$aI am looking for a positive continuous function $f$ such that for all positive $a,b>0$ 

$$a < b(b+2)\Longrightarrow f(a)<2f(b)$$ and $$a=b(b+2)\Longrightarrow f(a)=2f(b).$$

Does such a function exist?
I tried to constructing one using exponential functions, as they are positive, but I failed.
 A: Let $g:\mathbb{R}_{>1}\to\mathbb{R}_{>0}$ be defined by $g(x):=f(x-1)$ for all $x>1$.  Thus, if $a,b>0$ satisfies $a<b(b+2)$, or equivalently, $(a+1)<(b+1)^2$, then
$$g(a+1)=f(a) < 2\,f(b)=2\,g(b+1)\,.$$
If $a=b(b+2)$, which is the same as $(a+1)=(b+1)^2$, then
$$g(a+1)=f(a)=2\,f(b)=2\,g(b+1)\,.$$
Thus, if $h:\mathbb{R}\to\mathbb{R}$ is given by $$h(t):=\dfrac{\ln\Big(g\big(\exp(2^t)\big)\Big)}{\ln(2)}\text{ for all }t\in\mathbb{R}\,,$$
then
$$h(t+1)=h(t)+1\text{ for all }t\in\mathbb{R}\,.\tag{*}$$
That is,
$$f(x)=2^{h\left(\frac{\ln\big(\ln(x+1)\big)}{\ln(2)}\right)}\text{ for all }x>0\tag{#}\,.$$
In other words, you can start with any continuous function $h:\mathbb{R}\to\mathbb{R}$ that satisfies (*).  Then, any such a function $f:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ must take the form (#).  In particular, if $h(t)=t+\ln(c)$ for all $t\in\mathbb{R}$ and for a fixed $c>0$, then
$$f(x)=c\,\ln(x+1)\text{ for all }x>0\,.$$
There are, however, infinitely many other solutions.  For example, we can take $$h(t)=t+p(t)\text{ for all }t\in\mathbb{R}\,,$$
where $p:\mathbb{R}\to\mathbb{R}$ is an arbitrary continuous periodic function with period $1$.  Then,
$$f(x)=2^{p\left(\frac{\ln\big(\ln(x+1)\big)}{\ln(2)}\right)}\,\ln(x+1)\text{ for all }x>0\,.$$
For example, one can take $p(t)$ to be any function in the $\mathbb{R}$-span of
$$1,\sin(2\pi t),\cos(2\pi t),\sin(4\pi t),\cos(4\pi t),\sin(6\pi t),\cos(6\pi t),\ldots\,.$$
The only thing you may have to worry about is that $h$ should be a strictly increasing function.  However, that can be easily fixed by demanding that $p(t)$ be continuously differentiable almost everywhere with $p'(t)>-1$ for almost every $t\in[0,1)$ (this extra condition will remove some viable choices of $p$, though).  That is, something like $$p(t)=\frac{\cos(2\pi t)}{2\pi}\text{ for all }t\in\mathbb{R}$$
will also work.
