----- > [!proposition] Proposition. ([[condition for plane curve achieving minimum norm]]) > Let $\alpha:\mathbb{R} \to \mathbb{R}^{2}$ be a plane curve defined in the entire real line $\mathbb{R}$. Assume that $\alpha$ does not pass through the origin $0=(0,0)$ and that both $\lim_{ t \to +\infty } \|\alpha(t)\|=\infty \text{ and } \lim_{ t \to -\infty } \|\alpha(t)\|=\infty.$ > Then, there exists a point $t_{0} \in \mathbb{R}$ such that $\|\alpha(t_{0})\| \leq \|\alpha(t)\|$ for all $t \in \mathbb{R}$. > > This is in general false if we do not assume $\lim_{ t \to +\infty } \|\alpha(t)\|=\infty \text{ and } \lim_{ t \to -\infty } \|\alpha(t)\|=\infty$. Take, for example, the curve $\beta(t)=(x(t), x(t))=\left( \frac{1}{e^{t}}, \frac{1}{e^{t}} \right)$, $t \in \mathbb{R}$. The trace of $\beta(t)$ never touches the origin because $x(t)$ is a strictly-positive function. But as $t \to \infty$, we have $\beta(t) \to (0,0)$ and thus $\|\beta(t)\| \to (0,0)$ asymptotically. It follows that given any $t_{0} \in \mathbb{R}$, we can find $t'$ such that $\|\beta(t')\| < \|\beta(t_{0})\|$... for example, it suffices to choose $t':=t_{0}+59$. So $\|\beta(t)\|$ never achieves a minimum value. > [!proof]- Proof. ([[condition for plane curve achieving minimum norm]]) > Form the function $f(t)=\frac{1}{\|\alpha(t)\|}$; because we are guaranteed $\alpha \neq \b 0$, $f(t)$ is well-defined on all of $\mathbb{R}$. Moreover, $f(t)>0$ for all $t \in \mathbb{R}$. In light of this, take $t'=0$ (could be chosen to be any real number) and consider $\varepsilon_{}:=f(t')$. Using that $\lim_{ t \to \infty } f(t)=0=\lim_{ t \to -\infty }f(t)$, obtain $\delta_{a},\delta_{b} \in \mathbb{R}$ with $\delta_{a}<\delta_{b}$ such that for all $t<\delta_{a}$ we have $|f(t)-0|=|f(t)|=f(t)<\varepsilon$ and likewise for all $t>\delta_{b}$ we have $f(t)<\varepsilon$. So: $\begin{align} f(t) < & f(t') \text{ for all } t < \delta_{a} \\ f(t) < & f(t') \text{ for all } t < \delta_{b}. \end{align}$ Now consider the truncated continuum $[\delta_{a}, \delta_{b}] \subset (-\infty, \infty)$. This [[interval]] is [[compact]], thus the [[extreme value theorem]] guarantees the existence of $t_{0} \in [\delta_{a}, \delta_{b}]$ such that $f(t) \leq f(t_{0}) \text{ for all } t \in [\delta_{a}, \delta_{b}].$ Since this includes $f(t')$, we have $f(t) < t(t') \leq f(t_{0})$ for all $t$ outside of $[\delta_{a}, \delta_{b}]$, and thus we may conclude that $f$ attains a maximum value on $\mathbb{R}$. > A maximum value of $f(t)$ at $t_{0}$ corresponds to a minimum value of $\|\alpha(t)\|$ at $t_{0}$. So we are done. ![[CleanShot 2024-02-10 at [email protected]]] ![[CleanShot 2024-02-10 at 16.03.48.jpg]] ^d4fbd9 ----- #### ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag