The set $ C= C ( X ) $ of all continuous, real valued functions on a topological space$ X$ will be provided with an algebraic structure and an order struture.
Define:
$ Z ( f ) = f^{-1 } ( \{ 0 \} ) = \{ x \in X \vert f ( x ) = 0 \}$
I've encountered a problem solving these exercises . Can anyone help?
1: If $Z ( f )$ is a neighborhood of $ Z (g)$ , then $ f$ is a multiple of $ g$, that is $f = hg $ for some $ h \in C $ . furthermore , if $ X - Z(f)$ is compact , then $ h$ can be chosen to be bounded. [ Define $h( x) = f(x) / g(x)$ for$x \notin int Z (f )$, and $ h( x ) = 0 $ for some $x \in Z ( f) $]
2: If $\mid f \mid \leq \mid g \mid^{r}$ for some $r > 1$ ,then $ f $ is a multiple of $ g$. [ Define $h( x) = f( x ) / g ( x ) $ for some $x \notin int Z (g )$ and $h( x ) = 0$ otherwise.] Hence if $\mid f \mid \leq \mid g \mid $ , then $f^{r}$ is a multiple of $ g $ for every $r > 1$ for which $f^{r}$ is defined.
The hint is quite clear: we know that $Z(f)$ is a neighbourhood of $Z(g)$, so that $$Z(g) \subseteq \operatorname{int}(Z(f))$$
Then define $h(x) = {f(x) \over g(x)}$ for $x \notin \operatorname{int}(Z(f))$ which is a closed set, and $h(x)= 0$ for $x \in Z(f)$, another closed set.
If $x$ is in the overlap so in $Z(f) \cap (X\setminus \operatorname{int}(Z(f))$, i.e. the boundary of $Z(f)$, then we know $g(x) \neq 0$ (as otherwise $x \in Z(g)$ so $x \in \operatorname{int}(Z(f))$ contradiction) and $f(x) = 0, so both definitions agree.
So the pasting lemma (for functions defined on finitey many closed sets) says that $h$ is indeed continuous.
And $f(x) = h(x)g(x)$ on all of $X$. For $x \in Z(f)$ the left hand side is $0$ as is $h(x)$ and thus the right hand side too. Outside $\operatorname{int}(Z(f))$ we have $h(x) = {f(x) \over g(x)}$ so $h(x)g(x) = f(x)$ as well.
If $|f(x)| \le |g(x)|^r$ for some real $r>1$, we can again define $h(x) = {f(x) \over g(x)}$ for $x \notin Z(g)$ and $h(x) = 0$ for $x \in Z(g)$. This is continuous as $|{f(x) \over g(x)}| \le |{g(x)^r \over g(x)}| =|g^{r-1}(x)|$ outside of $Z(g)$ so that on the boundary of $Z(g)$ we have a common value $0$. As above one checks that $f(x) = h(x)g(x)$ everywhere.
The last part follows right away: suppose $|f| \le |g|$ then supposing $f^r$ is defined $|f^r| = |f|^r \le |g|^r$ and thus by the previous, $f^r$ is a multiple of $g$.