## 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 

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

This implies, setting $b = t_{i+1}$ and $a = t_i$, that

$$ 2^{-1-i} t_{i+1} \leq \left[ 2^{-i} t_i + (t_{i+1} - t_i) \sum_{j = i+1}^\infty 2^{-j} K_j \right] e^{K_{i}} $$

(here we rewrote $\int_{t_{i+1}}^\infty x(s) k(s) s^{-2} ~ds = \sum_{j = i+1}^\infty \int_{t_j}^{t_{j+1}} x(s)s^{-1} \cdot k(s) s^{-1} ~ds$, 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)

$$ \left[ e^{-K_{i}} - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j} \right] t_{i+1}  \leq \left[2 - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j} \right] t_i $$

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

$$ t_{i+1} \leq 2 t_i \cdot \frac{1}{e^{-K_{i}} - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j}} \leq \frac{2 t_i}{1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i + 1 + j}} $$

## The Estimates on $t$

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

$$ \prod_{i = i_0}^\infty (1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j}) $$

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

$$ \sum_{i = i_0}^\infty \ln (1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i+1+j}) \geq - 2\ln 2 \sum_{i = i_0}^\infty \left( K_i + \sum_{j = 0}^\infty 2^{-j} K_{i + 1 + j} \right) $$

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 

$$ \sum_{i = i_0}^\infty \sum_{j = 0}^\infty 2^{-j} K_{i + 1 + j} =  \sum_{j = 0}^\infty \sum_{i = i_0}^\infty 2^{-j} K_{i+1 + j} \leq 2 \sum_{i = i_0}^\infty K_{i+1} < \infty$$

and is also bounded. This concludes the proof.