If we define a continuous non-constant function $f:A\to [0,\infty)$ where $A\subset \mathbb{R^n}$ and is closed but not bounded, and if we also have that $\lim_{x\to \infty} f(x)=0$ then can we prove that $f$ has a max on $A$?
I know the extreme value theorem gives us that a continuous function attains a max provided the set it maps from is compact, but given the lack of boundeness of set $A$ i'm unsure of how to proceed.
On
First suppose there isn't a supremum for $\text{im}(f)$. Then for every $K>0$ there is some $x\in A$ such that $f(x)>K$. In particular for every $n\in\Bbb N$ there is some $x_n\in A$ such that $f(x_n)>n$.
Now consider the sequence $(x_n)_{n\in\Bbb N}$. If the sequence is unbounded then for every $L>0$ there is some $n\in\Bbb N$ such that $|x_n|>L$. Then we can't have $\lim_{x\to\infty}f(x)=0$, since that would mean that for every $\epsilon>0$ there is some $M>0$ such that $|f(x)|<\epsilon$ whenever $|x|>M$, but you can take $\epsilon<1$ and its corresponding $M$, and there also is some $n_0\in\Bbb N$ such that $|x_{n_o}|>M$ and thus $f(x_{n_0})>n_0\ge 1>\epsilon$. So $(x_n)_{n\in\Bbb N}$ is bounded.
Since it is bounded we can consider $s=\limsup_{n\in\Bbb N}x_n$ and $r=\liminf_{n\in\Bbb N}x_n$. Take $B=[r,s]\cap A$, which is a closed and bounded set, therefore compact.
We have that $(x_n)_{n\in\Bbb N}\subset[r,s]\cap A$, so there has to be some convergent subsequence $(x_{n_k})_{k\in\Bbb N}$. Let's say it converges to $l$. Then $l\in A$, since $A$ is closed and $(x_{n_k})_{k\in\Bbb N}\subset A$.
Knowing $f$ is continuous we can do $\lim_{k\to\infty}f(x_{n_k})=f(\lim_{k\to\infty}x_{n_k})=f(l)$, but $f(x_{n_k})\ge n_k$ so $\lim_{k\to\infty}f(x_{n_k})\ge\lim_{k\to\infty}n_k=+\infty$. This clearly can't happen, so there is a supremum for $\text{im}(f)$. Call it $S$.
If $f=0$ we have $S=0$ is indeed a maximum. If that's not the case, then there is some $x\in A$ such that $f(x)>0$, since $f:A\to[0,+\infty)$. Therefore the supremum $S\neq 0$, and since $\lim_{x\to\infty}f(x)=0$ we don't have $\lim_{x\to+\infty}f(x)=S$ or $\lim_{x\to-\infty}f(x)=S$. But since $S$ is the supremum, for every $\epsilon=\dfrac{1}{m}>0$ there is some $x_m\in A$ such that $|S-f(x_m)|<\dfrac{1}{m}$. Therefore the sequence $(f(x_m))_{m\in\Bbb N}$ converges to $S$.
The sequence $(x_m)_{m\in\Bbb N}$ has to be bounded as well, since otherwise we could consider an unbounded subsequence $(x_{m_k})_{k\in\Bbb N}$ and we would have $|S|=|S-0|\le|S-f(x_{m_k})|+|f(x_{m_k})-0|=$ $=|S-f(x_{m_k})|+|f(x_{m_k})|<\dfrac{1}{m_k}+|f(x_{m_k})|\xrightarrow{k\to\infty}0$, since $\lim_{x\to\infty}f(x)=0$, and then we would actually have $S=0$.
So again $(x_m)_{m\in\Bbb N}\subset A$ bounded and $A$ closed, and thus there is some convergent subsequence $(x_{m_k})_{k\in\Bbb N}$ converging to some $l\in A$, since $A$ is closed. Finally, using again the continuity of $f$ we get $S=\lim_{m\to\infty}f(x_m)=\lim_{k\to\infty}f(x_{m_k})=f(\lim_{k\to\infty}x_{m_k})=f(l)$, so $S$ is attained at $l\in A$. We conclude $S$ is the maximum of $f$.
If the function is non-constant, then there must exist some $x_0\in A$ such that $f(x_0)\neq 0.$ Since $\lim_{|x|\to\infty}f(x)=0$, there exists $M$ such that for all $x$ with $|x|>M$ we have $f(x)<f(x_0)/2$. Let $B=\{x\in \mathbb R^n: |x|\leq M\}$. Then $A\cap B$ is compact because it is closed and bounded, and it is nonempty because it contains $x_0$. Therefore, $f$ achieves a maximum on $A\cap B$, and that maximum is bigger than or equal to $f(x_0)$, so it is bigger than $f(x)$ for all $x$ outside of $A\cap B$, i.e., those $x$ with $|x|>M$.