Show that exp is not a tempered distribution I'm trying to show that $x \mapsto e^x$ is not a tempered distribution. For that I want to find a schwartz function $\varphi \in S(\mathbb{R}) $ such that the support of $\varphi$ is contained in $[-1,1]$ and
$$ \langle e^x, \tau_a\varphi\rangle = \int_\mathbb{R} e^x \varphi(x-a) dx = e^a \int_\mathbb{R} e^x \varphi(x) dx = e^a.
$$
Because then I can show by contradiction that $e^x \notin S'(\mathbb{R})$. Is it even possible to find such a $\varphi$?
 A: First, since integration-against-$e^x$ certainly is a distribution (dual to test functions $C^\infty_c$), if you have a Schwartz function supported on $[-1,1]$, it's actually a test function, and your displayed formula is correct. So I don't quite understand what contradiction you were getting.
Second, yes, the palpable fact that there are (e.g., positive, real-valued) Schwartz functions that integrate to $+\infty$ against $e^x$ is a compelling heuristic to believe that $e^x$ [sic] is not a tempered distribution. But it may be interesting to think about whether this is really a proof or not... in light of things like principal-value integrals, Hadamard's "finite part" integrals, and so on.
In particular, and perhaps connecting to your implicit argument-by-contradiction starting with the displayed formula, while there is a (not-necessarily-tempered) distribution $u$ given by integration against $e^x$, and with that displayed property under translation, we might imagine proving that $u$ has no extension to a tempered distribution with that same property under translation...?  Is this actually what you're thinking?
(As a silly comment: if there is no condition imposed on such an extension, then Hahn-Banach gives many... meaningless... extensions.)
If so, it might be technically more convenient to pose the condition (by differentiating) as $u'=a\cdot u$, with real $a$, to have the characterization be by a single equation.
Perhaps can you clarify?
