TITLE: How to show $e^{x}\cos[e^x]$ is a tempered distribution?
QUESTION [10 upvotes]: From Melrose, Lecture notes on Microlocal Analysis, Chapter 1. 
I was asked to show that the function
$$
u(x)=e^{x}\cos[e^{x}]
$$
is a tempered distribution. I tried to use the definition that there exist $k$ and $C_{k}$ such that
$$
|\int uv|\le C_{k}|v|_{k}
$$
where $$|v|_{k}=\sup_{\alpha+\beta\le k}|x^{\alpha}\partial_{x}^{\beta}v|$$
However I found I did know how to treat the oscillatory part
$$
\int^{\infty}_{M}uv
$$
where $M$ is a large enough number. Can someone give a hint?

REPLY [2 votes]: First note that $u$ is a ($C^\infty$) function and not a distribution. Hence, if we want to interpret $u$ as a distribution, we need to make some kind of identification.
Your book probably defines this identification as follows: If $f$ is a (locally integrable) function, we say that $f$ defines a tempered distribution if there is some tempered distribution $u$ such that
$$
u(\varphi)= \int f \cdot \varphi\, dx
$$
for all $\varphi \in C_c^\infty$ (not for all Schwartz functions).
This definition is sensible, because it is often possible to write down a Schwartz function such that the integral  $\int f \cdot \varphi$ does not exist (in the usual Lebesgue sense).
If you use that definition here, your problematic boundary terms will vanish and the distribution $u$ above will turn out to be $\frac{d}{dx}\sin(e^x)$.