Let $\{f_n\}$ be a sequence of (real-valued) smooth convex functions on $[0,1]$, with $f_n(0) = f_n(1) = 0$ for all $n$.

Let $t_n \in [0,1]$ be the minimizer of $f_n$ and assume that $M_n:= f_n(t_n) \to -\infty$.

Intuitively, it seems that that $V_n^2 := f''_n(t_n)$ should go to $\infty$. Can we obtain bounds on the relative growth of $V_n$ and $-M_n$? In particular, can we assure that $V_n/(-M_n) = o(1)$? (I mean, what further conditions one needs for this to hold?)

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    $\begingroup$ But $f_n(x)\geq -f'(0) x$, how may $f_n(t_n)$ tend to $-\infty$? $\endgroup$ Mar 2, 2016 at 17:52
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    $\begingroup$ I don't think that there is some simple control of this ratio. You could have some $f_n$ that approach a function with a non-differentiable kink at the minimum, even with fixed value at the minimum (some scaled and translated version of $\sqrt{x^2+1/n}$). $\endgroup$
    – Dirk
    Mar 2, 2016 at 18:29
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    $\begingroup$ Also, your intuition seems misleading: $f''_n(t_n)$ may very well vanish however large $-f_n(t_n)$ is... So I think you need to impose conditions, otherwise the ratio may be anything. $\endgroup$ Mar 3, 2016 at 15:31
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    $\begingroup$ For instance if $f_n(x)=nf(x)$, and if $f''$ vanishes at its minimum point, like e.g. $f(x)=\big(x-{1\over2}\big)^4-{1\over16}$, then of course for all $n$, $t_n$ is the same point, and $f''(t_n)=0$ . $\endgroup$ Dec 28, 2016 at 9:33
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    $\begingroup$ $V_n/(-M_n)$ may even diverge; check Dirk's example. $\endgroup$ Dec 30, 2016 at 2:22

1 Answer 1


$V_n$ can easily be bounded. Let $\varepsilon > 0$ fixed $$ f_n' (x) = \begin{cases} -1/2 + n(x - (1/2 - \varepsilon)) & x < 1/2 - \varepsilon\\ \varphi_n(x) & 1/2 - \varepsilon \le x \le 1/2 + \varepsilon \\ 1/2 + n(x - (1/2 + \varepsilon)) & 1/2 + \varepsilon < x \end{cases}$$ and $\varphi_n$ is a bounded increasing sequence such that $f_n$ is $C^\infty$, odd respect to $1/2$, $\varphi_n (1/2) = 0$ and not zero otherwise. Then $t_n = 1/2$ and $M_n \to - \infty$ and $f_n'' (t_n) = 0$.

  • $\begingroup$ @ D G, thanks. Not sure if I follow the argument. This seems like another counterexample. Are you saying that all cases are like this? Could you elaborate? How can $V_n$ be easily bounded in general. $\endgroup$
    – passerby51
    Dec 30, 2016 at 0:55
  • $\begingroup$ @passerby51 $V_n$ can be either bounded or unbounded. This was simply an example. There is no estimate in the fashion you ask. $\endgroup$
    – D G
    Dec 31, 2016 at 9:56

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