Suppose we have a function $x(t)$ that is analytic everywhere, $x(t)\geq 0$, and $x(0)=0$.
Suppose further that we know $x(t)$ is locally zero. ie. $x(t)=0$ for some small $t>0$.
Is there a way to conclude that $x(t)$ is identically zero for all $t>0$?
My appoach thus far is using a Taylor expansion:
$x(t) = x(0)+\dot{x}(0)t+\ddot{x}(0)t^2/2!+...$ with $t>0$
and considering the smallest k such that $x^k(0)\neq 0$. But I'm stuck.
Is it as simple as saying that $x(t)$ is analytic so it cannot have any "corners" hence it must be identically zero if it is locally zero? If this is the case how does it relate to the Taylor expansion (does it imply derivatives of any order at $t=0$ are zero?) Thanks.
It is a lot simpler that that. If $x(t)=0$ on some interval around $0$, then so is $\dot x(0)$, and then $\ddot x(0)$, etc. So all the coefficients in the Taylor expansion of $x$ are zero, implying that $x=0$ everywhere.