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$$ \varphi(x)=e^{1/(x^2-1)} $$ This function (on the interval $\quad]\!-1,1[ \,\,\, $, outside of it simply $\equiv0$) is used as the typical example of a test function / bump function, I have so far seen it it every book that covers $\mathcal{C}_0^\infty$ functions. But it's usually not called any specific name, though it does seem to have one, at least I heard it being called by some name recently, but forgot it.

I'd greatly like to know a name for this function, both for my computer functions library and for ease when writing proofs where a test function is required, and you can quickly reassure its existence with a simple "like the ...-function".

Friedrichs'sche Glättungsfunktion is in fact the name I was looking for!

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    $\begingroup$ In German it is sometimes called Friedrichs'sche Glättungsfunktion (roughly: Friedrichs's mollifying function) to honour its use in Friedrichs's work on differential equations. You can find a discussion and references on the Wikipedia-page on mollifiers. I don't know how "official" that name is, however. $\endgroup$
    – t.b.
    May 23, 2011 at 21:34
  • $\begingroup$ Usually you don't need an explicit formula for your bump function right? You just need to know that it exists. This functions allows you to prove that such $C^\infty$ bump functions exist. $\endgroup$
    – JT_NL
    May 23, 2011 at 23:41

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Bump function.

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  • $\begingroup$ Ahm... it would be a bit unfair to accept this answer now, considering Theo already gave the desired answer in the comments and bumb function refers to general functions of this type, not specifically $e^{1/(x^2-1)}$. $\endgroup$
    – leftaroundabout
    Sep 27, 2011 at 21:10
  • $\begingroup$ @leftaroundabout, probably a bit late to point this out. But that is not how StackExchange works. This answer answers your question. You should accept it to make it clear the question has been answered. People should write answers as answers and not comments or risk someone else getting their points (a risk I am sure they were also fine with or they would not have written their answer as a comment). $\endgroup$
    – Kvothe
    Sep 6 at 10:16

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