Suppose that $\Omega$ is a locally compact (but non-compact) Hausdorff space. Then $$C_{0}(\Omega)=\{x\in C(\Omega) \ | \ \forall\varepsilon>0, \ \{\omega\in\Omega \ | \ |x(\omega)|\geq\varepsilon\} \ \text{is compact}\}$$ supplied with the supremum norm is a Banach space. Let $x\colon\Omega\to\mathbb{R}$ be a function in the space $C_{0}(\Omega)$ such that $\|x\|=1$. Then $\{\omega\in\Omega \ | \ |x(\omega)|\geq1/2\}$ is non-empty and compact. Since $\Omega$ is Hausdorff, $K$ is closed in $\Omega$. And because $\Omega$ is non-compact, it follows that $K\neq\Omega$. So $U:=\Omega\setminus K$ is non-empty and open. Now I have another map $y\colon\Omega\to\mathbb{R}$ in $C_{0}(\Omega)$ that satisfies the following:
- $\forall\omega\in\Omega, \ 0\leq y(\omega)\leq1/2$,
- $\text{supp}(y)\subset U$, i.e. $y$ vanishes on $K$,
- $\exists\omega'\in U, \ y(\omega')=1/2.$
I'm working on a proof where I need the supremum norm of the continuous function $x\pm y$. I suspect (and hope) that $\|x\pm y\|=1$. By distinguishing the cases $\omega\in K$ and $\omega\in U$, it is not hard to prove that $\|x\pm y\|\leq 1$. However, I'm having trouble with the other inequality (assuming that it is true). Any suggestions are greatly appreciated.
Since $||x||=1$, there is $w' \in K$ such that $|x(w)|=1$. Hence, $$||x \pm y|| = \sup_{w \in \Omega} |(x \pm y)(w)| \geq |(x \pm y)(w')| = 1$$ as $y = 0$ on $K$.