I am reading a paper in which the following argument is made. We have two positive real valued functions $f(x)$ and $g(x)$. We know that $$\int\_0^x \int\_0^y f(z) \ dz \ dy \leq g(x).$$


It is then stated "and easy computation shows that, except for negligibly small intervals," $$\log f(x) = O(x + \log g(x)).$$


I am at a loss on how this comes about. $g(x)$ is convex, if that helps. Can anyone explain?



Estimating a quantity from an estimate in its integral
We begin with a lemma. Let $F$ be an increasing function tending to infty as $x\to\infty$.
Then $F'(x)\leq F^2(x)$, for $x\not\in E$, where $E$ is a set of finite measure.


Proof. Let $E$ be the set where $F'(x)\geq F^2(x)$. Then
$$|E|\leq\int\_{E\cap[1,\infty)}\frac{F'}{F^2}dx<\int\_1^\infty\frac{F'}{F^2}dx<\infty,$$
by the change of the variable $y=F(x)$.


Now your conditions imply that $g(x)\to\infty$; $g$ and $g'$ are increasing.


Suppose first that $g'(x)\to\infty$.
Your relation between $g$ and $f$ means that $g^{\prime\prime}=f$. Applying the lemma first
to $g$, and then to $g'$, we obtain the result $\log f(x)=O(\log g(x))$.
The case when $g'$ is bounded I leave as an exercise.


