Largest asymptotic growth for $2f(x)-f(2x)$

Question

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 $\Delta_f$?

Answer 1

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.