Does $\mathrm{e}^x$ belong to $\mathcal{S}'(\mathbb{R}^n)$? Starting with the two spaces of fuctions: $\mathcal{D}(\mathbb{R}^n)$ of smooth and compactly supported functions w/ its usual topology and $\mathcal{S}(\mathbb{R}^n)$ of smooth and fast decreasing funcitions (also named Schwartz space) w/ its usual topology too.
It is useful to think that $\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n)$.
Let $\mathrm{e}^{|x|} \in \mathcal{D}'(\mathbb{R}^n)$ defined by
$(\mathrm{e}^{|x|}, \varphi)=\int_{\mathbb{R}^n} \mathrm{e}^{|x|} \varphi(x)\,dx$.
I always used to think that $\mathrm{e}^{|x|} \notin \mathcal{S}'(\mathbb{R}^n)$. 
But, I recently noticed that I could apply the Hanh-Banach theorem to extend $\mathrm{e}^{|x|}$ for all test functions in $\mathcal{S}(\mathbb{R}^n)$. Is that true? Which is such extension?
 A: No. Yes, you can define $\Lambda:\mathcal D\to\Bbb C$ by $$\Lambda\phi=\int\phi(t)e^t\,dt.$$ But $\Lambda$ is not continuous in the subspace topology on $\mathcal D\subset\mathcal S$, so Hahn-Banch does not apply.
Edit: I'e been asked to prove that $\Lambda$ is not continuous in that topology. Of course, whether we prove this or not, the fact that we have not proved it is continuous means the argument from Hahn-Banach is at least incomplete.
Hmm. Recall that the topology on $\mathcal S$ is defined by the seminorms $$\rho_{N,\alpha}(f)=\sup_{|\beta|\le\alpha}\sup_{t\in\Bbb R}(1+|t|^N)|D^\beta f(t)|.$$
Fix $\phi\in\mathcal D$ with $\phi\ne0$ and $\phi\ge0$. Let $$\phi_n(t)=\phi(t-n).$$For every $N$, $\alpha$ it's clear that $$\rho_{N,\alpha}(\phi_n)=O(n^N),\quad(n\to\infty).$$But $$\Lambda\phi_n\ge ce^n.$$So there do not exist $N,\alpha,$ and $c$ with $$|\Lambda\phi|\le c\rho_{N,\alpha}(\phi)\quad(\phi\in\mathcal D).$$
A: $f \mapsto \int_{\mathbb{R}} e^x f(x) dx$ is certainly not in $\mathcal{S}'$ as defined, since it gives an infinite result when evaluated on a mollification of $e^{-|x|/2}$. Hahn-Banach does not offer an extension because there is no globally defined dominating function for this functional.
A: A slightly modification of @Ian answer: Find a smooth function $\varphi$ on ${\bf{R}}$ such that $\varphi=0$ on a neighbourhood of zero and $\varphi=1$ on the complement of the neighbourhood.
Then $\left<e^{x},\varphi(x)e^{-|x|/2}\right>$ simply does not exist as a real number.
