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Can somebody suggest a function that is (i) NOT bounded below, (ii) the second derivative is bounded above?

$f(x)=x^2$ is not a good example since $f(x) \geq 0$ (bounded below) and $f''(x)=2$ is bounded above.

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    $\begingroup$ How about $f(x)=-x$ ? $\endgroup$ May 17, 2018 at 13:36
  • $\begingroup$ @JohnSmith Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    May 31, 2018 at 22:02

3 Answers 3

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Starting from your idea let consider $f(x)=-x^2\le 0\implies f''(x)=-2.$

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It the bound on the second derivative is tight, we have

$$f''(x)=f''_{max}$$

and

$$f(x)=f''_{max}\frac{x^2}2+cx+c'$$

which is unbounded below for any $f''_{max}<0$ (or $f''_{max}=0$ and $c<0$).


Interestingly,

$$f(x)=-e^x=f''(x)$$ is unbounded below and bounded above.

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For $n\in\mathbb N$, consider $f_n(x)=-x^{2n}$ has second derivative $f_n''(x)=-2n(2n-1)x^{2n-2}$.

$f_n$ is not bounded below, and $f_n''$ is bounded above.

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