# Thread: Proof of Gronwall's Inequality

1. ## Proof of Gronwall's Inequality

I am trying to understand the proof to Gronwall's Inequality, but I have one major issue I was hoping someone could get me through. My issue actually makes the proof incorrect.

First, the statement of Gronwall's Inequality
Let

$\displaystyle \begin{matrix}u:[a,b]\rightarrow [0,\infty) \\v:[a,b]\rightarrow [0,\infty)\end{matrix}$
be continuous functions and let $\displaystyle C$ be a constant. Then if

$\displaystyle v(t) \le C + \int_a^t\! v(s) u(s)\, ds.$
For $\displaystyle t\in [a,b]$, it follows that

$\displaystyle v(t) \le C \exp\left(\int_a^t\! u(s)\, ds\right)$
for $\displaystyle t\in [a,b]$.

Proof:
$\displaystyle \dfrac{v(t)}{C + \int_a^t\! v(s) u(s)\, ds} \le 1$

How do we know $\displaystyle C + \int_a^t\! v(s) u(s)\, ds > 0$?

$\displaystyle \dfrac{v(t) u(t)}{C + \int_a^t\! v(s) u(s)\, ds} \le u(t)$
$\displaystyle \dfrac{d}{dt}\ln\left(C + \int_a^t\! v(s) u(s)\, ds\right) \le u(t)$
$\displaystyle \ln\left(C + \int_a^t\! v(s) u(s)\, ds\right) -\ln C \le \int_a^t\! u(s)\, ds$
$\displaystyle C + \int_a^t\! v(s) u(s)\, ds \le C exp\left(\int_a^t\! u(s)\, ds\right)$
How do we know $\displaystyle C > 0$?

EDIT: I guess both my questions boil down to how do we know $\displaystyle C > 0$? Since, we know $\displaystyle \int_a^t\! u(s) v(s)\, ds \ge 0$ by $\displaystyle \begin{matrix}u:[a,b]\rightarrow [0,\infty) \\v:[a,b]\rightarrow [0,\infty)\end{matrix}$

EDIT2: $\displaystyle v(t) \le C + \int_a^t\! v(s) u(s)\, ds \Rightarrow C + \int_a^t\! v(s) u(s)\, ds \ge 0$. However, I still don't see how we know $\displaystyle C > 0$.

2. Plug in u = 0. The result C > 0 pops right out.

3. I knew it was going to be that simple. Thank you.

4. You're welcome!