Let for a function $u: \mathbb R \rightarrow \mathbb R$ and $x_0 \in \mathbb R$, $r>0$: $$ w(u,x_0,r)=\sup_{B(x_0,r)} u-\inf_{B(x_0,r)}u, $$ where $B(x_0,r)=(x_0-r,x_0+r)$.
In Wikipedia (see here) is stated that if $u: \mathbb R \rightarrow \mathbb R$ satisfies the following condition $$ w(u,x_0, \frac{r}{2}) \leq \lambda w(u,x_0,r) $$ for fixed $0< \lambda <1$ and suficiently small values of $r$
then $u$ is Hoelder continuous.
How to understand this condition on $w$, is it indeed condition on the global Hoelder continuity?
How to prove this statement?
The quantity $w(u,x_0,r)$ is the oscillation of $u$ on the ball $B(x_0,r)$. Since this ball contains $x_0$, we have $$ |x-x_0|<r \implies |u(x)-u(x_0)|\le w(u,x_0,r) \tag{1} $$ Comparing (1) to Hölder continuity inequality $$ |x-x_0|<r \implies |u(x)-u(x_0)|\le Cr^\alpha \tag{2} $$ we see that the given condition $$w(u,x_0, r/2) \leq \lambda w(u,x_0,r)\tag3$$ should somehow be used to show $w(u,x_0,r)\le Cr^\alpha$ for some $\alpha\in (0,1)$.
###Proof that (3) implies Hölder continuity
Let $\alpha\in (0,1)$ be such that $2^{\alpha}\le 1/\lambda$ and consider the function $\phi(r) = r^{-\alpha}w(u,x_0,r)$. Since $$ \phi(r/2) =2^\alpha r^{-\alpha}w(u,x_0,r/2) \le 2^\alpha \lambda r^{-\alpha} w(u,x_0,r) \le \phi(r) $$ it follows that $$ \sup_{0<r\le R}r^{-\alpha}w(u,x_0,r) \le C:=\sup_{R/2<r\le R}r^{-\alpha}w(u,x_0,r) $$ where $R$ is a number such that (3) holds for $r\le R$. $\quad\Box$
It's not something people came up with for its own sake; this sort of condition naturally arises when one studies elliptic PDE. Roughly speaking: oscillation of $u$ on the ball $B(x_0,r)$ gives control on $\nabla u$ on compact subsets of the ball, which by the mean value theorem gives better control on the oscillation of $u$ on the ball $B(x_0,r/2)$. The factor $1/2$ is used simply because $1/2$ is the world's favorite number between $0$ and $1$.
Be careful with global when talking about Hölder continuity. One does not usually impose $|u(x)-u(y)|\le C|x-y|^\alpha$ for all $x,y$ in $\mathbb R$, because even the identity function $u(x)=x$ would fail that (because $\alpha < 1$). Note that $u(x)=x$ satisfies (3) with $\lambda=1/2$ and all $r$.
The part "suficiently small $r$" is also something of a concern when trying to make global conclusion. It could mean "sufficiently small, depending on $x_0$".