A non-negative function that is bounded by its integral

Question

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!

Answer 1

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 $u \equiv 0$.

Answer 2

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 $n\to\infty$.