Let assume that you have a sequence of twice differentiable functions $(f_n)_{n\in\mathbb{N}}\in\mathscr{C}^2(\mathbb{R})^{\mathbb{N}}$. Let suppose that for each $f_n$, it exists a $x_n\in\mathbb{R}$ such that:

  • $f_n$ is strictly convex on $(-\infty,x_n)$,
  • $f_n$ is strictly concave on $(x_n, +\infty)$.

Finally, let assume that the sequence uniformly converges to $f\in\mathscr{C}^2(\mathbb{R})$.

My feeling is that the sequence $x_n$ has to converge. Can one has an idea to show this ?

Thank you!


1 Answer 1


A counterexample is given by the formula $$f_n(x)=\frac1n\,\arctan(x-x_n)$$ for real $x$, where $(x_n)$ is any nonconvergent sequence of real numbers.

  • $\begingroup$ Thank you Iosif, I have however forget to precise that the limit cannot be linear. Do you think it is still incorrect ? The intuition behind the scene is that if the limit has strict convex and concave region, the concavity/convexity of these regions must fit the convexity of the sequence. $\endgroup$
    – NancyBoy
    Oct 3 at 20:42
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    $\begingroup$ You can modify the above example so that e.g. $f_n(x)= x^2$ for $x<0$, no? $\endgroup$ Oct 3 at 20:49
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    $\begingroup$ @Gaetano : Please finalize this matter. If you want to impose any additional conditions (such as not allowing the limit function to be affine, even on any nonempty interval), then please post such a modified question separately, after a careful preparation. If you keep moving the goalposts, especially after receiving a valid answer, people will be less willing to answer your questions. $\endgroup$ Oct 4 at 1:02
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    $\begingroup$ @Gaetano : The first name of Pietro Majer is Pietro, not Peter. $\endgroup$ Oct 4 at 1:05
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    $\begingroup$ @Gaetano : Concerning "finalizing the answer", I thought you were familiar with these guidelines. $\endgroup$ Oct 4 at 16:20

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