Properties of heat equation ** I simplified the question: **
On bounded domains, the maximum principle implies that the solution to the heat equation is (strictly) positive, if the initial and boundary data is positive. 
I would like to know whether this result is also true if applied on an unbounded domain:
Let $a>0$ and $g$ a continuous function with $g(a,t)>0$ for all $t \in [0,T]$ and $u(x,0) = u_0(x) >0$ for all $x \in (a,\infty)$ with $u_0 \in C^{\infty} \cap L^{\infty}.$
Let $u(x,t)$ be the solution to the heat equation $\left(\partial_t - \partial_{x}^2 \right)u=0$  with the above boundary data.
Does there exist then a version of the maximum principle saying that $u(x,t)>0$ for $(x,t) \in [a,\infty) \times [0,T]$?
 A: The answer is no. A counterexample has the form
$$u(x,t)=\sum_0^\infty \frac{g^{(n)}(t)}{((2n+1)!)^2}(r-R)^{2n+1},$$
where $r>R$ is the distance from the origin in $R^2$.
First one shows that this function satisfies the heat equation formally. Second,
there exists an infinitely differentiable $g\not\equiv 0$ such that $g^{(n)}(0)=0$
for all $n\geq 0$, and such that the above series converges. So our function is not zero,
satisfies the heat equation in the exterior of the ball,
and has zero boundary and initial conditions. For the existence of such $g$ one can refer to
http://www-bcf.usc.edu/~lototsky/MATH445/NonUniqueness-HeatEq.pdf
WLOG, $u$ is negavive somewhere, otherwise replace it by $-u$. Then add
a small positive constant to $u$, the boundary and initial conditions will be positive, but $u$ is still negative somewhere.
A: As a companion to Alexandre Eremenko's answer, the answer to your question is "yes" if we assume in addition that $u$ satisfies the growth condition $|u| < Ae^{M|x|^2}$ for some $A,\,M > 0$.
To see this assume for simplicity that $a = 0$ and $T = 1$. Let $h_R = -t^{-1/2}e^{-\frac{(x-R)^2}{4t}}$ and let $b_R = Ae^{MR^2}h_R$. By the comparison principle on bounded domains we have $u \geq b_R$ on $[0,\,R] \times [0,\,1]$. A short computation shows that $b_R \geq -8A\sqrt{M}e^{-3MR^2}$ on $[0,\,R/2] \times [0,\, (64M)^{-1}]$. Taking $R \rightarrow \infty$ we conclude that $u \geq 0$ for $t \in [0,\,(64M)^{-1}]$, and repeating the argument finitely many times we get $u \geq 0$ on $[0,\,T]$. That $u > 0$ follows from the strong maximum principle.
