## This is not a complete answer

The argument below doesn't give boundedness.

However, it gives that under the assumption $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, $x(t) = o(t^\lambda)$ for any $\lambda > 0$.

With stronger assumptions on $k$ the argument can also imply boundedness.

### Sublinear growth

First one observes that if the integral is finite, the growth rate is sublinear. This follows from the fact that if the integral $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x'(t) \to 0$ as $t\to\infty$, and hence $x(t) = o(t)$ by, e.g., L'Hopital.

### Upgrade the growth control

Integrating the bound for $x'(t)$ gives, for arbitrary $1 \leq t_1 < t_2$

$$ x(t_2) - x(t_1) \leq \int_{t_1}^{t_2} \int_t^\infty x(s) k(s) s^{-2} ds~dt $$

Integrating in parts in $t$, using that $dt = d(t - t_1)$

$$ x(t_2) - x(t_1) \leq (t_2 - t_1) \int_{t_2}^\infty x(s) k(s) s^{-2} ds + \int_{t_1}^{t_2} (t - t_1) x(t) k(t) t^{-2} dt $$

Now, for any $\delta, \beta> 0$, there exists $\tau_{\delta,\beta}$ such that for every $s \geq \tau_{\delta,\beta}$, it holds that

$$ x(s) \leq \delta s, \qquad \text{and} \qquad \int_{s}^\infty k(s)s^{-1} ds < \beta $$

since we know that $x$ grows sublinearly and the relevant integral converges. This implies that, for $t_2 > \tau_{\delta,\beta}$,

$$ x(t_2) \leq x(\tau_{\delta,\beta}) + (t_2 - \tau_{\delta,\beta}) \delta \beta + \int_{\tau_{\delta,\beta}}^{t_2} x(s) \cdot k(s) s^{-1} ds $$

By Gronwall's inequality this means that

$$ x(t_2) \leq [ \delta \tau_{\delta,\beta} + (t_2 - \tau_{\delta}) \delta \beta ] e^{\beta} $$

This final inequality can be used to estimate $\tau_{\delta/2,\beta}$. Consider the inequality

$$ [\delta \tau_{\delta,\beta} + (s - \tau_{\delta,\beta}) \delta\beta] e^{\beta} \leq \frac{\delta}{2} s \tag{A}$$

Solving this inequality we see that this is satisfied whenever

$$ \frac{e^{\beta} \tau_{\delta,\beta}}{\frac12 - \beta e^{\beta}} \leq s $$

For convenience write

$$ \frac{e^\beta}{\frac12 - \beta e^\beta} = 2^{1+\sigma(\beta)} $$

and note that $\sigma$ is continuous and $\sigma(0) = 0$.

This implies that $\tau_{\delta/2,\beta} \leq 2^{1 + \sigma(\beta)} \tau_{\delta}$.

Now, iterating this argument, starting from some sufficiently small $\delta$ that we fix, we see can construct an increasing sequence of times $\tau_{2^{-k} \delta,\beta}$ such that for $T \geq \tau_{2^{-k} \delta,\beta}$ it holds

$$ x(T) \leq 2^{-k} \delta T $$

We also have by iteration the bounds

$$ \tau_{2^{-k}\delta,\beta} \leq 2^{k + k\sigma(\beta)} \tau_{\delta,\beta} $$

So we have

$$ x(T) \lesssim_{\delta,\beta} T^{\sigma(\beta)} $$

### Boundedness

One can upgrade the estimate to get boundedness if one knows the decay rate of $\int_t^\infty k(s) s^{-1} ds$.

For example, if one knows that this integral is bounded by $C/\ln(t)^2$, then Gronwall will imply an estimate along the lines of

$$ (\tau_{\delta} + (s - \tau_\delta) \frac{C}{(\ln \tau_{\delta})^2}) e^{C/(\ln \tau_\delta)^2} < \frac12 s $$

replacing (A).

Now, by our estimates in the previous section we would've found that

$$ \tau_{2^{-k\delta}} \leq 2^{k(1 + \sigma)} \tau_{\delta} $$

for some $\sigma$. Bootstrapping from this we would get that

$$ \tau_{2^{-k} \delta} \leq 2^{1 + O(k^{-2})} \tau_{2^{-(k-1)} \delta} $$

The summability of the series $(k^{-2})$ then gives boundedness of $x(t)$.

[This argument can carry through as long as we know, for example, that $k(s) \lesssim (\ln (1+s))^{-2-\gamma}$ for some $\gamma > 0$; but seemingly fails for $k(s) = (\ln(1+s))^{-2}$.]