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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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up vote 2 down vote accepted

As it is written, it's certainly not correct because you use the unproven statement (1) to prove (3) which in turn proves (1), and logic does not work that way.

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.

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Now this is a good argument. Thank you! – Giuseppe Negro Dec 7 '11 at 18:01

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

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Thank you very much. – Giuseppe Negro Dec 7 '11 at 18:04

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