Suppose we have a $C^{1}(D)$ function $f : D \to \mathbb{R}$ where $D$ is an open subset of $\mathbb {R}^n$. I would like to claim $f$ has a unique global minimum. I have shown that $f$ is coercive on $D$, i.e., \begin{align*} \lim_{D \ni x_j \to \partial D} f(x_j) = +\infty, \end{align*} and $f$ has a unique stationary point in $D$. Now I would like to claim this stationary point is not saddle nor local maxima. Intuitively this is all very clear to me (maybe my intuition is completely wrong), if it is local maxima or saddle then to go to infinity on the boundary it must have other stationary points and this would contradict what I have shown. But I am having trouble to write a proof.
I think I have a proof. Please point out the errors if there were any.
Let $x^*$ be the stationary point, i.e., $\nabla f(x^*) = 0$. Let $S = \{x : f(x) \le 2 f(x^*)\}$. $S$ is nonempty since $x^* \in S$ and compact by coerciveness and continuity of $f$. Then $f$ achieves minimum on $S$. This minimum point must be stationary so it much coincide with $x^*$ by uniqueness of stationary point. Consequently, $\min \{f(x) : x \in D\} = f(x^*)$.