I asked a similar question here but did not specify all the details of the function I encountered. As that question has been answered, I ask the clarified version here.
Suppose $f: U \to \mathbb R$ is a $C^{\infty}$ smooth function where $U \in \mathbb R^n$ is an open simply connected set. Further, $f$ has following property: if $U \ni \|x_n\| \to \infty$, $f(x_n) \to +\infty$; if $\{x_n\} \subseteq U$ and $ \{x_n\} \to x \in \partial U$ (boundary of $U$), $f(x_n) \to + \infty$. This is equivalent to say all the sublevel sets of $f$ are compact. Now I know if $x_0 \in U$ is a critical point (that is, $\nabla f(x_0) = 0$), then $x_0$ is a strict local minimizer (that is, the Hessian $\nabla^2 f(x_0)$ is positive definite). My question is whether these information allows us to deduce that $f$ has only one critical point. A counterexample is certainly welcome.
