Let $X$ be a topological space (feel free to add some simplifying assumptions here, like “completely regular” provided at least the case of finite-dimensional manifolds is covered). Let $f,g \in C^*(X)$ where $C^*(X)$ denotes the ring of bounded continuous real-valued functions on $X$. Denote $f^\beta, g^\beta \colon \beta X\to \mathbb{R}$ the continuous functions corresponding to $f,g\colon X\to \mathbb{R}$ under the canonical isomorphism $C(\beta X) \cong C^*(X)$ where $\beta X$ is the Stone-Čech compactification of $X$, and let $Z(f^\beta),Z(g^\beta)$ be their zero-sets, i.e., $Z(f^\beta) = \{p\in\beta X : f^\beta(p)=0\}$.
Question: how can we tell whether $Z(f^\beta) \subseteq Z(g^\beta)$ merely by looking at $f,g$ (without looking at the Stone-Čech compactification)?
To give an idea of the flavor of criteria I'm looking for, let me observe that:
Fact: $Z(f^\beta) = \varnothing$ iff $f$ is bounded away from $0$ on $X$ (i.e. $\exists \varepsilon>0. |f|\geq\varepsilon$). (Proof: if $|f|\geq\varepsilon$ then clearly $|f^\beta|\geq \varepsilon$ so $Z(f^\beta)=\varnothing$. But conversely, if $Z(f^\beta)=\varnothing$, since $\beta X$ is compact, the image of $|f^\beta|$ is a compact subset of $\mathbb{R}_{>0}$, so it is lower bounded by some $\varepsilon>0$.) Equivalently, this means that $f$ is invertible in $C^*(X) \cong C(\beta X)$.
Based on the above fact, I first thought that $Z(f^\beta) \subseteq Z(g^\beta)$ meant $|g|\leq C|f|$ on $X$ for some constant $C$, but this is not correct (take $X=\mathbb{R}$ and $f\colon x\mapsto \min(1,x^2)$ and $g\colon x\mapsto \min(1,|x|)$: then $f^\beta$ and $g^\beta$ both vanish only at $0$ but we don't have $|g|\leq C|f|$). Maybe something like “$Z(f)\subseteq Z(g)$ and there exists $K\subseteq X$ compact and $C$ such that $|g|\leq C|f|$ outside $K$”?
Motivation: thinking about the PS in this answer made me ask whether, for $X$ a manifold and $f$ bounded continuous on $X$, we can find $g$ bounded and smooth on $X$ such that $Z(f^\beta) = Z(g^\beta)$ (and to find a criterion for the latter, looking at $Z(f^\beta) \subseteq Z(g^\beta)$ first seems natural).
Best Answer
Lemma: Suppose that $X$ is a compact Hausdorff space. Let $f,g:X\rightarrow[0,\infty)$ be continuous functions. Then the following are equivalent:
$Z(f)\subseteq Z(g)$.
For all $\epsilon>0$, there exists a $\delta>0$ where $f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$.
There exists a function $u:[0,\infty)\rightarrow[0,\infty)$ that is continuous at the point $0$ with $u(0)=0$ and where $g\leq u\circ f$.
There exists a continuous function $v:[0,\infty)\rightarrow[0,\infty)$ with $v(0)=0$ and where $g\leq v\circ f$.
There exists a continuous bijection $w:[0,\infty)\rightarrow[0,\infty)$ with $w(0)=0$ and where $g\leq w\circ f$ (the mapping $w$ is necessarily a homeomorphism and increasing).
Proof: $5\rightarrow 4,4\rightarrow 3.$ These are trivial.
$4\rightarrow 5.$ Suppose that $v:[0,\infty)\rightarrow[0,\infty)$ is a bijection with $v(0)=0$ and $g\leq v\circ f$. Then let $w:[0,\infty)\rightarrow[0,\infty)$ be the function defined by letting $w(x)=v(x)+x$. Then $g\leq w\circ f$ and $w(0)=0$.
$3\rightarrow 4.$ There is some continuous $v:[0,\infty)\rightarrow[0,\infty)$ where $v(0)=0$ and where either $\min(\max(g),u(y))\leq v(y)$ for all $y$. Therefore, for all $x$, we have $g(x)\leq u(f(x))$, so $g(x)\leq\min(\max(g),u(f(x)))\leq v(f(x)).$
$5\rightarrow 2$. Suppose that $\epsilon>0$. Then set $\delta=w^{-1}(\epsilon)$. If $x\in f^{-1}[0,\delta]$, then $f(x)\leq\delta$, so $g(x)\leq w(f(x))\leq w(\delta)=\epsilon.$ Therefore $x\in g^{-1}[0,\epsilon]$.
$2\rightarrow 1$. If $\epsilon>0$, then there is a $\delta>0$ where $Z(f)\subseteq f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$. Therefore, $Z(f)\subseteq\bigcap_{n=1}^{\infty}g^{-1}[0,1/n]=Z(g)$.
$1\rightarrow 3.$ Suppose that $Z(f)\subseteq Z(g)$. If $Z(f)=\emptyset$, then $\min(f)>0$, so just select a suitable function $u$ with $u(y)\geq\max(g)$ whenever $y\geq\min(f)$. Now, assume $Z(f)\neq\emptyset$.
Let $u:[0,\infty)\rightarrow[0,\infty)$ be the mapping such that $u(y)=\max\{g(x)\mid f(x)\leq y\}=\max g[f^{-1}(-\infty,y]].$ Then $u(0)=\max\{g(x)\mid f(x)\leq 0\}=0$. Furthermore, if $x_{0}\in X$, then $u(f(x_{0}))=\max\{g(x)\mid f(x)\leq f(x_{0})\}$. Therefore, $g(x_{0})\leq u(f(x_{0}))$. I claim that $u$ is upper semicontinuous (and therefore continuous at $0$).
Suppose that $y\in u^{-1}(-\infty,c)$. Then $\max g[f^{-1}(-\infty,y]]=\max\{g(x)\mid f(x)\leq y\}=u(y)<c$, so $g[f^{-1}(-\infty,y]]\subseteq(-\infty,c)$. Therefore, $f^{-1}(-\infty,y]\subseteq g^{-1}(-\infty,c)$. By compactness, there is some $z>y$ where $f^{-1}(-\infty,z]\subseteq g^{-1}(-\infty,c)$. However, if $s<z$, then $f^{-1}(-\infty,s]\subseteq g^{-1}(-\infty,c)$, so $g[f^{-1}(-\infty,s]]\subseteq(-\infty,c)$. Therefore, $u(s)=\max(g[f^{-1}(-\infty,s]])<c$ as well. Therefore, since there is a neighborhood $U$ of $y$ where $u(s)<c$ whenever $s\in U$, the function $u$ is upper-semicontinuous.
Q.E.D.
Theorem: Let $X$ be a completely regular space with compactification $C$. Suppose that $f,g:X\rightarrow[0,\infty)$ are continuous functions that extend to continuous maps $\overline{f},\overline{g}:C\rightarrow[0,\infty)$. Then the following are equivalent:
$Z(\overline{f})\subseteq Z(\overline{g})$.
For all $\epsilon>0$, there exists a $\delta>0$ where $f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$.
There exists a function $u:[0,\infty)\rightarrow[0,\infty]$ that is continuous at the point $0$ with $u(0)=0$ and where $g\leq u\circ f$.
There exists a continuous function $v:[0,\infty)\rightarrow[0,\infty)$ with $v(0)=0$ and where $g\leq v\circ f.$
There exists a continuous bijection $w:[0,\infty)\rightarrow[0,\infty)$ with $w(0)=0$ and where $g\leq w\circ f$.
Now suppose that $X$ is a locally compact regular space with compactification $C$, and suppose that $f,g:X\rightarrow[0,\infty)$ and $f,g$ extend to mappings $\overline{f},\overline{g}:C\rightarrow[0,\infty)$. The following result characterizes when $Z(\overline{f}|_{C\setminus X})\subseteq Z(\overline{g}|_{C\setminus X})$, so one can use the following result to produce more characterizations of when $Z(\overline{f})\subseteq Z(\overline{g})$ when $X$ is locally compact.
Theorem: Suppose that $X$ is a non-compact locally compact regular space with compactification $C$. Let $f,g:X\rightarrow[0,\infty)$ be bounded continuous functions, and let $\overline{f},\overline{g}:C\rightarrow[0,\infty)$ be the continuous extensions of $f,g$ to the domain $C$. Then the following are equivalent.
$Z(\overline{f}|_{C\setminus X})\subseteq Z(\overline{g}|_{C\setminus X})$.
For each $\epsilon>0$, there exists a $\delta>0$ and a compact $K\subseteq X$ such that if $x\in X\setminus K$, then $f(x)<\delta\rightarrow g(x)<\epsilon.$
There exists a continuous bijection $u:[0,\infty)\rightarrow[0,\infty)$ with $u(0)=0$ and a function $A:X\rightarrow[0,\infty)$ where $A^{-1}[\epsilon,\infty)$ is compact for each $\epsilon>0$ and where $g\leq A+(u\circ f)$.
Proof: $2\rightarrow 1$. Suppose that $c_{0}\in Z(f|_{C\setminus X})$. Therefore, let $(x_{d})_{d\in D}$ be a net that converges to $c_{0}$. Then for each $\epsilon>0$, there is some $\delta>0$ and compact set $K\subseteq X$ where if $x\in X\setminus K$ and $f(x)\leq\delta$, then $g(x)\leq\epsilon$. Then there is some $d_{0}\in D$ where if $d\leq d_{0}$, then $x_{d}\not\in K$ and where $f(x_{d})\leq\delta$. In this case, we have $g(x_{d})\leq\epsilon$. Therefore, we conclude that $g(x_{d})_{d\in D}\rightarrow 0$, so since $g$ is continuous, we know that $g(c_{0})=0$, and therefore $c_{0}\in Z(g|_{C\setminus X})$. Thus, $Z(f|_{C\setminus X})\subseteq Z(g|_{C\setminus X})$.
$1\rightarrow 3$. By the above results, we know that there is a continuous mapping $u:[0,\infty)\rightarrow[0,\infty)$ such that $\overline{g}|_{C\setminus X}\leq u\circ\overline{f}|_{C\setminus X}$. Therefore, let $A:X\rightarrow[0,\infty)$ be the mapping defined by $A=\max(0,g-(u\circ f))$. Then $g=g-(u\circ f)+(u\circ f)\leq A+(u\circ f),$ and $\overline{A}(c)=0$ whenever $c\in C\setminus X$, so $A^{-1}[\epsilon,\infty)$ is a compact subset of $X$ whenever $\epsilon>0$.
$3\rightarrow 2$. Suppose that $g\leq A+(u\circ f)$. Then for all $\epsilon>0$, there is a $\delta>0$ with $u(\delta)<\epsilon$. Therefore, suppose that $c\in C\setminus X$, and $\overline{f}(c)\leq\delta$. Then $$\overline{g}(c)\leq\overline{A}(c)+(u\circ\overline{f})(c)=u(\overline{f}(c))\leq u(\delta)<\epsilon.$$ In particular, there is no $c\in C\setminus X$ with $\overline{f}(c)\leq\delta$ and $\overline{g}(c)\geq\epsilon$. Therefore, if we set $K=\overline{f}^{-1}[0,\delta]\cap\overline{g}[\epsilon,\infty)$, then $K$ is a closed subset of $C$, so $K$ is compact, but $K$ is also a subset of $X$. Therefore, if $x\in X\setminus K$, then $f(x)\leq\delta\rightarrow g(x)<\epsilon$.
Q.E.D.