Let $(a,b) \subset (0,1)$. Consider the following transport equation
$$z\_t+z\_x=0, \ (t,x)\in(0,T)\times(0,1), \\z(t,0)=0, \ z(0,x)=z\_0(x).$$
It is clear that the solution to the above equation is given by $z(t,x)=z\_0(x-t),\ \text{if} \ x-t\in (0,1)$ and $0$ otherwise.
I want to prove the following observability inequality: There exists a positive constant $C$ such that
$$\int\_0^T \int\_a^b z\_0^2(x-t)dxdt\geq C\int\_0^1z\_0^2(x)dx.$$
I know that this inequality is satisfied if and only if $T \geq 1-b$ and $a=0$ but I don't see how to prove it. Any ideas or references?.
Thank you.



Observability inequality for the 1D transport equation
We have $0\le a\le b\le1$ and $T\in(0,\infty)$. We want to know when there is a positive constant $C$ such that
$$\int\_0^T dt\, \int\_a^b dx\, u^2(x-t)\geq C\int\_0^1 dx\,u^2(x) \tag{1}$$
for all measurable functions $u\colon\mathbb R\to\mathbb R$ such that $u(x)=0$ for $x\notin(a,b)$.


The answer is: never. Indeed, without loss of generality $a<b$. The left-hand side of (1) is
$$\int\_0^T dt\, \int\_a^b dx\, u^2(x-t) \\ 
=\int\_{\mathbb R} ds\, u^2(s)\int\_{\mathbb R} dt\,1\{s\in(0,1),t+s\in(a,b),t\in(0,T)\} \\
= 
\int\_0^1 ds\,u^2(s)w(s),$$
where
$$w(s):=\max[0,\min(T,b-s)-\max(0,a-s)].$$
Clearly, the weight function $w$ is continuous, so that $w(b-)=w(b)=0$.
Letting now $u:=1\_{(b-h,b)}$ with $h\downarrow0$, we see that the left-hand side of (1) is $\int\_{b-h}^b ds\,w(s)=o(h)$, whereas the right-hand side of (1) is $C\int\_{b-h}^b ds=Ch$, so that (1) fails to hols for any real $C>0$.


**Added in response to a comment by the OP:** Note that
$$w(s)=\max[0,\min(T,b-a,b-s,T-a+s)]\ge\min(b-1,T-a)=:m$$
for all $a\in[0,1]$ and $s\in[0,1]$.
So, (1) will hold with $C=m$. If you now want $C$ to be $>0$, just require that $m$ be $>0$; that is, (in addition to the condition $a\in[0,1]$) require that $b>1$ and $T>a$.


