# An application of Gronwall's lemma

I've come across a creative use of Gronwall's lemma which I would like to submit to the community. I suspect that the argument, while leading to a correct conclusion, is somewhat flawed.

We have a continuous mapping $g \colon \mathbb{R}\to \mathbb{R}$ such that

$$\tag{1} \forall \varepsilon>0\ \exists \delta(\varepsilon)>0\ \text{s.t.}\ \lvert x \rvert \le \delta(\varepsilon) \Rightarrow \lvert g(x) \rvert \le \varepsilon \lvert x \rvert$$

and a continuous trajectory $x\colon [0, +\infty) \to \mathbb{R}$ such that

$$\tag{2} e^{\alpha t}\lvert x(t)\rvert \le \lvert x_0\rvert+\int_0^t e^{\alpha s}\lvert g(x(s))\rvert\, ds.$$

Here $x_0=x(0)$ is the initial datum, which we may choose small as we wish, but $\alpha >0$ is a fixed constant that we cannot alter in any way.

Now comes the point. Fix $\varepsilon>0$. The lecturer says: Suppose we can apply (1) for all times $t \ge 0$. Then inserting (1) in (2) we get

$$e^{\alpha t}\lvert x(t) \rvert \le \lvert x_0\rvert + \varepsilon \int_0^t e^{\alpha s} \lvert x(s)\rvert \, ds$$

and from Gronwall's lemma we infer

$$\tag{3} \lvert x(t)\rvert \le e^{(\varepsilon - \alpha)t}\lvert x_0\rvert.$$

So if $\varepsilon <\alpha$ and $\lvert x_0 \rvert < \delta(\varepsilon)$, $\lvert x(s) \rvert$ is small at all times and our use of (1) is justified. We conclude that inequality (3) holds.

Does this argument look correct to you? I believe that the conclusion is correct, but that it requires more careful treatment.

Thank you.

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A correct argument is the following: Since $|x(0)|<\varepsilon$ and $x$ is continuous, $x$ is less than $\varepsilon$ in a small neighbourhood of 0. If $|x(t)|\ge \varepsilon$ for some time $t$, consider the infimum (call this $T$) of those $t$ where this happens, and $T>0$ by the above. But at $T$, the inequality (3) says that $|x(T)|<|x(0)|<\varepsilon$ and by continuity again this holds in a neighbourhood of $T$, contradicting the property that $T$ is the infimum.