# On property-$\bm{(R_{1})}$ and relative Chebyshev centers in Banach spaces-
II

Syamantak Das  and  Tanmoy Paul  Dept. of Mathematics
Indian Institute of Technology Hyderabad
India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We continue to study (strong) property-$(R_{1})$ in Banach spaces. As
discussed in [4], this study corresponds to a triplet $(X,V,\mathcal{F})$,
where $X$ is a Banach space, $V$ is a closed convex subset, and $\mathcal{F}$
is a subfamily of closed, bounded subsets of $X$. It is observed that if $X$
is a Lindenstrauss space then $(X,B_{X},\mathcal{K}(X))$ has strong
property-$(R_{1})$, $\mathcal{K}(X)$ represents the compact subsets of $X$.
This concludes that for any $F\in\mathcal{K}(X)$,
$\textrm{Cent}_{B_{X}}(F)\neq\emptyset$. This extends the well-known fact that
a compact subset of a Lindenstrauss space $X$ admits a nonempty Chebyshev
center in $X$. In this connection we extend our observation that
$\textrm{Cent}_{B_{X}}$ is Lipschitz continuous in $\mathcal{K}(X)$ if $X$ in
a Lindenstrauss space. It is observed that if $P$ is a bi-contractive
projection in $\ell_{\infty}$, then
$(\ell_{\infty},Range(P),\mathcal{K}(\ell_{\infty}))$ exhibits the strong
property-$(R_{1})$. Besides, if $V=span\\{(1,1\ldots,1\ldots)\\}$, then
$(\ell_{\infty},V,\mathcal{K}(\ell_{\infty}))$ exhibits the strong
property-$(R_{1})$, where $\mathcal{K}(\ell_{\infty})$ represents the set of
all compact subsets of $\ell_{\infty}$. Furthermore, stability results for
these properties are derived in continuous function spaces, which are then
studied for locally constrained subspaces with respect to a subspace of a
Banach space.

###### Key words and phrases:

Chebyshev center, restricted Chebyshev center, Property-$(R_{1})$,
Lindenstrauss space, $M$-ideal.

###### 2000 Mathematics Subject Classification:

Primary 46B20, 41A65

##  1\. Introduction

###  1.1. Prerequisites:

Some standard notations used in this study are introduced as follows: $X$
indicates a Banach space, whereas a subspace denotes a closed linear subspace.
For $x\in X$ and $r>0$, $B(x,r)$ and $B[x,r]$ denote the open and closed balls
with the center at $x$ and radius $r$, respectively. Furthermore, $B_{X}$ and
$S_{X}$ denote the closed unit ball and unit sphere of $X$, respectively.
Further, $\mathcal{B}(X),\mathcal{C}(X),\mathcal{K}(X),\mathcal{F}(X)$ denote
the set of all closed and bounded, closed and convex, compact, and finite
subsets of $X$, respectively. Real numbers are assumed to be the underlying
field for all spaces. For $x\in X,V\in\mathcal{C}(X)$, and
$B\in\mathcal{B}(X)$, the following are defined:

######  Notation.

  * (1)

$r(x,B)=\sup\\{\|x-b\|:b\in B\\}$

  * (2)

$\emph{rad}_{V}(B)=\inf\\{r(x,B):x\in V\\}$

  * (3)

$\emph{Cent}_{V}(B)=\\{v\in V:r(v,B)=\emph{rad}_{V}(B)\\}$

  * (4)

$\delta-\emph{Cent}_{V}(B)=\\{v\in V:r(v,B)\leq\emph{rad}_{V}(B)+\delta\\}$

  * (5)

For $B\subseteq X$, $B_{\eta}=\\{x\in X:d(x,B)\leq\eta\\}$ _for_ $\eta>0$.

  * (6)

$S_{\eta}(B)=\\{x\in X:r(x,B)\leq\eta\\}$ _for_ $\eta>0$.

Note that,

  * $($1$)$

$\textrm{Cent}_{V}(B)=\left\\{\bigcap\limits_{b\in
B}B[b,\textrm{rad}_{V}(B)]\right\\}\cap V$.

  * $($2$)$

$\delta-\textrm{Cent}_{V}(B)=\left\\{\bigcap\limits_{b\in
B}B[b,\textrm{rad}_{V}(B)+\delta]\right\\}\cap V$.

However, if $V\in\mathcal{C}(X)$ and $B\in\mathcal{B}(X)$, the set
$\textrm{Cent}_{V}(B)$ may be empty, although the set
$\delta-\textrm{Cent}_{V}(B)$ is nonempty for any $\delta>0$. A triplet
$(X,V,\mathcal{F})$, where $V\in\mathcal{C}(X)$ and
$\mathcal{F}(\subseteq\mathcal{B}(X))$, a subfamily of closed bounded subsets,
exhibits the restricted center property (rcp in short) if for all
$F\in\mathcal{F},\textrm{Cent}_{V}(F)\neq\emptyset$. Here,
$\textrm{rad}_{V}(F)$ represents the radius of the smallest ball (if it
exists) in $X$ centered at $V$ and containing $F$, $\textrm{Cent}_{V}(F)$
represents the possible points of the centers of these balls, and
$\delta-\textrm{Cent}_{V}(F)$ represents the intersection of $\delta$
perturbation of the radii of these balls. Authors in [3, 8, 11, 15]
investigated various properties (stated in the subsequent discussions) related
to the characters defined above, viz.
$\textrm{Cent}_{V}(F),\textrm{rad}_{V}(F)$, determined by various geometric
properties of the Banach space and also the type of the closed convex subset
$V$.

######  Definition 1.1.

[11] Let $V\in\mathcal{C}(X)$ and $\mathcal{F}\subseteq\mathcal{B}(X)$. We
call the triplet $(X,V,\mathcal{F})$ exhibits property-$(R_{1})$ if for $v\in
V,F\in\mathcal{F},r_{1},r_{2}>0$ the conditions $r(v,F)\leq r_{1}+r_{2}$ and
$S_{r_{2}}(F)\cap V\neq\emptyset$ imply that $V\cap B[v,r_{1}+\varepsilon]\cap
S_{r_{2}+\varepsilon}(F)\neq\emptyset$, for all $\varepsilon>0$.

Equivalently (see [3, Theorem 2.2, 2.4]), the triplet $(X,V,\mathcal{F})$
exhibits property-$(R_{1})$ if for $v\in V,F\in\mathcal{F},r_{1},r_{2}>0$ the
conditions $r(v,F)<r_{1}+r_{2}$ and $S_{r_{2}}(F)\cap V\neq\emptyset$ imply
that $V\cap B[v,r_{1}]\cap S_{r_{2}}(F)\neq\emptyset$.

The above is a set-valued analogue of the $1\frac{1}{2}$-ball property ([16])
and clearly $V$ has $1\frac{1}{2}$-ball property in $X$ if $(X,V,\mathcal{F})$
has property-$(R_{1})$ and $\mathcal{F}$ contains the singletons. A previous
study [11] reported that if $(X,V,\mathcal{F})$ exhibits the
property-$(R_{1})$, then it has rcp. However, [7] reported that the
$1\frac{1}{2}$-ball property is insufficient to ensure rcp for finite subsets.
In this context we recall [14, Proposition 2.2]. It is observed for a family
of bounded subsets $\mathcal{F}$, if for all $F\in\mathcal{F}$,
$\textrm{Cent}_{B_{Y}}(F)\neq\emptyset$ for a subspace $Y$ then for all
$F\in\mathcal{F}$ $\textrm{Cent}_{Y}(F)\neq\emptyset$. This concludes if
$(X,B_{X},\mathcal{F})$ exhibits rcp then $\textrm{Cent}_{X}(F)\neq\emptyset$
for all $F\in\mathcal{F}$.

Several characterization for the property-$(R_{1})$ are derived in [3]. It is
clear that for an arbitrary $V\in\mathcal{C}(X)$ and $v\in V$,
$r(v,F)\leq\textrm{rad}_{V}(F)+d(v,\textrm{Cent}_{V}(F))$.

######  Theorem 1.2.

[3, Theorem 2.4] Let $V$ be a closed convex subset of $X$. Then the triplet
$(X,V,\mathcal{F})$ exhibits property-$(R_{1})$ if and only if for $v\in V$
and $F\in\mathcal{F}$, $r(v,F)=\emph{rad}_{V}(F)+d(v,\emph{Cent}_{V}(F))$.

In [4], a stronger version of the property-$(R_{1})$: strong
property-$(R_{1})$, which is in fact a set valued version of strong
$1\frac{1}{2}$-ball property (see [7]), is studied.

######  Definition 1.3.

[4] Let $V\in\mathcal{C}(X)$ and $\mathcal{F}\subseteq\mathcal{B}(X)$. We call
the triplet $(X,V,\mathcal{F})$ exhibits strong property-$(R_{1})$ if for
$v\in V,F\in\mathcal{F},r_{1},r_{2}>0$ the conditions $r(v,F)\leq r_{1}+r_{2}$
and $S_{r_{2}}(F)\cap V\neq\emptyset$ imply that $V\cap B[v,r_{1}]\cap
S_{r_{2}}(F)\neq\emptyset$.

Several characterizations and examples of the strong property-$(R_{1})$ are
provided in [4]. Certain properties relevant to this study are listed.

######  Theorem 1.4.

Let $X$ be a Banach space, $V$ be a subspace of $X$, and $\mathcal{F}$ be a
subfamily of $\mathcal{B}(X)$. Then, the following are equivalent.

  * $(a)$

$(X,V,\mathcal{F})$ exhibits the strong property-$(R_{1})$.

  * $(b)$

$(X,V,\mathcal{F})$ exhibits the property-$(R_{1})$ and
$\forall~{}F\in\mathcal{F}$, $\emph{Cent}_{B_{V}}(F)\neq\emptyset$.

  * $(c)$

$\forall~{}v\in V$ and $F\in\mathcal{F}$, $r(v,F)=\emph{rad}_{V}(F)+\|v-z\|$,
for certain $z\in\emph{Cent}_{V}(F)$.

In all the above characterisations for (strong) property-$(R_{1})$ one can
choose $v=0$. [3] encountered that the space $C(K)$ where $K$ is a compact
Hausdorff space yields plenty of subspaces which satisfy (strong)
property-$(R_{1})$. In general, an $M$-ideal (see [6, Pg.1]) in a
Lindenstrauss space comes under the category of such subspaces (see [11]). For
instance, if $x^{*}$ is an extreme point of the dual unit ball of a
Lindenstrauss space $X$, then $\ker(x^{*})$ exhibits the strong
property-$(R_{1})$ for the set of all compact subsets of $X$. This study
examined the (strong) property-$(R_{1})$ and established various consequences,
stability properties, and examples thereof.

###  1.2. Summary of results:

The remainder of this paper is structured as follows.

Section 2 discusses several phenomena of the property-$(R_{1})$ and strong
property-$(R_{1})$. With respect to the finite subsets of a Lindenstrauss
space, the unit ball exhibits the strong property-$(R_{1})$. Moreover, if
$(X,Y,\mathcal{F}(X))$ exhibits the strong property-$(R_{1})$, then
$(X,B_{Y},\mathcal{F}(B_{X}))$ has property-$(P_{1})$. In addition, as
expected, the (strong) property-$(R_{1})$ is concluded to remain stable under
$c_{0}$ and $\ell_{\infty}$ sums.

A projection $P:X\rightarrow X$ such that $\|P\|\leq 1,\|I-P\|\leq 1$ is
referred to a bi-contractive projection on $X$. In Section 3, the range of any
bi-contractive projection in $\ell_{\infty}$ is derived as exhibiting the
strong property-$(R_{1})$ with respect to the compact subsets. Furthermore,
the $1$-dimensional subspace, $span\\{(1,1,\ldots,1,\ldots)\\}$ exhibits the
strong property-$(R_{1})$ with respect to the compact subsets of
$\ell_{\infty}$.

Section 4 demonstrates that the properties considered in this study remain
stable under continuous function spaces. For a compact Hausdorff space $K$,
$C(K,X)$ is considered to be the vector space of all continuous functions from
$K$ that take values in $X$. For an $f\in C(K,X)$, $\sup_{k\in K}\|f(k)\|$
defines a norm that makes the space complete. It is demonstrated, if
$(X,Y,\mathcal{F}(X))$ exhibits the property-$(R_{1})$, then
$(C(K,X),C(K,Y),\mathcal{F}(C(K,X)))$ has the property-$(R_{1})$ and vice
versa. This study adopted the technique used by Yost in [16, Theorem 2.1] to
confirm that a Weierstrass–Stone subspace of $C(K,X)$ exhibits the strong
property-$(R_{1})$ for the finite subsets. Strong property-$(R_{1})$ clearly
requires a routine verification for the subsequent results in [16].

In [10], Lindenstrauss provided an example of a Banach space $X$ and its
subspace $Y$ such that $Y$ is not one-complemented in $X$ although for any
subspace $Z,Y\subseteq Z\subseteq X$ $(Y\neq Z\neq X)$, $Y$ is one-
complemented in $Z$. We call a subspace constrained if it is in the range of a
norm-$1$ projection. Based on this example, Rao developed a local version of
constrained subspaces in Banach spaces: locally constrained subspace ([13]).
Let $Y,Z_{1}$ be subspaces of $X$ and let $Z_{2}=Y\cap Z_{1}$. Section 5
presents the results when $Z_{2}$ has the (strong) property-$(R_{1})$ in $Y$
and $Z_{2}$ is locally constrained with respect to $Z_{1}$ via $Y$ (Definition
5.1). In addition, the sufficient condition for a triplet
$(Y^{**},Y,\mathcal{B}(Y^{**}))$ exhibiting the property-$(R_{1})$ is derived,
where $Y$ is a subspace of $X$ and $(X^{**},X,\mathcal{B}(X^{**}))$ has the
property-$(R_{1})$.

##  2\. Various aspects of property-$(R_{1})$

The following theorem can be obtained from certain standard inequalities:
$|\textrm{rad}_{V}(F_{1})-\textrm{rad}_{V}(F_{2})|\leq d_{H}(F_{1},F_{2})$,
$|r(v_{1},F)-r(v_{2},F)|\leq\|v_{1}-v_{2}\|$, for $v_{1},v_{2}\in V$ and
$F_{1},F_{2},F\in\mathcal{B}(X)$. Here, $d_{H}$ represents the Hausdorff
metric defined over $\mathcal{B}(X)$.

######  Theorem 2.1.

Let $X$ be a Banach space and $V\in\mathcal{C}(X)$. If $(X,V,\mathcal{F}(X))$
exhibits the property-$(R_{1})$, then $(X,V,\mathcal{K}(X))$ exhibits the
property-$(R_{1})$.

Similar to Theorem 2.1, in certain cases, the strong property-$(R_{1})$ of a
$V\in\mathcal{C}(X)$ for finite subsets is sufficient for ensuring the same
for compact subsets.

######  Theorem 2.2.

Let $X$ be a Banach space and $V\in\mathcal{C}(X)$. Let $(X,\tau_{X})$ be a
lctvs, where $\tau_{X}$ represents a topology weaker than $(X,\|.\|)$. In
addition to that, we assume any (norm) bounded net in $X$ has a $\tau_{X}$
convergent subnet in $X$, and $\|.\|$ is lower semicontinuous in
$(X,\tau_{X})$. Then for a $\tau_{X}$-closed $V\in\mathcal{C}(X)$,
$(X,V,\mathcal{K}(X))$ exhibits the strong property-$(R_{1})$ whenever
$(X,V,\mathcal{F}(X))$ has the strong property-$(R_{1})$.

###### Proof.

Let $K\in\mathcal{K}(X)$. The aim is to prove that
$r(0,K)=\textrm{rad}_{V}(K)+\|z\|$, for certain $z\in\textrm{Cent}_{V}(K)$.

There exists a sequence $F_{n}\in\mathcal{F}(X)$ such that
$d_{H}(K,F_{n})\rightarrow 0$ as $n\rightarrow\infty$. Thus,
$r(0,F_{n})=\textrm{rad}_{V}(F_{n})+\|z_{n}\|$, where
$z_{n}\in\textrm{Cent}_{V}(F_{n})$ for all $n\in\mathbb{N}$. Because $(z_{n})$
is bounded in $X$, there exists a subnet $(z_{n_{i}})$ of $(z_{n})$ such that
$z_{n_{i}}\rightarrow z$ in $(X,\tau_{X})$ for certain $z\in V$. Hence,

$\displaystyle\|z\|\leq\liminf_{i}\|z_{n_{i}}\|=\liminf_{i}[r(0,F_{n_{i}})-\textrm{rad}_{V}(F_{n_{i}})]=r(0,K)-\textrm{rad}_{V}(K).$

Claim:  $r(z,K)=\textrm{rad}_{V}(K)$.

Because $z_{n_{i}}\rightarrow z$ in $(X,\tau_{X})$, the following is obtained.

$\displaystyle r(z,K)$ $\displaystyle\leq$
$\displaystyle\liminf_{i}r(z_{n_{i}},K)$ $\displaystyle\leq$
$\displaystyle\liminf_{i}[r(z_{n_{i}},F_{n_{i}})+d(F_{n_{i}},K)]$
$\displaystyle=$
$\displaystyle\liminf_{i}[\textrm{rad}_{V}(F_{n_{i}})+d(F_{n_{i}},K)]=\textrm{rad}_{V}(K)$

Hence, $r(z,K)=\textrm{rad}_{V}(K)$ and thus $z\in\textrm{Cent}_{V}(K)$.
Therefore, $d(0,\textrm{Cent}_{V}(K))\leq\|z\|\leq
r(0,K)-\textrm{rad}_{V}(K)$. The other inequality is evident. This completes
the proof. ∎

The following is obtained by applying Theorem 2.2.

######  Corollary 2.3.

Let $X$ be a dual (reflexive) Banach space and $V$ be a weak∗ (norm) closed
convex subset of $X$. If $(X,V,\mathcal{F}(X))$ exhibits the strong
property-$(R_{1})$, then $(X,V,\mathcal{K}(X))$ exhibits the strong
property-$(R_{1})$.

Let $V\in\mathcal{C}(X)$ be closed and convex. Let us recall the following.

######  Definition 2.4.

[11] For a subfamily $\mathcal{F}\subseteq\mathcal{B}(X)$, we call the triplet
$(X,V,\mathcal{F})$ exhibits property-$(P_{1})$ if for $\varepsilon>0$ and
$F\in\mathcal{F}$ there exists $\delta(\varepsilon,F)>0$ such that
$\delta-\textrm{Cent}_{V}(F)\subseteq\textrm{Cent}_{V}(F)+\varepsilon B_{X}$.

$(X,V,\mathcal{F})$ exhibits property-$(P_{1})$ if it has the
property-$(R_{1})$. In addition, a Banach space $X$ is considered a
Lindenstrauss space if $X^{*}$ is isometric with $L_{1}(\mu)$ for a certain
measure space $(\Omega,\Sigma,\mu)$. [9] is a standard reference for a
comprehensive study of Banach spaces of type Lindenstrauss spaces.
Furthermore, $X$ is a Lindenstrauss space if and only if for any finite family
of closed balls $\\{B_{i}\\}_{i=1}^{n}$ that are pairwise intersecting in $X$
actually intersects in $X$ ([9, Theorem 5.5, 6.1]). In this statement, $n$ can
be replaced by $4$. As discussed in Section 1, our next theorem extends the
fact that any finite subset of a Lindenstrauss space has a nonempty Chebyshev
center.

######  Theorem 2.5.

Let $X$ be a Lindenstrauss space. Then, $(X,B_{X},\mathcal{F}(X))$ has _rcp_.

###### Proof.

Let $F\in\mathcal{F}(X)$ be such that $F=\\{z_{1},z_{2},\ldots,z_{n}\\}$.
Consider the family of balls
$\mathcal{B}=\left\\{B_{X},B[z_{1},\textrm{rad}_{B_{X}}(F)],B[z_{2},\textrm{rad}_{B_{X}}(F)],\ldots,B[z_{n},\textrm{rad}_{B_{X}}(F)]\right\\}$.
Evidently, for any two $z_{i},z_{j}\in F$, $\|z_{i}-z_{j}\|\leq
2.\textrm{rad}_{B_{X}}(F)$.

Furthermore, for $z\in F$, $\|z-0\|\leq r(0,F)\leq r(s,F)+1$ for any $s\in
B_{X}$.

Hence, $\|z-0\|\leq\textrm{rad}_{B_{X}}(F)+1$ that follows
$B[z,\textrm{rad}_{B_{X}}(F)]\cap B_{X}\neq\emptyset$. Because $X$ is
$L_{1}$-predual, $\bigcap_{B\in\mathcal{B}}B\neq\emptyset$.

Hence, the result follows. ∎

######  Theorem 2.6.

Let $X$ be a Lindenstrauss space. Then, $(X,B_{X},\mathcal{F}(X))$ exhibits
property-$(P_{1})$.

###### Proof.

Let $F\in\mathcal{F}(X)$ be such that $F=\\{z_{1},\cdots,z_{n}\\}$ and
$\varepsilon>0$. Let $x\in\varepsilon-\textrm{Cent}_{B_{X}}(F)$.

Claim:  $x\in\textrm{Cent}_{B_{X}}(F)+\varepsilon B_{X}$.

Our assumption yields $\|x-z_{i}\|\leq\textrm{rad}_{B_{X}}(F)+\varepsilon$ for
all $i=1\cdots,n$. Thereafter, $B[x,\varepsilon]\cap
B[z_{i},\textrm{rad}_{B_{X}}(F)]\neq\emptyset$ for all $i=1\cdots,n$.
Additionally, $\|z_{i}-z_{j}\|\leq 2.\textrm{rad}_{B_{X}}(F)$. Further we have
$\|z_{i}-0\|\leq r(0,F)\leq r(s,F)+1$ for all $s\in B_{X}$ for all
$i=1,\cdots,n$. Hence, $\|z_{i}-0\|\leq\textrm{rad}_{B_{X}}(F)+1$. Since $X$
is a Lindenstrauss space, we have
$B[x,\varepsilon]\cap\left(\cap_{i=1}^{n}B[z_{i},\textrm{rad}_{B_{X}}(F)]\right)\cap
B_{X}\neq\emptyset$. Which in other words,
$B[x,\varepsilon]\cap\textrm{Cent}_{B_{X}}(F)\neq\emptyset$ and hence the
claim follows. ∎

######  Remark 2.7.

  * $(a)$

Let us recall that (see [3, Theorem 2.4]) the case $\delta=\varepsilon$ in the
definition of property-$(P_{1})$ ensures property-$(R_{1})$. Hence we have
$(X,B_{X},\mathcal{F}(X))$ has property-$(R_{1})$ when $X$ is a Lindenstrauss
space. In fact, we can draw a stronger conclusion, as derived in Theorem 2.8.

  * $(b)$

From [3, Theorem 2.5] it now follows that if $X$ is a Lindenstrauss space then
$\emph{Cent}_{B_{X}}:(\mathcal{F}(X),d_{H})\to(\mathcal{B}(X),d_{H})$ is
Lipschitz continuous.

######  Theorem 2.8.

Let $X$ be a Lindenstrauss space. Then, $(X,B_{X},\mathcal{F}(X))$ exhibits
strong property-$(R_{1})$.

###### Proof.

Let $x\in B_{X}$ and $F=\\{z_{1},z_{2},\ldots,z_{n}\\}\in\mathcal{F}(X)$.
Furthermore, let $r(x,F)\leq r_{1}+r_{2}$ and $S_{r_{2}}(F)\cap
B_{X}\neq\emptyset$ for certain $r_{1},r_{2}>0$.

The condition $r(x,F)\leq r_{1}+r_{2}$ ensures $B[x,r_{1}]\cap
B[z_{i},r_{2}]\neq\emptyset$ for $1\leq i\leq n$. In addition,
$S_{r_{2}}(F)\cap B_{X}\neq\emptyset$ ensures that $\|z_{i}-z_{j}\|\leq
2r_{2}$ for $i\neq j$ and $B[z_{i},r_{2}]\cap B_{X}\neq\emptyset$. However,
owing to $n.2.I.P.$ of $X$, $B_{X}\cap\bigcap_{i=1}^{n}B[z_{i},r_{2}]\cap
B[x,r_{1}]\neq\emptyset$ is obtained. This completes the proof. ∎

As discussed in Theorem 1.4, $\textrm{Cent}_{B_{Y}}(F)\neq\emptyset$ for
$F\in\mathcal{F}(X)$ if $(X,Y,\mathcal{F}(X))$ exhibits the strong
property-$(R_{1})$. In Theorem 2.9 it is derived that $B_{Y}$ admits
property-$(P_{1})$ for a suitable subfamily of finite subsets of $X$.

######  Theorem 2.9.

Let $(X,Y,\mathcal{F}(X))$ exhibits the strong property-$(R_{1})$. Then,
$(X,B_{Y},\mathcal{F}(B_{X}))$ has property-$(P_{1})$.

###### Proof.

We utilize the techniques used in [7, Theorem 2.9]. The detailed proof is
included for completeness.

Let $F\in\mathcal{F}(B_{X})$, $d=\textrm{rad}_{Y}(F)>0$, and $r(0,F)<1+d$.
Hence, there exists $\eta>0$ such that $r(0,F)-d=1-\eta$.

Considering $\varepsilon>0$, choose $0<\delta<1$ such that
$\delta+\frac{3\delta}{\delta+\eta}<\varepsilon$.

Now, $d=\textrm{rad}_{B_{Y}}(F)$ (see [4, Theorem 3.8]). Let
$y\in\delta-\textrm{Cent}_{B_{Y}}(F)$(that is,$\
r(y,F)<\textrm{rad}_{B_{Y}}(F)+\delta)$. Now, by the strong
property-$(R_{1})$,
$r(y,F)=\textrm{rad}_{Y}(F)+d(y,\textrm{Cent}_{Y}(F))=d+\underset{z\in\textrm{Cent}_{Y}(F)}{\inf}\|z-y\|$,
that is, $d(y,\textrm{Cent}_{Y}(F))=r(y,F)-d<\delta$. Hence, there exists $\
y_{0}\in\textrm{Cent}_{Y}(F)$ such that $\|y-y_{0}\|<\delta$. Clearly,
$\|y_{0}\|<\|y\|+\delta\leq 1+\delta$.

Claim: There exists $z\in\textrm{Cent}_{Y}(F)\cap B_{Y}$ such that
$\|y-z\|<\varepsilon$.

Now, $r(0,F)-d=1-\eta=d(0,\textrm{Cent}_{Y}(F))$ and there exists $\
z_{1}\in\textrm{Cent}_{Y}(F)$ with $\|z_{1}\|=1-\eta$.

Let $w_{\lambda}=\lambda y_{0}+(1-\lambda)z_{1}$ then
$\|w_{\lambda}\|\leq\lambda(1+\delta)+(1-\lambda)(1-\eta)=1+\delta\lambda-(1-\lambda)\eta$.

Now, $1+\delta\lambda-(1-\lambda)\eta=1\iff
1-\lambda=\frac{\delta}{\delta+\eta}\iff\lambda=\frac{\eta}{\delta+\eta}$.

Let $\lambda=\frac{\eta}{\delta+\eta}$ and $z=w_{\lambda}$, then
$0<\lambda<1$.

$\|y_{0}-z\|=(1-\lambda)\|y_{0}-z_{1}\|\leq\frac{3\delta}{\delta+\eta}$
because $\|y_{0}-z_{1}\|\leq 2+1=3$.

Additionally, $z\in\textrm{Cent}_{Y}(F)$ because $\textrm{Cent}_{Y}(F)$ is
convex and $\|z\|\leq 1+\delta\lambda-(1-\lambda)\eta=1$.

Hence, we have $z\in\textrm{Cent}_{B_{Y}}(F)$ and
$\|y-z\|\leq\|y-y_{0}\|+\|y_{0}-z\|<\delta+\frac{3\delta}{\delta+\eta}<\varepsilon$.
This proves the claim.

Hence,
$\delta-\textrm{Cent}_{B_{Y}}(F)\subseteq\textrm{Cent}_{B_{Y}}(F)+\varepsilon
B_{X}$. ∎

For a Banach space $X$ we introduce the following notations.

$X_{0}=\bigoplus_{c_{0}}X=\\{(x_{n}):x_{n}\in X,\lim_{n}\|x_{n}\|=0\\}$,

$X_{\infty}=\bigoplus_{\ell_{\infty}}X=\\{(x_{n}):x_{n}\in
X,\sup_{n}\|x_{n}\|<\infty\\}$

$X_{1}=\bigoplus_{\ell_{1}}X=\\{(x_{n}):x_{n}\in
X,\sum_{n}\|x_{n}\|<\infty\\}$

$\bigoplus_{\ell_{1}(m)}X=\oplus_{i=1}^{m}X$ with norm
$\sum_{i=1}^{m}\|x_{i}\|$ and

$\bigoplus_{\ell_{\infty}(m)}X=\oplus_{i=1}^{m}X$ with norm
$\max_{i=1}^{m}\|x_{i}\|$.

######  Theorem 2.10.

Let $Y$ be a subspace of $X$. Then, $(X,Y,\mathcal{F}(X))$ has the (strong)
property-$(R_{1})$ if and only if $(X_{0},Y_{0},\mathcal{F}(X_{0}))$ has the
(strong) property-$(R_{1})$.

###### Proof.

First, the result for the property-$(R_{1})$ is derived.

It is sufficient to prove that $(X_{0},Y_{0},\mathcal{F}(X_{0}))$ exhibits the
property-$(R_{1})$ if $(X,Y,\mathcal{F}(X))$ has the property-$(R_{1})$. Proof
of this fact is outlined below.

Let $F\in\mathcal{F}(X_{0})$ and $F(n)\subseteq X$ be the corresponding
component, that is, $F(n)\in\mathcal{F}(X)$. Suppose
$F=\\{x_{1},\ldots,x_{k}\\}$, $y_{0}=(y_{0}(n))\in Y_{0}$ and $r_{1},r_{2}>0$
be such that $S_{r_{2}}(F)\cap Y_{0}\neq\emptyset$ and
$r(y_{0},F)<r_{1}+r_{2}$.

Claim: $S_{r_{2}}(F)\cap B[y_{0},r_{1}]\cap Y_{0}\neq\emptyset$

It is clear that there exists $N$ sufficiently large, such that $0\in
S_{r_{2}}(F(n))\cap B[y_{0}(n),r_{1}]\cap Y$, for all $n>N$. Now, for $1\leq
n\leq N$, we choose $y(n)\in S_{r_{2}}(F(n))\cap B[y_{0}(n),r_{1}]\cap Y$. As
a result, an element $(y(n))$, a member of the set as specified in the claim
above, exists.

Nevertheless, regarding the remaining part, it is sufficient to prove that
$(X_{0},Y_{0},\mathcal{F}(X_{0}))$ exhibits a strong property-$(R_{1})$ if
$(X,Y,\mathcal{F}(X))$ exhibits a strong property-$(R_{1})$.

This study only shows that $\textrm{Cent}_{B_{Y_{0}}}(F)\neq\emptyset$, for
$F\in\mathcal{F}(X_{0})$.

Let $F\in\mathcal{F}(X_{0})$ and $F(n)\in\mathcal{F}(X)$ be as defined above.
Based on the assumption, $\textrm{Cent}_{B_{Y}}(F(n))\neq\emptyset$ for all
$n\in\mathbb{N}$. Let $y(n)\in\textrm{Cent}_{B_{Y}}(F(n))$ be obtained for all
$n\in\mathbb{N}$.

It is clear that, $y=(y(n))\in\bigoplus_{c_{0}}Y$, because
$\textrm{rad}_{B_{Y}}(F(n))\rightarrow 0$.

Since for any $z=(z(1),z(2),\cdots)\in\bigoplus_{c_{0}}B_{Y}$, $r(y,F)\leq
r(z,F)$ and because of $\|(y(n))\|_{\infty}\leq 1$,
$y\in\textrm{Cent}_{\bigoplus_{c_{0}}B_{Y}}(F)=\textrm{Cent}_{B_{Y_{0}}}(F)$,
the result follows from Theorem 1.4. ∎

A similar argument also yields the following.

######  Theorem 2.11.

Let $Y$ be a subspace of $X$. Then, $(X,Y,\mathcal{B}(X))$ exhibits the
(strong) property-$(R_{1})$ if and only if
$(X_{\infty},Y_{\infty},\mathcal{B}(X_{\infty}))$ exhibits the (strong)
property-$(R_{1})$.

######  Remark 2.12.

  * $(a)$

If $Y$ is a finite co-dimensional proximinal subspace of $c_{0}$, then there
exists $n\in\mathbb{N}$, such that $Y=F\oplus_{\infty}Z$, where $F$ is a
subspace of $\ell_{\infty}(n)$ and $Z=\\{(x_{i})\in c_{0}:x_{i}=0,1\leq i\leq
n\\}$. From Theorem 2.11, it is clear that $Y$ has property-$(R_{1})$ in
$c_{0}$ if and only if $F$ has property-$(R_{1})$ in $\ell_{\infty}(n)$. We do
not know the characterization of $\alpha_{i}\in\ell_{1}(n)$ where $1\leq i\leq
m$, considering $dim(c_{0}/Y)=m$, satisfying the condition that
$\cap_{i}\ker(\alpha_{i})$ has property-$(R_{1})$ in $\ell_{\infty}(n)$.

  * $(b)$

If $Y$ is a finite co-dimensional proximinal subspace of $c_{0}$, then from
the decomposition $Y=F\oplus_{\infty}Z$, it is also clear that $Y$ has the
property-$(R_{1})$ in $c_{0}$ if and only if $Y$ has the property-$(R_{1})$ in
$\ell_{\infty}$.

We now consider the derivation of the result for the $\ell_{1}$-sum. The
following proposition has been ignored in this study owing to its simple
proof.

######  Proposition 2.13.

Let $X$ be a Banach space and $Y_{1},Y_{2}$ be two subspaces of $X$. Then, for
$F_{1},F_{2}\in\mathcal{B}(X)$ the following can be obtained:

  * $(a)$

$r((y_{1},y_{2}),F_{1}\times F_{2})=r(y_{1},F_{1})+r(y_{2},F_{2})$ $\forall$
$(y_{1},y_{2})\in Y_{1}\oplus_{1}Y_{2}$.

  * $(b)$

$\emph{rad}_{Y_{1}\oplus_{1}Y_{2}}(F_{1}\times
F_{2})=\emph{rad}_{Y_{1}}(F_{1})+\emph{rad}_{Y_{2}}(F_{2})$.

  * $(c)$

$\emph{Cent}_{Y_{1}\oplus_{1}Y_{2}}(F_{1}\times
F_{2})=\emph{Cent}_{Y_{1}}(F_{1})\oplus_{1}\emph{Cent}_{Y_{2}}(F_{2})$.

  * $(d)$

$d(0,\emph{Cent}_{Y_{1}\oplus_{1}Y_{2}}(F_{1}\times
F_{2}))=d(0,\emph{Cent}_{Y_{1}}(F_{1}))+d(0,\emph{Cent}_{Y_{2}}(F_{2}))$.

Let $X$ be a Banach spaces. For a fixed $n$ define
$\mathcal{H}=\\{\Pi_{i=1}^{n}F_{i}:F_{i}\in\mathcal{F}(X)\\}$. The following
obtained as a consequence of Proposition 2.13.

######  Theorem 2.14.

Let $X$ be a Banach spaces and $Y$ be a subspace of $X$. Then
$(X,Y,\mathcal{F}(X))$ exhibits the property-$(R_{1})$ if and only if
$(\bigoplus_{\ell_{1}(n)}X,\bigoplus_{\ell_{1}(n)}Y,\mathcal{H})$ exhibits the
property-$(R_{1})$.

For a Banach space $X$, recall the notations defined before section 1.2. For
$F\in\mathcal{F}(X)$, we identify $F\times F\times\ldots\times
F(n-\mbox{times})$ with $\\{(x_{1},x_{2},\ldots x_{n},0,0\ldots):x_{i}\in
F\\}$. Let
$\mathfrak{F}=\\{\Pi_{i=1}^{n}F_{i}:F_{i}=F,F\in\mathcal{F}(X),n\in\mathbb{N}\\}$.
$\mathfrak{F}$ is now identified with a subfamily of $\mathcal{B}(X_{1})$,
more precisely a subfamily of $\mathcal{F}(X_{1})$.

######  Theorem 2.15.

Let $Y$ be a subspace of $X$. Then $(X,Y,\mathcal{F}(X))$ exhibits the
property-$(R_{1})$ if and only if $(X_{1},Y_{1},\mathfrak{F})$ exhibits the
property-$(R_{1})$.

###### Proof.

Here, proving that the condition is sufficient concludes the proof.

Let $\mathcal{W}\in\mathfrak{F}$, $y\in Y_{1}$, and $r_{1},r_{2}>0$ be such
that $r(y,\mathcal{W})<r_{1}+r_{2}$ and $S_{r_{2}}(\mathcal{W})\cap
Y_{1}\neq\emptyset$. Then clearly $\mathcal{W}=\Pi_{i=1}^{N}F_{i}$ and we get
a large $l\in\mathbb{N}(l>N)$ such that $\|y_{i}\|<r_{1}+r_{2}$, for all
$i\geq l$. Let
$\mathcal{W}_{l}=\\{(x_{1},x_{2},\ldots,x_{l}):\exists~{}(w_{i})\in\mathcal{W},x_{i}=w_{i},1\leq
i\leq l\\}$ and $\Lambda=(y_{1},y_{2},\ldots,y_{l})$.

It is clear that, $r(\Lambda,\mathcal{W}_{l})<r_{1}+r_{2}$ and
$S_{r_{2}}(\mathcal{W}_{l})\cap\bigoplus_{\ell_{1}(l)}Y\neq\emptyset$.

Now, from Theorem 2.14, $S_{r_{2}}(\mathcal{W}_{l})\cap
B[\Lambda,r_{1}]\cap\bigoplus_{\ell_{1}(l)}Y\neq\emptyset$ is obtained.

Let $(z_{1},\cdots,z_{l})$ be in the intersection above. Let
$z=(z_{1},\cdots,z_{l},0,\ldots)$. Then, $z\in S_{r_{2}}(\mathcal{W})\cap
B[y,r_{1}]\cap Y_{1}$. Hence, the conclusion follows. ∎

However, it is not known whether the $\ell_{1}$-sum remains stable for
$(X_{1},Y_{1},\mathcal{F}(X_{1}))$. We conclude this section by establishing a
sufficient condition for a triplet to possess property-$(R_{1})$.

######  Theorem 2.16.

Let $Y$ be a subspace of a Banach space $X$. If
$(X^{**},Y^{\perp\perp},\mathcal{F}(X^{**}))$ has property-$(R_{1})$ then
$(X,Y,\mathcal{F}(X))$ has property-$(R_{1})$.

###### Proof.

The proof is straightforward and follows from the extended version of
Principle of local reflexivity as stated in [1, Theorem 3.2]. ∎

##  3\. Examples from subspaces of $\ell_{\infty}$

It is well-known that in $\ell_{\infty}(2)$ the subspace $span\\{(1,1)\\}=Z$
(say) has $1\frac{1}{2}$-ball property. The simplest manner of observing this
is that for any $(x,y)\in\ell_{\infty}(2)$, $d((x,y),Z)=\frac{|x-y|}{2}$ and
the unique best approximation from $(x,y)$ to the subspace $Z$ is
$(\frac{x+y}{2},\frac{x+y}{2})$. The identity
$|x|\vee|y|=\frac{|x-y|}{2}+\frac{|x+y|}{2}$ for an arbitrary pair of real
numbers $x,y$ ensures the equivalent formulation ([5]) that characterizes
$1\frac{1}{2}$-ball property in a Banach space.

Consequently, using similar arguments it can be easily observed that
$span\\{(1,-1)\\}$ (=$W$ say) has $1\frac{1}{2}$-ball property in
$\ell_{\infty}(2)$. This study claims that both the subspaces viz. $Z,W$
exhibit the property-$(R_{1})$ for finite subsets of $\ell_{\infty}(2)$.

######  Proposition 3.1.

Let $X=\ell_{\infty}(2)$ and $Z=\textrm{span}\\{(1,1)\\}$. Then,
$(X,Z,\mathcal{F}(X))$ exhibits the strong property-$(R_{1})$ in $X$.

###### Proof.

Let $F\in\mathcal{F}(X)$. Furthermore, let $(z,z)\in Z$ and $r_{1},r_{2}>0$ be
such that $r((z,z),F)\leq r_{1}+r_{2}$ and $S_{r_{2}}(F)\cap Z\neq\emptyset$.

Claim: $B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z\neq\emptyset$.

If $card(F)=1$, then the claim follows from the fact that $Z$ has
$1\frac{1}{2}$-ball property in $X$. Suppose the assertion holds for
$card(F)=n$. The following proof is when $card(F)=n+1$. Assume that
$F=\\{(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{n+1},y_{n+1})\\}$.

Let $(p,p)\in S_{r_{2}}(F)\cap Z=\cap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z$. Because $Z$ exhibits the strong property-$(R_{1})$ when $F$ has $n$
elements, the following is obtained $(s_{i},s_{i})\in
B[(z,z),r_{1}]\cap\cap_{\begin{subarray}{c}{j=1}\\\ {j\neq
i}\end{subarray}}^{n+1}B[(x_{j},y_{j}),r_{2}]$ for certain
$(s_{i},s_{i})\in\ell_{\infty}(2)$ for $i=1,2,\ldots,n$. Here, $s_{1},s_{2}$
are chosen and the arguments are as stated below.

Case 1:  When $p\leq s_{1}\leq s_{2}$.

Then, $-r_{2}\leq p-x_{1}\leq s_{1}-x_{1}\leq s_{2}-x_{1}\leq r_{2}$ and
$-r_{2}\leq p-y_{1}\leq s_{1}-y_{1}\leq s_{2}-y_{1}\leq r_{2}$, and thus
$(s_{1},s_{1})\in B[(x_{1},y_{1}),r_{2}]$.

Thus, $(s_{1},s_{1})\in
B[(z,z),r_{1}]\cap\bigcap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z=B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z$.

Case 2:  When $p\leq s_{2}\leq s_{1}$.

Then $-r_{2}\leq p-x_{2}\leq s_{2}-x_{2}\leq s_{1}-x_{2}\leq r_{2}$ and
$-r_{2}\leq p-y_{2}\leq s_{2}-y_{2}\leq s_{1}-y_{2}\leq r_{2}$, and thus
$(s_{2},s_{2})\in B[(x_{2},y_{2}),r_{2}]$.

Thus, $(s_{2},s_{2})\in
B[(z,z),r_{1}]\cap\bigcap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z=B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z$.

Case 3:  When $s_{1}\leq s_{2}\leq p$.

Then $-r_{2}\leq s_{1}-x_{2}\leq s_{2}-x_{2}\leq p-x_{2}\leq r_{2}$ and
$-r_{2}\leq s_{1}-y_{2}\leq s_{2}-y_{2}\leq p-y_{2}\leq r_{2}$, and thus
$(s_{2},s_{2})\in B[(x_{2},y_{2}),r_{2}]$.

Thus, $(s_{2},s_{2})\in
B[(z,z),r_{1}]\cap\bigcap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z=B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z$.

Case 4:  When $s_{2}\leq s_{1}\leq p$.

Then $-r_{2}\leq s_{2}-x_{1}\leq s_{1}-x_{1}\leq p-x_{1}\leq r_{2}$ and
$-r_{2}\leq s_{2}-y_{1}\leq s_{1}-y_{1}\leq p-y_{1}\leq r_{2}$, and thus
$(s_{1},s_{1})\in B[(x_{1},y_{1}),r_{2}]$.

Thus, $(s_{1},s_{1})\in
B[(z,z),r_{1}]\cap\bigcap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z=B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z$.

Case 5:  When $s_{1}\leq p\leq s_{2}$.

Then $z\leq s_{1}+r_{1}\leq p+r_{1}$ and $p-r_{1}\leq s_{2}-r_{1}\leq z$, and
thus $|p-z|\leq r_{1}$.

Thus, $(p,p)\in
B[(z,z),r_{1}]\cap\bigcap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z=B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z$.

Case 6:  When $s_{2}\leq p\leq s_{1}$.

Then $z\leq s_{2}+r_{1}\leq p+r_{1}$ and $p-r_{1}\leq s_{1}-r_{1}\leq z$, and
thus $|p-z|\leq r_{1}$.

Thus, $(p,p)\in
B[(z,z),r_{1}]\cap\bigcap_{i=1}^{n+1}B[(x_{i},y_{i}),r_{2}]\cap
Z=B[(z,z),r_{1}]\cap S_{r_{2}}(F)\cap Z$.

Hence, the result follows when $card(F)=n+1$ and this completes the proof. ∎

Moreover, a similar conclusion can be derived for the subspace
$span\\{(1,-1)\\}$.

######  Proposition 3.2.

Let $X=\ell_{\infty}(2)$ and $W=\textrm{span}\\{(1,-1)\\}$. Then,
$(X,W,\mathcal{F}(X))$ has the strong property-$(R_{1})$ in $X$.

We now establish that the range of any bi-contractive projection in
$\ell_{\infty}$ exhibits the property-$(R_{1})$. Let us recall that, if
$P:\ell_{\infty}\rightarrow\ell_{\infty}$ is a bi-contractive projection, then
either $Px(n)=\frac{x(n)+x(\tau(n))}{2}$ or $Px(n)=\frac{x(n)-x(\tau(n))}{2}$
for $x\in\ell_{\infty}$ and $n\in\mathbb{N}$ (see [2, Theorem 3.9]). Here,
$\tau:\mathbb{N}\rightarrow\mathbb{N}$ is a permutation such that $\tau^{2}=I$
(identity).

Thus, if $P$ is a bi-contractive projection on $\ell_{\infty}$ and $\tau$ is
the corresponding permutation on $\mathbb{N}$, then, for $n\in\mathbb{N}$,
either $\tau(n)=n$ or $\tau(n)=m$ for certain $m\in\mathbb{N}$ and in the
second case $\tau(m)=n$. This concludes, for a bi-contractive projection
$P:\ell_{\infty}\rightarrow\ell_{\infty}$,

  * $($1$)$

When $\tau(n)=n$: either $Px(n)=x(n)$ or $Px(n)=0$.

  * $($2$)$

When $\tau(n)=m$ ($n\neq m$): either $(Px(n),Px(m))=\alpha(1,1)$ or
$(Px(n),Px(m))=\alpha(1,-1)$ for some scalar $\alpha$.

Hence, if $P\neq 0$ then $P(\ell_{\infty})$ is isometrically isomorphic with
either $\left(\Pi_{n\in
A}(\alpha_{n},\alpha_{n})\right)_{\infty}\oplus_{\infty}\left(\Pi_{m\in
B}(\beta_{m},-\beta_{m})\right)_{\infty}\oplus_{\infty}W$ or $\ell_{\infty}$.
Here, $W$ is an $M$-summand of $\ell_{\infty}$ and $A,B\subseteq\mathbb{N}$.
Based on [4, Theorem 3.5] $W$ exhibits the strong property-$(R_{1})$ in
$\ell_{\infty}$. Furthermore, the subspaces $\left(\Pi_{n\in
A}(\alpha_{n},\alpha_{n})\right)_{\infty}$ and $\left(\Pi_{m\in
B}(\beta_{m},-\beta_{m})\right)_{\infty}$ exhibit the strong
property-$(R_{1})$ in $\ell_{\infty}$ when considering Propositions 3.1, 3.2
and Theorem 2.11. Finally, according to Theorem 2.11, the subspace $range(P)$
has the strong property-$(R_{1})$ in $\ell_{\infty}$. Hence, the following is
obtained.

######  Theorem 3.3.

Let $P$ be a bi-contractive projection in $\ell_{\infty}$ and $Y=range(P)$.
Then, $(\ell_{\infty},Y,\mathcal{F}(\ell_{\infty}))$ has the strong
property-$(R_{1})$ in $\ell_{\infty}$.

We now consider the derivation for the subspace
$span\\{(1,1,\ldots,1,\ldots)\\}$, which exhibits the strong
property-$(R_{1})$ in $\ell_{\infty}$ for finite subsets.

######  Theorem 3.4.

Let $X=\ell_{\infty}$ and $Y=\textrm{span}\\{(1,1,\ldots)\\}$. Then,
$(X,Y,\mathcal{F}(X))$ exhibits the strong property-$(R_{1})$.

###### Proof.

Let $F\in\mathcal{F}(X)$, $(x,x,\ldots)\in Y$, $r_{1},r_{2}>0$ be such that
$r((x,x,\ldots),F)\leq r_{1}+r_{2}$ and $S_{r_{2}}(F)\cap Y\neq\emptyset$.

Claim: $B[(x,x,\ldots),r_{1}]\cap S_{r_{2}}(F)\cap Y\neq\emptyset$.

Step 1: Let $card(F)=1$. Let $F=\\{(x(1),x(2),\ldots)\\}$. Then, we have
$B[(x,x,\ldots),r_{1}]\cap B[(x(1),x(2),\ldots),r_{2}]\neq\emptyset$ and
$B[(x(1),x(2),\ldots),r_{2}]\cap Y\neq\emptyset$. Let $(z(1),z(2),\ldots)\in
B[(x,x,\ldots),r_{1}]\cap B[(x(1),x(2),\ldots),r_{2}]$ and $(y,y,\ldots)\in
B[(x(1),x(2),\ldots),r_{2}]$.

Let $\alpha=\underset{i\in\mathbb{N}}{\inf}z(i)$ and
$\beta=\underset{i\in\mathbb{N}}{\sup}z(i)$. Then, $\alpha\leq z(i)\leq\beta$
for all $i\in\mathbb{N}$.

Case 1: When $y\leq\alpha$.

Then, $y\leq\alpha\leq z(i)$ for all $i\in\mathbb{N}$. Then, $-r_{2}\leq
y-x(i)\leq\alpha-x(i)\leq z(i)-x(i)\leq r_{2}$ for all $i\in\mathbb{N}$.
Hence, $|\alpha-x(i)|\leq r_{2}$ for all $i\in\mathbb{N}$. Let
$\varepsilon>0$. Then, there exists $N\in\mathbb{N}$ such that
$z(N)-\varepsilon\leq\alpha$. Hence, $x-\varepsilon\leq
z(N)+r_{1}-\varepsilon\leq\alpha+r_{1}$ and $\alpha-r_{1}\leq z(N)-r_{1}\leq
x$. Thus, $|\alpha-x|\leq r_{1}+\varepsilon$.

Therefore, $(\alpha,\alpha,\ldots)\in B[(x,x,\ldots),r_{1}]\cap
B[(x(1),x(2),\ldots),r_{2}]$.

Case 2: When $\alpha\leq y\leq\beta$.

Let $\varepsilon>0$. There exists $N,N^{\prime}\in\mathbb{N}$ such that
$z(N)-\varepsilon\leq\alpha$ and $\beta\leq z(N^{\prime})+\varepsilon$.
Subsequently, $x-\varepsilon\leq z(N)-\varepsilon+r_{1}\leq\alpha+r_{1}\leq
y+r_{1}$ and $y-r_{1}\leq\beta-r_{1}\leq z(N^{\prime})+\varepsilon-r_{1}\leq
x+\varepsilon$. Then, $|y-x|\leq r_{1}+\varepsilon$.

Hence, $(y,y,\ldots)\in B[(x,x,\ldots),r_{1}]\cap
B[(x(1),x(2),\ldots),r_{2}]$.

Case 3: When $\beta\leq y$.

Then, $z(i)\leq\beta\leq y$ for all $i\in\mathbb{N}$. Furthermore, $-r_{2}\leq
z(i)-x(i)\leq\beta-x(i)\leq y-x(i)\leq r_{2}$ for all $i\in\mathbb{N}$. Hence,
$|\beta-x(i)|\leq r_{2}$ for all $i\in\mathbb{N}$. Let $\varepsilon>0$. Then,
there exists $N\in\mathbb{N}$ such that $\beta\leq z(N)+\varepsilon$. Hence,
$x\leq z(N)+r_{1}\leq\beta+r_{1}$ and $\beta-r_{1}\leq
z(N)-r_{1}+\varepsilon\leq x+\varepsilon$. Thus, $|\beta-x|\leq
r_{1}+\varepsilon$.

Thus, $(\beta,\beta,\ldots)\in B[(x,x,\ldots),r_{1}]\cap
B[(x(1),x(2),\ldots),r_{2}]$.

Hence, $(X,Y,\mathcal{F}(X))$ has the strong property-$(R_{1})$ when
$card(F)=1$.

Step 2: Suppose the assertion holds for all $F\in\mathcal{F}(X)$ when
$card(F)=n$. Let $card(F)=n+1$. Let $F=\\{x_{1},\ldots,x_{n+1}\\}$, where
$x_{i}=(x_{i}(1),x_{i}(2),\ldots)$ for $i=1,\ldots,n+1$. Then, we have
$B[(x,x,\ldots),r_{1}]\cap B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]\neq\emptyset$
for all $i=1,\ldots,n+1$ and
$\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]\cap Y\neq\emptyset$. Now,
because $(X,Y,\mathcal{F}(X))$ exhibits the property-$(R_{1})$ when
$card(F)=n$, the following is obtained:
$B[(x,x,\ldots),r_{1}]\cap\cap_{\begin{subarray}{c}{i=1}\\\ {i\neq
2}\end{subarray}}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]\cap Y\neq\emptyset$
and
$B[(x,x,\ldots),r_{1}]\cap\cap_{i=2}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]\cap
Y\neq\emptyset$. Let $(p,p,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{\begin{subarray}{c}{i=1}\\\ {i\neq
2}\end{subarray}}^{n+1}B[(x_{i}(1),x_{i}(2)\ldots),r_{2}]$, $(q,q,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=2}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$
and $(s,s,\ldots)\in\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Case 1: When $s\leq p\leq q$.

Then, $-r_{2}\leq s-x_{2}(i)\leq p-x_{2}(i)\leq q-x_{2}(i)\leq r_{2}$ for all
$i\in\mathbb{N}$. Thus, $(p,p,\ldots)\in B[(x_{2}(1),x_{2}(2),\ldots),r_{2}]$.

Hence, $(p,p,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Case 2: When $s\leq q\leq p$.

Then, $-r_{2}\leq s-x_{1}(i)\leq q-x_{1}(i)\leq p-x_{1}(i)\leq r_{2}$ for all
$i\in\mathbb{N}$. Thus, $(q,q,\ldots)\in B[(x_{1}(1),x_{1}(2),\ldots),r_{2}]$.

Hence, $(q,q,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Case 3: When $p\leq q\leq s$.

Then, $-r_{2}\leq p-x_{1}(i)\leq q-x_{1}(i)\leq s-x_{1}(i)\leq r_{2}$ for all
$i\in\mathbb{N}$. Thus, $(q,q,\ldots)\in B[(x_{1}(1),x_{1}(2),\ldots),r_{2}]$.

Hence, $(q,q,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Case 4: When $q\leq p\leq s$.

Then, $-r_{2}\leq q-x_{2}(i)\leq p-x_{2}(i)\leq s-x_{2}(i)\leq r_{2}$ for all
$i\in\mathbb{N}$. Thus, $(p,p,\ldots)\in B[(x_{2}(1),x_{2}(2),\ldots),r_{2}]$.

Hence, $(p,p,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Case 5: When $p\leq s\leq q$.

Then, $x\leq p+r_{1}\leq s+r_{1}$ and $s-r_{1}\leq q-r_{1}\leq x$. Thus,
$|s-x|\leq r_{1}$.

Hence, $(s,s,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Case 6: When $q\leq s\leq p$.

Then, $x\leq q+r_{1}\leq s+r_{1}$ and $s-r_{1}\leq p-r_{1}\leq x$. Thus,
$|s-x|\leq r_{1}$.

Hence, $(s,s,\ldots)\in
B[(x,x,\ldots),r_{1}]\cap\cap_{i=1}^{n+1}B[(x_{i}(1),x_{i}(2),\ldots),r_{2}]$.

Thus, the assertion holds for all $F\in\mathcal{F}(X)$. ∎

It is clear that the subspaces of type
$\left(\Pi_{n}(\alpha_{n},\alpha_{n})\right)_{\infty}$ or
$\left(\Pi_{m}(\beta_{m},-\beta_{m})\right)_{\infty}$ stated before Theorem
3.3, are $w^{*}$-closed, so is the subspace in Theorem 3.4. Hence, by
Corollary 2.3, the conclusions in Theorems 3.3 and 3.4 remain valid for
$\mathcal{K}(X)$.

##  4\. Subspaces of $C(K,X)$ with property-$(R_{1})$

In this section, the following fact is proven. By $K$ and $C(K,X)$ we denote a
compact Hausdorff space and the Banach space of $X$-valued continuous
functions over $K$ as discussed in section $1$.

######  Theorem 4.1.

Let $Y$ be a subspace of $X$. Then, $(X,Y,\mathcal{F}(X))$ has the
property-$(R_{1})$ if and only if $(C(K,X),C(K,Y),\mathcal{F}(C(K,X)))$ has
the property-$(R_{1})$.

Before proving Theorem 4.1, a few supporting results must be derived.

######  Proposition 4.2.

Let $f_{1},f_{2},\ldots,f_{k}\in C(K)$ and $\varepsilon>0$. Then, there exists
a finite family $(\varphi_{i})_{i=1}^{m}\subseteq C(K)$ where
$(\varphi_{i})_{i=1}^{m}$ forms a partition of unity and there exists
$h_{1},h_{2},\ldots,h_{k}\in span\\{\varphi_{i}:1\leq i\leq m\\}$ such that
$\|f_{i}-h_{i}\|_{\infty}<\varepsilon$ for $1\leq i\leq k$.

###### Proof.

Case 1: When $k=1$.

Let $\\{V_{i}:1\leq i\leq n\\}$ be a finite open cover of $K$ such that
$|f(z)-f(w)|<\varepsilon$ for $z,w\in V_{i}$ and $1\leq i\leq n$. Let
$(\varphi_{i})_{i=1}^{n}$ be a partition of unity such that
$0\leq\varphi_{i}\leq 1$, $1\leq i\leq n$ and $S(\varphi_{i})\subseteq V_{i}$
([12, Theorem 2.13]). Choose $v_{i}\in V_{i}$ and define
$h=\sum_{i=1}^{n}f(v_{i})\varphi_{i}$. Then,
$|f(x)-h(x)|=|\sum_{i=1}^{n}f(x)\varphi_{i}(x)-\sum_{i=1}^{n}f(v_{i})\varphi_{i}(x)|\leq\sum_{j}|\varphi_{i_{j}}(x)||f(x)-f(v_{i_{j}})|$.
The last sum is taken over all those $j$’s for which $x\in V_{i_{j}}$.
Clearly, $\sum_{j}\varphi_{i_{j}}(x)|f(x)-f(v_{i_{j}})|\leq\varepsilon$.

Case 2: When $k>1$.

Without loss of generality, this study assumes that $k=2$; no new ideas are
involved in other values of $k$.

Let $(\varphi_{i})_{i=1}^{n}$ and $(\varrho_{i})_{i=1}^{m}$ be two partitions
of unity in $C(K)$ such that there exists $h_{1},h_{2}$ where $h_{1}\in
span\\{\varphi_{i}:1\leq i\leq n\\}$ and $h_{2}\in span\\{\varrho_{i}:1\leq
i\leq m\\}$ where $\|f_{i}-h_{i}\|<\varepsilon$ for $i=1,2$. Then,
$\\{\varphi_{i}\varrho_{j}:1\leq i\leq n,1\leq j\leq m\\}$ is a partition of
unity and $h_{i}\in span\\{\varphi_{i}\varrho_{j}:1\leq i\leq n,1\leq j\leq
m\\}$. This completes the proof. ∎

Let $(\varphi_{i})_{i=1}^{N}$ be a finite partition of unity in $C(K)$
corresponding to an open cover $\mathcal{U}$ of $K$ obtained as in [12,
Theorem 2.3], then we call $(\varphi_{i})_{i=1}^{N}$ is a partition of unity
in $C(K)$ subordinate to the cover $\mathcal{U}$. In this case,
$(\varphi_{i})_{i=1}^{N}$ corresponds to a subspace $Z$ of $C(K,X)$:
$Z=\\{\sum_{i=1}^{N}x_{i}\varphi_{i}:x_{i}\in X,1\leq i\leq n\\}$. It is clear
that $Z\cong\bigoplus_{\ell_{\infty}(N)}X$.

######  Proposition 4.3.

Let $X$ be a Banach space and $K$ be a compact Hausdorff space and let
$f_{1},f_{2},\ldots,f_{n}\in C(K,X)$. Then, for $\varepsilon>0$, there exists
a subspace $Z$ of $C(K,X)$, where $Z\cong\bigoplus_{\ell_{\infty}(m)}X$ for
some $m$, and $d(f_{i},Z)\leq\varepsilon$, $1\leq i\leq n$.

###### Proof.

This study followed Proposition 4.2 to construct a subspace $Z\subseteq
C(K,X)$. If $S_{i}=f_{i}(K)$, then $S_{i}\subseteq X$ is a compact set. Let us
fix $i$. For every $s\in S_{i}$, let $B(s,\varepsilon)\cap S_{i}$ be a ball in
$S_{i}$. For a finite sub-cover $\mathcal{U}_{i}$ of
$\\{f_{i}^{-1}(B(s,\varepsilon)\cap S_{i}):s\in S_{i}\\}$, we may choose a
finite partition of unity subordinate to the cover $\mathcal{U}_{i}$; say,
$(\varphi_{j})$. If $(s_{j})_{j=1}^{n}\subseteq S_{i}$ is a finite set of
points corresponding to the cover $\mathcal{U}_{i}$, then
$\|f_{i}-\sum_{j}\varphi_{j}s_{j}\|\leq\varepsilon$.

We now continue the process for other values of $i$. Following the arguments
for $n$ functions as derived in Proposition 4.2, it is concluded for a finite
dimensional subspace $Z$ of $C(K,X)$ where
$Z\cong\bigoplus_{\ell_{\infty}(m)}X$, for some $m\in\mathbb{N}$. Clearly,
$d(f_{i},Z)\leq\varepsilon$, $1\leq i\leq n$, and this completes the proof. ∎

We are now ready to prove Theorem 4.1.

######  Theorem 4.4.

Let $Y$ be a subspace of $X$. Then, $(X,Y,\mathcal{F}(X))$ has the
property-$(R_{1})$ if and only if $(C(K,X),C(K,Y),\mathcal{F}(C(K,X)))$ has
the property-$(R_{1})$.

###### Proof.

Let $F\in\mathcal{F}(C(K,X))$ and $g\in C(K,Y)$. Let $r_{1},r_{2}>0$ such that
$S_{r_{2}}(F)\cap C(K,Y)\neq\emptyset$ and $r(g,F)\leq r_{1}+r_{2}$.
Consequently, the following is developed.

Claim: $S_{r_{2}+\varepsilon}(F)\cap B(g,r_{1}+\varepsilon)\cap
C(K,Y)\neq\emptyset$ for all $\varepsilon>0$.

Suppose that $h\in S_{r_{2}}(F)\cap C(K,Y)$. From Proposition 4.3 there exists
$Z\subseteq C(K,X)$ where $Z\cong\bigoplus_{\ell_{\infty}(k)}X$ for certain
$k$ such that for all $f\in F$, $d(f,Z)<\varepsilon$. Let
$F=\\{f_{1},f_{2},\ldots,f_{n}\\}$ and
$F^{\prime}=\\{f_{1}^{\prime},f_{2}^{\prime},\ldots,f_{n}^{\prime}\\}$ where
$f_{i}^{\prime}\in Z$ and $\|f_{i}-f_{i}^{\prime}\|<\varepsilon$.
Additionally, there exists $W\subseteq C(K,Y)$ where
$W\cong\bigoplus_{\ell_{\infty}(m)}Y$ such that there exist
$g^{\prime},h^{\prime}\in W$ and
$\|g-g^{\prime}\|<\varepsilon,\|h-h^{\prime}\|<\varepsilon$. Subsequently,
without a loss of generality, $m=k$ may be assumed. Furthermore, from the
assumption, $S_{r_{2}+2\varepsilon}(F^{\prime})\cap W\neq\emptyset$ and
$r(g^{\prime},F^{\prime})\leq r_{1}+r_{2}+2\varepsilon$ is obtained. In
addition, from Theorem 2.11
$(\bigoplus_{\ell_{\infty}(k)}X,\bigoplus_{\ell_{\infty}(k)}Y,\mathcal{F}(\bigoplus_{\ell_{\infty}(k)}X))$
with the property-$(R_{1})$ is obtained; hence,
$S_{r_{2}+2\varepsilon}(F^{\prime})\cap B(g^{\prime},r_{1}+2\varepsilon)\cap
W\neq\emptyset$.

Let $h$ be in the intersection of the abovementioned set. After identifying
$h$ with an element in $C(K,Y)$, we obtain $h\in S_{r_{2}+3\varepsilon}(F)\cap
B(g,r_{1}+3\varepsilon)\cap C(K,Y)$. Moreover, because $\varepsilon>0$ is
arbitrary, the proof follows. ∎

However, it is not yet known whether analogous results such as Theorem 4.4 are
true for the spaces of the form $L_{1}(\mu,X)$ or not. Nevertheless, if the
triplet $(L_{1}(\mu,X),L_{1}(\mu,Y),\mathcal{F}(L_{1}(\mu,X)))$ has the
property-$(R_{1})$, then $(X,Y,\mathcal{F}(X))$ has the property-$(R_{1})$.

######  Theorem 4.5.

Let $E$ be a real Lindenstrauss space, $K$ and $S$ be compact Hausdorff
spaces, and $\psi:K\rightarrow S$ be a continuous onto map. Let
$\psi^{*}:C(S,E)\rightarrow C(K,E)$ be the natural isometric embedding
expressed as $\psi^{*}f=f\circ\psi$. Then,
$(C(K,E),\psi^{*}C(S,E),\mathcal{F}(C(K,E)))$ exhibits the strong
property-$(R_{1})$.

###### Proof.

Let $F\in\mathcal{F}(C(K,E))$ such that $F=\\{f_{1},\ldots,f_{n}\\}$. Suppose
$r>0$ such that $r(0,F)\leq 1+r$ and
$S_{r}(F)\cap\psi^{*}C(S,E)\neq\emptyset$. Furthermore, define
$\eta:S\rightarrow\mathcal{B}(E)$ by

$\displaystyle\eta(y)$ $\displaystyle=$ $\displaystyle
B[0,1]\cap\left(\underset{k\in\psi^{-1}(y)}{\bigcap}\cap^{n}_{i=1}B[f_{i}(k),r]\right)$
$\displaystyle=$ $\displaystyle B[0,1]\cap\left(\cap^{n}_{i=1}\\{a\in
E:f_{i}(\psi^{-1}(y))\subseteq B[a,r]\\}\right).$

Each $\eta(y)$ is closed and convex.

Claim: $\eta(y)\neq\emptyset$ for all $y\in S$.

Let $\psi^{*}g\in\psi^{*}C(S,E)\cap
S_{r}(F)=\psi^{*}C(S,E)\cap\left(\cap^{n}_{i=1}B[f_{i},r]\right)$.

For  $k_{1},k_{2}\in\psi^{-1}(y)$, we have

$\displaystyle\|f_{i}(k_{1})-f_{j}(k_{2})\|$ $\displaystyle\leq$
$\displaystyle\|f_{i}(k_{1})-g(y)\|+\|f_{j}(k_{2})-g(y)\|$ $\displaystyle\leq$
$\displaystyle\|f_{i}-\psi^{*}g\|+\|f_{j}-\psi^{*}g\|\leq 2r$

for $i,j=1,\ldots,n$. Hence, $B[f_{i}(k_{1}),r]\cap
B[f_{j}(k_{2}),r]\neq\emptyset$. Because $\|f_{i}\|\leq r+1$, $B[0,1]\cap
B[f_{i}(k),r]\neq\emptyset$ for all $i=1,\ldots,n$. Thus, the entire family of
balls defining $\eta(y)$ has a pairwise non-empty intersection property.
Moreover, owing to the collection of centers
$\\{0\\}\cup\cup^{n}_{i=1}f_{i}(\psi^{-1}(y))$ being compact,
$\eta(y)\neq\emptyset$ is obtained.

Claim: $\eta$ is lower semicontinuous.

Let $G\subseteq E$ be open, and $y_{0}\in\\{y:\eta(y)\cap G\neq\emptyset\\}$
and $a\in\eta(y_{0})\cap G$. Then, $\|a\|\leq
1,f_{i}(\psi^{-1}(y_{0}))\subseteq B[a,r]$ and $B[a,\varepsilon]\subseteq G$
for certain $\varepsilon>0$ and for all $i=1,\ldots,n$. As $K$ is compact, the
map $y\rightarrow\psi^{-1}(y)$ is upper semicontinuous. Hence,
$N=\cap^{n}_{i=1}\\{y:f_{i}(\psi^{-1}(y))\subseteq\textrm{int}B[a,r+\varepsilon]\\}$
is an open set containing $y_{0}$. If $y\in N$, then $B[a,\varepsilon]\cap
B[f_{i}(k),r]\neq\emptyset$ for all $k\in\psi^{-1}(y)$ and for all
$i=1,\ldots,n$. Additionally, $B[a,\varepsilon]\cap B[0,1]\neq\emptyset$.
Furthermore, because $E$ is a real Lindenstrauss space, $\eta(y)\cap
B[a,\varepsilon]\neq\emptyset$ for all $y\in N$. Thus,
$N\subseteq\\{y:\eta(y)\cap G\neq\emptyset\\}$, where $\\{y:\eta(y)\cap
G\neq\emptyset\\}$ is open and $\eta$ is lower semicontinuous.

Now, applying Michael’s selection theorem, a continuous selection
$h:S\rightarrow E$ is obtained, such that $h(y)\in\eta(y)$ for all $y\in S$.
Accordingly, $\psi^{*}h\in\cap^{n}_{i=1}B[f_{i},r]\cap
B[0,1]\cap\psi^{*}C(S,E)$. This completes the proof. ∎

The following is obtained as a consequence of Theorem 4.5.

######  Corollary 4.6.

Let $K,S,E,\psi$ be as in Theorem 4.5. Furthermore, $y_{0}\in S$ is set and
let $M=\\{\psi^{*}f:f\in C(S,E)~{}\textrm{and}~{}f(y_{0})=0\\}$. Then,
$(C(K,E),M,\mathcal{F}(C(K,E)))$ has a strong property-$(R_{1})$.

###### Proof.

Let $f,r,\eta$ be similar to that in the proof of Theorem 4.5. If
$\psi^{*}g\in M\cap\cap^{n}_{i=1}B[f_{i},r]$,
$\|f_{i}(x)\|=\|f_{i}(x)-\psi^{*}g(x)\|\leq r$ for all $x\in\psi^{-1}(y_{0})$
and for all $i=1,\ldots,n$. Hence, $0\in\eta(y_{0})$. If we define
$\eta_{0}:S\rightarrow\mathcal{B}(E)$ by

$\eta_{0}(y)=\begin{cases}\eta(y)\quad&\text{if}\,y\neq y_{0}\\\
\\{0\\}\quad&\text{if}\,y=y_{0}\\\ \end{cases}$

This $\eta_{0}$ is clearly lower semicontinuous. Subsequently, on applying
Michael’s selection theorem, a continuous selection $h:S\rightarrow E$ is
obtained, such that $h(y)\in\eta_{0}(y)$ for all $y\in K$. Hence, the
assertion follows. ∎

##  5\. Property-$(R_{1})$ for constrained subspaces

Recall the following from [13].

######  Definition 5.1.

Let $Z_{1},Y$ be closed subspaces of $X$ and $Z_{2}=Z_{1}\cap Y$. $Z_{2}$ is
said to be locally constrained with respect to $Z_{1}$ via $Y$, provided for
all $z\in Z_{1}$, there are onto linear projections
$P:\textrm{span}\\{z,Z_{2}\\}\longrightarrow Z_{2}$ and
$Q:\textrm{span}\\{z,Y\\}\longrightarrow Y$ of norm one, such that
$P(z)=Q(z)$.

######  Remark 5.2.

Consider $Y\subseteq X$ as a constrained subspace with a projection $P$. Let
$Z_{1},Z_{2}$ be closed subspaces of X such that $P(Z_{1})\subseteq Z_{1}$ and
$Z_{2}=Z_{1}\cap Y$. Now, for any $z\in Z_{1}$,
$P|_{\textrm{span}\\{z,Z_{2}\\}}:\textrm{span}\\{z,Z_{2}\\}\longrightarrow
Z_{2},Q=P|_{\textrm{span}\\{z,Y\\}}:\textrm{span}\\{z,Y\\}\longrightarrow Y$
are projections of norm one and clearly $P(z)=Q(z)$.

######  Theorem 5.3.

Let $Z_{1},Y\subseteq X$ be closed subspaces of $X$ and $Z_{2}=Z_{1}\cap Y$.
Suppose that $Z_{2}$ is locally constrained with respect to $Z_{1}$ via $Y$.
Then, if $(X,Z_{1},\mathcal{B}(X))((X,B_{Z_{1}},\mathcal{B}(X)))$ has rcp,
then $(Y,Z_{2},\mathcal{B}(Y))((Y,B_{Z_{2}},\mathcal{B}(Y)))$ has rcp.

###### Proof.

Let $F\in\mathcal{B}(Y)$. As per the assumption, there exists a $z\in Z_{1}$
such that $r(z,F)=\text{rad}_{Z_{1}}(F)$. Let $P$ and $Q$ be the projections
for $z$ as provided in the definition for a locally constrained space. Then,
$P(z)\in Z_{2}$ and

$\displaystyle r(P(z),F)$ $\displaystyle=$ $\displaystyle\underset{y\in
F}{\sup}\|P(z)-y\|$ $\displaystyle=$ $\displaystyle\underset{y\in
F}{\sup}\|Q(z)-Q(y)\|$ $\displaystyle\leq$ $\displaystyle\underset{y\in
F}{\sup}\|z-y\|$ $\displaystyle=$ $\displaystyle\textrm{rad}_{Z_{1}}(F)$
$\displaystyle\leq$ $\displaystyle\textrm{rad}_{Z_{2}}(F).$

Thus, the proof for $B_{Z_{2}}$ is now clear and follows owing to
$\|P\|=\|Q\|=1$. This completes the proof. ∎

Rao observed that if $Z_{2}$ is locally constrained with respect to $Z_{1}$
via $Y$ and $Z_{1}$ has (strong) $1\frac{1}{2}$-ball property in $X$, then
$Z_{2}$ exhibits a (strong) $1\frac{1}{2}$-ball property in $Y$ ([13]). A
similar analog is derived for the (strong) property-$(R_{1})$.

######  Theorem 5.4.

Let $Z_{1},Y$ be subspaces of $X$ and $Z_{2}=Z_{1}\cap Y$. Suppose $Z_{2}$ is
locally constrained with respect to $Z_{1}$ via $Y$. If
$(X,Z_{1},\mathcal{B}(X))$ has the (strong) property-$(R_{1})$, then
$(Y,Z_{2},\mathcal{B}(Y))$ exhibits the (strong) property-$(R_{1})$.

Furthermore, we have the following.

######  Theorem 5.5.

Suppose the triplet $(X^{**},X,\mathcal{B}(X^{**}))$ has the
property-$(R_{1})$, and $Y$ is a closed subspace such that for all $x\notin
Y$, Y is constrained in $\textrm{span}\\{x,Y\\}$. Then,
$(Y^{**},Y,\mathcal{B}(Y^{**}))$ has the property-$(R_{1})$.

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