Holder Continuous Functions on $[0,1]$ are complete + Banach space

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 $f$, $$ \left\|\sum_{n=0}^{N}f_{n}-f\right\|_{C^{\alpha}} = \left\|\sum_{n=N+1}^{\infty}f_{n}\right\|_{C^{\alpha}} \le \sum_{n=N+1}^{\infty}\|f_{n}\|_{C^{\alpha}}\rightarrow 0 \mbox{ as } N\rightarrow\infty. $$

I personally find it easier to work with an $f$ that has been defined differently. Let $(f_n)_{n=1}^\infty$ be Cauchy in $\Lambda_\alpha([0,1])$. You can use the Cauchy criterion to show $f_n(x)$ converges pointwise to some $f(x)$ for any $x \in [0,1]$ (I'll let you work through the details). So define $f(x)$ to be the pointwise limit in $(\mathbb{R}, |\cdot |)$ as $n \to \infty$ of $f_n(x)$ for all $x \in [0,1]$.

Now we need to show:

1) $f_n \rightarrow f$ in the Hölder norm

2) $f$ is Hölder continuous on $[0,1]$

Task 1 is a bit technical. Consider each term separately. Fix $\epsilon > 0$. Since $f_n$ is Cauchy there exists $N$ such that for all $m,n \geq N$, $|f_n(0) -f_m(0)| < \epsilon/2$. Fix $n$ and let $m \rightarrow \infty$ to get $|f_n(0) -f(0)| < \epsilon/2$.

For the other term we have for all $x,y \in [0,1]$, $m,n \geq N$ as above, $$\frac{|f_n(x) - f_n(y) - (f_m(x) - f_m(y))|}{|x-y|^\alpha} < \epsilon/2$$ Again, fix $m$ and take $n$ to $\infty$. Some rearranging will yield $$ \lim_{n \to \infty} \frac{f_n(x) - f_n(y)}{(x-y)^\alpha} = \frac{f(x) - f(y)}{(x-y)^\alpha}$$ in $(\mathbb{R}, |\cdot |)$. But that implies for $n \geq N$, $$\sup_{x,y \in [0,1], x\neq y} \frac{|f_n(x) - f_n(y) - (f(x) - f(y))|}{|x-y|^\alpha} < \epsilon/2$$
Combining the two terms and by the triangle inequality we get convergence in $\Lambda_\alpha([0,1])$.

Task 2 is pretty quick. By the above results and the reverse triangle inequality, for some $N$, $x,y \in [0,1]$, $n \geq N$, $$\frac{|f(x) - f(y)|}{|x-y|^\alpha} - \frac{|f_n(x) - f_n(y)|}{|x-y|^\alpha} < \epsilon$$ Let $\epsilon = 1$. Since $f_n(x) \in \Lambda_\alpha([0,1])$ there exists a $C>0$ such that $|f_n(x) - f_n(y)| \leq C|x-y|^\alpha$. Use this to show $$\frac{|f(x) - f(y)|}{|x-y|^\alpha} < 1 + C$$ and so $f \in \Lambda_\alpha([0,1])$.

I will use the symbol $||\cdot ||$ for the $\Lambda_\alpha$-norm and $||\cdot ||_\infty$ for the supremum norm.

Suppose $\{f_n\}$ is a Cauchy sequence, and without loss of generality we can assume that $||f_n - f_{n+1}|| < 2^{-n}$ by taking a subsequence. This implies that $$ (*)\qquad|f_{n+1}(x) - f_{n}(x) - f_{n+1}(y) + f_n(y)| \leq 2^{-n} |x-y|^\alpha, \quad \forall x,y\in [0,1]. $$ But $|f_{n+1}(0) - f_n(0)| < 2^{-n}$ as well, so for $m>n$, $$ |f_n(x) - f_m(x)|\leq |f_n(0) - f_m(0)| + |f_n(x) - f_m(x) -f_n(0) + f_m(0)| \leq (1+|x|^\alpha)\sum_{k=n+1}^m \frac{1}{2^k}. $$ This last estimate comes from applying $(*)$ and the triangle inequality for each integer between $n$ and $m$. This shows that $\{f_n(x)\}$ is Cauchy for each $x$, hence we can define $f(x) = \lim_{n\to \infty} f_n(x)$ for each $x$.

Now consider the functions $$ g_n(x) = f_1(x) + \sum_{k=1}^n(f_{k+1}(x)-f_k(x)) = f_{n+1}(x), $$ where the last equality follows by telescoping. We then see that $g_n(x) \to f(x)$ as $n\to \infty$. Observe that $$\sum_{k=1}^\infty |f_{k+1}(x)-f_k(x)|\leq \sum_{k=1}^\infty ||f_{k+1}-f_k||_\infty \leq \sum_{k=1}^\infty ||f_{k+1}-f_k||<\infty,$$ so the series $\sum_{k=1}^\infty (f_{k+1}(x)-f_{k}(x))$ converges absolutely. (We can then rearrange the terms without care.) Then, for any $x,y$, $$\begin{align} \frac{\left|\sum_n(f_{n+1}(x)-f_n(x)) - \sum_n(f_{n+1}(y)-f_n(y))\right|}{|x-y|^\alpha} &\leq \sum_n \frac{|(f_{n+1}(x)-f_n(x)) - (f_{n+1}(y) - f_n(y)) |}{|x-y|^\alpha}\\ &\leq \sum_n ||f_{n+1} - f_n|| < \infty. \end{align}$$

Thus, by taking a supremum over all $x\neq y$ in $[0,1]$, we arrive at $||\sum_{n} f_{n+1} - f_n|| < \infty$, hence it lies in $\Lambda_\alpha$. This implies that $f=f_1 + \sum_n (f_{n+1}-f_n) \in \Lambda_\alpha$.