I have a question about the proof of Gronwall's inequality as given in Chicone: Ordinary Differential Equations with Applications.

Gronwall: Suppose that $a<b$ and let $\alpha, \phi,$ and $\psi$ be nonnegative continuous functions defined on the interval $[a,b]$. Moreover, suppose that $\alpha$ is differentiable on $(a,b)$ with nonnegative continuous derivative $\dot\alpha$. If, for all $t \in [a,b]$, $$\phi(t) \leq \alpha(t) + \int_{a}^{t}\psi(s)\phi(s)ds,$$ then $$\phi(t) \leq \alpha(t)e^{\int_{a}^{t}\phi(s)ds}$$ for all $t \in [a,b]$.

Proof: Assume for the moment that $\alpha(a)>0$. In this case $\alpha(t) \geq \alpha(a)>0$ on the interval $[a,b]$. The function on the interval $[a,b]$ defined by $t \mapsto \alpha(t) + \int_{a}^{t}\psi(s)\phi(s)ds$ is positive and exceeds $\phi$.

The last sentence doesn't make sense. I understand the function is positive. But how does it exceed $\phi$?

  • $\begingroup$ uhm -- this is the assumption. $\endgroup$ – Thomas Apr 16 '14 at 17:44
  • 1
    $\begingroup$ Right, but the proof statement makes it sound like it is verifiable. $\endgroup$ – user85362 Apr 16 '14 at 17:45
  • 3
    $\begingroup$ It just repeats the assumption made in the theorem. If a theorem states an assumption as a prerequisite, it is of course something which you can use to prove it. $\endgroup$ – Thomas Apr 16 '14 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy