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.


1 Answer 1


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 $T>a$.

  • $\begingroup$ Thank you Mr. Pinelis for this clarification. Does the inequality any chance to hold for some $T, a, b$ ?. Because in control theory, this is a classical result but I can not find it in any book. Thank you again sir. $\endgroup$
    – Gustave
    Jul 16, 2020 at 12:34
  • 1
    $\begingroup$ @Gustave : As my answer shows, the inequality cannot hold for any $T,a,b$ satisfying your conditions $0\le a\le b\le1$ and $T\in(0,\infty)$. The most probable reason why you cannot find this "classical result" in any book is that you remember the result incorrectly. However, as now shown in the addition to my answer, your inequality will hold if $a\in[0,1]$,$b>1$, and $T>a$. $\endgroup$ Jul 16, 2020 at 17:04
  • $\begingroup$ Thank you sir for the great answer. I'm very thankful for you. $\endgroup$
    – Gustave
    Jul 17, 2020 at 4:38

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