \sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})}$$ is finite.

Also from page 254,

Definitions.(i) If $u : U \to \mathbb{R}$ is bounded and continuous, we write $$\|u\|_{C(\bar{U})}:=\sup_{x\in U}|u(x)|.$$(ii) The $\gamma$th-Hölder seminorm of $u : U \to \mathbb{R}$ is $$[u]_{C^{0,\gamma}(\bar{U})}:=\sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|u(x)-u(y)|}{|x-y|^\gamma} \right\},$$ and the $\gamma$th-Hölder norm is $$\|u\|_{C^{0,\gamma}(\bar{U})}:=\|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})}.$$

This is all I got so far:

\begin{align}

\|u_k-u\|_{C^{k,\gamma}(\bar{U})}&=\sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})} \\

&= \sum_{|\alpha|\le k} \sup_{x\in U} |u_k(x)-u(x)|+ \sum_{|\alpha|=k} \sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|[u_k(x)-u(x)]-[u_k(y)-u(y)]|}{|x-y|^\gamma} \right\}.

\end{align}

Now, where can I go from here, to show that $\lim_{k \to \infty}\|u_k-u\|_{C^{k,\gamma}(\bar{U})}=0$? The sequence is Cauchy, and I have to use that fact somehow.

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We have to show the following: given a Cauchy sequence $(u_n)_{n\in \mathbb N}$ in $C^{k,\gamma}(\overline U),$ there is a $u \in C^{k,\gamma}(\overline U)$ such that $\lim_{n\rightarrow\infty}u_n = u$ in $C^{k,\gamma}(\overline U).$

Note that the Hölder norm is the “sum” of the $C^k$ norm (i.e. sup-norm up to the $k$-th derivatives) and the Hölder condition with parameter $\gamma \in (0,1)$ for the $k$-th derivatives. From the $C^k$ part of the Hölder norm and the completeness of $C^k(\overline U),$ we get a $u \in C^k(\overline U)$ with $\lim_{n\rightarrow\infty}u_n = u$ in $C^k(\overline U).$ Now let’s look at the Hölder condition. Let’s fix a multi-index $\alpha$ with $|\alpha| = k$ and write

$$

v := D^\alpha u\qquad and \qquad v_n := D^\alpha u_n

$$

for short. Then we have for fixed $x\neq y$

$$

\begin{align}

\frac{|v(x) – v(y)|}{|x-y|^\gamma} & = \frac{|v(x) – v_m(x) + v_m(x) – v_m(y) + v_m(y) – v(y)|}{|x-y|^\gamma} \\

& \leq \frac{|v(x) – v_m(x)|}{|x-y|^\gamma} + \frac{|v_m(x) – v_m(y)|}{|x-y|^\gamma} + \frac{|v_m(y) – v(y)|}{|x-y|^\gamma}

\end{align}

$$

for arbitrary $m \in \mathbb N.$ Here, the first and third summands on the r.h.s. can be made arbitrarily small by choosing $m$ large enough, since $v_m \rightarrow v$ uniformly, and the second summand on the r.h.s. is bounded independent of $m$ and $x\neq y$ since the sequence $u_n$ is Cauchy, hence bounded, in the Hölder norm. From all this we get a constant $M_\alpha$ such that

$$

\frac{|v(x) – v(y)|}{|x-y|^\gamma} \leq M_\alpha \qquad independent\ of\ x\neq y.

$$

This shows that we have

$$

u\in C^{k,\gamma}(\overline U).

$$

So far so good. Finally, we have to show that $u_n \rightarrow u$ in $C^{k,\gamma}(\overline U).$ From the construction of $u,$ we already have $u_n \rightarrow u$ in $C^k(\overline U),$ so we only have to show convergence in the Hölder seminorms of the $k$-th derivatives. We use the notation $\alpha,v,v_n$ from above. We have

$$

\begin{align}

\frac{|(v-v_m)(x) – (v-v_m)(y)|}{|x-y|^\gamma} & = \frac{|v(x) – v_m(x) – v(y) + v_m(y)|}{|x-y|^\gamma} \\

& = \frac{|(\lim_{k\rightarrow\infty}v_k(x)) – v_m(x) – (\lim_{k\rightarrow\infty}v_k(y)) + v_m(y)|}{|x-y|^\gamma} \\

& = \lim_{k\rightarrow\infty}\frac{|v_k(x) – v_m(x) – v_k(y) + v_m(y)|}{|x-y|^\gamma} \\

& = \lim_{k\rightarrow\infty}\frac{|(v_k – v_m)(x) – (v_k – v_m)(y)|}{|x-y|^\gamma}.

\end{align}

$$

Here, the r.h.s. can be made arbitrarily small *independent of* $x\neq y$ by choosing $m$ large enough, since the sequence $u_n$ is Cauchy in the Hölder norm. So we find $v_n \rightarrow v$ in the Hölder seminorm. Since there are only finitely many multi-indices $\alpha$ (or equivalently, $k$-th derivatives) to consider, we can conclude

$$

u_n \rightarrow u\qquad in\ C^{k,\gamma}(\overline U),

$$

as desired.

Since your comment to my first answer is tricky to read, let me restate it. The follow-up question is: how does

$$

\frac{|v(x)-v(y)|}{|x-y|^\gamma} \le M_\alpha < \infty \qquad independent\ of\ x\neq y \qquad\qquad\qquad(*)$$

for every multi-index $\alpha$ of order $k$ imply

$$

u \in C^{k,\gamma}(\bar{U})?

$$

The answer is as follows: in order to show $u \in C^{k,\gamma}(\bar{U}),$ we have to show that all the ingredients of the Hölder norm of $u$ can be calculated, and the value of the Hölder norm thus obtained is finite.

By construction of $u,$ we have

$$

\sum_{|\beta|\leq k} \|D^\beta u\|_{C(\overline U)} < \infty.

$$

Moreover, (*) above implies that for a fixed multi-index $\alpha$ of order $k,$ we have

$$

[D^\alpha u]_{C^{,\gamma}(\overline U)} \leq M_\alpha < \infty.

$$

From this we get

$$

\sum_{|\alpha| = k}[D^\alpha u]_{C^{,\gamma}(\overline U)} \leq \sum_{|\alpha| = k} M_\alpha < \infty.

$$

So we have all in all

$$

\|u\|_{C^{k,\gamma}(\bar{U})} = \sum_{|\beta|\leq k} \|D^\beta u\|_{C(\overline U)} + \sum_{|\alpha| = k}[D^\alpha u]_{C^{,\gamma}(\overline U)} < \infty

$$

and thus

$$

u \in C^{k,\gamma}(\bar{U}),

$$

as desired.

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