$u(x)$ is an absolutely continuous function on $[0,1]$ with $u'(x)=a(x)u(x), u(0)=0$ and $a(x)$ bounded. Show that $u(x)\equiv 0$.

Question

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 $[0,1]$.

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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.

Answer 1

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 $x$.

Answer 2

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 $[0,1]$.

Answer 3

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.