# A smooth function $f(x)$ has a unique minimum. It $f$ also varies smoothly in time, does the location of its minimum vary smoothly in time?

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 minimum in $x$ for every fixed $t\in[0,1]$.

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?

• If the minimums are nondegenerate (meaning the second derivative is positive at the minimum) then you can use the implicit function theorem to get that indeed the the points where the mínima are attained is a smooth curve. Otherwise I suspect your "curve" could be very pathological. Mar 29, 2017 at 4:16
• Unique relative minimum, or absolute (global)? I don't know if it matters for the result but am not sure exactly what the hypothesis is. Mar 29, 2017 at 4:19
• @JonasMeyer I meant for it to be a unique global minimum, though as you say I am not sure this matters for the result. Mar 29, 2017 at 5:05
• @Jose27 you are right, that is a good idea to work in the implicit function theorem. Unfortunately I am not assuming nondegenerate minima. Mar 29, 2017 at 5:10
• I asked a modified version of this question at math.stackexchange.com/questions/2208776/… Mar 29, 2017 at 15:12

The function $f(x,t)=(tx-1)^2 - \exp\left(\dfrac{1}{x(x+2)}\right)1_{(-2,0)}(x)$ seems to be a counterexample, where $1_A(x)$ is the indicator function for $A$. The second term is a bump function with min at $x=-1$. Here $\chi(0)=-1$, but $\chi(t)=\frac1t$ when $t>0$.

• Well done! Interestingly, your counterexample does not appear to work in the "unqiue local minimum" version of my question. Perhaps that distinction between unique global minimum and unique local minimum does matter for this question after all. Mar 29, 2017 at 5:56
• FWIW I think you get the same type of counterexample but without as neat a form for $\chi$ by taking $(tx-1)^2-e^{-(x+1)^2}$, which is even real analytic. But this is qualitatively the only type of counterexample I know of, where the global min goes off to infinity and disappears to be replaced by a local min that was waiting in the wings. Mar 29, 2017 at 12:23
• @DarrenOng: I saw your other question, which is interesting. I would be very surprised if $\chi$ is not at least continuous in that case although I don't have proof even of that let alone differentiability. Mar 30, 2017 at 18:03