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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?

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  • $\begingroup$ $e^{|x|}$ or $e^{|x|^{2}}$? Perhaps you may have to assume $n=1$. $\endgroup$ – user284331 Jan 22 '18 at 15:38
  • $\begingroup$ Thanks. I edited it. $\endgroup$ – user418048 Jan 22 '18 at 16:40
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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 $$|\Lambda\phi|\le c\rho_{N,\alpha}(\phi)\quad(\phi\in\mathcal D).$$

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  • $\begingroup$ At first I wanted to argue $\Lambda$ is not continuous in the subspace topology but I fail. So how to argue that formally? $\endgroup$ – user284331 Jan 22 '18 at 15:58
  • $\begingroup$ @user284331 Take $f_n$ which are mollifications of $e^{-|x|/2} 1_{[-n,n]}(x)$, then $f_n$ are "trying to converge" to $e^{-|x|/2}$ in the subspace topology of $\mathcal{D}$ inherited from $\mathcal{S}$ while $\Lambda f_n$ is blowing up. (It is important here that the subspace topology of $\mathcal{D}$ inherited from $\mathcal{S}$ and the usual topology of $\mathcal{D}$ are totally different.) $\endgroup$ – Ian Jan 22 '18 at 16:07
  • $\begingroup$ @Ian Of course you need to show that $(f_n)$ is convergent, or at least bounded, in $\mathcal S$... $\endgroup$ – David C. Ullrich Jan 22 '18 at 16:11
  • $\begingroup$ @DavidC.Ullrich Quite right, but you can't quite iron out the details of that proof until you specify the details of the mollifications. $\endgroup$ – Ian Jan 22 '18 at 16:13
  • $\begingroup$ @user284331 See edit. $\endgroup$ – David C. Ullrich Jan 22 '18 at 16:23
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$f \mapsto \int_{\mathbb{R}} e^x f(x) dx$ 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.

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  • $\begingroup$ Is there any problem if we simply put $e^{-x/2}$, I mean, why do we need a mollification of $e^{-|x|/2}$? $\endgroup$ – user284331 Jan 22 '18 at 15:50
  • $\begingroup$ @user284331 Just $e^{-x/2}$ is not in $\mathcal{S}(\mathbb{R})$ because of bad behavior on the left. But $e^{-|x|/2}$ isn't either because it's not smooth. $\endgroup$ – Ian Jan 22 '18 at 15:51
  • $\begingroup$ Ah... I forgot the left side. $\endgroup$ – user284331 Jan 22 '18 at 15:51
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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.

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