What is the relation between Lipschitz and $BMO$ spaces? Let $0 < \alpha < 1$. The space $\text{Lip}_{\alpha}(\mathbb{R}^n)$ of the Lipschitz continuous functions with order $\alpha$ is
$$
\text{Lip}_{\alpha}(\mathbb{R}^n)=\{f: |f(x)-f(y)|\le C|x-y|^{\alpha} \quad\text{  for a.e. }\,x, y \in \mathbb{R}^n\}.
$$
The smallest such constant $C$ is called the $\text{Lip}_{\alpha}(\mathbb{R}^n)$ norm of $f$ and is denoted by $\|f\|_{\text{Lip}_{\alpha}(\mathbb{R}^n)}$.
This norm has the following integral representation:
$$
\|f\|_{\text{Lip}_{\alpha}(\mathbb{R}^n)}\thickapprox\sup_{B}\frac{1}{|B|^{1+\alpha/n}}\int_{B}|f(x)-f_{B}|dx,
$$
where $f_{B}=\frac{1}{|B|}\int_{B}f(y)dy$. It is clear that for $\alpha=0$ the right hand side is the definition of $BMO$ norm of $f$.
According to these facts can we say that Lipschitz space includes $BMO$ space? What is the relation between Lipschitz and $BMO$ spaces?
 A: I'm going to define a bunch of spaces which all describe functions of "regularity $\alpha$" in some sense.
Hölder spaces: Here $\alpha$ will be in $[0,1]$. Define $Lip_{\alpha}(\Bbb T^n)$ to be the space of all functions $f:\Bbb T^n \to \Bbb R$ such that $|f(x)-f(y)| \leq C|x-y|^{\alpha}$  for some $C>0$ independent of $x,y \in \Bbb T^n$. The smallest such constant $C$ is called the Holder seminorm, denoted by $[f]_{\alpha}$. The Banach space norm on $Lip_{\alpha}(\Bbb T^n)$ is defined by $\|f\|_{L^{\infty}(\Bbb T^n)}+[f]_{\alpha}.$ Note that when $\alpha = 0$ we just get $L^{\infty}(\Bbb T^n)$. Equivalently one may describe $Lip_{\alpha}(\Bbb T^n)$ as the set of functions $f$ such that $\sup_{x\in Q}|f(x)-f_Q| \leq C|Q|^{\alpha/n},$ for all cubes $Q \subset \Bbb T^n$, where $f_{Q} := \frac1{|Q|} \int_Q f$, and $|Q|$ is the Lebesgue measure of $Q$. (Proving this equivalence is difficult.)
Besov spaces: Here $\alpha$ can be any real number. Any function $f:\Bbb T^n \to \Bbb R$ admits a canonical decomposition called the Littlewood-Paley decomposition $f = \sum_{j\ge 0} f_j$. The Besov space $B^{\alpha}_{\infty,\infty}(\Bbb T^n)$ consists of those functions $f$ such that $\|f_j\|_{L^{\infty}(\Bbb T^n)} \leq C2^{-\alpha j}$ for some $C$ which is independent of $j$. The smallest constant $C$ for which the inequality holds is called the Besov norm. This induces a Banach space structure on $B^{\alpha}_{\infty,\infty}$. The space $B^1_{\infty,\infty}$ is called the Zygmund class and is equivalently described as the set of all functions $f$ such that $$|f(x+h)+f(x-h)-2f(x)| \leq C|h|,$$ and $B^0_{\infty,\infty}$ consists of the distributional derivatives of functions from the Zygmund class.
BMO spaces: Here $\alpha$ will be in $[0,1]$. Let us define the space $BMO_{\alpha}(\Bbb T^n)$ to be the space of all functions $f:\Bbb T^n \to \Bbb R$ such that $\sup_Q \frac{1}{|Q|^{1+\alpha/n}}\int_{Q} |f-f_Q|dx <\infty$, where the sup is over all cubes $Q\subset \Bbb T^n$, and $f_{Q} := \frac1{|Q|} \int_Q f$, and $|Q|$ is the Lebesgue measure of $f$. The norm on $BMO_{\alpha}$ is defined to be that supremum, which makes it a Banach space.
Continuous function spaces: Here $\alpha=:k$ must take values in $\Bbb N$. Then $C^{k}(\Bbb T^n)$ is defined to be the set of all functions $f:\Bbb T^n \to \Bbb R$ such that all partial derivatives of order up to $k$ are continuous. The norm is defined to be sum of the uniform norms of all of the partial derivatives up to order $k$. Again, we get a Banach space.

So now the question is: how are all of these spaces related?
Theorem 1: If $\alpha \in (0,1)$ then $$ Lip_{\alpha}(\Bbb T^n) = B^{\alpha}_{\infty,\infty} (\Bbb T^n)= BMO_{\alpha}(\Bbb T^n).$$ All of the norms are equivalent.
Theorem 2: For $\alpha = 0$ we have the following inclusions: $$C^0(\Bbb T^n) \subsetneq L^{\infty}(\Bbb T^n) \subsetneq BMO_0(\Bbb T^n) \subsetneq B^0_{\infty,\infty}(\Bbb T^n).$$ So none of the norms are equivalent. For $\alpha=1$ we have the corresponding sequence of proper inclusions.
Basically the equivalences in Theorem 1 always boil down to a computation on dyadic blocks. They fail for $\alpha=0$ due to the fact that the series $\sum 2^{-\alpha n}$ diverges for $\alpha=0$.
Sorry if this was unclear. Will try to update with references.
