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I'm reading some papers that utilize functional analysis. I came across the following definition: Let $\mathcal{X}$ be a metric space and let $g:\mathcal{X}\rightarrow\mathbb{R}$ be a continuous, strictly positive function. We define $C_g(\mathcal{X})$ to be the space of all continuous $f:\mathcal{X}\rightarrow \mathbb{R}$ such that $\|f\|_g<\infty$, where $$\|f\|_g := \sup_{x\in\mathcal{X}} \,\frac{\vert f(x)\vert}{g(x)}.$$

I am looking for a reference that talks about this space in more detail. I couldn't find anything with a quick google search. (I'm not even sure if this space has a name.) Apparently $C_g(\mathcal{X})$ is a Banach space w.r.t. the norm $\|\cdot\|_g$. But I would like to understand this space better.

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Your space $C_g(\mathcal{X})$ is isometric to a more standard Banach space $C_b(\mathcal{X})$, the space of bounded continuous functions on $\mathcal{X}$, where the correspondence is \begin{align*} f\in C_g(\mathcal{X})&\longrightarrow \frac{f}{g}\in C_b(\mathcal{X})\\ gf\in C_g(\mathcal{X})&\longleftarrow f\in C_b(\mathcal{X}). \end{align*}

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  • $\begingroup$ Thank you. Is there any reason one may want to operate in $C_g(\mathcal{X})$ versus $C_b(\mathcal{X})$? $\endgroup$
    – Satana
    May 26, 2019 at 21:11
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    $\begingroup$ There are times when you want to put the weight $g$ to, e.g., enforce a decay constraint, when $f$ rather than $gf$ looks more natural. But it is really just a matter of taste in this case. $\endgroup$ May 26, 2019 at 21:18
  • $\begingroup$ Makes sense. Just to clarify, do you mean when $f\in C_g(\mathcal{X})$ looks more natural than $gf\in C_g(\mathcal{X})$ with $f\in C_b(\mathcal{X})$. $\endgroup$
    – Satana
    May 26, 2019 at 21:23
  • $\begingroup$ It really depends on the problem at hand. For example, $f\in C_b$ gives $f^2\in C_b$, but the corresponding statement in $C_g$ would introduce $g$ that you may need to carry around. $\endgroup$ May 26, 2019 at 21:50

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