I am trying to find smooth functions $f : \mathbb{R}\_+ \to \mathbb{R}\_+$ such that the quantity $$\Delta\_f(x) := 2f(x)-f(2x)$$
is positive for $x$ large enough and has the **greatest asymptotic growth**.




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It seems clear that one cannot go beyond a linear growth, since $\Delta\_f$ vanishes for linear functions and will likely be negative for super-linear ones. However, one can construct many examples of almost linear asymptotic growths.


For example, plugging $f(x) := x^a$ for some $a\in(0,1)$ yields $\Delta\_f(x) = (2-2^a) x^a$.


Choosing $f(x) := \frac{x}{\ln x}$ yields $\Delta\_f(x) \sim 2 \ln 2 \frac{x}{(\ln x)^2}$ which has a larger asymptotic growth.




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What would be your candidates for even larger asymptotic growths? And is there a way to prove a (sublinear) a priori upper bound on $\Delta\_f$?



Largest asymptotic growth for $2f(x)-f(2x)$
Let us discretise the problem by setting $a\_n=2^{-n}f(2^n)$, $b\_n=2^{-n-1}\Delta\_f(2^n)$. Then your relation becomes,
$$b\_n=a\_n-a\_{n+1}.$$
since $a\_n,b\_n$ are non-negative, we conclude that
$$\sum\_{n=1}^\infty b\_n<\infty.$$
This is a necessary and sufficient condition. Indeed, take any summable sequence $b\_n$ of positive numbers, then we can define
$$a\_n=\sum\_{k=n}^\infty b\_k$$ as the sequence of partial sums, and obtain your equation on the sequence $x\_n=2^n$. Then you can interpolate by choosing $f(x)$ arbitrarily on the interval $(1,2)$.


Returning to your original notation, the growth condition becomes
$$\int\frac{\Delta\_f(x)}{x^2}dx<\infty.$$
This is a necessary and sufficient condition.
For example, we cannot have $\Delta\_f(x)\sim x/\log x$, but can have
$$\Delta\_f(x)\sim\frac{x}{\log x(\log\log x)^{1+\epsilon}},$$
and so on.


