Let $f(x,t)$ be a smooth function $\mathbb R^2\to\mathbb R$ such that $F_t(x):=f(x,t)$ has a unique local minimum in $x$ for every fixed $t\in[0,1]$. Further assume that this local minimum of $F_t(x)$ is also the unique global minimum of $F_t(x)$.

How regularly does the location of this unique minimum vary with respect to $t$? In other words, if $x=\chi(t)$ is the $x$-value where $F_t(x)$ attains its unique minimum, can we say that $\chi(t)$ is a smooth function of $t$? If not, is $\chi(t)$ differentiable or continuous?

I asked a similar version of this question here. The answer was correct and very clever, but I was wondering what happens if we insist that the unique global minimum was also a unique local minimum.

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    $\begingroup$ For fixed $t$, $f(x, t) = (x^3 - t)^2$ has its unique local and global minimum at $x = t^{1/3}$. This is not differentiable at $t = 0$. $\endgroup$ Apr 1, 2017 at 19:13
  • $\begingroup$ @RaviFernando Make it an answer! $\endgroup$ Apr 1, 2017 at 21:05
  • $\begingroup$ Very good answers! But this is a bit of an awkward situation for me. Both answers taken together answer my question, and I would prefer to accept both answers rather than choose one. However I don't think there is a way for me to accept two answers. What should I do? $\endgroup$
    – Darren Ong
    Apr 3, 2017 at 0:55
  • $\begingroup$ You can award 250 to each I think. $\endgroup$
    – zhw.
    Apr 4, 2017 at 2:59
  • $\begingroup$ @zhw I did not put up the bounty, Jonas Meyer did. $\endgroup$
    – Darren Ong
    Apr 4, 2017 at 3:05

2 Answers 2


For fixed $t$, $f(x,t) = (x^3−t)^2$ has its unique local and global minimum at $x = t^{1/3}$. This is not differentiable at $t=0$.

  • $\begingroup$ I wonder if this is the highest ratio of bounty to character count of any answer in MSE history $\endgroup$
    – zhw.
    Apr 4, 2017 at 17:30
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    $\begingroup$ @zhw.: math.stackexchange.com/a/74383 $\endgroup$ Apr 4, 2017 at 22:42
  • $\begingroup$ @JonasMeyer Hilarious, thanks. $\endgroup$
    – zhw.
    Apr 5, 2017 at 18:26

I am not sure about the differentiability of $\chi$, but unless I missed something it is really easy to show that $\chi$ is continuous if $f$ is.

Indeed, consider $t_0\in [0,1]$ and let $k_0=\chi(t_0)$. Let $\varepsilon>0$, and $g(t)= \min(f(k_0-\varepsilon,t)-f(k_0,t),f(k_0+\varepsilon,t)-f(k_0,t))$. Then $g$ is continuous since $f$ is, and $g(t_0)>0$, so there must be a $\eta>0$ such that $g(t)>0$ for any $t\in[t_0-\eta,t_0+\eta]\cap [0,1]$. Consider now, for such $t$, the restriction of $F_t(x)=f(x,t)$ to $[k_0-\varepsilon,k_0+\varepsilon]$ ; by compactness, there is a $z\in [k_0-\varepsilon,k_0+\varepsilon]$ where $F_t$ reaches its minimum. Since $z\leq F_t(k_0)$, $z$ must be strictly inside $[k_0-\varepsilon,k_0+\varepsilon]$. Because of the hypotheses on $f$, this forces $z$ to coincide with $\chi(t)$. This shows that $|\chi(t)-\chi(t_0)|<\varepsilon$ whenever $|t-t_0|<\eta$, and this is the classic definition of continuity.

  • $\begingroup$ Why? He helped fix an (admittedly minor) issue in the answer. His comment should remain to show his contribution, that's not how the delete function is supposed to be used. $\endgroup$
    – MM8
    Apr 1, 2017 at 22:18
  • $\begingroup$ Not to remove legitimate contributions from users, no matter how minor you may think they are. $\endgroup$
    – MM8
    Apr 2, 2017 at 22:13
  • $\begingroup$ SE is a gameified system. The person posting the answer still gets reputation for their better attempts and deleting old attempts does not rob them of being recognized for their contribution via rep. Having other people delete their contribution in comments removes their chance at reputation increases despite their contribution, directly harming the gamification incentive system that SE has going. This belongs on meta, not here but I still really think you are subverting the system by trying to make people delete their contributions. $\endgroup$
    – MM8
    Apr 3, 2017 at 16:21
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    $\begingroup$ @TimonG.: There is no reputation for comments. Comments that are no longer relevant can be deleted with no harm. Sometimes they're worth keeping for providing lasting insight and giving credit to the person providing it, but for typos it doesn't serve a purpose I'm aware of. $\endgroup$ Apr 3, 2017 at 16:57
  • $\begingroup$ Happy to delete my comment, including this one, although it's been fun. $\endgroup$
    – zhw.
    Apr 4, 2017 at 2:58

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