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Is there a practical example of a real, one-variable function with compact support that is not defined piecewise? Most computer algebra systems have a hard time dealing with such expressions, so a bump function that isn't piecewise defined would be very helpful.

Xander Henderson
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Johnny
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    If by "isn't piecewise" you mean that you want some "nice formula" involving the usual functions $\sin,\cos,\tan,\arctan,\exp,\ln$ etc, then unfortunately, this is not possible, because such functions are analytic (locally expressible as power series). If an analytic function vanishes on a non-trivial open subset of the real line, then it must vanish identically (uniqueness of analytic continuation). – peek-a-boo Feb 11 '22 at 10:28
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    Does $f(x) = \max(0, 1-|x|)$ count? – daw Feb 11 '22 at 10:32
  • with bump function you are meaning any continuous function of finite-extension? or you are using the definition of bump-function from the smooth functions with compact-support $\in C_c^\infty$?... answers are going to be completely different for both kind of meanings, so be aware of it. – Joako Feb 14 '22 at 18:51

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Take$$f(x)=1-\sqrt{x^2}+\sqrt{\left(1-\sqrt{x^2}\right)^2},$$whose support is $[-1,1]$. If $x\in[-1,1]$, $f(x)=2\bigl(1-|x|\bigr)$.

José Carlos Santos
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    Huh, very smart. At first glance I thought this was only defined on $[-1,1]$ (which wouldnt answer op's question), but the square takes care of that. – Just dropped in Feb 11 '22 at 10:37
  • I was looking for at least C^2 functions, but I guess those can't be created from analytical functions. – Johnny Feb 11 '22 at 11:23
  • This solution, being a compact-support function not defined piece-wise, it is a bump function under the meaning of finite extent functions sometime used for teaching convolutions, but if the OP is referring to the class of bump functions $\in C_c^\infty$ (it is didn´t specified, but it looks for me is the spirit of the question), the function will fail to be a bump-function since it is not a smooth function. – Joako Feb 14 '22 at 18:48
  • José Carlos Santos: I use your example $f(x)$ to made this function $q(x) = e^{-\frac{x^2}{f(x)}}$ which I believe it is an example of a bump function $\in C_c^\infty$ not defined piece-wise, but I couldn´t prove it formally, so I left the question here if you are interested to review it. – Joako Feb 15 '22 at 00:12
  • @José Carlos Santos: In my last comment there is a typo: the function was $q(x)=e^{-\frac{x^2}{f(x^2)}}$. – Joako Feb 15 '22 at 15:08
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Using a slight modification of the answer given by @José Carlos Santos above, I believe is been "informally proved" here that the following function is a bump function $\in C_c^\infty$ not defined piecewise: $$f(x) = \left(1-x^2+ \sqrt{\left(1-x^2\right)^2}\right)\exp\left(-\frac{x^2}{1-x^2}\right)$$

Joako
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  • Can you also create sort of the opposite of a bump function, i.e. a function that is flat in some finite interval in a similar manner? (See https://math.stackexchange.com/questions/4764011/non-piecewise-definition-of-a-function-that-is-flat-in-some-interval-a-b-and) – Kvothe Sep 05 '23 at 15:28
  • @Kvothe Now I better understand the topic I think my answer is wrong in the strict mathematical sense since the function is not defined at $x={-1;\ 1}$ even when I have "hiddenly" defined piecewise using a polynomyal/square_root form of the Heaviside unitary step function - Nevertheless, it still continuous since the limits exists and are equal in both sides of the problematic points. Since having a non-zero measure interval of constant value is forbidden for power series due the Identity Theorem, I think is hardly possible to find a proper one (as mentioned by EthanBolker in your question) – Joako Sep 05 '23 at 16:31