Let
- $(M,d)$ be a metric space;
- $\Lambda\subseteq M$;
- $E$ be a $\mathbb R$-Banach space.
Let $$\left\|f\right\|_{B(\Lambda,\:E)}:=\sup_{x\in\Lambda}\left\|f(x)\right\|_E\;\;\;\text{for }f:\Lambda\to E.$$ It's easy to show that $$B(\Lambda,E):=\left\{f:\Lambda\to E\mid\left\|f\right\|_{B(\Lambda,\:E)}<\infty\right\},$$ equipped with $\left\|\;\cdot\;\right\|_{B(\Lambda,\:E)}$ is a $\mathbb R$-Banach space. Now, let $$[f]_{C^{0+\alpha}(\Lambda,\:E)}:=\sup_{\substack{x,\:y\:\in\:\Lambda\\x\:\ne\:y}}\frac{\left\|f(x)-f(y)\right\|_E}{d(x,y)^\alpha}\;\;\;\text{for }f:\Lambda\to E$$ for $\alpha\in(0,1]$. Moreover, let $$C^{0+\alpha}(\Lambda,E):=\left\{f\in B(\Lambda,E):[f]_{C^{0+\alpha}(\Lambda,\:E)}<\infty\right\}$$ and $$\left\|f\right\|_{C^{0+\alpha}(\Lambda,\:E)}:=\left\|f\right\|_{B(\Lambda,\:E)}+[f]_{C^{0+\alpha}(\Lambda,\:E)}\;\;\;\text{for }f:\Lambda\to E.$$
How can we show that $C^{0+\alpha}(\Lambda,E)$ equipped with $\left\|\;\cdot\;\right\|_{C^{0+\alpha}(\Lambda,\:E)}$ is a $\mathbb R$-Banach space?
Let $(f_n)_{n\in\mathbb N}\subseteq C^{0+\alpha}(\Lambda, E)$ be Cauchy. There is a unique $f\in B(\Lambda,E)$ with $$\left\|f-f_n\right\|_{B(\Lambda,\:E)}\xrightarrow{n\to\infty}0.\tag1$$ Now, $$\frac{\left\|f(x)-f(y)\right\|_E}{d(x,y)^\alpha}=\lim_{n\to\infty}\frac{\left\|f_n(x)-f_n(y)\right\|_E}{d(x,y)^\alpha}\le\lim_{n\to\infty}\left\|f_n\right\|_{C^{0+\alpha}(\Lambda,\:E)}<\infty$$ for all $x,y\in\Lambda$ with $x\ne y$ and hence $f\in C^{0+\alpha}(\Lambda,E)$. Moreover, $$\frac{\left\|(f-f_n)(x)-(f-f_n)(y)\right\|_E}{d(x,y)^\alpha}=\lim_{m\to\infty}\frac{\left\|(f_m-f_n)(x)-(f_m-f_n)(y)\right\|_E}{d(x,y)^\alpha}\le\limsup_{m\to\infty}\left\|f_m-f_n\right\|_{C^{0+\alpha}(\Lambda,\:E)}\xrightarrow{n\to\infty}0$$ for all $x,y\in\Lambda$ with $x\ne y$ and hence $f\in C^{0+\alpha}(\Lambda,E)$ with $$[f-f_n]_{C^{0+\alpha}(\Lambda,\:E)}\xrightarrow{n\to\infty}0.$$
We should be done. However, I've seen other people proving the claim in a different way. First of all, they assume that $\Lambda$ is open. Why should this be necessary?
Moreover, they claim that the $f$ in my proof above can be chosen from $B(\overline\Lambda,E)$ and that $(1)$ holds with respect to $\left\|\;\cdot\;\right\|_{B(\overline\Lambda,\:E)}$.
It's clear to me that (without the openness assumption), each $g\in C^{0+\alpha}(\Lambda,E)$ admits a unique extension $\tilde g\in C^{0+\alpha}(\overline\Lambda,E)$ with $$[\tilde g]_{C^{0+\alpha}(\overline\Lambda,\:E)}=[g]_{C^{0+\alpha}(\Lambda,\:E)}.$$ But I don't see why (if $f$ is chosen to be its unique extension $\tilde f$) the claimed convergence should hold (since we should only know that $\left\|f-f_n\right\|_{B(\Lambda,\:E)}\le\left\|f-f_n\right\|_{B(\overline\Lambda,\:E)}$ and we only know that the left-hand side converges as $n\to\infty$).
However, as I said before, I don't see why this should be of any use at all. So, am I missing something?