The relation of the norms of Lipschitz functions Let $\mbox{Lip}_\alpha X$ and $\mbox{lip}_\alpha X$ over metric space $X$ denote the algebra of Lipschitz functions and its subspace. I know that for each $\alpha<1$, $\mbox{Lip}_1 X \subset \mbox{lip}_\alpha X$. But the first question is that what is the relation between their norms? For example, is it correct that $\|\cdot\|_{\mbox{lip}_\alpha X} \leq \|\cdot\|_{\mbox{Lip}_1 X}$?
The second question is about the compact supported elements in $\cal A$ if $\cal A$ is one of $\mbox{lip}_\alpha X$ or $\mbox{Lip}_\alpha X$ for some $\alpha$: 
Can we say that ${\cal A} \cap C_c(X)$ is dense in $\cal A$? Why?
 A: I'm assuming that $X$ is a bounded metric space (on unbounded spaces Lipschitz functions do not form an algebra, and natural inclusions between Lipschitz/Hölder spaces fail). The (semi-)norm on $\mathrm{Lip}_\alpha X$ is $$ \|u\| = \sup_{x,y\in X, \ x\ne y}\frac{|u(x)-u(y)|}{d(x,y)^\alpha}.$$ A function $u\in \mathrm{Lip}_\alpha X$ belongs to the "little" Hölder space $\mathrm{lip}_\alpha X$ if $$\sup_{x,y\in X, \ 0<d(x,y)<\delta}\frac{|u(x)-u(y)|}{d(x,y)^\alpha}\to 0 \quad \text{ as }\ \delta\to 0.$$ It's not hard to check that $\mathrm{lip}_\alpha X$ is a closed subspace of $\mathrm{Lip}_\alpha X$, and thus it is a Banach space of its own, with the same norm (subspace norm). 
The principal difference between the two spaces is that $\mathrm{lip}_\alpha X$ is separable (unless $X$ is huge) while $\mathrm{Lip}_\alpha X$ is usually non-separable. An example may help to illustrate this difference: for each $a\in (0,1)$ the function $f_a(x)=\sqrt{(x-a)^+}$ belongs to $\mathrm{Lip}_{1/2}(0,1)$ but not to $\mathrm{lip}_{1/2}(0,1)$. Since $\|f_a-f_b\|=1$ whenever $a\ne b$, the space $\mathrm{Lip}_{1/2}(0,1)$ is nonseparable. The definition of the "little" space directly precludes such constructions. 
The following analogy may be helpful: $\mathrm{lip}_{\alpha}$ is to $\mathrm{Lip}_{\alpha}$ what $c_0$ is to $\ell_\infty$ (or VMO to BMO).
To answer your question about norm relation: yes, such an inequality holds but it requires a multiplicative constant: $\|\cdot \|_{\mathrm{Lip}_\alpha} \le (\mathrm{diam}\, X)^{1-\alpha}\|\cdot \|_{\mathrm{Lip}_1}$.  
Concerning density: compactly supported functions are not dense in $\mathrm{Lip}_{1/2}(0,1)$, because $\sqrt{x}$ is at distance at least $1$ from any such function. They should be dense in $\mathrm{lip}_{\alpha}X$, at least for reasonable spaces $X$, but I don't expect the proof to be enjoyable. 
