# Proving that on a plane curve defined on the reals there exists $t_0$ such that $\vert \alpha(t_0) \vert \leq \vert \alpha(t) \vert \ \forall t$

Let $\alpha: \mathbb{R} \to \mathbb{R}^2$ be a closed plane curve defined on the real line. Suppose that $\alpha$ doesn't go through the origin $\textbf{0} = (0,0)$ and that: $$\displaystyle{\lim_{t \to \pm \infty}} \vert\alpha(t)\vert = \infty$$

a) Prove that there exists $t_0 \in \mathbb{R}$ such that $\vert \alpha(t_0) \vert \leq \vert \alpha(t) \vert \ \forall t \in \mathbb{R}$.

b) Show, with an example, that the affirmation made in (a) is false if we don't suppose that both limits (as $t$ goes to $\infty$ and $-\infty$) go to $\infty$.

I think b) is easy, but I'm having some trouble with a). I had an idea to do it by contradiction but it led me nowhere. Any help would be appreciated.

• define closed plane curve precisely please – zhw. May 16 '18 at 17:25

I'll assume that $\alpha$ is continuous.
Since $\lim_{t\to+\infty}\bigl|\alpha(t)\bigr|=+\infty$, there is a $t_+\in[0,+\infty)$ such that $t\geqslant t_+\implies\bigl|\alpha(t)\bigr|\geqslant\bigl|\alpha(0)\bigr|$ and since $\lim_{t\to-\infty}\bigl|\alpha(t)\bigr|=+\infty$, there is a $t_-\in(\infty,0]$ such that $t\leqslant t_-\implies\bigl|\alpha(t)\bigr|\geqslant\bigl|\alpha(0)\bigr|$. Since the function$$\begin{array}{ccc}[t_-,t_+]&\longrightarrow&\mathbb{R}_+\\t&\mapsto&\bigl|\alpha(t)\bigr|\end{array}$$is continuous and its domain is a closed and bounded interval, it attains a minimum at some $t_0\in[t_-,t_+]$. And if $t\in\mathbb{R}\setminus[t_-,t_+]$, then $\bigl|\alpha(t)\bigr|\geqslant\bigl|\alpha(0)\bigr|\geqslant\bigl|\alpha(t_0)\bigr|$. Therefore, your function attains a minimum at $t_0$.