# Some remarks on contractive and existence sets

Maciej Ciesielski and Grzegorz Lewicki

###### Abstract

Let $X$ be a real or complex Banach space and let $F\subset X$ be a non-empty
set. $F$ is called an existence set of best coapproximation (existence set for
brevity), if for any $x\in X$, $R_{F}(x)\neq\emptyset,$ where

$R_{F}(x)=\\{d\in F:\|d-c\|\leq\|x-c\|\hbox{ for any }c\in F\\}.$

It is clear that any existence set is a contractive subset of $X.$ The aim of
this paper is to present some conditions on $F$ and $X$ under which the
notions of exsistence set and contractive set are equivalent.

AMS Classification: 47B37, 46E30, 47H09.
Key Words and Phrases: Banach spaces, reflexivity, strict convexity,
contractive and existence sets, one complemented spaces.

##  1 Introduction

Let $X$ be a real or complex Banach space and let $F\subset X$ be a non-empty
set. A continuous mapping $P:X\rightarrow F$ is called a projection onto $F,$
whenever $P|_{F}=Id,$ that is $P^{2}=P.$ Setting

$Min(F)=\\{z\in X:\hbox{ for every }c\in F,x\in X,\hbox{ if
}\|z-c\|\geq\|x-c\|\hbox{ then }x=z\\},$

we say that $F\subset X$ is optimal if $Min(F)=F.$ Observe that for any
$F\subset X,$ $F\subset Min(F).$ This notion has been introduced by Beauzamy
and Maurey in [5], where basic properties concerning optimal sets can be
found.

A set $F\subset X$ is called an existence set of best coapproximation
(existence set for brevity), if for any $x\in X$, $R_{F}(x)\neq\emptyset,$
where

$R_{F}(x)=\\{d\in F:\|d-c\|\leq\|x-c\|\hbox{ for any }c\in F\\}.$

Notice that any contractive set is an existence set. Indeed, if
$P:X\rightarrow F$ is a contractive projection, then $Px\in R_{F}(x)$ for any
$x\in X.$ Also it is clear that any existence set is an optimal set. The
converse, in general, is not true. However, the following result is satisfied.

######  THEOREM 1.

([5], Prop. 2) If $X$ is one-complemented in $X^{**}$ and strictly convex,
then any optimal subset of $X$ is an existence set in $X,$ which, in
particular, holds true for strictly convex spaces $X,$ such that $X=Z^{*}$ for
some Banach space $Z.$

Existence and optimal sets have been studied by many authors from different
points of view, mainly in the context of approximation theory (see e.g. [2],
[3], [[10] \- [16]], [19], [20], [24], [22], [[27] \- [36]]).
Recall that a closed subspace $F$ of a Banach space $X$ is called one-
complemented if there exists a linear projection of norm one from $X$ onto
$F.$ It is obvious that any one-complemented subspace is an existence set. The
converse, in general, is not true. By a deep result of Lindenstrauss, [24]
there exists a Banach space $X$ and $F$ a linear subspace of $X,$
codim$(F)=2,$ such that:

a). $F$ is one-complemented in any containing it hyperplane $Y$ of $X;$

b). $F$ is not one-complemented in $X.$
This gives you immediately an example of a subspace being an existence set
which is not one-complemented. However, in [5] (see also [22] p.121), the the
following result has been proven.

######  THEOREM 2.

(see [5], Prop. 5). Let $V$ be a linear subspace of a smooth, reflexive and
strictly convex Banach space. If $V$ is an optimal set then $V$ is one-
complemented in $X.$ If $X$ is a smooth Banach space, then any subspace of $X$
which is an existence set is one-complemented. Moreover, in both cases a norm-
one projection from $X$ onto $V$ is uniquely determined.

The aim of this paper is to present some conditions on a Banach space $X$ and
a convex and closed set $F\subset X$ under which the notions of existence set
and contractive set are equivalent. In other words, we will study the problem
of existence of a non-expansive selection $P:X\rightarrow F$ such that $Px\in
R_{F}(x)$ for any $x\in X.$
In Section 1 some preliminary results will be demonstrated.
The main results of the paper will be presented in Section 2. In Section 2 we
assume that $F$ has a nonempty interior in $X.$
In Section 3 we consider the general case.
All the results will be demonstrated for real Banach spaces. However, at the
end of Section 3, we show how to apply the results obtained in the real case
to the case of complex Banach spaces (see Lemma 8, Lemma 9 and Theorem 16).
Now we present some notions which will be used in this paper.
In the sequel by $S(X)$ we denote the unit sphere in a Banach space X and by
$S(X^{*})$ the unit sphere in its dual space. A functional $f\in S(X^{*})$ is
called a supporting functional for $x\in X,$ if $f(x)=\|x\|.$ Analogously, a
point $x\in S(X)$ is called a norming point for $f\in X^{*}$ if $f(x)=\|f\|.$
A point $x\in X$ is called a smooth point if it has exactly one supporting
functional. A Banach space $X$ is called smooth if any $x\in S(X)$ is a smooth
point.
By $ext(X)$ we denote the set of all extreme points of $S(X).$ A Banach space
$X$ is called strictly convex if $ext(X)=S(X).$
If $F$ is a linear subspace of a Banach space $X,$ by ${\cal P}(X,F)$ we will
denote the set of all linear, continuous projections from $X$ onto $F.$ If $X$
is a Banach space and $Z\subset X$ is an affine subspace of $X,$ for $F\subset
Z$ by $int_{Z}(F)$ ($int(F)$ if $Z=X$), we denote the interior of $F$ with
respect to $Z.$
To the end of this paper, unless otherwise stated, all Banach spaces are real.
Also we will need

######  Definition 1.

It is said that a Banach space $X$ has (CFPP) if and only if for any nonempty,
convex, closed and bounded set $F\subset X$ and any nonexpansive mapping
$T:X\rightarrow X$ such that $T(F)\subset F,$ $T$ has a fixed point in $F.$

## Section 1

We start with a lemma presenting basic properties of existence sets.

######  LEMMA 1.

Let $X$ be a Banach space and let $F\subset X$ be an existence set. Then for
any $d\in X,$ $F+d$ is an existence set. Also, for any $t\in\mathbb{R},$ $tF$
is an existence set. If $F$ is a linear subspace then $F$ is an existence set
if and only if for any $x\in X\setminus F$ there exists $P_{x}\in{\cal
P}(F_{x},F),$ $\|P_{x}\|=1,$ where $F_{x}=span[x]\oplus F.$

Proof.  Fix $d\in X$ and $x\in X.$ Let $p\in R_{F}(x-d).$ Let $c\in F+d,$
$c=c_{1}+d,$ wher $c_{1}\in F.$ Then

$\|c-(p+d)\|=\|c_{1}-p\|\leq\|x-d-c_{1}\|=\|x-c\|,$

which shows that $p+d\in R_{F+d}(x).$ Now fix $t\in\mathbb{R}.$ If $t=0,$ then
$F=\\{0\\}$ is obviously an existence set. If $t\neq 0,$ fix $p\in
R_{F}(x/t).$ Let $c\in tF,$ $c=tc_{1}.$ Observe that

$\|p-c_{1}\|\leq\|p-x/t\|$

and consequently

$|t|\|p-c_{1}\|\leq|t|\|p-x/t\|,$

which implies that $\|tp-c\|\leq\|tp-x\|.$ Hence $tp\in R_{tF}(x).$ Now assume
that $F$ is a linear subspace and fix $x\in X\setminus F.$ Then any $y\in
F_{x}$ can be represented in the unique way as $y=tx+v_{o},$ where $v_{o}\in
F.$ Fix $p\in R_{F}(x).$ We show that $tp+v_{o}\in R_{F}(tx+v_{o}).$ This is
obvious if $t=0.$ So assume that $t\neq 0.$ Fix $v\in F.$ Since
$\frac{v-v_{o}}{t}\in F,$

$\|p-\frac{v-v_{o}}{t}\|\leq\|x-\frac{v-v_{o}}{t}\|$

and consequently

$|t|\|p-\frac{v-v_{o}}{t}\|\leq|t|\|x-\frac{v-v_{o}}{t}\|,$

which imples that $\|tp+v_{o}-v\|\leq\|tx+v_{o}-v\|=\|y-v\|.$ Notice that a
linear mapping $P_{x}:F_{x}\rightarrow F$ defined by
$P_{x}(tx+v_{o})=tp+v_{o}$ belongs to ${\cal P}(F_{x},F).$ By the above
reasoning, for any $y\in F_{x},$ $\|P_{x}y\|=\|P_{x}y-0\|\leq\|y\|,$ which
shows that $\|P_{x}\|=1.$ Conversely, if there exists $P_{x}\in{\cal
P}(F_{x},F),$ $\|Px\|=1,$ then for any $v\in F,$ and $y\in F_{x},$

$\|P_{x}y-v\|=\|P_{x}(y-v)\|\leq\|y-v\|$

which shows our claim.

######  LEMMA 2.

(compare with [8], Lemma 1). Let $X$ be a Banach space and let $C\subset X$ be
a locally compact and convex set. For $F\subset C,$ $F\neq\emptyset,$ define

$N(F)=\\{f:C\rightarrow C:\|y-f(x)\|\leq\|y-x\|\hbox{ for any }y\in F,x\in
C\\}.$

Then $N(F)$ is compact in the topology of pointwise-weak convergence.

Proof.  Fix $x_{o}\in F.$ For $x\in X$ set $C_{x}=\\{y\in
C:\|y-x_{o}\|\leq\|x-x_{o}\|\\}.$ Notice that for $f\in N(F)$

$\|f(x)-x_{o}\|\leq\|x-x_{o}\|.$

Hence $N(F)\subset P=\Pi_{x\in C}C_{x}.$ Since $C$ is convex and locally
weakly compact, by the Mazur Theorem, $C_{x},$ as a bounded subset of $C,$ is
weakly compact. By the Tychonoff Theorem, $P$ is a compact set in the topology
of pointwise-weak convergence, which we denote by $\tau.$ Hence to show that
$N(F)$ is compact, we need to demonstrate that $N(F)$ is a $\tau$ closed
subset of $P.$ Let $\\{f_{\gamma}\\}\subset N(F)$ be a net $\tau$-converging
to $f\in P.$ Hence for any $y\in F,f_{\gamma}(y)\rightarrow f(y)$ in the weak
topology. Since $f_{\gamma}(y)=y,$ for any $\gamma,$ $f(y)=y.$ Moreover, for
$y\in F,x\in C,$

$\|y-f(x)\|\leq\liminf_{\gamma}\|y-f_{\gamma}(x)\|\leq\|y-x\|.$

Hence $f\in N(F),$ which completes the proof.

######  LEMMA 3.

(compare with [8], Lemma 2) Let $F,C,X$ and $N(F)$ be as in Lemma 2. Then
there exists $r\in N(F)$ such that for any $f\in N(F),$ $x\in C$ and $y\in F$

$\|y-f(r(x))\|=\|y-r(x)\|.$

Proof.  First we define a partial ordering in $N(F).$ For any $f,g\in N(F)$ it
is said that $f<g$ if and only if for any $x\in C$ and $y\in F$

$\|f(x)-y\|\leq\|g(x)-y\|\hbox{ and }\|f(x_{o})-y_{o}\|<\|g(x_{o})-y_{o}\|$

for some $y_{o}\in F,$ $x_{o}\in C.$ It is said that $f\leq g$ if and only if
$f<g$ or $f=g.$ It is easy to see that $(N(F),\leq)$ is a partially ordered
set. Observe that $N(F)\neq\emptyset,$ since $id_{C}\in N(F).$ Now we show
that there exists a minimal element in $(N(F),\leq).$ First notice that for
any $f\in N(F)$ the set $A_{f}=\\{g\in N(F):g\leq f\\}$ is $\tau$-closed in
$N(F),$ which follows easily from definition of our partial ordering. Hence by
Lemma 2, $A_{f}$ is a compact subset of $N(F).$ Now let $G\subset N(F)$ be a
chain. We show that there exists a smallest element in $G.$ Notice that if
$f_{1},...,f_{n}\in G$ then without loss of generality we can assume that
$f_{n}\leq f_{n-1}\leq...\leq f_{1}.$ Hence

$\bigcap_{j=1}^{n}A_{f_{j}}=A_{f_{n}}\neq\emptyset.$

Since the sets $A_{f}$ are closed and $N(F)$ is compact, $\bigcap_{f\in
G}A_{f}\neq\emptyset.$ Hence any $g\in\bigcap_{f\in G}A_{f}$ is the smallest
element in $G.$ By the Kuratowski-Zorn lemma, there exist $r\in N(F)$ a
minimal element in $(N(F),\leq).$ Let $f\in N(F).$ Hence for any $x\in C$ and
$y\in F,$

$\|y-(f\circ r)(x)\|\leq\|y-r(x)\|\leq\|y-x\|,$

which shows that $f\circ r\leq r.$ If for some $x_{o}\in C$ and $y_{o}\in F,$
$\|y_{o}-(f\circ r)(x_{o})\|<\|y_{o}-r(x_{o})\|,$ then $f\circ r\neq r$ and
$f\circ r\leq r,$ a contradiction with the minimality of $r.$

######  THEOREM 3.

(compare with [8], Theorem 1) Let $F,C,X$ and $N(F)$ be as in Lemma 2. Assume
that for any $x\in C$ there exists $h\in N(F)$ such that $h(x)\in F.$ Then
there exists $g\in N(F)$ such that for any $x\in C$ $g(x)\in F.$ In
particular, $r(x)\in R_{F}(x)$ for any $x\in C,$ where $r$ is a maximal
element from Lemma 3. If $C=X,$ then $F$ is an existence set in $X.$

Proof.  Let $r\in N(F)$ be as in Lemma 3. We show that for any $x\in X,$
$r(x)\in F.$ Notice that, since $r(x)\in C,$ there exists $h\in N(F)$ such
that $h(r(x))\in F.$ Let $y=h(r(x)).$ Observe that by Lemma 3,

$0=\|h(r(x))-y\|=\|y-r(x)\|.$

Hence $r(x)=y\in F,$ as required.

######  THEOREM 4.

Let $X$ be a reflexive space. For $f\in X^{*}\setminus\\{0\\}$ define

$G=\\{x\in X:f(x)\leq 0\\}.$

If $G$ is an existence set then $F=ker(f)$ is one-complemented in $X.$

Proof.  Since $G$ is an existence set, for any $x\in S(X),$ we can select
$P_{o}x\in R_{G}(x).$ Define a mapping $P:X\rightarrow G$ by $P0=0,$
$Px=\|x\|P_{o}(x/\|x\|)$ for $x\neq 0.$ Notice that $Px\in R_{G}(x)\subset G$
for any $x\in X.$ This is obvious if $x=0.$ If $x\neq 0,$ for any $d\in G,$

$\|P_{o}(x/\|x\|)-d/\|x\|)\leq\|(x-d)/\|x\|\|$

and consequently

$\|Px-d\|=\|\|x\|P_{o}(x/\|x\|)-d)\leq\|x-d\|.$

In particular, for any $d\in F=ker(f),$ $\|Px-d\|\leq\|x-d\|$ and
$P|_{F}=id_{F}.$ This shows that $P\in N(F)$ (in our case $C=X).$ Now we show
that for any $x\in X$ there exists $Q\in N(F)$ such that $Qx\in F.$ Fix $x\in
X.$ If $f(x)=0,$ then $Id_{F}(x)\in F.$ Now assume that $f(x)>0.$ If
$f(Px)=0,$ then $Px\in F.$ If $f(Px)<0,$ then there exists $0<\alpha<1$ such
that $f(\alpha x+(1-\alpha)Px)=0.$ Since $N(F)$ is convex, $Q=\alpha
Id_{X}+(1-\alpha)P\in N(F)$ and $Qx\in F.$ Now assume that $f(x)<0$ and set
$G_{1}=\\{x\in X:f(x)\geq 0\\}.$ By Lemma 1 for any $x\in X,$
$P_{1}x=-P(-x)\in R_{G_{1}}(x)\subset G_{1}.$ Since $F\subset G_{1},$ for any
$d\in F,$ $\|P_{1}x-d\|\leq\|x-d\|$ and $P_{1}|_{F}=id_{F}.$ This shows that
$P_{1}\in N(F).$ If $f(P_{1}x)=0,$ then $P_{1}x\in F.$ If $f(P_{1}x)>0,$ then
there exists $0<\alpha<1$ such that $f(\alpha x+(1-\alpha)P_{1}x)=0.$ Since
$N(F)$ is convex, $Q_{1}=\alpha Id_{X}+(1-\alpha)P_{1}\in N(F)$ and $Q_{1}x\in
F.$ By Theorem 3 applied to $C=X,$ there exists $Q:X\rightarrow ker(f)$ such
that $Q|_{ker(f)}=id|_{ker(f)}$ and and $\|Qx-d\|\leq\|x-d\|$ for any $x\in X$
and $d\in ker(f).$ Hence $ker(f)$ is an existence set. By Lemma 1, $ker(f)$ is
one-complemented in $X.$

######  THEOREM 5.

Let $X$ be a Banach space and let $f\in X^{*}\setminus\\{0\\}.$ If $ker(f)$ is
one-complemented in $X,$ then for any $d\in\mathbb{R}$ the set $G_{d}=\\{x\in
X:f(x)\leq d\\}$ is a contractive subset of $X.$

Proof.  By Lemma 1, we can assume that $d=0.$ Fix $P\in{\cal P}(X,ker(f)),$
$\|P\|=1.$ Define $Q:X\rightarrow G_{o}$ by $Qx=x$ if $f(x)\leq 0$ and $Qx=Px$
if $f(x)>0.$ We show that for any $x,z\in X$ $\|Qx-Qz\|\leq\|x-z\|.$ If
$f(x)\leq 0$ and $f(z)\leq 0,$ then

$\|Qx-Qz\|=\|x-z\|.$

If $f(x)\geq 0$ and $f(z)\geq 0,$ then

$\|Qx-Qz\|=\|Px-Pz\|\leq\|x-z\|.$

Now assume that $f(x)>0$ and $f(z)<0.$ Let $y\in X$ be so chosen that $Py=0$
and $f(y)=1.$ Then, it is easy to see that for any $w\in X,$ $Pw=w-f(w)y.$
Hence if $w=\alpha y+v$ where $v\in ker(f)$ and $\alpha\in\mathbb{R},$ then

$\|Pw\|=\|\alpha Py+Pv\|=\|v\|\leq\|w\|=\|\alpha y+v\|.$

This means that for any $v\in ker(f)$ and $t\geq 0$ the function
$g_{v}(t)=\|ty+v\|$ satisfies

$g_{v}(t)\geq g_{v}(0)=\|v\|.$

Sinec $g_{v}$ is a convex function, $g_{v}$ is increasing in $[0,+\infty).$
Since $f(x)>0,$ if $x=\alpha y+v,$ then $f(x)=\alpha>0.$ Analogoulsy, since
$f(z)<0,$ if $z=\beta y+w,$ then $f(z)=\beta<0.$ Notice that, since the
function $g_{v-w}$ is increasing in $[0,+\infty),$

$\|Qx-Qz\|=\|Px-z\|=\|v-(\beta y+w\|=\|-\beta y+(v-w)\|$ $=g_{v-w}(-\beta)\leq
g_{v-w}(-\beta+\alpha)=$ $=\|(-\beta+\alpha)y+v-w\|=\|\alpha y+v-(\beta
y+w)\|=\|x-z\|,$

as required. Since $Q|_{G_{o}}=id_{G_{o}},$ $G_{o}$ is a contractive set. The
proof is complete.

######  LEMMA 4.

Let $X$ be a Banach space and let $C\subset X$ be a closed, convex and bounded
set such that $0\in int(C).$ Define for $x\in X,$

$f(x)=\inf\\{t>0:x/t\in C\\}.$

Then $f$ is a convex and continuous function.

Proof.  First we show that for any $x_{1},x_{2}\in X,$ $f(x_{1}+x_{2})\leq
f(x_{1})+f(x_{2}).$ Fix $x_{1},x_{2}\in X$ and $\epsilon>0.$ Then there exists
$t_{1}>0$ and $t_{2}>0$ satisfying $t_{1}<f(x_{1})+\epsilon$ and
$t_{2}<f(x_{2})+\epsilon$ such that $\frac{x_{i}}{t_{i}}\in C$ for $i=1,2.$
Notice that

$\frac{x_{1}+x_{2}}{t_{1}+t_{2}}=\frac{t_{1}}{t_{1}+t_{2}}\frac{x_{1}}{t_{1}}+\frac{t_{2}}{t_{1}+t_{2}}\frac{x_{2}}{t_{2}}.$

Since $C$ is convex, by definiton of $t_{1}$ and $t_{2},$
$\frac{x_{1}+x_{2}}{t_{1}+t_{2}}\in C.$ This shows that

$f(x_{1}+x_{2})\leq f(x_{1})+f(x_{2})+2\epsilon$

and consequently $f$ is a subadditive function. Moreover, it is easy to see
that $f(ax)=af(x)$ for any $a\geq 0$ and $x\in X.$ Hence for any $a,b\geq 0,$
$a+b=1,$ and $x,y\in X$

$f(ax+by)\leq f(ax)+f(by)\leq af(x)+bf(y),$

which shows that $f$ is a convex function. Now we prove that $f$ is
continuous. Let $\|x_{n}-x\|\rightarrow 0.$ Fix $\epsilon>0$ and $t>0$ such
that $f(x)<t<f(x)+\epsilon.$ Since $0\in int(C),$ there exists $r>0$ such that
$B_{o}(x/t,r)\subset C.$ Hence for $n\geq n_{o},$ $x_{n}/t\in
B_{o}(x/t,r)\subset C.$ Hence for any $n\geq n_{o},$ $f(x_{n})\leq t\leq
f(x)+\epsilon$ and conseqently,

$\limsup_{n}f(x_{n})\leq f(x).$

Now, assume on the contrary that

$0\leq d=\liminf_{n}f(x_{n})<f(x).$

Fix $d<t<f(x).$ Hence $f(x)>0,$ so $x\neq 0.$ Then there exists a subsequence
$x_{n_{k}}$ such that $f(x_{n_{k}})<t$ and consequently $x_{n_{k}}/t\in C.$
Since $C$ is closed, $x/t\in C.$ Hence $f(x)\leq t,$ which is a contradiction.
Finally we get

$\limsup_{n}f(x_{n})\leq f(x)\leq\liminf_{n}f(x_{n})$

which shows that $f(x)=\lim_{n}f(x_{n}),$ as required.

######  REMARK 1.

If $C=-C$ then the function $f$ from Lemma 4 is the Minkowski functional of
$C.$

######  LEMMA 5.

Let $X$ be a Banach space and let $C\subset X$ be a closed, convex and bounded
set such that $0\in int(C).$ Let $\\{x_{n}\\}\subset\sigma(C)$ be a dense
subset of $\sigma(C)$ such that for any $n\in\mathbb{N}$ there exists exactly
one supporting functional $f_{n}$ for $x_{n},$ with $\|f_{n}\|=1.$ Assume that
$f_{n}(x_{n})=d_{n}$ and $f_{n}(v)\leq d_{n}$ for any $v\in V.$ Put

$D_{n}=\\{x\in X:f_{n}(x)\leq d_{n}\\}.$

Then $C=\bigcap_{n\in\mathbb{N}}D_{n}.$

Proof.  By definition of $f_{n}$ and $D_{n},$
$C\subset\bigcap_{n\in\mathbb{N}}D_{n}.$ Now assume on the contrary that there
exists $y_{o}\in\bigcap_{n\in\mathbb{N}}D_{n}\setminus C.$ Since $0\in
int(C),$ there exists $t\in(0,1)$ such that $ty_{o}\in\sigma(C).$ Fix
$\\{y_{n}\\}\subset\\{x_{n}\\}$ such that

$\|y_{n}-ty_{o}\|\rightarrow 0.$ (1)

This is possible, since $\\{x_{n}\\}$ is a dense subset of $\sigma(C).$ Let
$d_{n}$ and $f_{n}$ be so chosen that $f_{n}(y_{n})=d_{n}$ and $V\subset
D_{n}=\\{x\in X:f_{n}(x)\leq d_{n}\\}.$ Now we show that $\\{d_{n}\\}$ is
bounded. Assume on the contrary that this is not true. Without loss of
generality, passing to a subsequence, if necessary, we can assume that
$d_{n}\rightarrow+\infty.$ Let $R>0$ be so chosen that $C\subset B(0,R).$
Since $f_{n}(y_{n})=d_{n},$ and $y_{n}\in C,$ for $n\geq n_{o},$

$\sup\\{f_{n}(z):z\in B(0,R)\\}\geq f_{n}(y_{n})\geq 2R.$

Hence

$\sup\\{f_{n}(z):z\in B(0,1)\\}\geq 2$

and consequently, $\|f_{n}\|>2$ for $n\geq n_{o}$ which is a contradiction.
Now we show that $e_{o}=\liminf_{n}d_{n}>0.$ Assume on the contrary that
$\liminf_{n}d_{n}=0.$ Without loss of generality, passing to a subsequence, if
necessary, we can assume that $d_{n}\rightarrow 0.$ Fix $r>0$ such that
$B(0,r)\subset int(C).$ Fix $n_{o}\geq 2$ such that $\frac{r}{n_{o}}<1.$ Fix
$k_{o}$ such that $0\leq d_{k}<\frac{r}{n_{o}}$ for $k\geq k_{o}.$ Fix $x\in
B(0,r).$ If $f_{k}(x)\geq 0,$ then

$0\leq f_{k}(x)\leq d_{k}<\frac{r}{n_{o}}.$

If $f_{k}(x)<0,$ then $f_{k}(-x)\geq 0$ and $f_{k}(-x)\leq
d_{k}<\frac{r}{n_{o}}.$ Hence

$\sup\\{|f_{k}(x)|:x\in B(0,r)\\}\leq\frac{r}{n_{o}}$

and consequenntly $\|f_{k}\|\leq\frac{1}{n_{o}}<1$ which is a contradiction.
Without loss of generality, passing to a subsequence if necessary, by (1), we
can assume that $\|y_{n}-ty_{o}\|\rightarrow 0$ and $d_{n}\rightarrow
e_{o}>0.$ Notice that

$f_{n}(ty_{o})=f_{n}(y_{n})-f_{n}(y_{n}-ty_{o})=d_{n}-f_{n}(y_{n}-ty_{o})\geq
d_{n}-\|y_{n}-ty_{o}\|.$

Hence

$\liminf_{n}f_{n}(y_{o})=\frac{\liminf_{n}f_{n}(ty_{o})}{t}\geq\frac{\lim_{n}d_{n}}{t}=\frac{e_{o}}{t}.$

Since $t\in(0,1)$ and $e_{o}>0,$ $f_{n}(y_{o})>d_{n}$ for $n\geq n_{o}.$ Hence
$y_{o}\notin\bigcap_{n\in\mathbb{N}}D_{n},$ which is a contradiction. The
proof is complete.

######  LEMMA 6.

Let $X$ be a reflexive Banach space. Let $\\{F_{t}\\}_{t\in T}$ be a family of
convex existence sets directed by $\subset.$ Then $F=cl(\bigcup_{t\in
T}F_{t})$ is also an existence set.

Proof.  Since $\\{F_{t}\\}_{t\in T}$ is directed by $\subset,$ for any
$t_{1},t_{2}\in T$ there exists $t_{3}\in T$ such that $F_{t_{1}}\cup
F_{t_{2}}\subset F_{t_{3}}.$ Define a partial ordering $\leq$ in $T$ by
$t_{1}\leq t_{2}$ provided $F_{t_{1}}\subset F_{t_{2}}.$ Fix
$F_{t_{o}}=F_{o}\in\\{F_{t}\\}_{t\in T}.$ Since $\\{F_{t}\\}_{t\in T}$ is
directed by $\subset,$ $F=cl(\bigcup_{t\geq t_{o}}F_{t}).$ By the axiom of
choice, for any $t\geq t_{o}$ we can select $P_{t}\in N(F_{t})$ (see Lemma 2)
such that $P_{t}(x)\in R_{F_{t}}(x)$ for any $x\in X.$ Observe that
$\\{P_{t}\\}_{t\geq t_{o}}\subset N(F_{o}).$ Set for any $t\geq t_{o}$
$A_{t}=cl(\\{P_{s}:s\geq t\\})$ where the closure is taken with respect to the
topology in $N(F_{o})$ defined in Lemma 2. Since $\\{F_{t}\\}_{t\in T}$ is
directed by $\subset,$ for any $n\in\mathbb{N}$ and $t_{1},...t_{n}\geq t_{o}$
there exists $u\in T$ such that

$\bigcap_{i=1}^{n}A_{t_{i}}\supset A_{u}\neq\emptyset.$

By the proof of Lemma 2, $\bigcap_{t\geq t_{o}}A_{t}\neq\emptyset.$ Fix
$P\in\bigcap_{t\geq t_{o}}A_{t}.$ We show that for any $x\in X,$ $Px\in
R_{F}(x).$ Fix $x\in X$ and $d\in\bigcup_{t\geq t_{o}}F_{t}.$ Then there
exists $s\geq t_{o}$ such that $d\in F_{s}$ and consequently $d\in F_{u}$ for
$u\geq s.$ Since $P\in A_{s},$

$\|Px-d\|\leq\liminf_{u\geq s}\|P_{u}x-d\|\leq\|x-d\|.$

If $d\in F,$ then there exists a sequence $\\{d_{n}\\}\subset\bigcup_{t\geq
t_{o}}F_{t}$ such that $\|d_{n}-d\|\rightarrow 0.$ By the previous reasoning,

$\|Px-d\|=\lim_{n}\|Px-d_{n}\|\leq\lim_{n}\|x-d_{n}\|=\|x-d\|,$

as required.

######  COROLLARY 1.

Let $X$ be a reflexive Banach space and let $F$ be a convex existence set.
Define for any $v\in F$ and $t>0$ $F_{t}=(1-t)v+tF$ and let
$C_{v}=cl(\bigcup_{t>0}F_{t}).$ Then $F$ is an existence set.

Proof.  By Lemma 1, $F_{t}$ is an existence set for any $t>0.$ Since
$F_{t}\subset F_{s}$ for $t\leq s,$ by Lemma 6 $C_{v}$ is an existence set
too.

######  LEMMA 7.

Let $X$ be a reflexive Banach space. Let $\\{F_{t}\\}_{t\in T}$ be a family of
convex existence sets directed by $\supset.$ If $F=\bigcap_{t\in
T}F_{t}\neq\emptyset,$ then $F$ is also an existence set.

Proof.  Since $\\{F_{t}\\}_{t\in T}$ is directed by $\supset,$ for any
$t_{1},t_{2}\in T$ there exists $t_{3}\in T$ such that $F_{t_{1}}\cap
F_{t_{2}}\supset F_{t_{3}}.$ Define a partial ordering $\leq$ in $T$ by
$t_{1}\leq t_{2}$ provided $F_{t_{1}}\supset F_{t_{2}}.$ By the exiom of
choice, for any $t\in T$ we can select $P_{t}\in N(F_{t})$ such that
$P_{t}(x)\in R_{F_{t}}(x)$ for any $x\in X.$ Observe that $\\{P_{t}\\}_{t\in
T}\subset N(F).$ Set for any $t\geq t_{o}$ $A_{t}=cl(\\{P_{s}:s\geq t\\})$
where the closure is taken with respect to the topology in $N(F)$ defined in
Lemma 2. Since $\\{F_{t}\\}_{t\in T}$ is directed by $\supset,$ for any
$n\in\mathbb{N}$ and $t_{1},...t_{n}\geq t_{o}$ there exists $u\in T$ such
that

$\bigcap_{i=1}^{n}A_{t_{i}}\supset A_{u}\neq\emptyset.$

By the proof of Lemma 2, $\bigcap_{t\in T}A_{t}\neq\emptyset.$ Fix
$P\in\bigcap_{t\in T}A_{t}.$ We show that for any $x\in X,$ $Px\in R_{F}(x).$
Fix $x\in X$ and $d\in F.$ Hence $d\in F_{s}$ for any $s\in T$ and
consequently

$\|Px-d\|\leq\liminf_{s}\|P_{s}x-d\|\leq\|x-d\|.$

To finish the proof we need to show that $Px\in F$ for any $x\in X.$ Assume on
the contrary that there exists $x_{o}\in X$ such that $Px_{o}\notin F.$ Then
there exists $t_{o}\in T$ such that $Px_{o}\notin F_{t_{o}}.$ Hence
$Px_{o}\notin F_{t}$ for any $t\geq t_{o}.$ Since $P_{t}x_{o}\in F_{t}\subset
F_{t_{o}}$ for $t\geq t_{o},$ by the Mazur Theorem $Px_{o}\in F_{t_{o}},$
which is a contradiction.

## Section 2

Now we state the main result of this paper.

######  THEOREM 6.

Let $X$ be a reflexive, separable Banach space. Let $F\subset X$ be a
nonempty, closed, bounded and convex existence set with nonempty interior in
$X.$. Then $F$ is an intersection of a countable family of contractive half-
spaces.

Proof.  By Lemma 1 we can assume that $0\in int(F).$ Define a function
$f:X\rightarrow\mathbb{R}$ by

$f(x)=\inf\\{t>0:x/t\in C\\}.$

By Lemma 4 $f$ is a convex and continuous function. By the Mazur Theorem $f$
is Gateaux differentiable on a countable, dense subset of $X.$ Since
$f(tx)=tf(x)$ for any $x\in X$ and $t>0,$ there exists a dense countable
subset $Z=\\{z_{n}\\}$ of $\sigma(C)$ such that $f$ is is Gateaux
differentiable at any $z\in Z.$ Let for $n\in\mathbb{N}$
$D_{n}=cl(\bigcup_{t>0}F_{t,n}),$ where $F_{t,n}=\\{(1-t)z_{n}+tF\\}.$ By
Corollary 1, $D_{n}$ is an existence set for any $n\in\mathbb{N}.$ Since $f$
is is Gateaux differentiable at $z_{n},$ there exists $f_{n}\in S(X^{*})$ and
$d_{n}\in\mathbb{R}$ such that $D_{n}=\\{x\in X:f_{n}(x)\leq d_{n}\\}$ and
$f_{n}(x_{n})=d_{n}.$ (Since $0\in int(C),$ $d_{n}>0.)$ By Theorem 4 and Lemma
1, $ker(f_{n})$ is one-complemented in $X.$ By Theorem 5, $D_{n}$ is a
contractive half-space. By Lemma 5, $F=\bigcap_{n=1}^{\infty}D_{n},$ as
required. Now we will show a sufficient condition under which intersections of
countable families of contractive sets are contractive.

######  THEOREM 7.

Let $(X,\|\cdot\|)$ be a reflexive Banach space. Let $\|\cdot\|_{n}$ be a
sequence of strictly convex norms on $X$ such that there exists a sequence
$\\{s_{n}\\}$ of nonegative numbers, $s_{n}\rightarrow 0$ satisfying for any
$n\in\mathbb{N}$ and $x\in X$

$(1-s_{n})\|x\|_{n}\leq\|x\|\leq(1+s_{n})\|x\|_{n}$ (2)

Assume that $\\{F_{k}\\}$ is a countable family of convex sets such that for
any $k,n\in\mathbb{N}$ $F_{k}$ is a contractive set with respect to
$\|\cdot\|_{n}.$ Assume that $F=\bigcap_{k\in\mathbb{N}}F_{k}\neq\emptyset.$
Then $F$ is a contractive subset of $X$ with respect to $\|\cdot\|.$

Proof.  By Lemma 1, we can asume that $0\in F.$ Fix $n\in\mathbb{N}.$ Let for
$k\in\mathbb{N}$ $P_{k,n}:X\rightarrow F_{k}$ be a contractive mapping with
respect to $\|\cdot\|_{n}.$ Fix a sequence of positive numbers $\\{c_{k}\\}$
such that $\sum_{k=1}^{\infty}c_{k}=1.$ Define $P_{n}:X\rightarrow X$ by

$P_{n}x=\sum_{k=1}^{\infty}c_{k}P_{k,n}x.$

(Since $\sum_{k=1}^{\infty}c_{k}=1,$ $P_{n}$ is well-defined.) Now we show
that $Fix(P_{n})=F.$ By definition of $P_{k,n},$ $F\subset Fix(P_{n}).$ Now
asume that $x\in Fix(P_{n})\setminus\\{0\\}.$ Since $0\in F,$ for any
$k\in\mathbb{N}$ $\|P_{k,n}(x)\|_{n}\leq\|x\|_{n}.$ Fix $f_{n}\in S(X^{*})$
(with respect to $\|\cdot\|_{n})$ such that $f_{n}(x)=\|x\|_{n}.$ Notice that

$\|x\|_{n}=f(x)=f_{n}(\sum_{k=1}^{\infty}c_{k}P_{k,n}x)=\sum_{k=1}^{\infty}c_{k}f_{n}(P_{k,n}x)\leq\sum_{k=1}^{\infty}c_{k}\|P_{k,n}x\|_{n}\leq\|x\|_{n}.$

Hence for any $k\in\mathbb{N}$ $\|P_{k,n}x\|_{n}=\|x\|_{n}$ and
$f_{n}(P_{k,n}x)=\|x\|_{n}.$ Hence for any $k\in\mathbb{N}$
$\frac{P_{k,n}x}{\|P_{k,n}x\|_{n}}$ is a norming point for $f_{n}.$ Since
$(X\|\cdot\|_{n})$ is strictly convex, $f_{n}$ has exactly one norming point
with respect to $\|\cdot\|_{n}.$ Hence for any $k\in\mathbb{N},$ $P_{k,n}x=x,$
which shows that $x\in F.$ By ([8], Theorem 2 and Lemma 3), $F=Fix(P_{n})$ is
a contractive subset of $(X,\|\cdot\|_{n}).$ Let $Q_{n}:X\rightarrow F$ be a
contractive mapping with respect to $\|\cdot\|_{n}.$ Define for any $M\geq 1$

$N_{M}(F)=\\{f:X\rightarrow X:\|y-f(x)\|\leq M\|y-x\|\hbox{ for any }y\in
F,x\in X\\}.$

Reasoning as in Lemma 2, we can show that $N_{M}(F)$ is a compact set with
respect to the topology of weak pointwise convegence. Put

$M=sup\\{\frac{1+s_{n}}{1-s_{n}}:n\in\mathbb{N}\\}.$

Notice that for any $x\in X$ and $y\in F,$

$\|Q_{n}x-y\|\leq(1+s_{n})\|Q_{n}x-y\|_{n}\leq(1+s_{n})\|x-y\|_{n}\leq\frac{1+s_{n}}{1-s_{n}}\|x-y\|\leq
M\|x-y\|.$

Hence $Q_{n}\in N_{M}(F)$ for any $n\in\mathbb{N}.$ Renasoning as in Lemma 3,
we can show that $\bigcap_{n\in N}A_{n}\neq\emptyset,$ where
$A_{n}=cl(\\{Q_{k}:k\geq n\\}).$ Fix $Q\in\bigcap_{n\in N}A_{n}.$ We show that
$Q$ is a contractive mapping from $X$ onto $F$ with respect to $\|\cdot\|.$
Notice that for any $x,y\in X$

$\|Q_{n}x-Q_{n}y\|_{n}\leq\|x-y\|_{n}\leq\frac{\|x-y\|}{1-s_{n}}.$

Since $s_{n}\rightarrow 0,$

$\liminf_{n}\frac{\|Q_{n}x-Q_{n}y\|_{n}}{1-s_{n}}\leq\|x-y\|.$

Since $Q\in\bigcap_{n\in N}A_{n},$ this shows that

$\|Qx-
Qy\|\leq\liminf\|Q_{n}x-Q_{n}y\|\leq\liminf_{n}\frac{\|Q_{n}x-Q_{n}y\|_{n}}{1-s_{n}}\leq\|x-y\|,$

as required. Moreover, since for any $n\in\mathbb{N}$ $Q_{n}x\in F$ and $F$ is
convex, by the Mazur Theorem, $Qx\in F.$ Since for any $n\in\mathbb{N}$ and
$x\in F,$ $Q_{n}x=x,$ $Qx=x.$ The proof is complete.

######  REMARK 2.

Observe that, in general, the countable intersection of contractive sets need
not to be even an existence set. In ([19], Example 2.10), it was shown that
$F=ker(f_{1})\cap ker(f_{2})\subset l_{\infty}^{(4)}$ is not an existence set.
Here $f_{1}=(1,0,0,0)$ and $f_{2}=(1/2,1/6,1/6,1/6).$ (We understand that
$f_{i}(x)=\sum_{j=1}^{4}x_{j}f_{j},$ for $x\in l_{\infty}^{(4)}$ and $i=1,2.)$
Observe that $F=\bigcap_{n=1}^{\infty}F_{n},$ where

$F_{n}=\\{x\in l_{\infty}^{(4)}:|f_{i}(x)|\leq 1/n,i=1,2\\}.$

By Lemma 7 and Theorem 1 for any $n\in\mathbb{N}$ $F_{n}$ is not an existence
set. Observe that for any $n\in\mathbb{N}$
$F_{n}=\bigcup_{k=1}^{\infty}F_{n,k},$ where

$F_{n,k}=\\{x\in F_{n}:|x_{i}|\leq k\hbox{ for }k=1,...,4\\}.$

By Lemma 6 and Theorem 1, for any $k,n\in\mathbb{N},$ $F_{n,k}$ is not an
existence set too.

An immediate consequence of Theorem 6 and Theorem 7 is

######  THEOREM 8.

Let $(X,\|\cdot\|)$ and $F$ be as in Theorem 6. Assume that $(X,\|\cdot\|)$
satisfies the assumptions of Theorem 7. Then the following conditions are
equivalent:

a. $F$ is an existence set;

b. $F$ is a contractive set;

c. $F$ is an intersection of a countable number of contractive half-spaces.

Proof.  Assume that $F$ is an existence set. By Theorem 6, $F$ is an
intersection of a countable number of contractive half-spaces. By Theorem 7,
$F$ is a contractive set. Since each existence set is contractive, the proof
is complete. Now we present two applications of Theorem 8.

######  THEOREM 9.

Let $X=l_{1}^{(n)}$ and let $F\subset X$ be a convex bounded set with nonempty
interior in $X.$ Then the following conditions are equivalent

a. $F$ is an existence set;

b. $F$ is a contractive set;

c. $F$ is an intersection of a countable number of contractive half-spaces.

Proof.  Assume that $F$ is a convex existence set in $X.$ By Theorem 6, $F$ is
an intersection of a countable family $\\{F_{k}\\}$ of half-spaces. By Theorem
4, each $F_{k}$ is determined by $f_{k}\in S(X^{*})$ such that $ker(f_{k})$ is
one-complemented in $X.$ By[4], each $f_{k}$ has at most two coordinates
different from zero. By [7] and [1], for each $k,$ $ker(f_{k})$ is also one-
complemented in $l_{p}^{(n)}$ for $p\geq 1.$ Fix a sequence $p_{l}>1$
$p_{l}\rightarrow 1.$ By Theorem 5, for any $k,l$ $F_{k}$ is a contractive
subset of $l_{p_{l}}^{(n)}.$ Applying Theorem 8 to
$\|\cdot\|_{l}=\|\cdot\|_{p_{l}}$ and $\\{F_{k}\\}$ we get the result.

######  THEOREM 10.

Let $X=l_{\infty}^{(n)}$ and let $F\subset X$ be a convex set. Assume that
$F=\bigcap_{k=1}^{\infty}F_{k},$ where for each $k$ $F_{k}$ is a half space
determined by $f_{k}\in S(X^{*})$ having at most two coordinates different
from zero. Then the following conditions are equivalent

a. $F$ is an existence set;

b. $F$ is a contractive set;

Proof.  Fix a sequnece $p_{l}\rightarrow+\infty.$ By [7] and [1], and Theorem
5, for any $k,l$ $F_{k}$ is a contractive subset of $l_{p_{l}}^{(n)}.$
Reasoning as in Theorem 9 we get the result. Now we present a sufficient
condition (in which strict convexity is not assumed), under which the
intersection of a countable family of contractive half-spaces is a contractive
set.

######  THEOREM 11.

Let $X$ be a reflexive Banach space satisfying (CFPP) (see Def. 1). Assume
that $F=\bigcap_{k=1}^{\infty}F_{k},$ where for each $k$ $F_{k}$ is a
contractive half-space determined by $f_{k}\in S(X^{*}).$ (By Theorem 4, for
each $k$ $ker(f_{k})$ is one complemented in $X.)$ Let $P_{k}$ be a norm one
projection from $X$ onto $ker(f_{k}).$ (By [4], for any $x\in X$
$P_{k}x=x-f_{k}(x)y_{k}$ where $y_{k}\in X$ satisfies $f_{k}(y_{k})=1.)$
Assume furthermore that there exists a sequence $\\{d_{k}\\}$ of nonegative
numbers such that $F_{k}=\\{x\in X:f_{k}(x)\leq d_{k}\\}.$ If $0\notin
cl(conv(\\{y_{k}\\}_{k\in\mathbb{N}})),$ then $F$ is a contractive set.

Proof.  We follow the idea included in [8], Lemma 3). Fix a sequence $a_{n}$
of positive numbers such that $\sum_{n=1}^{\infty}a_{n}=1.$ By the proof of
Theorem 5 and Lemma 1 the mapping $Q_{n}:X\rightarrow F_{n}$ defined by
$Q_{n}x=x-f_{n}(x-z_{n})y_{n}$ if $f_{n}(x)>d_{n}$ and $Q_{n}x=x$ in the
opposite case, is a contractive projection from $X$ onto $F_{n}.$ (Here for
each $n\in\mathbb{N}$ $z_{n}$ is so chosen that $f_{n}(z_{n})=d_{n}.)$ Define
$Q:X\rightarrow X$ by

$Qx=\sum_{n=1}^{\infty}a_{n}Q_{n}.$

It is easy to see that $Q$ is a nonexpansive mapping. We show that
$Fix(Q_{n})=F.$ It is easy to see that $F\subset Fix(Q).$ Assume on the
contrary that there exists $x\in Fix(Q)\setminus F.$ Since
$F=\bigcap_{k=1}^{\infty}F_{k},$ the set

$Z=\\{n\in\mathbb{N}:Q_{k}(x)\neq x\\}\neq\emptyset.$

By defintion of $Q$

$x=\sum_{k\notin Z}a_{k}x+\sum_{k\in Z}a_{k}Q_{k}x.$

(We assume that $\sum_{k\notin Z}=0$ if $Z=\mathbb{N}.)$ By defintion of
$Q_{k},$

$0=\sum_{k\in Z}a_{k}f_{k}(x-z_{k})y_{k}.$

Since $a_{k}>0$ and $f_{k}(x-z_{k})>0$ for $k\in Z,$ $0\in
cl(conv(\\{y_{k}\\}_{k\in\mathbb{N}})),$ which is a contradiction.

######  REMARK 3.

In [8], Theorem 4, sufficient conditions for a Banach space $X$ satisfying
(CFPP) are presented. In particular, any finite-dimensional Banach space
satisfies (CFPP).

######  EXAMPLE 1.

Let $X=l_{\infty}^{(n)}.$ Let for $k\in\mathbb{N},$ $f^{k}\in S(X^{*}),$
$f^{k}=(f^{k}_{1},...,f^{k}_{n})$ be so chosen that

$|f^{k}_{j}|\geq\sum_{i\neq j}|f^{k}_{i}|$ (3)

for $j\in\\{1,...,\\}$ depending on $k.$ By [4], for any $k,$ $ker(f^{k})$ is
one-complemented in $X.$ Fix a sequence of nonegative numbers $\\{d_{k}\\}.$
By Theorem 5 the half spaces $F_{k}=\\{x\in X:f^{k}(x)\leq d_{k}\\}$ are
contractive subsets of $X.$ Assume that $F=\bigcap_{k=1}^{\infty}F_{k},$ is a
nonempty set . Let for any $j\in\\{1,...,n\\},$
$Z_{j}=\\{k\in\mathbb{N}:|f^{k}_{j}|\geq\sum_{i\neq j}|f^{k}_{i}|\\}.$ Assume
that for some $j\in\\{1,...,n\\}$ such that $Z_{j}\neq\emptyset,$
$f^{k}_{j}>0$ for any $k\in Z_{j}.$ By [4] and the proof of Theorem 11, for
each $k\in\mathbb{N},$ $Q_{k}x=x-f^{k}(x-z_{k})y_{k},$ where
$y_{k}=(0,(\frac{1}{f^{k}_{j}})_{j},...,0)$ where $j\in\\{1,...,n\\}$
satisfies (3). Hence by our assumption on $Z_{j},$ $0\notin
cl(conv(\\{y_{k}\\}_{k\in\mathbb{N}})).$ By Theorem 11, $F$ is a contractive
subset of $X.$ Observe that if for some $k$ $f^{k}$ has more than two
coordinates different from zero, this result cannot be deduced from Theorem
10.

######  REMARK 4.

The assumption that $F$ is a bounded set in Theorem 8 can be weakend. If we
assume that $F=cl(\bigcup_{t\in T}F_{t})$ where $\\{F_{t}\\}_{t\in T}$ is a
directed by $\subset$ family of bounded and convex existence sets such that
$int(F_{t_{o}})\neq\emptyset$ for some $t_{o}\in T,$ by Lemma 6 the conditions
(a) and (b) from Theorem 8 are equivalent. If $T=\mathbb{N}$ and
$int(F)\neq\emptyset,$ then by the Baire Property
$int(F_{t_{o}})\neq\emptyset$ for some $t_{o}\in\mathbb{N}.$

In particular, we can prove

######  THEOREM 12.

Let $(X,\|\cdot\|)$ be a strictly convex, reflexive and separable Banach space
and let $F\subset X$ be a bounded set with nonempty interior (compare with
Remark 4.) Then the following conditions are equivalent

a. $F$ is an existence set;

b. $F$ is a contractive set;

c. $F$ is an intersection of a countable number of contractive half-spaces.

d. $F$ is an optimal set.

Proof.  If $F$ is an existence set, then, by definition, $F$ is an optimal
set. Since $X$ is strictly convex, if $F$ is an optimal set, by [5], $F$ is a
convex set. By Theorem 1, $F$ is an existence set. Hence by Theorem 8, the
proof is complete.

######  REMARK 5.

There exists Banach spaces such that there is no bounded, convex sets with
nonempty interior being existence sets. For example, if we take $X=L_{p}[0,1]$
for $1<p<+\infty,$ then there is no $F\subset X,$ $F$ convex with
$int(F)\neq\emptyset$ being an existence set. Indeed, if such an $F$ exists,
then by Theorem 6 $F$ would be an intersection of a countable family of
contractive half-spaces $\\{D_{n}\\}.$ By Theorem 4, each $D_{n}$ is
determined by $f_{n}\in S(X^{*})$ with $ker(f_{n})$ being one-complemented.
But by [5], (see also [14], [21], [28] and [35]),each hyperplane in $X$ is not
one complemented.

## Section 3

Now, we prove some results on existence sets $F$ without assumptions that they
have nonempty interior in the whole space $X.$ To the end of this section, if
is not otherwise stated, we assume that $int_{Z}(F)\neq\emptyset$ where
$Z=cl(span(F))$ and $int_{Z}(F)$ means the interior with respect to $Z.$

######  REMARK 6.

It may happen that $int_{Z}(F)=\emptyset.$ For example, if we take
$X=L_{p}[0,1]$ for $1\leq p<+\infty,$ and $F=\\{f\in X:f\geq 0\\}$ then it is
easy to see that $Span(F)=X$ and $int_{X}(F)=\emptyset.$ Observe that $F$ is a
contractive subset of $X.$ Indeed, it is easy to see that the mapping
$Pf=f\chi(D_{f,+})$ is a contractive projection onto $F.$ Here $\chi(D_{f,+})$
denote the characteristic function of the set $D_{f,+}=\\{t\in[0,1]:f(t)\geq
0\\}.$ If $X$ is finite-dimensional, then $int_{Z}(F)\neq\emptyset$ for any
$F\subset X,$ $F\neq\emptyset.$

We start with

######  THEOREM 13.

Let $X$ be a Banach space and let $F\subset X$ be a convex and bounded
(compare with Remark 4) existence set. Assume that $Z=cl(span(F))$ satisfies
the assumption of Theorem 8. If $Z$ is a contractive subset of $X$ then $F$ is
a contractive subset of $X.$

Proof.  Since $F$ is an existence set in $X,$ $F$ is an existence set in $Z.$
By Theorem 8 applied to $Z$ there exists a contractive projection
$Q:Z\rightarrow F.$ Let $P:X\rightarrow Z$ be a contractive projection. The
$Q\circ P$ is a contractive projection from $X$ anto $F.$ The proof is
complete. Now we present examples of Banach spaces $X$ in which any subspace
being an existence set is a contractive set. In [22], the folowing result was
shown.

######  THEOREM 14.

Let $X$ be a Banach space and let $Z\subset X,$ be a linear subspace, which is
an existence set. Put

$G_{Z}=\\{z\in Z\setminus\\{0\\}:\hbox{ there exists exactly one }f\in
S(X^{*}):f(z)=\|z\|\\}.$ (4)

Assume that the norm closure of $G_{Z}$ in $X$ is equal to $Z.$ Then there
exists exactly one projection $P\in{\cal P}(X,Z)$ such that $\|P\|=1,$ which
means that $Z$ is a contractive subset of $X.$

######  REMARK 7.

In particular, if $X$ is a smooth space, then (4) is satisfied. Applying
Theorem 14 it was shown in [22], that in $c_{o}$, $l_{1}$ and some Musielak-
Orlicz sequence spaces any subspace $Z$ which is an existence set is one-
complemented. Also it was shown in [20] that any subspace $Z$ of the Lorentz
sequence space $l_{1,w}$ which is an existence set, is contractive.

######  COROLLARY 2.

Let $X$ be a reflexive Banach space. Let $F\subset X$ be a bounded and convex
existence set such that $dim(Z=span(F))<\infty.$ If $X$ is strictly convex and
smooth then $X$ is contractive. If $X$ is finite-dimensional, strictly convex
and and smooth Banach space then any bounded optimal set is contractive.

Proof.  If $dim(Z=span(F))<\infty,$ then $int_{Z}(F)\neq\emptyset.$ By Theorem
13 and Theorem 14 $F$ is contractive set. If $X$ is strictly convex and
reflexive and $F$ is a bounded optimal set then by [5], $F$ is a convex
existence set. If $X$ is finite-dimensional, then $int_{Z}(F)\neq\emptyset$
for any $F\subset X.$ The proof is complete. The folowing result can be
applied to sets satisfying $int_{Z}(F)=\emptyset.$ The proof of it is the same
as in [19].

######  THEOREM 15.

Let $X$ be a reflexive smooth Banach space and let $F\subset X$ be a convex
existence set such that $F=cl(\bigcup_{t\in T}^{\infty}F_{t}),$ where
$\\{F_{t}\\}_{t\in T}$ is a family of convex and compact existence sets
ordered by $\subset.$ Then $F$ is a contractive set. If $F$ is a convex,
compact existence set then the assumption of reflexivity can be omitted.

Proof.  Let $x\in X$ and fix $t\in T.$ By ([19],Th. 3.3) there exists
$P_{t}x\in R_{F_{t}}(x)$ such that for any $d\in F_{t}$ and $s\geq 0$

$\|Px-d\|\leq\|sx+(1-s)P_{t}x-d).$

Since $X$ is smooth, by ([9], Lemma 1), $P_{t}x$ is uniquely determined. By
([9], Th. 1) $F_{t}$ is a contractive set. (Up to now the reflexivity is not
needed.) By ([8]), Lemma 4, $F$ is a contractive set.

######  REMARK 8.

In general, we do not know under which conditions on a Banach space $X$ any
existence and convex set can be represented as $F=cl(\bigcup_{t\in
T}^{\infty}F_{t}),$ where $\\{F_{t}\\}_{t\in T}$ is a family of convex and
compact existence sets ordered by $\subset.$ Such a result has been proven
[19], Lemma 3.7 for reflexive Köthe sequence spaces. For $l_{p}$ spaces such a
result has been demonstrated in [10].

######  REMARK 9.

Notice that in general a contractive set need not to be convex. Let, for
example $X=l_{\infty}^{(2)}.$ Define

$F=\\{(x,y)\in X:|y|\leq|x|\\}.$

Then after elementary, but tedious calculations, one can see that the mapping
$P:X\rightarrow F$ defined by

$P(x,y)=\begin{pmatrix}(x,sgn(y)x)&\text{ for $x\geq 0,|y|>|x|$}\\\
(x,-sgn(y)x)&\text{ for $x<0,|y|>|x|$}\\\ (x,y)&\text{ for
$|y|\leq|x|$}\end{pmatrix}$

is a contractive projection from $X$ onto $F.$ It would be be interesting to
obtain a relation between existence and contractive sets in nonconvex case.

Notice that the following two well-known lemmas permit us to adopt the results
proved for real Banach spaces in the case of complex Banach spaces. We present
the proofs of them for a conveniece of the reader.

######  LEMMA 8.

Let $X$ be a complex Banach space with a Hamel basis $H=\\{h_{t}\\}_{t\in T}.$
Let $X_{R}$ be a real linear space spanned by $\\{h_{t},ih_{s}\\}_{s,t\in T}.$
Then $H_{R}=\\{h_{t},ih_{s}\\}_{s,t\in T}$ is a Hamel basis of $X_{R}$ over
$\mathbb{R}.$ Let us equipp $X_{R}$ with a norm induced from $X,$ i.e.,

$\|\sum_{j=1}^{n}a_{j}h_{t_{j}}+\sum_{j=1}^{n}b_{j}ih_{t_{j}}\|=\|\sum_{j=1}^{n}(a_{j}+ib_{j})h_{t_{j}}\|$

Then the mapping $Iz:X\rightarrow X_{R}$ defined by

$Iz(\sum_{j=1}^{n}a_{j}h_{t_{j}})=\sum_{j=1}^{n}re(a_{j})h_{t_{j}}+\sum_{j=1}^{n}im(a_{j})ih_{t_{j}}$

is a linear surjective isometry over $\mathbb{R}$ which means that
$\|Iz(x)\|=\|x\|,$ $Iz(x+y)=Iz(x)+Iz(y),$ $Iz(ax)=aIz(x)$ for
$a\in\mathbb{R},$ and $x,y\in X.$ In particular, $X_{R}$ is a real Banach
space.

Proof.  Since $H$ is a Hamel basis over $\mathbb{C},$ the set
$\\{h_{t},ih_{s}\\}_{s,t\in T}$ is linearly independent over $\mathbb{R}.$ The
fact that $Iz$ is a linear isometry over $\mathbb{R}$ follows immediately from
the defintion of the norm in $X_{R}.$ To prove surjectivity, fix $x\in X_{R}.$
Then $x=\sum_{j=1}^{n}a_{j}y_{t_{j}}+\sum_{k=1}^{m}b_{k}iy_{s_{k}}$ with
$a_{j}\neq 0$ and $b_{k}\neq 0.$ Let $S=\\{t_{1},...,t_{n},s_{1},..,s_{m}\\}.$
Let $z=\sum_{j\in S}c_{j}y_{j},$ where $c_{j}=a_{j}+ib_{j}.$ (We put $a_{j}=0$
if $j\notin\\{s_{1},...,s_{m}\\}$ and $b_{j}=0$ if
$j\notin\\{t_{1},...,t_{n}\\}.$ It is clear that $Iz(z)=x,$ which completes
the proof.

######  LEMMA 9.

Let $X$ be a complex Banach space. Then $X$ is reflexive if and only if
$X_{R}$ is reflexive, $X$ is strictly convex if and only if $X_{R}$ is
strictly convex, $X$ is smooth if and only if $X_{R}$ is smooth. Moreover $X$
satisfies (CFPP) if and only if $X_{R}$ satisfies (CFPP).

Proof.  The proofs of all above properties follows from Lemma 8. Before
presenting them, fix $f\in X^{*}.$ Then $f=re(f)+i(im(f)).$ Put $g=re(f).$
Since $f(ix)=if(x)$ for any $x\in X,$ we easily get that $f(x)=g(x)-ig(ix).$
Moreover, $\|f\|=\|g\|.$ Indeed, it is immediate that $\|g\|\leq\|f\|.$ To
prove a converse, fix $x\in X,$ $\|x\|=1.$ Then $f(e^{it}x)=g(e^{it}x)$ for
some $t\in[0,2\pi].$ Hence

$|f(x)|=|f(e^{it}x)|=|g(e^{it}x)|\leq\|g\|\|e^{it}x\|=\|g\|$

which shows our claim. Moreover, by the above reasoning if $g\in(X_{R})^{*}$
then the mapping $f:X\rightarrow\mathbb{C}$ given by $f(x)=g(x)-ig(ix)$
belongs to $X^{*}$ and $\|f\|=\|g\|.$ Hence, in particular, any $f\in
S(X^{*})$ attains its norm at some $x\in S(X)$ if and only if any
$g\in(X_{R})^{*}$ attains its norm in some $z\in S(X_{R}).$ Consequently, by
the James Theorem $X$ is reflexive if and only if $X_{R}$ is reflexive.
Analogously, for any $x\in S(X)$ there exists exactly one $f\in S(X^{*})$
satisfying $f(x)=\|x\|=1$ if and only if for any $z\in S(X_{R})$ there exists
exactly one $g\in S((X_{R})^{*})$ satisfying $g(z)=\|z\|=1.$ Hence $X$ is
smooth if and only if $X_{R}$ is smooth. The equivalence of strict convexity
for $X$ and $X_{R}$ and the equivalence of (CFPP) for $X$ and $X_{R}$ follows
immediately from Lemma 8, so we omit the proofs.

######  THEOREM 16.

Theorem 6, Theorem 7, Theorem 8 (By a half-space determined by $f\in S(X^{*})$
we understand the set $\\{x\in X:re(f)(x)\leq d\\}.),$ Theorem 9, Theorem 10,
Theorem 11, Theorem 12, Theorem 13, Theorem 14 and Theorem 15 holds true for
any complex Banach space $X$.

Proof.  Let $X$ be a complex Banach space and let $F\subset X.$ Then by Lemma
8, $F$ is a convex set if and only if $Iz(F)$ is a convex set, $F$ is an
existence set in $X$ if and only if $Iz(F)$ is a and existence set in $X_{R}$
and $F$ is a contractive subset of $X$ if and only if $Iz(F)$ is a contractive
subset of $X_{R}.$ By Lemma 9, we can adopt the proofs of the above mentionned
theorems given in the real case to the complex case.

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$\begin{array}[]{l}\textnormal{\small Maciej CIESIELSKI}\\\ \textnormal{\small
Institute of Mathematics}\\\ \textnormal{\small Pozna\'{n} University of
Technology}\\\ \textnormal{\small Piotrowo 3A, 60-965 Pozna\'{n}, Poland}\\\
\textnormal{\small email: maciej.ciesielski@put.poznan.pl;}\\\ \textnormal{
}\end{array}$

$\begin{array}[]{l}\textnormal{\small Grzegorz LEWICKI}\\\ \textnormal{\small
Department of Mathematics and Computer Science}\\\ \textnormal{\small
Jagiellonian University}\\\ \textnormal{\small 30-348 Krak\'{o}w, \L
ojasiewicza 6, Poland}\\\ \textnormal{\small email:
grzegorz.lewicki@im.uj.edu.pl;}\end{array}$


# The Steinitz Realization Problem

Sameera Vemulapalli  Department of Mathematics, Stanford University
<EMAIL_ADDRESS>
###### Abstract.

Let $K$ be a number field and let $n\in\mathbb{Z}_{>1}$. The _Steinitz
realization problem_ asks: does every element of $\operatorname{Cl}(K)$ occur
as the Steinitz class of a degree $n$ extension of $K$? In this article, we
give an affirmative answer to the Steinitz realization problem for all $n$ and
$K$.

###### Key words and phrases:

Algebraic numbers; Polynomials; Class groups; Geometry of numbers

###### 2010 Mathematics Subject Classification:

11R04 (primary), 11R09, 11R29, 14H05 (secondary)

##  1\. Introduction

Fix $n\in\mathbb{Z}_{>1}$ a positive integer. To any degree $n$ extension of
number fields $L/K$, we may canonically associate an element of the ideal
class group of $K$, namely $[\mathfrak{a}]$ where
$\mathcal{O}_{L}\simeq\mathcal{O}_{K}^{n-1}\oplus\mathfrak{a}$ as an
$\mathcal{O}_{K}$-module. The class $[\mathfrak{a}]$ is called the _Steinitz
class_ of $L/K$ and is denoted $\operatorname{St}(L/K)$. By a theorem of
Hecke, the square of the Steinitz class $\operatorname{St}(L/K)$ is the class
of the relative discriminant $\operatorname{Disc}(L/K)$. The _Steinitz
realization problem_ asks: does every element of $\operatorname{Cl}(K)$ occur
as the Steinitz class of a degree $n$ extension of $K$?

When $n=2,3,4,5$, there is an affirmative answer to the Steinitz realization
problem; in these cases the Steinitz classes of $S_{n}$-extensions are
_equidistributed_ in $\operatorname{Cl}(K)$ when degree $n$ number fields are
ordered by discriminant. The case $n=2,3$ was proven by Kable and Wright [kab]
and the case $n=4,5$ was proven by Bhargava, Shankar, and Wang [bsw]. The
proofs rely on the parametrizations of degree $n$ number fields; for $n>6$,
there are no such parametrizations. The purpose of this article is to give an
affirmative answer to the Steinitz realization problem for all $n$.

######  Theorem 1.1.

Every element of $\operatorname{Cl}(K)$ occurs as the Steinitz class of a
degree $n$ extension of $K$ with squarefree discriminant.

We give a sketch of the proof. A _rank n ring_ over $\mathcal{O}_{K}$ is a
ring that is locally free rank of $n$ as an $\mathcal{O}_{K}$-module. By work
of Wood [wood], a binary $n$-ic form

$f(x,y)=f_{0}x^{n}+\dots+f_{n}y^{n}\in\operatorname{Sym}^{n}(\mathcal{O}_{K}\oplus\mathcal{O}_{K})$

gives rise to a rank $n$ ring $R_{f}$. If $\mathfrak{a}$ is an integral ideal
of $\mathcal{O}_{K}$ such that $f_{n-1}\in\mathfrak{a}$ and
$f_{n}\in\mathfrak{a}^{2}$, we construct a rank $n$ ring $R_{f}(\mathbf{a})$
and an inclusion $R_{f}\xhookrightarrow{}R_{f}(\mathbf{a})$. When $f$ is
irreducible, both rings are orders in a number fields. The Steinitz class of
$R_{f}(\mathfrak{a})$ is $[\mathfrak{a}^{-1}]$ and
$\operatorname{Disc}(R_{f}(\mathbf{a}))=\mathfrak{a}^{-2}\operatorname{Disc}(R_{f})$.

If $\operatorname{Disc}(R_{f}(\mathfrak{a}))$ is squarefree as an ideal and
$f(x,y)$ is irreducible, then $R_{f}(\mathfrak{a})$ is the maximal order in a
degree $n$ extension of $K$ with Steinitz class $[\mathfrak{a}^{-1}]$. Thus,
Theorem 1.1 immediately follows from the existence of an appropriate $f(x,y)$.

######  Theorem 1.2.

For every nonzero proper prime ideal $\mathfrak{a}$ of $\mathcal{O}_{K}$
coprime to $(n!)$, there are infinitely many monic degree $n$ polynomials

$f(x)\coloneqq x^{n}+f_{1}x^{n-1}\dots+f_{n}$

with coefficients in $\mathcal{O}_{K}$ such that:

  1. (1)

$f$ is irreducible over $K$;

  2. (2)

$f_{n-1}\in\mathfrak{a}$ and $f_{n}\in\mathfrak{a}^{2}$;

  3. (3)

and $\operatorname{Disc}(R_{f}(\mathfrak{a}))$ is squarefree as an ideal in
$K$.

The proof of Theorem 1.2 is a slight generalization of the proof given by
Kedlaya when proving that there exist infinitely many irreducible monic
polynomials over $\mathbb{Z}$ with squarefree discriminant [ked].

###  1.1. Acknowledgements

The author would like to thank Ravi Vakil and Kiran Kedlaya for valuable
conversations. She would also like to thank the NSF and Stanford University,
where this work was completed.

##  2\. Construction of $R_{f}(\mathfrak{a})$

By work of Wood [wood], a binary $n$-ic form
$f_{0}x^{n}+\dots+f_{n}y^{n}\in\operatorname{Sym}^{n}(\mathcal{O}_{K}\oplus\mathcal{O}_{K})$
gives rise to a rank $n$ ring $R_{f}$ as follows. As an
$\mathcal{O}_{K}$-module, put $R_{f}\coloneqq\mathcal{O}_{K}^{n}$ and label
the copies of $\mathcal{O}_{K}$ so we write:

$R_{f}=(\mathcal{O}_{K})_{0}\oplus\dots\oplus(\mathcal{O}_{K})_{n-1}.$

Put $\zeta_{1},\dots,\zeta_{n-1}\coloneqq 1$ and $\zeta_{n}\coloneqq-f_{n}$.
Define $1\in(\mathcal{O}_{K})_{0}$ to be the multiplicative identity in
$R_{f}$. For $1\leq i\leq j\leq n-1$ and $1\leq k\leq n$, define the
multiplication maps
$(\mathcal{O}_{K})_{i}\otimes(\mathcal{O}_{K})_{j}\rightarrow(\mathcal{O}_{K})_{k}$
by:

  1. (1)

$x\otimes y\rightarrow-f_{i+j-k}\zeta_{k}xy$ if $\max(i+j-n,1)\leq k\leq i$;

  2. (2)

$x\otimes y\rightarrow f_{i+j-k}\zeta_{k}xy$ if $j<k\leq\min(i+j,n)$;

  3. (3)

and $x\otimes y\rightarrow 0$ otherwise.

In the notation above, $(\mathcal{O}_{K})_{n}$ refers to
$(\mathcal{O}_{K})_{0}$.

For any nonzero integral ideal $\mathfrak{a}$ of $K$, define
$R_{f}(\mathfrak{a})$ to be the locally free $\mathcal{O}_{K}$-module given
by:

$R_{f}(\mathfrak{a})\coloneqq(\mathcal{O}_{K})_{0}\oplus\dots\oplus\mathfrak{a}^{-1}(\mathcal{O}_{K})_{n-1}.$

There is a natural inclusion of $\mathcal{O}_{K}$-modules given by
$R_{f}\subseteq R_{f}(\mathfrak{a})$.

######  Lemma 2.1.

$R_{f}(\mathfrak{a})$ has a $\mathcal{O}_{K}$-algebra structure extending that
of $R_{f}$ if and only if $f_{n-1}\in\mathfrak{a}$ and
$f_{n}\in\mathfrak{a}^{2}$. When $f_{n-1}\in\mathfrak{a}$ and
$f_{n}\in\mathfrak{a}^{2}$, there is a unique $\mathcal{O}_{K}$-algebra
structure on $R_{f}(\mathfrak{a})$ extending that of $R_{f}$, the Steinitz
class of $R_{f}(\mathfrak{a})$ is $[\mathfrak{a}^{-1}]$ and
$\mathfrak{a}^{-2}(\operatorname{Disc}(R_{f}))=(\operatorname{Disc}(R_{f}(\mathfrak{a})))$.

###### Proof.

The first statement follows directly from the multiplication table The second
is immediate from the construction. ∎

##  3\. Proof of Theorem 1.2

The proof given in this section is essentially that given by Kedlaya [ked].
The main differences are that we work over a number field and impose a few
additional congruence conditions needed for our application. We repeat much of
Kedlaya’s construction verbatim, and encourage the reader to refer to
Kedlaya’s paper for more detail. Fix a class
$[\mathfrak{a}]\in\operatorname{Cl}(K)$, and pick a representative prime ideal
$\mathfrak{a}$ coprime to $(n!)$. For
$a_{1},a_{2},\dots,a_{n-1},b\in\mathcal{O}_{K}$, put:

$Q_{a}(x)\coloneqq n(x-a_{1}/n)(x-a_{2})\dots(x-a_{n-1})$ $P_{a,b}(x)\coloneqq
b+\int_{0}^{x}Q_{a}(t)dt$

The integral in the definition of $P_{a,b}(x)$ should be interpreted formally;
$Q_{a}(x)$ is a polynomial. The discriminant of $P_{a,b}$, up to sign, is the
resultant of $P_{a,b}(x)$ and $Q_{a}(x)$. So, we obtain the following equality
of ideals:

$\Delta_{a,b}=(n^{n}P_{a,0}(a_{1}/n)+n^{n}b)\prod_{i=2}^{n-1}(P_{a,0}(a_{i})+b).$

Suppose $a_{1}\in\mathfrak{a}$ and $b\in\mathfrak{a}^{2}$. If we write
$P_{a,b}(x)=x^{n}+f_{1}x^{n-1}+\dots+f_{n}$ and assume that $P_{a,b}(x)$ has
coefficients in $\mathcal{O}_{K}$, then $f_{n}=b\in\mathfrak{a}^{2}$ and

$f_{n-1}=(-1)^{n-1}\prod_{i=1}^{n-1}a_{i}\in\mathfrak{a}.$

Furthermore, $\mathfrak{a}^{2}$ divides $(n^{n}P_{a,0}(a_{1}/n)+n^{n}b)$ and
hence divides $\Delta_{a,b}$. Therefore, we wish to show that there exists a
choice of $a_{1},\dots,a_{n-1},b$ with $a_{1}\in\mathfrak{a}$ and
$b\in\mathfrak{a}^{2}$ such that $P_{a,b}(x)$ has integral coefficients, is
irreducible, and the ideal $\mathfrak{a}^{-2}\Delta_{a,b}$ is squarefree. We
will do this by fixing an appropriate choice of $a_{1},\dots,a_{n-1}$ and then
applying a squarefree sieve to choose $b$.

We will first show that there exist $a_{1},\dots,a_{n-1}\in\mathcal{O}_{K}$
such that:

  1. (1)

$a_{1}\in\mathfrak{a}$ and $a_{i}\equiv 1\pmod{\mathfrak{a}}$ for
$i=2,\dots,n-1$;

  2. (2)

$P_{a,0}$ has integral coefficients;

  3. (3)

for every prime ideal $\mathfrak{p}\neq\mathfrak{a}$, there exists $b$ such
that $\Delta_{a,b}$ is not divisible by $\mathfrak{p}^{2}$;

  4. (4)

there exists $b_{2}\in\mathfrak{a}^{2}$ such that $\Delta_{a,b_{2}}$ is not
divisible by $\mathfrak{a}^{3}$;

  5. (5)

there exists a prime ideal $\mathfrak{p}_{1}$ and $b_{1}\in\mathcal{O}_{K}$
such that $P_{a,b_{1}}$ is irreducible modulo $\mathfrak{p}_{1}$;

  6. (6)

and the polynomials
$(n^{n}P_{a,0}(a_{1}/n)+n^{n}b),(P_{a,0}(a_{2})+b),\dots,(P_{a,0}(a_{n-1})+b)$
are pairwise coprime.

To do this, we’ll need the following two lemmas.

######  Lemma 3.1 (Kedlaya, Lemma 2.3 [ked]).

Let $\gamma\in\mathcal{O}_{K}$ be such that $(\gamma)$ is a principal prime
ideal coprime to $(n(n-1))$. Then there exist infinitely many primes
$\mathfrak{p}_{1}$ modulo which the polynomial
$R(x)=x^{n}-\gamma^{n-1}x+\gamma$ is irreducible and its derivative
$R^{\prime}(x)=nx^{n-1}-\gamma^{n-1}$ splits into linear factors.

###### Proof.

The polynomial $R^{\prime}(x)$ has splitting field
$L=K(\zeta_{n-1},n^{1/(n-1)})$, in which $(\gamma)$ does not ramify because
$(\gamma)$ does not divide $(n(n-1))$, where here $\zeta_{n-1}$ is an
$(n-1)$–th root of unity. Thus $R$ is an Eisenstein polynomial with respect to
any prime above $(\gamma)$ in $L$; in particular, $R$ is irreducible over $L$.
By the Chebotarev density theorem, there exist infinitely many totally split
prime ideals of $L$ modulo which $R$ is irreducible; the norm to $K$ of any
such prime ideal is a prime ideal of the desired form. ∎

######  Lemma 3.2 (Kedlaya, Lemma 2.4 [ked]).

For any field $F$ of characteristic zero, there exist $a_{1},\dots,a_{n-1}$
such that

$P_{a,0}(a_{1}/n),P_{a,0}(a_{2}),\dots,P_{a,0}(a_{n-1})$

are distinct.

Let $\gamma$, $\mathfrak{p}_{1}$, and $R(x)$ be as in Lemma 3.1 and ensure
that $\mathfrak{p}_{1}\nmid\mathfrak{a}(n!)$. Apply Lemma 3.2 to obtain
$a^{\prime}_{1},\dots,a^{\prime}_{n-1}$ so that

$P_{a^{\prime},0}(a^{\prime}_{1}/n),P_{a^{\prime},0}(a^{\prime}_{2}),\dots,P_{a^{\prime},0}(a^{\prime}_{n-1})$

have distinct values. Choose a prime ideal
$\mathfrak{p}_{2}\nmid\mathfrak{a}(n!)\mathfrak{p}_{1}$ such the reductions
have well-defined and distinct values modulo $\mathfrak{p}_{2}$. Now use the
Chinese remainder theorem to choose $a_{1},\dots,a_{n-1}\in\mathcal{O}_{K}$
subject to the following congruence conditions:

  * •

$a_{1}$ is coprime to $(n)$ and is contained in $\mathfrak{a}$ and the ideal
generated by $(n-1)!$;

  * •

the integers $a_{2},\dots,a_{n-1}$ are contained in $(n!)$ and $a_{i}\equiv-1$
modulo $\mathfrak{a}$ for $i=2,\dots,n-1$;

  * •

the quantities $a_{1}/n,\dots,a_{n-1}$ are congruent to the roots of
$R^{\prime}(x)$ modulo $\mathfrak{p}_{1}$;

  * •

and $a_{i}\equiv a^{\prime}_{i}$ modulo $\mathfrak{p}_{2}$.

We’ll now show that this choice of $a_{1},\dots,a_{n-1}$ satisfies conditions
$(1)-(6)$.

  1. (1)

Follows directly from the choice of $a_{i}$.

  2. (2)

Observe that:

$Q_{a}(x)\equiv(nx-a_{1})x^{n-2}=nx^{n-1}-a_{1}x^{n-2}\pmod{(n!)},$

so $P_{a,0}$ has integer coefficients.

  3. (3)

Suppose that $\mathfrak{p}\neq\mathfrak{a}$ is a prime ideal dividing $(n!)$.
Then for $b\in\mathcal{O}_{K}$, we have $P_{a,0}(a_{i})+b\equiv
b\pmod{\mathfrak{p}}$ for $i=2,\dots,n-1$ because $a_{i}\in\mathfrak{p}$. If
$\mathfrak{p}\mid(n)$, then

$n^{n}P_{a,0}(a_{1}/n)+n^{n}b\equiv a_{1}^{n}\pmod{\mathfrak{p}}$

and if $\mathfrak{p}\nmid(n)$, then

$n^{n}P_{a,0}(a_{1}/n)+n^{n}b\equiv n^{n}b\pmod{\mathfrak{p}}$

Consequently, $\Delta_{a,1}$ is not divisible by $\mathfrak{p}$. Now suppose
that $\mathfrak{p}$ is a prime ideal coprime to $\mathfrak{a}(n!)$, so
$\lvert\mathcal{O}_{K}/\mathfrak{p}\rvert>n$. There are at least
$\lvert\mathcal{O}_{K}/\mathfrak{p}\rvert-n$ choices of
$b\in\mathcal{O}_{K}/\mathfrak{p}$ for which $\Delta_{a,b}$ is indivisible by
$p$, because each linear factor rules out exactly one choice of $b$.

  4. (4)

Choose $b\in\mathfrak{a}^{2}$ so that $P_{a,0}(a_{1}/n)+b\not\equiv
0\pmod{\mathfrak{a}^{3}}$. To conclude it suffices to show that
$P_{a,0}(a_{i})\not\equiv 0\pmod{\mathfrak{a}}$ for all $i=2,\dots,n-1$.
Because $a_{i}\equiv-1$ modulo $\mathfrak{a}$ for $i=2,\dots,n-1$ and
$P_{a,0}(0,-a_{2},\dots,-a_{n-1})$ is a nonzero polynomial with positive
coefficients in $a_{2},\dots,a_{n-1}$, we can ensure this by choosing
$\mathfrak{a}$ to have sufficiently large norm.

  5. (5)

Ensured by the fact that the quantities $a_{1}/n,\dots,a_{n-1}$ are congruent
to the roots of $R^{\prime}(x)$ modulo $\mathfrak{p}_{1}$.

  6. (6)

Ensured by the fact that $a_{i}\equiv a^{\prime}_{i}$ modulo
$\mathfrak{p}_{2}$.

Given such a choice of the $a_{1},\dots,a_{n-1}$ above, let $T$ be the set of
all $b\in\mathfrak{a}^{2}$ such that $b\equiv b_{1}\pmod{\mathfrak{p}_{1}}$
amd $b\equiv b_{2}\pmod{\mathfrak{a}^{3}}$. For any $b\in T$, the polynomial
$P_{a,b}(x)=x^{n}+f_{1}x^{n-1}+\dots+f_{n}$ is irreducible with integral
coefficients such that $f_{n}\in\mathfrak{a}^{2}$, $f_{n-1}\in\mathfrak{a}$,
$\mathfrak{p}_{1}\nmid\Delta_{a,b}$, and $\mathfrak{a}^{3}\nmid\Delta_{a,b}$.
For $b\in K$, put $\lvert
b\rvert=\sqrt{\sum_{i=1}^{\deg(K)}\lvert\sigma_{i}(b)\rvert}$ where
$\sigma_{1},\dots,\sigma_{\deg(K)}$ are the embeddings of $K$ into
$\mathbb{C}$. The proof of Theorem 1.2 is concluded by the following lemma.

######  Lemma 3.3.

Let $T$ be a set in $\mathcal{O}_{K}$ defined by congruence conditions at a
finite set of primes $S$. Let
$c_{1},\dots,c_{k},d_{1},\dots,d_{k}\in\mathcal{O}_{K}$ and put
$A(x)=\prod_{i=1}^{k}(c_{i}x+d_{i})$. Suppose that $A(x)$ is squarefree away
from $S$. For $\mathfrak{p}\neq\mathfrak{a}$, let

$a(\mathfrak{p})=\frac{\\#\\{b\in\mathcal{O}_{K}/\mathfrak{p}^{2}\;\mid\;A(b)\not\equiv
0\pmod{\mathfrak{p}^{2}}\\}}{\\#(\mathcal{O}_{K}/\mathfrak{p}^{2})}$

Then

$\lim_{N\rightarrow\infty}\frac{\\#\\{b\in T\mid\lvert b\rvert\leq N\text{ and
}A(b)\text{ is squarefree away from }S\\}}{\\#\\{b\in T\mid\lvert b\rvert\leq
N\\}}=\prod_{\mathfrak{p}\notin S}a(\mathfrak{p}).$

If $a(\mathfrak{p})>0$ for all $\mathfrak{p}\notin S$, then
$\prod_{\mathfrak{p}\neq\mathfrak{a}}a(\mathfrak{p})>0$.

###### Proof.

If $a(\mathfrak{p})>0$ for all $\mathfrak{p}\notin S$, then there exists a $k$
such that:

$\prod_{\mathfrak{p}\notin S}a(\mathfrak{p})\geq\prod_{\mathfrak{p}\notin
S}\bigg{(}1-\frac{k}{N_{K/\mathbb{Q}}(\mathfrak{p})^{2}}\bigg{)}>0,$

which proves the second claim. To prove the first claim it suffices to prove
the following tail estimate:

$\\#\\{b\in T\mid\lvert b\rvert\leq N,\exists\;\mathfrak{p}\text{ s.t.
}\operatorname{Nm}(\mathfrak{p})>\sqrt{N}\text{ and }A(b)\equiv
0\pmod{\mathfrak{p}^{2}}\\}=o(N^{\deg(K)}).$

We upper bound the tail estimate by the sum of the following expressions:

$\sum_{i=1}^{k}\sum_{\operatorname{Nm}(\mathfrak{p})>\sqrt{N}}\\#\\{b\in
T\mid\lvert b\rvert\leq N\text{ and }c_{i}b+d_{i}\equiv
0\pmod{\mathfrak{p}^{2}}\\}$ $\sum_{1\leq i<j\leq
k}^{k}\sum_{\operatorname{Nm}(\mathfrak{p})>\sqrt{N}}\\#\\{b\in T\mid\lvert
b\rvert\leq N\text{ and }c_{i}b+d_{i}\equiv c_{j}b+d_{j}\equiv
0\pmod{\mathfrak{p}}\\}$

Both sums are upper bounded by $O(N^{\deg(K)-1/2})$, completing the proof. ∎

## References