Suppose we have a bounded metric space $(X,d)$ and a countable measurable partition $P$ of $X$ (with respect to some probability measure $\mu$). We know that the Holder space with exponent $\alpha$, denoted $C^\alpha(X)$, consisting of those functions $f:X\to \mathbb{R}$ with $\|f\|_\alpha<\infty$, where $\|f\|_\alpha=\|f\|_\infty +|f|_\alpha$ (here $\|f\|_\infty=\sup\limits_{x\in X}|f(x)|$ and $|f|_\alpha=\sup\limits_{x\neq y}\frac{|f(x)-f(y)|}{d(x,y)^\alpha}$ ) is a Banach space.
Consider the space $C^{\alpha,loc}(X)$ consisting of those functions $f:X\to\mathbb{R}$ such that $f|_p\in C^\alpha(X)$ for all $p\in P$, where $f|_p$ denotes the restriction of $f$ to $p$. I want to make this into a Banach space with respect to a suitable norm. Furthermore, I want this norm, $\|\cdot\|_{\alpha,loc}$, say, to satisfy $\|f\|_{\alpha,loc}=\|f\|_\alpha$ for all $f\in C^\alpha(X)$. Does anyone have any ideas on what norm I can define? Is this even possible? Thanks!
Edit: Perhaps the following works? $\|f\|_{\alpha,loc}:=\sum\limits_{p\in P}\left\|f|_p\right\|_\alpha \mu(p)$.
There is no natural way to make $C^{\alpha, \rm loc}(X)$ into a Banach space, let alone in a way that has the norm agreeing with $C^\alpha(X)$ norm. You could try the weighted sum (as in the question), or the supremum of $\|f_{|p}\|_\alpha$. Both fail to even be finite on $C^{\alpha, \rm loc}(X)$, since a function may be in $C^\alpha$ on each element of partition, yet have infinite weighted sum (or supremum) of the terms $\|f_{|p}\|_\alpha$. And to expect a piecewise expression to magically turn into $\|f\|_\alpha$ when $f\in C^\alpha(X)$ is entirely unrealistic.
The space $C^{\alpha, \rm loc}(X)$ is a metrizable locally convex space with the topology induced by the countable family of seminorms $\|f_{|p}\|_\alpha$. That's all the useful structure that is has.