I have that $f(x) : \mathbb{R}_+ \rightarrow \mathbb{R}$ is a uniformly continuous function s.t. $f(0) = \alpha \in \mathbb{R}_+$ and $lim_{x \rightarrow \infty} f(x)=0.$
I believe that if I can show that $\mathbb{R}_n$ is bounded, then I can use the extreme value theorem to show that the maximum must exist and be in $\mathbb{R}_+$. My idea is that I can define my distance function $d(x,y) = |x-y|$ and create a complete metric space. Thus my uniformly continuous function maps from a complete metric space to $\mathbb{R}$, making my function bounded. Since my function is uniformly continuous and bounded on a closed set $\mathbb{R}_+$, there exists $c$ and $d$ in $[0, \infty)$ such that:
$f(c) \leq f(x) \leq f(d) \forall x \in [0, \infty)$ with $d$ being my maximum.
Split into three cases.
Case 1. $\alpha>0$. As $\lim_{x\rightarrow \infty} f(x) =0, \exists A >0$ s.t. $\forall x > A, |f(x)| < \alpha$.
Thus, we have a closed and bounded interval $[0,A]$, and so by the extreme value theorem we know that there must exist $c,d \in [0,A]$ s.t. $f(c) \leq f(x) \leq f(d) \,\forall x \in [0,A]$.
Thus, $d$ is the maximum of $f(x)$ on $[0,\infty)$.
Case 2. $\alpha =0$. Let $\exists x' \in \mathbb{R}_+$ s.t. $f(x')>\alpha$. As $\lim_{x\rightarrow \infty} f(x) =0, \exists B >0$ s.t. $\forall x > B, |f(x)| < f(x')$.
Thus, we have a closed and bounded interval $[0,B]$, and and we can apply the extreme value theorem.
Case 3. $\alpha =0$. Let $x'$ s.t. $f(x')> \alpha$ not exist $\in \mathbb{R}_n$. Then $\alpha \geq x \forall x \in \mathbb{R}_+$ and thus is our maximum.
Is this correct?