Existence of minimizer for a function on an open set

Question

Let $f : \mathbb{R}^d \setminus \{0 \} \to \mathbb{R} \cup \{+\infty \}$ be a continuous function such that $f(k . x) = f(x)$ for $k > 0, k \in \mathbb{R}$ and $x \neq 0, x \in \mathbb{R}^d$. Show that f has a minimizer.

I am stuck:

We can restrict our domain to $D = \{x \in \mathbb{R}^d : x \neq 0, \big|\big|x \big|\big| \leq 1 \}$ and show that f has a minimizer over D.

I think the problem is equivalent to whether a continuous function over an open bounded set is guaranteed to have a minimizer.

Answer 1

Hint: $f$ has a minimum on $\{x:\|x\|=1\}$. Show that this minimum is actually the minimum of $f$ on the entire domain.

Continuous functions on open sets need not have minimizers.

Answer 2

Instead, restrict to $\{\,x\in\Bbb R^d: \|x\|=1\,\}$, which is compact.