Vtyjkyuk

I 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.

Let $g:mathbbR_>1tomathbbR_>0$ 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 $h:mathbbRtomathbbR$ is given by $$h(t):=dfraclnBig(gbig(exp(2^t)big)Big)ln(2)text for all tinmathbbR,,$$
then
$$h(t+1)=h(t)+1text for all tinmathbbR,.tag*$$
That is,
$$f(x)=2^hleft(fraclnbig(ln(x+1)big)ln(2)right)text for all x>0tag#,.$$

In other words, you can start with any continuous function $h:mathbbRtomathbbR$ that satisfies (*). Then, any such a function $f:mathbbR_>0tomathbbR_>0$ must take the form (#). In particular, if $h(t)=t+ln(c)$ for all $tinmathbbR$ 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 $$h(t)=t+p(t)text for all tinmathbbR,,$$
where $p:mathbbRtomathbbR$ is an arbitrary continuous periodic function with period $1$. Then,
$$f(x)=2^pleft(fraclnbig(ln(x+1)big)ln(2)right),ln(x+1)text for all x>0,.$$
For example, one can take $p(t)$ to be any function in the $mathbbR$-span of
$$1,sin(2pi t),cos(2pi t),sin(4pi t),cos(4pi t),sin(6pi t),cos(6pi t),ldots,.$$
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 $tin[0,1)$ (this extra condition will remove some viable choices of $p$, though). That is, something like $$p(t)=fraccos(2pi t)2pitext for all tinmathbbR$$
will also work.