Sufficient and necessary conditions to get an infinite fiber $g⁻©ö(w)$
I want to verify the proof of this result and get some start ideas to
overcome the different steps of this proof.
Lemma: Let $g$ be a real analytic function. Then we have the equivalence
$((a)¡ü(b))¢¢(c)$, where the statements $(a),(b)$ and $(c)$ are given by:
(a) $g$ has infinitely many real zeros.
(b) $g$ assumes arbitrarily large and arbitrarily small values, i.e., for
all $K>0$, there are $s©û,s©ü$ with $g(s©û)<-K$ and $g(s©ü)>K$,
(c) The fiber $g⁻©ö(w)$ is infinite for all $w¡ôℝ$.
Proof: (1) To prove that $(a)¡ü(b)¢¡(c)$, assume by contraduction that the
equation $g(s)=w$ has only finitely many solutions $(b_{j})_{1¡Âj¡Âi}$.
That means there are $z©û<z©ü$ such that $g(s)¡Áw$ for $w<z©û$ or $w>z©ü$.
Let $z©ý$ be the largest zero of $g$ smaller than $z©û$, and $z©þ$ the
smallest zero larger than $z©ü$. The exitence of $z©ý$ and $z©þ$ is
guaranteed by the premises that $g$ has infinitely many zeros and the fact
that $g$ is analytic, so $g$ has only finitely many zeros in the compact
interval $[z©û,z©ü]$ (or $g$ is identically $0$, but that has been ruled
out by the premise that it take arbitrarily large values), hence there
must be infinitely many outside the interval. Let
$K=max({|g(s)|:z©ý¡Âs¡Âz©þ})+|w|$. By assumption, there are $s©û,s©ü$ with
$ g(s©û)<-K$ and $g(s©ü)>K$. By the intermediate value theorem, there is
an $s_{w}$ between $z©ý$ or $z©þ$ and $s©û$ or $s©ü$ with $g(s_{w})=w$.
Contradiction.
We note that if $g$ is not analytic then it satisfies conditions (a) and
(b) but does not satisfy (c).
(2) To prove that $(c)¢¡(a)¡ü(b)$, we note first that $g$ has infinitely
many zeros because the fiber $g⁻©ö(0)$ is infinite. This proves item
(a). To prove item (b), let $K>0$ and $w=K+¥å,¥å>0$. The fiber
$g⁻©ö(K+¥å)$ is nonempty because it is infinite, hence there is
$s©ü$ such that $g(s©ü)=w=K+¥å>K$. By the same method, $g⁻©ö(-K-¥å)$
is nonempty, so there is $s©û$ such that $g(s©û)=-K-¥å<-K$.
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