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I'm trying to find a smooth function $f$ on the real line with compact support, which has a unique minimum (at some $x_{0}\in\mathbb{R}$) and such that its third derivative there is negative, i.e. $f'''(x_{0})<0$. I played around with exponential, ("cut offs" of) trigonometric functions etc., but I couldn't find any. Any help is appreciated.

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By smooth, do you mean infinitely differentiable at every point (including the boundaries of support)? – cardinal May 11 '11 at 16:49
up vote 3 down vote accepted

Consider the function $f$ defined by $f(x)=(x^2-x^3-1)\exp(1/(x^2-1))$ if $|x|<1$ and $f(x)=0$ if $|x|\ge1$, and the point $x_0=0$.

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How about $-10+10x^2-x^3+x^4$, then "round off the corners" where it crosses the x axis to make it smooth and compact support?

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The fourth derivative of this function is the constant 24 everywhere. How do you plan on "rounding the corners" to make it smooth? (I'm quite likely just being dense.) – cardinal May 11 '11 at 16:56
@cardinal: I would multiply it by a bump function that is zero outside $[-.8,.8],$ one over $[-.75,.75]$ and infinitely differentiable. That will zero out the fourth derivative outside $[-.8,.8]$ as well. – Ross Millikan May 11 '11 at 17:08
I'll check that, as I was trying something similar. – cardinal May 11 '11 at 17:09

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