Embedding of weighted Holder space into $L^p$ Suppose we have a bounded metric space $(X,d)$ and a countable measurable partition $Q$ of $X$ (with respect to some probability measure $\mu$) such that $\mu(q)>0$ for all $q\in Q$. 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 $\|f\|_\infty=\sup\limits_{x}|f(x)|$ and $|f|_\alpha=\sup\limits_{x\neq y}\frac{|f(x)-f(y)|}{d(x,y)^\alpha}$ ) is a Banach space.
We define $\mathcal{C}$ to consist of those functions $f:X\to \mathbb{R}$ for which $\|f\|=\sum\limits_{q\in Q}\left\|f|_q\right\|_\alpha\mu(q)<\infty$. This can be shown to be a Banach space.
What I want to show is that this Banach space embeds in $\mathcal{L}^p(X)$. However, this is as far as I can get:
For $f\in \mathcal{C}$, $\|f\|_p=\left\|\sum\limits_{q\in Q}f|_q\right\|_p\leq \sum\limits_{q\in Q}\left\|f|_q\right\|_p\leq \sum\limits_{q\in Q}\left\|f|_q\right\|_{\infty}\mu(q)^{1/p}\leq\sum\limits_{q\in Q}\left\|f|_q\right\|_{\alpha}\mu(q)^{1/p}$.
This is not quite $\|f\|$ (due to the presence of $\frac{1}{p}$). Does anyone know how to show this is $\leq c\|f\|$ for some $c>0$? Is this even possible?
Thanks!
 A: The key point is that since $Q$ is a partition, the following equality holds: $\left\lvert f(x)\right\rvert^p=\sum_{q\in Q}\left\lvert f(x)\right\rvert^p\mathbf 1_q\left(x\right)$. Using the fact that 
$$
\left\lvert f(x)\right\rvert\mathbf 1_q\left(x\right) \leqslant 
\left\|f|_q\right\|_\infty \mathbf 1_q\left(x\right)\leqslant \left\|f|_q\right\|_\alpha\mathbf 1_q\left(x\right)$$
we obtain that 
$$
\left\lvert f(x)\right\rvert^p\leqslant \sum_{q\in Q}\left\|f|_q\right\|_\alpha^p\mathbf 1_q\left(x\right).
$$
Integrating ver $X$, we derive that 
$$
\left\lVert f \right\rVert^p\leqslant \sum_{q\in Q}\left\|f|_q\right\|_\alpha^p\mu\left(q\right)$$
which is not exactly what we want. However, we cannot do much better, for example in the case where $f$ is constant equal to a positive $c_q$ on $q$, we have 
$$
\left\lVert f\right\rVert_p=\left(\sum_{q\in Q}c_q^p\mu\left(q\right)\right)^{1/p}\mbox{ and }\left\lVert f\right\rVert = \sum_{q\in Q}c_q \mu\left(q\right) 
$$
hence if $c_q$ are chosen big enough, finiteness of $\left\lVert f\right\rVert$ can happen without that of $\left\lVert f\right\rVert_p$, unless $p=1$.
