Let $E$ be a locally compact Hausdorff space and $$C_0(E):=\left\{f\in C(E):\left\{|f|\ge\varepsilon\right\}\text{ is compact for all }\varepsilon>0\right\}.$$
Let $f\in C_0(E)$ with $$\inf_{x\in E}f(x)<0\tag1.$$ How can we show that there is a $x_0\in E$ with $$f(x_0)\le\min(f(x),0)\;\;\;\text{for all }x\in E\tag2?$$
I guess we can simply apply the fact$^1$ that if $g\in C_0(E)$, there is a $x_0\in E$ with $|g(x_0)|=\sup_{x\in E}|g(x)|$ to the function $g:=\min(f,0)$ or am I missing something?
$^1$ Assume $E$ is a normed $\mathbb R$-vector space and $g\not\equiv0$. Then there is a $x_1\in E$ with $\varepsilon:=|g(x_1)|>0$. Since $\left\{|g|\ge\varepsilon\right\}$ is compact, there is a $r>0$ with $$|g(x)|<\varepsilon\;\;\;\text{for all }\left\|x\right\|_E>r.\tag1$$ By continuity of $g$ and compactness of the closed ball around $0\in E$ with radius $r$, there is a $x_0\in E$ with $$|g(x_0)|=\sup_{\left\|x\right\|_E\:\le\:r}|g(x)|\ge\varepsilon\tag4.$$ Maybe we can use a similar argumentation even when $E$ is a general locally compact Hausdorff space.
Indeed, for $g \in C_0(E)$ there exists $x_0 \in E$ such that $|g(x_0)| = \sup _{x\in E}|g(x)|$:
The set $K = \left\{|g| \ge \frac12 \sup _{x\in E}|g(x)| \right\}$ is compact so $|g|\big|_K$ attains its maximum, i.e. there exists $x_0 \in E$ such that $$|g(x_0)| = \sup_{x \in K}|g(x)|$$ But we have $|g(x)| < \frac12 \sup _{x\in E}|g(x)|$ for $x \notin K$ so $$|g(x_0)| = \sup_{x \in K}|g(x)| = \sup_{x \in E}|g(x)|$$
Applying this to $g = \min(f,0)$ we have $|g| = |f|\chi_{\{f < 0\}}$ so $f(x_0) \le 0$.
Also, if $f(x) < 0$ then $|g(x)| = |f(x)|$ so for such $x$ we have $$f(x_0) = -|g(x_0)| \le -|g(x)| = f(x)$$ Therefore $f(x_0) \le g(x)$ for all $x \in E$.