1
$\begingroup$

I have some doubts about my solution, since I am not familiar with measure theory.

This is the question:

Let $a > 0, u\colon[0,a] \to [0,+\infty)$ continous. Show:

$ \exists L \geq 0\ \forall t \in [0,a]: u(t) \le \int_0^t Lu(s)\;ds \implies \forall t \in [0,a]: u(t) = 0 $

My attempt is:

We show the contraposition:

$ \exists t \in [0,a]: u(t) \neq 0 \implies \forall L \geq 0 \;\exists j \in [0,a]: u(j) > \int_{0}^{j}Lu(s) ds $

If $ u(0) > 0 $ choose $ j = 0 $.

Now suppose $ u(0) = 0 $. We can define by assumption some $ 0 < s := \inf\{t \in [0,a]\ |\ \forall j < t: u(j) = 0\ \text{and}\ u(t) \neq 0 \} $

Hence, we use the integrability of $u$ and choose $j := s$.

\begin{align*} \int_{0}^{j}Lu(s) \,ds = 0 < u(j) \end{align*}

But I am not sure if this integral exists, since the upper bound is some infimum, and even if it converges, that it is $0$.

I would be happy about any advice.

Thank you!

$\endgroup$
  • $\begingroup$ Actually, you may want to choose $j=0$ if $u(0)>0$. Also, the set you take the infimum of is empty because $u$ is continuous. $\endgroup$ – Hagen von Eitzen Jun 30 '18 at 13:07
  • $\begingroup$ Actualy you are right. I was mistaken there. This surely, makes only sense if $u(0) > 0$, but we have have a problem if $u(0) = 0$, since then $0 > 0$ for $j=0$. I'll correct this. $\endgroup$ – Slyder Jun 30 '18 at 13:18
2
$\begingroup$

Say $u\le c$ on $[0,a]$. Then $$u(t)\le\int_0^t Lc=Lct.$$

Hence $$u(t)\le\int_0^t L(Lcs)\,ds=L^2c\frac{t^2}2.$$

And so on; in fact $$u(t)\le L^nc\frac{t^n}{n!}$$ for $n=1,2\dots$, by induction. But $L^nt^n/n!\to0$ as $n\to\infty$.

$\endgroup$
  • $\begingroup$ Thank you! Very simple :) $\endgroup$ – Slyder Jul 1 '18 at 8:00
1
$\begingroup$

A completely different approach motivated by Grönwall's lemma:

Define $V(t) = L\int_0^t u(s)\,ds$. Notice that $V \ge 0$ and $V \in C^1[0,a]$ and we have

$$V'(t) = Lu(t) \le L^2\int_0^tu(s)\,ds = LV(t)$$

Multiply this inequality with $e^{-Lt}$ to get $V'(t)e^{-Lt} \le LV(t)e^{-Lt}$ so

$$\frac{d}{dt}\big(V(t)e^{-Lt}\big) = e^{-Lt}(V'(t) - LV(t)) \le 0$$

Integrating this on $[0,x]$ gives

$$V(x)e^{-Lx} = V(x)e^{-Lx} - V(0) = \int_0^x \frac{d}{dt}\big(V(t)e^{-Lt}\big)\,dt \le 0$$

so $V(x) \le 0, \forall x \in [0,a]$. Therefore $V \equiv 0$ so $u \equiv 0$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.