Let $x(t),t\in [1,\infty)$ be a nondecreasing positive function satisfying the following inequality:
$$
x'(t) \le \int\_t^{+\infty} x(s)\frac{k(s)}{s^2}\,ds,
$$
for any $t \ge 1$, where $k(t),t\in [1,\infty)$ is a nonincreasing positive function such that
$$
\int\_1^{+\infty}\frac{k(s)}{s}\,ds <\infty.
$$


Can we prove that $x(t)$ is a bounded function?



A Riccati type integral inequality
The boundedness statement is true
---------------------------------


The general argument is similar to what I gave [in my previous answer](https://mathoverflow.net/a/323578/3948), which was essentially off by a log due to certain inefficiencies in the estimates. Here I rewrite the argument to get rid of the log loss. 


To start, as explained in the previous answer, we immediately have that $x(t)$ is sublinear. Since the inequality concerning $x(t)$ is linear, we can assume, without loss of generality, that $\sup x(t)/t \leq 1$ by a simple rescaling of $x$. This means we can find a sequence of times $1 = t\_0, t\_1, t\_2, \ldots$ defined by 


$$ t\_i = \inf \{ t\in [1,\infty) : \forall s \geq t, x(s) \leq 2^{-i} s \}. $$


(Note that implicitly since $x$ is differentiable it is also continuous, so $x(t\_i) = 2^{-i} t\_i$.)


Our goal is to estimate $t\_i$. Specifically, we want to show that $2^{-i}t\_i$ is bounded. 


We will also denote by 


$$ K(t) = \int\_t^\infty \frac{k(s)}{s} ~ds, \quad K\_i = \int\_{t\_i}^{t\_{i+1}} \frac{k(s)}{s} ~ds.$$


We note that $K\_i \searrow 0$ and the numbers are in fact summable by assumption. 


### Gronwall


Integrating by parts the differential inequality for $x'$ we get (as I argued in the previous answer) for $1 \leq a < b$


$$ x(b) - x(a) \leq (b-a) \int\_b^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds + \int\_a^b x(s) \frac{s - a}{s} \frac{k(s)}{s} ~ds $$


Estimating $(s-a)/s \leq 1$ we have that, by Gronwall's inequality 


$$ x(b) \leq \left[ x(a) + (b-a) \int\_a^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)} $$


This implies, setting $b = t\_{i+1}$ and $a = t\_i$, that


$$ 2^{-1-i} t\_{i+1} \leq \left[ 2^{-i} t\_i + (t\_{i+1} - t\_i) \sum\_{j = i}^\infty 2^{-j} K\_j \right] e^{K\_{i}} $$


(here we rewrote $\int\_{t\_{i}}^\infty x(s) k(s) s^{-2} ~ds = \sum\_{j = i}^\infty \int\_{t\_j}^{t\_{j+1}} x(s)s^{-1} \cdot k(s) s^{-1} ~ds$, and used the decaying bound on $x(s)s^{-1}$ above $t\_j$, and the fact that all functions involved are positive.)


Simplify (by the summability of $K\_j$ we can assume from here on the indices $i$ are always larger than some sufficiently large $i\_0$ such that the two terms in the brackets below are guaranteed to be positive)


$$ \left[ e^{-K\_{i}} - \sum\_{j = 0}^\infty 2^{-j} K\_{i+j} \right] t\_{i+1} \leq \left[2 - \sum\_{j = 0}^\infty 2^{-j} K\_{i+j} \right] t\_i $$


So for all sufficiently large $i$ we have the bound (using the convexity of the exponential function)


$$ t\_{i+1} \leq 2 t\_i \cdot \frac{1}{e^{-K\_{i}} - \sum\_{j = 0}^\infty 2^{-j} K\_{i+j}} \leq \frac{2 t\_i}{1 - K\_i - \sum\_{j = 0}^\infty 2^{-j} K\_{i + j}} $$


The Estimates on $t$
--------------------


To show our desired conclusion it suffices to show that the infinite product


$$ \prod\_{i = i\_0}^\infty (1 - K\_i - \sum\_{j = 0}^\infty 2^{-j} K\_{i+j}) $$


is bounded below away from zero. Now, the summability of $K\_i$ implies that we can choose $i\_0$ sufficiently large that $\sum\_{i = i\_0}^\infty K\_i < \frac12$. This implies


$$ \sum\_{i = i\_0}^\infty \ln (1 - K\_i - \sum\_{j = 0}^\infty 2^{-j} K\_{i+j}) \geq - 2\ln 2 \sum\_{i = i\_0}^\infty \left( K\_i + \sum\_{j = 0}^\infty 2^{-j} K\_{i + j} \right) $$


the first term in the sum is obviously bounded by the summability of $K\_i$. For the second term we interchange the order of summation 


$$ \sum\_{i = i\_0}^\infty \sum\_{j = 0}^\infty 2^{-j} K\_{i + j} = \sum\_{j = 0}^\infty \sum\_{i = i\_0}^\infty 2^{-j} K\_{i + j} \leq 2 \sum\_{i = i\_0}^\infty K\_{i+1} < \infty$$


and is also bounded. This concludes the proof. 


