How to prove that Holder space is normed linear space

Question

Can you some one please tell how to prove Holder Space is Normed Linear Space

The Holder Space $C^{k,\gamma}(\bar{U})$ consisting of the all $u \in C^k(\bar{U})$ for which the norm

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

is finite

Definition 1:

If $u:U\to \mathbb{R}$ is bounded and continuous , we write

$$\|u\|_{C(\bar{U})}:=\sup_{x\in U}|u(x)|.$$

Definition 2

The $\gamma^{th} -$ Holder 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 $\gamma^{th} -$ Holder Norm is

$$\|u\|_{C^{0,\gamma}(\bar{U})}:=\|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})}.$$

and please explain those norms ..I was trying to understand things but i can't thank you very much

Answer 1

Hint: \begin{align} \frac{|u(x)-u(y)+v(x)-v(y)|}{|x-y|} \leq \frac{|u(x)-u(y)|}{|x-y|}+\frac{|v(x)-v(y)|}{|x-y|} \end{align} and \begin{align} \sup\{a_n+b_n\} \leq \sup\{a_n\}+\sup\{b_n\}. \end{align}

Answer 2

One way to show completeness of a normed linear space $X$ is to show that every absolutely summable sequence is summable. That is, assume $\{ x_n \} \subset X$ satisfies $\sum_{n}\|x_n\| < \infty$, and show that $\sum_{n}x_n$ converges to some $x\in X$. IF this is true for every summable sequence, then $X$ is complete. This equivalent of completeness is handy in cases like yours.