Closure of a set and extrema values of a function

Given the subset of $\mathbb{R}^2$

$$\mathcal{C}=\left \{ e^{-t}\left ( \cos t, \sin t \right ) \; \text{such that} \; t \geq 0 \right \}$$

we define the function $f:\overline{\mathcal{C}} \rightarrow \mathbb{R}$ as

$$f(x, y)= \left\{\begin{matrix} - \left ( x^2+y^2 \right )^{-1} &,  & (x, y) \in \mathcal{C} \\   0& , &\text{elsewhere}  \end{matrix}\right.$$

1. Find the closure $\overline{\mathcal{C}} \subset \mathbb{R}^2$ of $\mathcal{C}$ and prove that is compact.
2. Find all (local) extrema of $f$ and characterize them.

Solution

We have seen what $\mathcal{C}$ is in the topic here . It is actually a spiral.

1. Its outermost point is $(1,0)$ and it spirals in toward the origin. $\overline{\mathcal{C}}=\mathcal{C}\cup\{(0,0)\}.$ Since $\mathcal{C}$ is bounded, its closure is compact.
2. Define $(x(t),y(t))$ as in the representation of $\mathcal{C}$ (as a parameterized curve). Then $f(x(t),y(t))=e^{2t}.$ Since $e^{2t}$ is increasing as $t$ increases, there are no local maxima for $f$ anywhere, and $f$ is unbounded above. On the other hand $f(0,0)=0,$ and that is a local minimum for $f.$ The hypotheses of the Extreme Value Theorem are not met; $f$ is not continuous on $\overline{\mathcal{C}}$ and (as a related fact), it is not uniformly continuous on $\mathcal{C}.$