Let $\mu\in\mathcal{S}'(R)$ be a Schwartz distribution. The solution of a heat equation with $\mu$ as the initial data is


$$
u(t,x)= \int\_R \frac{e^{-\frac{(x-y)^2}{2t}}}{\sqrt{2\pi t}} \mu(d y)
$$


You can assume that $\mu$ is non-negative, i.e., a measure on $R$.


The problem is how $u(t,x)$ behaves for $x$ fixed as $t\rightarrow\infty$. I guess that it might not increase too fast for large $t$, e.g., it does not increase like $e^t$. Do anyone have any idea? 


Thank you very much in advance!


EDIT:
Here is one try: 
If we smooth $\mu$ by a test function to get say $\mu\_n$, then $\mu\_n$ is a smooth function with at most certain polynomial increase. However, the degree of the domination polynomial might depend on $n$...



Long time behavior of the heat equation on R
Denote $\Gamma(x,t)$ the fundamental solution of the heat equation form the integral. By the theorem of L. Schwarz for any $\mu\in S'(S)$ there is a number $m\in \mathbb N$ and $C>0$ such that $$|u(x,t)|=|(\mu,\Gamma(t,\cdot))|\le C\|\Gamma(t,\cdot)\|\_m,$$ where
$
\|\varphi\|\_m=$


$\sup\_{\alpha \le m,\ x \in \mathbb R}(1+|x|)^m |\partial^\alpha \varphi(x)|.
$ 


It is straightforward to obtain, as it is said in the comments, that the $u$ increases at most polynomially. Namely, it is known that $$|\partial\_x^k \Gamma(t,x)|\le C\_k t^{-(k+1)/2}e^{-c\_k x^2/t}.\ $$ From here it is easy to get $$|x^m \partial\_x^k \Gamma(t,x)|\le C\_{k,m} t^{(m-k-1)/2}e^{-c\_{k,m} x^2/t},$$ since $|y|^m e^{-c y} \le C\_m$. Putting $y=x^2/t\ $ we have $|x|^m e^{-c x^2/t}\le t^{m/2}e^{-c x^2/2t}$.


