Space of continuous functions versus $L^\infty$

Question

Let $(S,\Sigma,\mu)$ be a measure space. The Riesz-Fischer theorem shows that the Lebesgue spaces $L^p(S,\Sigma,\mu)$ are Banach spaces with the metric induced by the $p$-norm. In particular $$L^\infty(S,\Sigma,\mu) $$ is a Banach space, with the metric $$d(f,g)=\operatorname{ess sup}_{x \in S} |f(x)-g(x)|.$$ In the case where $S$ is a subset of the real line, one can also discuss the space of continuous functions $C(S,w)$, with the weighted supremum norm $$\|f\|=\sup_{x \in S} |w(x) f(x)|.$$

I have several questions regarding these spaces:

1. Are the two concepts related somehow? Perhaps by choosing a particular measure $\mu$?
2. Is there a standard notation for what I've called $C(S,w)$?
3. Under what conditions on $S$ and $w$ is $C(S,w)$ a Banach space?

Thank you!

Answer 1

(2) Both $C(S, w)$ and $C_w(S)$ are used; the latter may be more common and it appears in Encyclopedia of Mathematics.

(1) First, an observation: $L^\infty(S, \Sigma, \mu)$ depends only on the ideal of null-sets with respect to measure $\mu$, not on the measure itself. It's all about what sets we can drop when taking $\operatorname{ess\,sup}$.

Let $\mu$ be the counting measure on $S$. The weighted space $C(S, w)$ is isometrically isomorphic to the subspace $M$ of $L^\infty(S, \Sigma, \mu)$ that consists of functions $g$ such that $g/w$ is continuous. Indeed, using the counting measure means that $\operatorname{ess\,sup}$ is the same as $\sup$ and so the $L^\infty$ norm of $g$ is the weighted supremum norm of $f=g/w$.

If $S$ is an interval and $w$ is continuous, one can use the Lebesgue measure instead of the counting measure, because then $g$ itself is continuous whenever $g/w$ is continuous, and for continuous functions $\operatorname{ess\,sup} = \sup$ for any measure with the property that null sets have empty interior.

(3) If $w$ is bounded below by a positive constant, then $C(S, w)$ is complete. For this we only need to show $M$ is closed in $L^\infty$. But uniform convergence $g_n\to g$ implies uniform convergence $g_n/w\to g/w$, hence $g/w$ is continuous.

If $w$ is not bounded away from $0$, we should not expect completeness. For example, on $(-1,1)$ with $w(x)=|x|$ the functions $$ f_n(x) = x/|x|^{1-1/n} $$ converge to $x/|x|$ in the norm of $C(S, w)$, because $$ \left(\frac{x}{|x|^{1-1/n}} - \frac{x}{|x|}\right)|x| = x|x|^{1/n} - x \to 0 \text{ uniformly} $$