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We had the following version of Gronwall's Lemma in our lecture:

Let $a<b, c \in \mathbb{R}$, $v: [a,b] \mapsto \mathbb{R}$ measurable and bounded and $u:[a,b] \mapsto [0, \infty)$ continuous st $v(t) \leq c + \int_{a}^t u(s)v(s)ds$ for all $t \in [a,b]$. Then we have $v(t) \leq c \cdot \exp( \int_{a}^t u(s)ds)$.

In the proof, we differentiated between the case that $c=0$ and $c \neq 0$.

In the first case, we claimed that $v(t) \leq 0\quad \forall t \in [a,b]$ and assumed that there's a $t \in [a,b]$ st $v(t) >0$. We then set $t_0=\sup \{t \in [a,b] | v( \tau) \leq 0\, \forall \tau \in [a,t] \} \in [a,b].$ One subcase now was that $t_0=b$ and the other one that $t_0 < b$. Here we stated that for $h>0$ small enough, we get

$$0< \sup_{0 \leq t \leq t_0+h} v(t) = \sup_{t_0 \leq t \leq t_0+h} v(t) \leq ...$$ Question: Unfortunately, I neither understand the first inequality nor the second equality. Could someone please explain this to me?

In the second case ($c \neq 0$), we defined $w(t):= C \cdot \exp(\int_a^t u(s)ds) \forall t \in [a,b]$ and then worked with $\alpha(t):= v(t)-w(t)$ in order to apply the first case.

Question: I was wondering if $w$ is measurable and bounded? Plus whether I really need this property in order to apply the first case?

Thank you for your help!

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  • $\begingroup$ what is exactly the second inequality? What did you want to write instead of $\dots$ in the displayed inequality? $\endgroup$ Aug 23, 2017 at 9:39
  • $\begingroup$ @uniquesolution Thank you for your edits first of all! I talked about the second equality (not inequality) and by $...$ I just wanted to imply that the proof goes on (but that I do not have any questions about this part) :) $\endgroup$
    – SallyOwens
    Aug 23, 2017 at 9:48

2 Answers 2

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  1. The first inequality: If it were false for some $h>0$, then you could replace $t_0$ by $t_0+h$, contradicting the definition of $t_0$.

  2. The second comes from the fact that for $t\leq t_0$: $v(t)\leq 0$ and using the first inequality.

  3. Measurability and boundedness imply that the integral of $u v$ is well-defined and finite. Otherwise the very first inequality would not make sense.

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For the first inequality:

If there is a small $h$ such that the inequality is not true, then we could choose a bigger $t_0$ (which is defined via supremum).

For the second equality: You just "forget" the values $t<t_0$ when taking the supremum. Those dont matter, because $v(t)\leq 0$ for these values.

Yes, the function $w$ is bounded and measurable, it can be shown that it is monotone (non-decreasing).

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