# A Riccati type integral inequality

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?

## The boundedness statement is true

The general argument is similar to what I gave in my previous answer, 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.

• It is not clear to me why $x(b) \leq \left[ x(a) + (b-a) \int_b^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)}$ is true. I guess that it should be$x(b) \leq \left[ x(a) + (b-a) \int_a^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)}$ and the next line is $2^{-1-i} t_{i+1} \leq \left[ 2^{-i} t_i + (t_{i+1} - t_i) \sum_{j = i}^\infty 2^{-j} K_j \right] e^{K_{i}}$. Then the rest arguments work as before. Feb 21, 2019 at 16:16
• @Totoro: that part is exactly the same as my previous answer. Do you see how I got the bound for $x(t_2) - x(t_1)$ through integration by parts in the previous part? There are two boundary terms corresponding to the points $t_2 (=b)$ and $t_1 (=a)$. The $t_1$ term vanish because I chose to integrate by parts against $t - t_1$ which vanishes there. Feb 21, 2019 at 20:36
• Oh, wait, I see what you meant; your question is with my application of Gronwall. Yes, you are correct. Feb 21, 2019 at 20:42

## 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.

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 $$k(s) = (\ln(1+s))^{-2}$$.]

• Sharpening the arguments above a bit more, a sufficient condition for guaranteeing boundedness can be written as $$\int_1^\infty \frac{1}{t} \int_t^\infty \frac{k(s)}{s} ds~dt < \infty$$ Feb 19, 2019 at 18:21

Without requiring that the integral $$$$\int_1^\infty x(s)\frac{k(s)}{s^2}\,ds \tag{1}$$$$ be finite, the answer is no. Indeed, then one can take e.g. $$k(s)=1/s$$ and $$x(s)=e^s$$.

I don't know whether the condition that the integral in (1) be finite changes the answer.