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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

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dezdichado
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  • A little problem: those truncated versions of exponentials are not smooth, so are not Schwartz... – paul garrett Apr 19 '16 at 21:06
  • @paulgarrett Oh, I see. So I just need to make them smooth by modifying them at $|x| = n$? – dezdichado Apr 19 '16 at 21:21
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    It is not immediately clear to me that modifying a function slightly in an effort to smooth it will not change the derivatives greatly. – Aaron Apr 19 '16 at 21:27
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    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! :) – paul garrett Apr 19 '16 at 21:47
  • I have been playing with the bump functions but to no avail. Can you think of a different approach than this to show that a simple distribution on $\mathbb{R}$ is not a tempered one? – dezdichado Apr 19 '16 at 21:56
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    In this case, you can just name a Schwartz function $\varphi$ such that $$\int_{-\infty}^{+\infty} \varphi(x)e^x,dx = +\infty.$$ For example $\varphi(x) = \exp( - \sqrt{1+x^2})$. – Daniel Fischer Apr 22 '16 at 19:09
  • @paulgarrett Interesting time to come across a comment by you. I just finished an algebra assignment out of your Abstract Algebra text, now starting my analysis homework... and there you are! – Prince M Mar 13 '18 at 03:43
  • @PrinceM, :) ...... – paul garrett Mar 13 '18 at 05:41

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Some comments:

  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).

orangeskid
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