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 $n$...

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 decrease not increase too fast for large $t$, at least e.g., it does not increase like $e^t$. Do anyone have any idea? Thank you very much in advance!

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 decrease for large $t$, at least it does not increase like $e^t$. Do anyone have any idea? Thank you very much in advance!