Non isolated minimum Consider the $C^2$ function $F:\mathbb{R}^k \rightarrow \mathbb{R}^k$
is it possible for it to be such that $x_n$ is a strict local minimizer for all $n$ and $x_n \rightarrow x$ where $x$ is a strict local minimizer itself.
I know that $\sin(\frac{1}{x})$ has a somewhat similar property, but it is not continuous at zero.
Thanks!
 A: Here is an example: $F(x) = x^6 (\sin \frac{1}{x})^2 + x^8$.
Establishing that this satisfies your criteria takes a little work:
First, let us establish the relevant smoothness properties:
Note that $F$ is smooth for $x \ne 0$, and near $x=0$ we have some $K_0$ such that $F(x) \le K_0x^6$, we have $|F(x)-F(0) - 0.x| \le K x^6$, hence $F'(0) = 0$. Some differentiation gives $F'(x) = 8{x}^{7}+6 {\mathrm{sin}\frac{1}{x}}^{2}{x}^{5}-2\mathrm{cos} \frac{1}{x} \mathrm{sin} \frac{1}{x} {x}^{4}$ for $x \neq 0$, and since near $x=0$ we have some $K_1$ such that $|F'(x)| \le K_1 x^4$, hence $F'$ is continuous.
Since $|F'(x)-F'(0)| \le K_1 x^4$, we see that $F''(0) = 0$.
Some monotonous work shows that for $x \neq 0$ we have $|F''(x)| \le K_2 x^2$, from which it follows that $F$ is $C^2$. Finally, we note that $F(z) = z^6 (\sin \frac{1}{z})^2 + z^8$ is an analytic function on $\operatorname{re}z > 0$.
Since $F(x) \ge x^8 \ge 0 = F(0)$, we see that $0$ is a strict local minimizer.
Let $\mu_n = \frac{1}{\frac{\pi}{2}+n\pi}$, $\zeta_n = \frac{1}{n\pi}$. We see that $\mu_{n+1} < \zeta_n < \mu_n$ and
$F(\mu_n) = (\frac{1}{\frac{\pi}{2}+n\pi})^6 (1+ (\frac{1}{\frac{\pi}{2}+n\pi})^2) \ge (\frac{1}{\frac{\pi}{2}+n\pi})^6 = (\frac{1}{n\pi})^6 \left( \frac{n \pi}{\frac{\pi}{2}+n\pi} \right)^6$,
$F(\zeta_n) = (\frac{1}{n\pi})^6 (\frac{1}{n\pi})^2$, and 
$F(\mu_{n+1}) = (\frac{1}{\frac{3\pi}{2}+n\pi})^6 (1+ (\frac{1}{\frac{3\pi}{2}+n\pi})^2) \ge (\frac{1}{\frac{3\pi}{2}+n\pi})^6 = (\frac{1}{n\pi})^6 \left( \frac{n \pi}{\frac{3\pi}{2}+n\pi} \right)^6$.
Since $\frac{n \pi}{\frac{\pi}{2}+n\pi} > \frac{n \pi}{\frac{3\pi}{2}+n\pi} > \frac{1}{n\pi}$ for $n \ge 1$, we see that $F(\mu_n) > F(\zeta_n), F(\mu_{n+1}) > F(\zeta_n)$, and so $F$ has a minimizer $x_n$ in $(\mu_{n+1}, \mu_n)$. To see that this is a strict minimizer, suppose $y_k \to x_n$ is a sequence such that $F(y_k) = F(x_n)$, then since $F$ is analytic, this would imply that $F$ is constant on $\operatorname{re}z > 0$, which is a contradiction. Hence $x_n$ is a strict local minimizer.
Since $x_n \to 0$, we have the desired result.
