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.

As you presented it, this is completely bogus: it is an example of the logical fallacy called "begging the question".
