$u(x)$ is an absolutely continuous function on $[0,1]$ with $u'(x)=a(x)u(x), u(0)=0$ and $a(x)$ bounded. Show that $u(x)\equiv 0$. 
Suppose that $a(x)$ is a bounded measurable function on $[0,1]$, and $u(x)$ is an absolutely continuous function on $[0,1]$, which satisfies $u'(x)=a(x)u(x)$ for $a.e.\, x\in [0,1]$. Further suppose that $u(0)=0$. Prove that $u(x)\equiv 0$ on $[0,1]$.


My attempt:
Since absolute continuity implies differentiability a.e. and 
$$ u(x)=u(x)-u(0)=\int_0^x u'(t)dt=\int_0^xa(t)u(t)dt $$
but I cannot deduce anything useful from the above identity. 
Indeed, we have $$|u(x)|\le\int_0^x|a(t)||u(t)|dt$$
and if we assume that $|a(x)|\le M$, we have:
\begin{align}
|u(x)|\le M|u(\xi)|x
\end{align}
where $\xi\in (0,x)$. But this does not lead us to everywhere close to the statement that $u(x)\equiv 0$ on $[0,1]$. Can someone give me a hint? Thanks.
 A: Hint: $g(x) = u(x)e^{-\int_0^x a(t) \, dt}$ is absolutely continuous, and $g'(x) = 0$ a.e. in $[0, 1]$.
Some details:


*

*$a(x)$ is bounded and measurable, and therefore integrable on the (bounded) interval $I = [0, 1]$. It follows that $A(x) = \int_0^x a(t) \, dt $ is absolutely continuous on $I$, and $A'(x) = a(x)$ a.e.

*The composition of a Lipschitz-continuous function with an absolutely continuous function is absolutely continuous. It follows that $v(x) = \exp(-A(x))$ is absolutely continuous on $I$.

*The product of two absolutely continuous functions on a compact interval is absolutely continuous. Therefore $g(x) = u(x)v(x) =  u(x)e^{-\int_0^x a(t) \, dt}$ is absolutely continuous on $I$. It follows that $g$ is differentiable a.e., and
$$
g(x) = g(0) + \int_0^x g'(t) \, dt \, .
$$

*For almost all $x \in I$, $u$ and $A$ are differentiable at $x$ with $A'(x) = a(x)$, so that $g'(x) = 0$ a.e. in $I$. It follows that $g(x) = g(0) = 0$ for all $x$.
A: If you call $F(x)=\int\limits_0^x|u(t)|\,dt$ and you estimate $|u|$ as you did, using the fact that $|a(t)|\le M$ you get for $F$:
$$
F'(x)\le MF(x)
$$
on $[0,1]$. Clearly, $F\ge 0$ and $F(0)=0$. By plain comparison, $F$ will be dominated by the solution of the IVP
$$
G'(x)=MG(x);\qquad G(0)=0,
$$
which is $G(x)=0$. Therefore $0\le F(x)\le G(x)=0$ and $F=0$. Being continuous and non-negative, from this it follows that $|u(x)|=0$ on $[0,1]$.
A: Choose $\epsilon$ sufficiently small such that $\epsilon\sup_{t\in[0,1]} |a(t)|<1$. From the first equation you wrote and the triangle inequality, we have that for all $x\in[0,\epsilon]$
$$
|u(x)|\leq \epsilon \sup_{t\in[0,1]}|a(t)|\sup_{x\in [0,\epsilon]}|u(x)|.
$$
Taking the supremum on the left side shows that $\sup_{x\in[0,\epsilon]}|u(x)|=0$, or in other words $u(x)=0$ for all $x\in [0,\epsilon]$. Now repeat the argument but shifted over by $\epsilon$ until the entire interval is covered.
