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.
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.