Examples of strict inclusion of function classes Consider the following three classes of Lipschitz functions $f:\mathbb{R}^n\to\mathbb{R}^n$:

*

*$A=\{\exists\beta,R>0:\,\langle f(x)-f(y),x-y\rangle\leq-\beta|x-y|^2\,\forall|x|,|y|>R\}$

*$B=\{\exists\alpha,\beta>0:\,\langle f(x)-f(y),x-y\rangle\leq\alpha-\beta|x-y|^2\,\forall x,y\in\mathbb{R}^n\}$

*$C=\{\exists\alpha,\beta>0:\,\langle f(x),x\rangle\leq\alpha-\beta|x|^2\,\forall x\in\mathbb{R}^n\}$
We certainly have that $A\subset B\subset C$. I'm interested in explicit examples (preferably in dimension $n=1$) which show that these inclusions are strict.
 A: I will give functions which serve as counterexamples, but will only provide a sketch
for why these counterexamples work.
First, consider the function
$$
f(x) = 2\sin(x) - x.
$$
Then this function is in $B$, but not in $A$.
One can show that $f$ is not in $A$ using derivatives.
One can see that there are arbitrarily large points $x$
with $f'(x)>0$, which causes a problem in the definition of $A$.
Some intuition why the function is in $B$:
If $|x-y|$ is small, then any problems caused by the $\sin$ term
can be dealt with the constant $\alpha$.
If $|x-y|$ is large, then the $\sin$ term is irrelevant relative to the $x$ term.
For the second counterexample, the function is more complicated to describe.
We require that
$$
f(0)=0,\;
f(-n) = n,\;
f(2^{2n}) = -2^{2n},\;
f(2^{2n+1}) = -2^{2n}
$$
holds for all $n\in\Bbb N$ and we interpolate linearly between these points.
Then this function is in $C$, but not in $B$.
The function has a linear growth, but has plateaus, and the size of the plateaus grows
as $|x|$ grows.
Because of $f(x)\leq -x/2$ one can show that the function is in $C$.
However, if one chooses $x=2^{2n}$, $y=2^{2n+1}$
then one can see that there are no constants $\alpha,\beta$
such that the definition in $B$ can be satisfied.
