# Prove that $e^x$ is not a tempered distribution on $\mathbb{R}$

Consider the following sequence of functions $\psi_n(x) = e^{-(1+\varepsilon)x} \dfrac{1_{|x|\leq n}}{n}$. Clearly, $|\psi_n^{(m)}(x)|\leq\dfrac{(1+\varepsilon)^m}{n}$. Hence, the $\psi_n$-s are convergent to $0$ in $\mathscr{S}(\mathbb{R})$. However, it is easily computed that $\int_{\mathbb{R}} \psi_n(x)e^xdx = \int_{-n}^{n} e^{-\varepsilon x}dx = \dfrac{1}{\varepsilon}\dfrac{e^{n\varepsilon} - e^{-n\varepsilon }}{n}\geq\dfrac{e^\varepsilon - 1}{\varepsilon}$. Therefore, $v(x) = e^x$ is not a tempered distrubition.

Can anybody check if my attempt at proving the claim correct? My idea was based on this discussion

• A little problem: those truncated versions of exponentials are not smooth, so are not Schwartz... Commented Apr 19, 2016 at 21:06
• @paulgarrett Oh, I see. So I just need to make them smooth by modifying them at $|x| = n$? Commented Apr 19, 2016 at 21:21
• It is not immediately clear to me that modifying a function slightly in an effort to smooth it will not change the derivatives greatly. Commented Apr 19, 2016 at 21:27
• As @Aaron speculates, smoothing functions at discontinuous cut-offs can change the Schwartz semi-norms ... significantly. I guess the operational question is whether it's simpler (in whatever context you find yourself) to see whether you can be sufficiently subtle in doing smooth truncations... versus taking a somewhat different approach to the question. It's true that we seem not to collectively have well-known examples of Schwartz functions that decay (much) more slowly than exponentials! :) Commented Apr 19, 2016 at 21:47
• In this case, you can just name a Schwartz function $\varphi$ such that $$\int_{-\infty}^{+\infty} \varphi(x)e^x\,dx = +\infty.$$ For example $\varphi(x) = \exp( - \sqrt{1+x^2})$. Commented Apr 22, 2016 at 19:09

1. if $$\phi$$ is a distribution of function type, $$\phi\ge 0$$, and $$\phi$$ is also a tempered distribution, then $$\phi$$ has polynomial growth ( our $$\phi(x) = e^x$$ is therefore not a tempered distribution).
2. There exist $$\phi$$ tempered distribution of function type not of polynomial growth. Example: consider $$\psi(x) = \sin e^x$$, of function type, bounded, so tempered, and $$\phi= \psi'$$, of function type, $$\phi(x) = \cos e^x \cdot e^x$$, tempered, of function type, but not of polynomial growth ( notice that the sign of $$\phi$$ is not constant).