$\def\R{\mathbb{R}}$
Consider a space $C^{1,\beta}$ of all continuous differentiable functions $f\colon\R\to\R$ such that their derivative $f'$ is Holder continuous with exponent $\beta$.
A standard way to define a norm on this space is to put
$$ \|f\|_{1,\beta}:=\sup_{x\in\R} |f(x)|+\sup_{x\in\R} |f'(x)|+\sup_{x\neq y}\frac{|f'(x)-f'(y)|}{|x-y|^\beta}. $$
My question is why do we need here the middle term $\sup_{x\in\R} |f'(x)|$? If we just put $$ \|f\|:=\sup_{x\in\R} |f(x)|+\sup_{x\neq y}\frac{|f'(x)-f'(y)|}{|x-y|^\beta}. $$ would this be a norm? Would the space equipped with this norm be a Banach space? Would this norm be equivalent to the standard norm $\|\cdot\|_{1,\beta}$?
Thanks!
Yes, if you omit the middle term you get an equivalent norm. (The reason the middle term is typically included is just to make various things simpler.)
This is a generalization of the classical Landau inequality, which says that a bound on $f$ and a bound on $f''$ imply a bound a $f'$. One can give a simple proof like so: Assume $f'$ is "large" at a point. The hypothesis on continuity of $f'$ shows that $f'$ is large on an interval of a certain length, which implies that $f$ must be large somewhere. In fact:
Theorem If $|f|\le a$ and $|f'(t)-f'(s)|\le b|s-t|^\beta$ then $$|f'|\le c_\beta a^{\frac{\beta}{\beta+1}}b^{\frac{1}{\beta+1}}.$$
Proof: Say $f'(0)=m>0$. Then $$f'(t)\ge f'(0)-bt^\beta\ge m/2, \quad(0<t<t_0=(m/2b)^{1/\beta}).$$Hence $$2a\ge f(t_0)-f(0)\ge\frac{m}{2}t_0,$$and the claimed inequality follows on inserting the definition of $t_0$ and unravelling things.