I am studying for an Analysis prelim and I am stuck on how to show that the following space of Holder continuous functions is complete:
$$\Lambda_{\alpha}([0,1]) = \{f: [0,1] \rightarrow \mathbb{R} \mid \sup_{x,y \in [0,1], x\neq y} \frac{|f(x) – f(y)|}{|x-y|^{\alpha}} < \infty \text{ and } 0< \alpha \leq 1\}$$
with the norm $\| \cdot \|_{\Lambda_{\alpha}}$ defined by
$$\|f\|_{\Lambda_{\alpha}} = |f(0)| + \sup_{x,y \in [0,1], x\neq y} \frac{|f(x) – f(y)|}{|x-y|^{\alpha}}$$
I think its easy to see that $\Lambda_{\alpha}([0,1]) \subset B([0,1])$, i.e., it is a subset of bounded functions on $[0,1]$ which is complete. My thought is that if I let $\{f_n\}$ be a Cauchy sequence in $\Lambda_{\alpha}([0,1])$ and let $f \in B([0,1])$, such that $f_n \rightarrow f$ in the uniform norm, then we can show $f \in \Lambda_{\alpha}([0,1])$ and that $f_n \rightarrow f$ in the Holder norm defined above. Not sure how to show this though. Perhaps I could also somehow show that any absolutely convergent series converges in $\Lambda_\alpha$, but I'm not sure how to do this either. Any help would be appreciated.
Best Answer
An equivalent way to prove that a normed space is complete is to show that every absolutely convergent series converges. That is, suppose $\{ f_{n} \}_{n=1}^{\infty}$ satisfies $\sum_{n=1}^{\infty}\|f_{n}\| < \infty$, and show that $\sum_{n=1}^{\infty}f_{n}$ converges in norm to some $f$ in the space.
In this case, assume $\{ f_{n} \}\subset C^{\alpha}[0,1]$ and $$ \sum_{n=0}^{\infty}\|f_{n}\|_{C^{\alpha}} < \infty. $$ As mentioned $\|f\|_{C} \le \|f\|_{C^{\alpha}}$. Therefore, because $C[0,1]$ is complete, then $\sum_{n=0}^{\infty}f_{n}$ converges in $C$ to some $f$. Furthermore, $f \in C^{\alpha}$ because \begin{align} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}} & = \frac{|\sum_{n}f_{n}(x)-\sum_{n}f_{n}(y)|}{|x-y|^{\alpha}} \\ & \le \frac{\sum_{n}|f_{n}(x)-f_{n}(y)|}{|x-y|^{\alpha}} \\ & = \sum_{n=1}^{\infty}\frac{|f_{n}(x)-f_{n}(y)|}{|x-y|^{\alpha}} \\ & \le \sum_{n=1}^{\infty}\|f_{n}\|_{C^{\alpha}} = M < \infty. \end{align} Hence, $f=\sum_{n=1}^{\infty}f_{n} \in C^{\alpha}$ and $$\left\|\sum_{n=1}^{\infty}f_{n}\right\|_{C^{\alpha}} \le \sum_{n=1}^{\infty}\|f_{n}\|_{C^{\alpha}}.$$ To see that $\sum_{n}f_{n}$ converges in $C^{\alpha}$ to $f$, $$ \left\|\sum_{n=0}^{N}f_{n}-f\right\|_{C^{\alpha}} = \left\|\sum_{n=N+1}^{\infty}f_{n}\right\|_{C^{\alpha}} \le \sum_{n=N+1}^{\infty}\|f_{n}\|_{C^{\alpha}}\rightarrow 0 \mbox{ as } N\rightarrow\infty. $$