# Uniqueness of Hahn–Banach extensions and some of its variants

Soumitra Daptari  and  Tanmoy Paul  Dept. of Mathematics
Indian Institute of Technology Hyderabad
, India<EMAIL_ADDRESS>&<EMAIL_ADDRESS>
###### Abstract.

In this study, we analyze the various strengthening and weakening of the
uniqueness of the Hahn–Banach extension. In addition, we consider the case in
which $Y$ is an ideal of $X$. In this context, we study the
property-$(U)/(SU)/(HB)$ and property-$(k-U)$ for a subspace $Y$ of a Banach
space $X$. We obtain various new characterizations of these properties. We
discuss various examples in the classical Banach spaces, where the
aforementioned properties are satisfied and where they fail. It is observed
that a hyperplane in $c_{0}$ has property-$(HB)$ if and only if it is an
$M$-summand. Considering $X,Z$ as Banach spaces and $Y$ as a subspace of $Z$,
by identifying $(X\widehat{\otimes}_{\pi}Y)^{*}\cong\mathcal{L}(X,Y^{*})$, we
observe that an isometry in $\mathcal{L}(X,Y^{*})$ has a unique norm-
preserving extension over $(X\widehat{\otimes}_{\pi}Z)$ if $Y$ has
property-$(SU)$ in $Z$. It is observed that a finite dimensional subspace $Y$
of $c_{0}$ has property-$(k-U)$ in $c_{0}$, and if $Y$ is an ideal, then
$Y^{*}$ is a $k$-strictly convex subspace of $\ell_{1}$ for some natural $k$.

###### Key words and phrases:

property-$(wU)$, property-$(U)$, property-$(SU)$, property-$(HB)$,
$L_{1}$-predual, $M$-ideal, $k$-Chebyshev subspace

###### 2020 Mathematics Subject Classification:

Primary 46A22, 46B10, 46B25; Secondary 46B20, 46B22.

dedicated to the memory of Professor Ashoke K. Roy

##  1\. Introduction

The Hahn–Banach extension theorem is one of the most long-standing topics in
functional analysis. Many illustrious mathematicians including Lawrence
Narici, Edward Beckenstein, A. R. Halbrook, Mart Poldvere, Dirk Werner, and
Eve Oja have contributed to the understanding of this continuously-discussed
theorem. The reader can follow the link in [MO] to obtain an overview of the
importance of this theorem. This paper follows the roadmap of the unique
existence of norm-preserving extensions. For a subspace $Y$ of $X$, if the
norm-preserving extension over $X$ is unique for every $y^{*}\in Y^{*}$, the
subspace is said to have property-$(U)$ (see [P]). A slightly weaker property
than that stated above is given as follows.

We say $Y$ has property-$(wU)$ whenever $y^{*}\in Y^{*}$ satisfying
$\|y^{*}\|=y^{*}(y_{0})$ for some $y_{0}\in S_{Y}$ has a unique norm-
preserving extension to $X$. The property-$(wU)$ is known as the weak
Hahn–Banach extension property (see [LA]) in the literature; we maintain
similarities with this property while referring to the other properties along
with it. We continue to study the uniqueness of the Hahn–Banach extension. In
a recent paper ([DPR]), the property-$(SU)$ was extensively studied. A
potpourri presentation of the said property led us to further strengthen this
property. In this paper, we focus on property-$(wU)$, property-$(HB)$, and
property-$(k-U)$.

We now list a few notations and definitions that are used throughout this
manuscript.

###  1.1. Notations and Definitions

$X$ denotes a Banach space over real scalars. We consider $Y$ to be a closed
subspace of $X$. $B_{X}$ and $S_{X}$ represent the closed unit ball and the
unit sphere of $X$, respectively.

  1. (1)

$Y^{\\#}=\\{x^{*}\in X^{*}:\|x^{*}\|=\|x^{*}|_{Y}\|\\}$.

  2. (2)

$HB(y^{*})=\\{x^{*}\in X^{*}:\|x^{*}\|=\|y^{*}\|~{}\&~{}x^{*}|_{Y}=y^{*}\\}$,
for $y^{*}\in Y^{*}$.

  3. (3)

For $A\subseteq X$, $\dim A=\dim(\textrm{span}\\{A-a\\})$, where $a\in A$.

  4. (4)

For $x\in X\setminus\\{0\\}$, $S(X^{*},x)=\\{x^{*}\in
S_{X^{*}}:x^{*}(x)=\|x\|\\}$.

  5. (5)

$Y^{\perp}=\\{x^{*}\in X^{*}:x^{*}|_{Y}=0\\}$.

  6. (6)

$d(x,Y)=\inf_{y\in Y}\|x-y\|$ and $P_{Y}(x)=\\{y\in Y:\|x-y\|=d(x,Y)\\}$.

  7. (7)

For Banach spaces $X,Z$, $X\cong Z$ means that $X$ is isometrically isomorphic
to $Z$.

We now provide a few important definitions that are relevant to the central
theme of this investigation.

A projection $P:X\rightarrow X$ is a linear map with $\|P\|<\infty$, unless
otherwise stated, such that $P^{2}=P$. By a contractive projection $P$ on $X$,
we mean $P:X\rightarrow X$ such that $\|P\|=1$. A bi-contractive projection
$P:X\rightarrow X$ is a projection where $\|P\|=1=\|I-P\|$.

######  Definition 1.1.1.

A subspace $Y$ is called an ideal in $X$ if there exists a contractive
projection $P$ on $X^{*}$ with $\ker(P)=Y^{\perp}$.

A few stronger versions of property-$(U)$ are the following.

######  Definition 1.1.2.

Let $Y$ be a subspace of $X$, and suppose there exists a projection
$P:X^{*}\rightarrow Y^{\perp}$ with $\|P\|=1$. Let $G=(I-P)(X^{*})$.

  * $(a)$

[O] $Y$ is said to have property-$(SU)$ in $X$ if for each $x^{*}\in X^{*}$,

$\|x^{*}\|>\|y^{\\#}\|,$

$\text{ whenever }x^{*}=y^{\\#}+y^{\perp}\text{ with }y^{\\#}\in
G,y^{\perp}(\neq 0)\in Y^{\perp}.$

  * $(b)$

[H] $Y$ is said to have property-$(HB)$ in $X$ if for each $x^{*}\in X^{*}$,

$\|x^{*}\|>\|y^{\\#}\|~{}\&~{}\|x^{*}\|\geq\|y^{\perp}\|,$ $\text{ whenever
}x^{*}=y^{\\#}+y^{\perp}\text{ with }y^{\\#}\in G,y^{\perp}(\neq 0)\in
Y^{\perp}.$

######  Definition 1.1.3.

[HWW] A subspace $J$ is called an $M$-ideal in $X$ if there exists a subspace
$Z$ of $X^{*}$ such that $X^{*}=J^{\perp}\oplus_{\ell_{1}}Z$.

Every $M$-ideal has property-$(HB)$. A subspace $J$ is said to be semi
$M$-ideal if $X^{*}=J^{\perp}\oplus_{\ell_{1}}Z$ for some closed set $Z$ in
$X^{*}$. The obvious choice for $Z$ for the (semi) $M$-ideal is $J^{\\#}$. A
semi $M$-ideal with a linear $J^{\\#}$ is an $M$-ideal. $J$ is a (semi)
$M$-ideal if and only if it satisfies the $(2)3$-ball property in $X$ (see [L,
Theorem 6.9, 6.10]).

From [LH] and [HWW], we recall that a Banach space $X$ is called
$L_{1}$-predual if $X^{*}$ is isometrically isomorphic to $L_{1}(\mu)$ for
some measure space $(\Omega,\Sigma,\mu)$, and $X$ is called $M$-embedded if
its canonical image in $X^{**}$ is an $M$-ideal in $X^{**}$.

By a hyperplane in $X$, we mean a subspace of type $\\{x\in X:x^{*}(x)=0\\}$
for some $x^{*}\in X^{*}$. A finite co-dimensional subspace is a finite
intersection of the hyperplanes in $X$.

######  Definition 1.1.4.

[X] A subspace $Y$ is said to have property-$(k-U)$ in $X$, for some
$k\in\mathbb{N}$ if $\dim HB(y^{*})\leq k-1$ for all $y^{*}\in S_{Y^{*}}$.

Let us recall that $Y$ is said to be a proximinal subspace of $X$ if
$P_{Y}(x)\neq\emptyset$ for all $x$. Subspace $Y$ is said to be a Chebyshev
subspace if $P_{Y}(x)$ is a singleton for all $x\in X$. In general, the map
$P_{Y}:X\rightarrow 2^{Y}$ is closed convex valued and not necessarily linear
if $Y$ is Chebyshev. We call $P_{Y}$ the metric projection of $Y$.

######  Definition 1.1.5.

[S] Let $k\in\mathbb{N}$.

  * $(a)$

A proximinal subspace $Y$ of $X$ is said to be a $k$-Chebyshev subspace if
$0\leq\dim(P_{Y}(x))\leq k-1$ for all $x\in X$.

  * $(b)$

$X$ is said to be $k$-strictly convex if for any $k+1$ linearly independent
elements $x_{1},x_{2},\dots,x_{k+1}\in S_{X}$, we have
$\|\sum_{i=1}^{k+1}x_{i}\|<k+1$.

For a closed and bounded set $B$ of a normed space $X$, by $\textrm{ext}(B)$,
we denote the set of all extreme points of $B$.

###  1.2. A short background of this work

The smoothness of a (non-zero) vector $x$ in a Banach space is related to the
uniqueness of the functional in its state space $S(X^{*},x)$. Property-$(U)$
(and property-$(wU)$) extends this notion to subspaces. If there exists a
linear Hahn–Banach extension operator $S:Y^{*}\rightarrow X^{*}$ (i.e., $Y$ is
an ideal in $X$), property-$(U)$ ensures the uniqueness of the existence of
such an $S$ and also establishes an embedding of $Y^{*}$ into $X^{*}$; in this
case, $X^{*}=Y^{\\#}\bigoplus Y^{\perp}$, a vector space decomposition. Note
that property-$(HB)$ ensures that the corresponding projection is bi-
contractive. It is relevant that property-$(U)$ is not invariant under
isometry.

We also extend our study to cases in which a subspace $Y$ fails to have
property-$(U)$, but the dimension of the set $HB(y^{*})$ has a finite upper
bound for $y^{*}\in Y^{*}$, known as property-$(k-U)$, as defined in
Definition 1.1.4. If $Y$ is a subspace of $X$, then

$Y$ has property-$(U)$ in $X$ $\Longleftrightarrow$ $Y^{\perp}$ is a Chebyshev
subspace of $X^{*}$ [P].

$X^{*}$ is a strictly convex $\Longleftrightarrow$ every subspace of $X$ has
property-$(U)$ [T, F].

The $k$-valued analogues of these cases are as follows.

######  Theorem 1.2.1.

[X] Let $k\in\mathbb{N}$ and $Y$ be a subspace of $X$.

  * $(a)$

$Y$ has property-$(k-U)$ in $X$ if and only if $Y^{\perp}$ is a $k$-Chebyshev
subspace of $X^{*}$.

  * $(b)$

$X^{*}$ is $k$-strictly convex if and only if every subspace of $X$ has
property-$(k-U)$ in $X$.

In this case, (property-$(k-U)$), the corresponding notion of smoothness is
multismooth, as discussed in [LR]. It is clear from the definition that an $n$
co-dimensional subspace $Y$ has property-$((n+1)-U)$; consequently, the James
space has property-$(k-U)$ in its bi-dual. By contrast, no subspace of
$\ell_{\infty}/c_{0}$ (and similarly for $B(\ell_{2})/K(\ell_{2})$) has
property-$(k-U)$ in their respective superspaces; a fortiori, if $u$ is
unitary in a $C^{*}$-algebra $\mathcal{A}$, then
$\textrm{span}(S(\mathcal{A}^{*},u))=\mathcal{A}^{*}$.

###  1.3. Major outcomes of this work

We organize this work in the following manner.

In Section $2$, we discuss property-$(wU)$ in Banach spaces. In Theorem 2.1.1,
we provide new characterizations of property-$(wU)$ in Banach space. The
$3$-space problem for property-$(wU)$ is partially proven in this section.

Section 3 discusses the stability properties of various types of uniqueness of
the Hahn–Banach extensions stated earlier, namely property-$(SU),(HB)$. Some
new characterizations are given for these properties along with property-$(U)$
in Banach spaces in Theorems 3.1.1, 3.1.3, and 3.1.4. We partially solve the
3-space problem for property-$(HB)$. Some examples are given at the end,
establishing that property-$(wU)$($(SU)$) is strictly weaker than
property-$(U)$($(HB)$).

In Section 4, we discuss property-$(k-U)$ in Banach spaces. In Theorem 4.1.5,
it is characterized when a reflexive subspace $Y$ has property-$(k-U)$ in a
Banach space $X$. In Theorem 4.2.1, we discuss a class of finite-dimensional
subspaces of $L_{1}(\mu)$ that does not have property-$(k-U)$ for any $k$.
Some examples are also given with this property.

##  2\. On weakly Hahn–Banach smoothness

In this section, we study property-$(wU)$. We begin with some
characterizations of property-$(wU)$. Let us recall that for $x^{*}\in X^{*}$,
a hyperplane $(x^{*})^{-1}\left(0\right)$ is proximinal if and only if $x^{*}$
attains its norm in the unit sphere of $X$.

###  2.1. Characterizations

######  Theorem 2.1.1.

Let $Y$ be a subspace of the Banach space $X$. Then, the following are
equivalent.

  * $(a)$

$Y$ has property-$(wU)$ in $X$.

  * $(b)$

Every proximinal subspace $Z\subseteq Y$, $Y/Z$ has property-$(wU)$ in $X/Z$.

  * $(c)$

$P_{Y^{\perp}}(x^{*})=\\{0\\}$ for $x^{*}\in X^{*}$ satisfying
$x^{*}(y)=\|x^{*}\|$, for some $y\in S_{Y}$.

###### Proof.

$(a)\Rightarrow(b)$. Suppose $Y$ has property-$(wU)$ in $X$, and $Z$ is
proximinal in $Y$. Let $F\in(Y/Z)^{*}$ be a norm attaining functional. That
is, there exists $y+Z\in Y/Z,\|y+Z\|=1$ such that $F(y+Z)=\|F\|$. Considering
$F$ as an element in $Z^{\perp}_{Y^{*}}$, $y^{*}$, say, is clearly norm-
attaining at some $y+z\in S_{X}$, for some $z\in Z$. Now, $y^{*}$ has a unique
norm-preserving extension to $X$; let us call it $x^{*}$. Since $x^{*}\in
Z^{\perp}_{X^{*}}$, we may identify $x^{*}$ to some
$\widetilde{F}\in(X/Z)^{*}$ with
$\|F\|=\|y^{*}\|=\|x^{*}\|=\|\widetilde{F}\|$. To prove the uniqueness of
$\widetilde{F}$, let us assume $G\in(X/Z)^{*}$ such that $G|_{Y/Z}=F$.
Clearly, $G\in Z^{\perp}_{X^{*}}$, which we call $w^{*}$. Then,
$w^{*}|_{Y}=y^{*}$ and $\|w^{*}\|=\|y^{*}\|$. Hence, we have $w^{*}=x^{*}$,
which proves the necessity of the condition.

$(b)\Rightarrow(a)$. Let us assume that $Y$ does not have property-$(wU)$ in
$X$. Then, there exists $y^{*}\in S_{Y^{*}}$ such that $y^{*}$ attains its
norm on $B_{Y}$ and distinct $x_{1}^{*},x_{2}^{*}\in HB(y^{*})$. Let
$Z=\ker(y^{*})$; then, $Z$ is proximinal in $Y$. Suppose $y_{0}\in B_{Y}$ is
such that $y^{*}(y_{0})=\|y^{*}\|$, then
$\|y_{0}+Z\|=d(y_{0},Z)=|y^{*}(y_{0})|=1$. We can now identify
$y^{*},x_{1}^{*},x_{2}^{*}$ in the corresponding quotient spaces, i.e.,
$Z^{\perp}_{Y^{*}}$ and $Z^{\perp}_{X^{*}}$, respectively. If we define
$F=y^{*}$ and $G_{1},G_{2}$ as $x_{1}^{*},x_{2}^{*}$, respectively, then
clearly it is a contradiction to the assumption that $Y/Z$ has property-$(wU)$
in $X/Z$.

$(a)\Rightarrow(c).$ Let $0\neq y^{\perp}\in P_{Y^{\perp}}(x^{*})$ for some
$x^{*}\in X^{*}$, as stated in $(c)$. Then, $x^{*},x^{*}-y^{\perp}\in
HB(x^{*}|_{Y})$. This contradicts the fact that $Y$ has property-$(wU)$ in
$X$.

$(c)\Rightarrow(a).$ If $y^{*}\in S_{Y^{*}}$ such that $|y^{*}(y)|=1$, for
some $y\in S_{Y}$, then, any $x^{*}\in HB(y^{*})$, we have that $x^{*}(y)=1$.
Hence, for distinct $x_{1}^{*},x_{2}^{*}\in HB(y^{*})$, we obtain
$x_{1}^{*}-x_{2}^{*}\in P_{Y^{\perp}}(x_{1}^{*})$, which contradicts our
assumption on $(c)$. ∎

###  2.2. A variant of the 3-space problem

We now focus on the main result of this section. Let us recall the
characterization [LA, Theorem 2.2] for property-$(wU)$ of a subspace $Y$ in
$X$.

######  Theorem 2.2.1.

Let $Y,Z$ be the subspaces of $X$ such that $Z\subseteq Y\subseteq X$. If $Z$
is a semi-M-ideal in $X$ and $Y/Z$ has property-$(wU)$ in $X/Z$, then $Y$ has
property-$(wU)$ in $X$.

###### Proof.

Let us choose $x_{0}\in S_{X}$, $y_{0}\in S_{Y}$ and $\varepsilon>0$.

Claim:  $\exists~{}R\geq 1$ such that $Y\cap B(x_{0}+Ry_{0},R+\varepsilon)\cap
B(x_{0}-Ry_{0},R+\varepsilon)\neq\emptyset.$

Now, if $x_{0}\in Z$, then the aforementioned intersection holds true for any
$R>0$ and clearly contains $x_{0}$. Moreover, if $y_{0}\in Z$, then the
aforementioned intersection is non-empty for $R=1$ (see [LA, Pg.98]). Hence,
without loss of generality, we assume that $x_{0},y_{0}\notin Z$.

Define $p=\frac{1}{\|x_{0}+Z\|}$, $q=\frac{1}{\|y_{0}+Z\|}$ and choose
$\eta>0$ such that $\eta<\frac{\varepsilon p}{q+2p}$.

From [LA, Theorem 2.2], there exists $r\geq 1$ such that,

$Y/Z\cap B(px_{0}+Z+rqy_{0}+Z,r+\eta)\cap
B(px_{0}+Z-rqy_{0}+Z,r+\eta)\neq\phi.$

That is,

$Y/Z\cap B(px_{0}+rqy_{0}+Z,r+\eta)\cap B(px_{0}-rqy_{0}+Z,r+\eta)\neq\phi.$

Let $r^{\prime}>\max\\{\frac{p}{q},r\\}$. Since $r^{\prime}>r$, we have

$Y/Z\cap B(px_{0}+r^{\prime}qy_{0}+Z,r^{\prime}+\eta)\cap
B(px_{0}-r^{\prime}qy_{0}+Z,r^{\prime}+\eta)\neq\phi.$

Hence, $Y/Z\cap
B(x_{0}+\frac{r^{\prime}q}{p}y_{0}+Z,\frac{r^{\prime}}{p}+\frac{\eta}{p})\cap
B(x_{0}-\frac{r^{\prime}q}{p}y_{0}+Z,\frac{r^{\prime}}{p}+\frac{\eta}{p})\neq\phi$.

As $q\geq 1$, we have

$Y/Z\cap
B(x_{0}+\frac{r^{\prime}q}{p}y_{0}+Z,\frac{qr^{\prime}}{p}+\frac{q\eta}{p})\cap
B(x_{0}-\frac{r^{\prime}q}{p}y_{0}+Z,\frac{qr^{\prime}}{p}+\frac{q\eta}{p})\neq\phi$.

Let $y+Z\in Y/Z\cap
B(x_{0}+\frac{r^{\prime}q}{p}y_{0}+Z,\frac{qr^{\prime}}{p}+\frac{q\eta}{p})\cap
B(x_{0}-\frac{r^{\prime}q}{p}y_{0}+Z,\frac{qr^{\prime}}{p}+\frac{q\eta}{p})$,
i.e.,
$\|x_{0}\pm\frac{r^{\prime}q}{p}y_{0}-y+Z\|\leq\frac{qr^{\prime}}{p}+\frac{q\eta}{p}$.
Let $z_{i}\in Z$, $i=1,2,$ such that
$\|x_{0}\pm\frac{r^{\prime}q}{p}y_{0}-y+z_{i}\|\leq\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+\eta$.

Consider the balls
$B_{1}=B(x_{0}+\frac{r^{\prime}q}{p}y_{0}-y,\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+\eta)$
and

$B_{2}=B(x_{0}-\frac{r^{\prime}q}{p}y_{0}-y,\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+\eta)$.

Because $B_{1}\cap B_{2}\neq\phi$ and $B_{i}\cap Z\neq\phi$ $(i=1,2)$, from
the $2$-ball property of $Z$, we obtain

$\exists z\in Z\cap
B\left(x_{0}+\frac{r^{\prime}q}{p}y_{0}-y,\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+2\eta\right)\cap
B\left(x_{0}-\frac{r^{\prime}q}{p}y_{0}-y,\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+2\eta\right).$

Thus, $y+z\in Y\cap
B(x_{0}+\frac{r^{\prime}q}{p}y_{0},\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+2\eta)\cap
B(x_{0}-\frac{r^{\prime}q}{p}y_{0},\frac{qr^{\prime}}{p}+\frac{q\eta}{p}+2\eta)$.
As $\frac{r^{\prime}q}{p}>1$ and $\frac{q\eta}{p}+2\eta<\varepsilon$, the
claim follows for $R\geq\frac{r^{\prime}q}{p}$.

This completes the proof. ∎

We now discuss the cases of vector-valued functions. A contra-positive
argument leads to the following hypothesis. Let $(\Omega,\Sigma,\mu)$ be a
probability space and $Y$ be a subspace of $X$ such that $X^{*}$ has the
Radon–Nikod$\acute{y}$m property (see [DU, pg. 61] for the definition). We
denote this property by RNP in the remainder of this paper.

######  Theorem 2.2.2.

If $L_{p}(\mu,Y)$ has property-$(wU)$ in $L_{p}(\mu,X)$, then $Y$ has
property-$(wU)$ in $X$, where $1<p<\infty$.

Furthermore, we have the following.

######  Theorem 2.2.3.

Let $Y$ be a subspace of $X$, and let $X^{*}$ has the RNP. If $C(K,Y)$ has
property-$(wU)$ in $C(K,X)$, then $Y$ has property-$(wU)$ in $X$.

We do not know the validity of the converse of Theorem 2.2.2, 2.2.3.

##  3\. On property-$(SU)$ and property-$(HB)$

###  3.1. Characterizations

We begin this section with the following characterizations of property-$(U)$,
$(SU)$, and $(HB)$.

######  Theorem 3.1.1.

Let $Y$ be a subspace of $X$; then, the following are equivalent:

  * $(a)$

$Y$ has property-$(U)$ in $X$.

  * $(b)$

For every subspace $Z(\neq 0)$ of $Y$, $Y/Z$ has property-$(U)$ in $X/Z$.

  * $(c)$

For every hyperplane $Z$ in $Y$, $Y/Z$ has property-$(U)$ in $X/Z$.

###### Proof.

From [DPR, Theorem 4.1$(a)$], we have that $(a)\Rightarrow(b)$ and
$(b)\Rightarrow(c)$ are obvious. We only need to show that
$(c)\Rightarrow(a)$.

Let us assume that $Y$ does not have property-$(U)$ in $X$. Hence, there
exists $y^{*}\in Y^{*}$ and distinct $x_{1}^{*},x_{2}^{*}\in HB(y^{*})$. Then,
there exists $z\in X$ such that $x_{1}^{*}(z)\neq x_{2}^{*}(z)$. Define
$Z=\ker(y^{*})$, then $Z\subseteq Y$ is a hyperplane in $Y$ and
$(Y/Z)^{*}=\textrm{span}\\{y^{*}\\}$. The spanning vector $y^{*}$ possesses
two distinct norm-preserving extensions over $(X/Z)$. Hence, the result
follows. ∎

######  Remark 3.1.2.

  * $(a)$

Note that the condition $(c)$ in Theorem 3.1.1 can be restated as follows: for
a hyperplane $Z$ in $Y$ and $y\in Y\setminus Z$, $y+Z$ is a smooth point of
$X/Z$.

  * $(b)$

From the previous remark and Theorem 3.1.1 it follows that if $c_{0}\subseteq
Y\subseteq\ell_{\infty}$ where $\dim(Y/c_{0})=1$ then $Y$ can not have
property-$(U)$ in $\ell_{\infty}$.

Our next theorem characterizes property-$(SU)$ in Banach spaces.

######  Theorem 3.1.3.

Let $Y$ be a subspace of $X$: then, the following are equivalent:

  * $(a)$

$Y$ has property-$(SU)$ in $X$.

  * $(b)$

$Y$ has property-$(U)$ in $X$ and $Y$ is an ideal in $X$.

  * $(c)$

$Y^{\perp}$ is Chebyshev in $X^{*}$, and the metric projection
$P_{Y^{\perp}}:X^{*}\rightarrow Y^{\perp}$ is a linear projection.

###### Proof.

$(a)\Longleftrightarrow(b)$ follows from [O].

$(a)\Rightarrow(c)$. Let us recall from [O] that, $(a)$ holds if and only if
$Y^{\\#}$ is a linear subspace. Let $x_{1}^{*},x_{2}^{*}\in X^{*}$. It is
sufficient to show that
$P_{Y^{\perp}}(x_{1}^{*}+x_{2}^{*})=P_{Y^{\perp}}(x_{1}^{*})+P_{Y^{\perp}}(x_{2}^{*})$.
Let $y_{1}^{\perp},y_{2}^{\perp}\in Y^{\perp}$ such that
$P_{Y^{\perp}}(x_{1}^{*})=y_{1}^{\perp}$ and
$P_{Y^{\perp}}(x_{2}^{*})=y_{2}^{\perp}$. Consequently, for $i=1,2$
$\|x_{i}^{*}-y_{i}^{\perp}\|=d(x_{i}^{*},Y^{\perp})=\|x_{i}^{*}|_{Y}\|=\|(x_{i}-y_{i}^{\perp})|_{Y}\|$.
Thus, $x_{1}^{*}-y_{1}^{\perp},x_{2}^{*}-y_{2}^{\perp}\in Y^{\\#}$ and hence
by $(a)$, $x_{1}^{*}+x_{2}^{*}-y_{1}^{\perp}-y_{2}^{\perp}\in Y^{\\#}$ and
$\|x_{1}^{*}+x_{2}^{*}-y_{1}^{\perp}-y_{2}^{\perp}\|\leq\|x_{1}^{*}+x_{2}^{*}-y_{1}^{\perp}-y_{2}^{\perp}+y^{\perp}\|$
for all $y^{\perp}\in Y^{\perp}$. Hence,
$\|x_{1}^{*}+x_{2}^{*}-y_{1}^{\perp}-y_{2}^{\perp}\|\leq
d(x_{1}^{*}+x_{2}^{*}-y_{1}^{\perp}-y_{2}^{\perp},Y^{\perp})=d(x_{1}^{*}+x_{2}^{*},Y^{\perp})$.
Therefore,
$P_{Y^{\perp}}(x_{1}^{*}+x_{2}^{*})=y_{1}^{\perp}+y_{2}^{\perp}=P_{Y^{\perp}}(x_{1}^{*})+P_{Y^{\perp}}(x_{2}^{*})$.

To prove $(c)\Rightarrow(b)$, we show that $Y^{\perp}$ is the kernel of a
contraction. The proof of $\|I-P_{Y^{\perp}}\|=1$ remains. Let $x^{*}\in
X^{*}$ and $x^{*}=y^{\perp}+z^{*}$. Then,
$\|z^{*}\|=\|x^{*}-y^{\perp}\|=d(x^{*},Y^{\perp})\leq\|x^{*}\|$, and hence,
the result follows. ∎

Focusing on property-$(HB)$, we have the following.

######  Theorem 3.1.4.

Let $Y$ be a subspace of $X$. Then, the following are equivalent.

  * $(a)$

$Y$ has property-$(HB)$ in $X$.

  * $(b)$

$Y$ has property-$(U)$ in $X$, and there exists a projection
$P:X^{*}\rightarrow Y^{\perp}$ such that $\|P\|=1=\|I-P\|$.

  * $(c)$

$Y^{\perp}$ is Chebyshev in $X^{*}$, and the metric projection
$P_{Y^{\perp}}:X^{*}\rightarrow Y^{\perp}$ is a linear projection of norm-$1$.

###### Proof.

$(a)\Longleftrightarrow(b)$. Clearly, we have $(a)\Rightarrow(b)$. By
contrast, it is sufficient to show that $\|(I-P)(x^{*})\|<\|x^{*}\|$ whenever
$(I-P)(x^{*})\neq x^{*}$. Suppose that is not the case, then there exists
$x_{0}^{*}\in X^{*}$ such that $(I-P)(x_{0}^{*})\neq x_{0}^{*}$ and
$\|(I-P)(x_{0}^{*})\|=\|x_{0}^{*}\|$. Clearly,
$\|x_{0}^{*}|_{Y}\|=\|(I-P)x_{0}^{*}|_{Y}\|\leq\|(I-P)x_{0}^{*}\|$. Let
$\widetilde{x_{0}^{*}}$ be a norm-preserving extension of $x_{0}^{*}|_{Y}$.
Then $x_{0}^{*}-\widetilde{x_{0}^{*}}\in Y^{\perp}$; hence,
$(I-P)x_{0}^{*}=(I-P)\widetilde{x_{0}^{*}}$. Consequently,
$\|(I-P)x_{0}^{*}\|=\|(I-P)\widetilde{x_{0}^{*}}\|\leq\|\widetilde{x_{0}^{*}}\|=\|x_{0}^{*}|_{Y}\|$.
Thus $(I-P)(x_{0}^{*}),x_{0}^{*}\in HB(x_{0}^{*}|_{Y})$, which is a
contradiction.

$(c)\Rightarrow(b).$ If $(c)$ holds, then for any $x^{*}\in X^{*}$, there
exists $y^{\perp}\in Y^{\perp}$ and $y^{*}\in X^{*}$ such that
$x^{*}=y^{\perp}+y^{*}$. Now, we have
$\|y^{*}\|=\|x^{*}-y^{\perp}\|=d(x^{*},Y^{\perp})\leq\|x^{*}\|$. Hence, both
$P_{Y^{\perp}}$ and $I-P_{Y^{\perp}}$ are of norm-$1$ and $(b)$ follows.

$(a)\Rightarrow(c).$ Clearly, $Y$ has property-$(U)$ in $X$. Now, if $(a)$
holds, then $X^{*}=Y^{\perp}\bigoplus Y^{\\#}$. Now for every $x^{*}\in
X^{*}$, there exists a unique $y^{\perp}\in Y^{\perp}$ and $y^{\\#}\in
Y^{\\#}$ such that $x^{*}=y^{\perp}+y^{\\#}$ where
$\|y^{\perp}\|\leq\|x^{*}\|$. Hence,
$\|x^{*}-y^{\perp}\|=\|y^{\\#}\|=\|y^{\\#}|_{Y}\|=\|x^{*}|_{Y}\|=d(x^{*},Y^{\perp})$.
Therefore, $y^{\perp}=P_{Y^{\perp}}(x^{*})$, which completes the proof. ∎

Our next observation serves a sufficient condition for property-$(HB)$.

######  Proposition 3.1.5.

Let $Y$ be a subspace of $X$. If $Y^{\perp\perp}$ has property-$(HB)$ in
$X^{**}$, then $Y$ has property-$(HB)$ in $X$.

###### Proof.

Clearly, $Y$ has property-$(U)$ in $X$ if $Y^{\perp\perp}$ has property-$(U)$
in $X^{**}$. In fact, if $y^{*}\in Y^{*}$ and distinct $x_{1}^{*},x_{2}^{*}\in
HB(y^{*})$, then distinct $J_{X^{*}}(x_{1}^{*}),J_{X^{*}}(x_{2}^{*})\in
HB(J_{Y^{*}}(y^{*}))$, where $J_{X^{*}}:X^{*}\hookrightarrow X^{***}$ is the
canonical embedding. It has been shown in [DPR, pg. 11] that $Y$ is ideal in
$X$ if and only if $Y^{\perp\perp}$ is ideal in $X^{**}$. Hence, $Y$ has
property-$(SU)$ in $X$ if $Y^{\perp\perp}$ has property-$(HB)$ in $X^{**}$,
and we have the decomposition $X^{*}=Y^{\perp}\bigoplus Y^{\\#}$.

Owing to the property-$(HB)$ of $Y^{\perp\perp}$ in $X^{**}$, there exists
$Q:X^{***}\rightarrow X^{***}$, which is a bi-contractive projection with
$\ker(Q)=Y^{\perp\perp\perp}$. As $Y$ has property-$(SU)$ in $X$, there exists
a projection $P:X^{*}\rightarrow Y^{\\#}$ with $\|P\|=1$. Clearly,
$P^{**}:X^{***}\rightarrow Y^{***}$ is a contractive projection with
$\ker(P^{**})=Y^{\perp\perp\perp}$, which leads to $Q=P^{**}$.

Now, if there exists $x^{*}\in X^{*}$ such that $\|(I-P)x^{*}\|>\|x^{*}\|$
then $\|(I-Q)x^{*}\|=\|(I-P^{**})x^{*}\|>\|x^{*}\|$. Property-$(HB)$ of
$Y^{\perp\perp}$ in $X^{**}$ is contradicted. ∎

###  3.2. On subspaces of tensor product spaces

We now identify a few cases mainly in the spaces of vector valued functions
and tensor product spaces, where property-$(HB)$ remains stable. By
$(\Omega,\Sigma,\mu)$ we mean a probability space, and let $L_{p}(\mu,X)$ be
the Banach space consisting of all $p$th Bochner integrable functions (see
[DU] for details) as defined in Section 2. It is well known that
$L_{p}(\mu,X)^{*}\cong L_{q}(\mu,X^{*})$, $\frac{1}{p}+\frac{1}{q}=1$ if and
only if $X^{*}$ has the RNP.

A routine verification of the property-$(HB)$ in the proofs of [DPR, Theorem
3.4, 3.6] allows us to prove the following Theorems 3.2.1 and 3.2.2.

######  Theorem 3.2.1.

Let $Y$ be a subspace of $X$, and let $X^{*}$ has the RNP. Then,
$L_{p}(\mu,Y)$ has property-$(HB)$ in $L_{p}(\mu,X)$ if and only if $Y$ has
property-$(HB)$ in $X$, for $1<p<\infty$.

We denote the injective (projective) tensor product for Banach spaces $X$ and
$Z$ by $X\widehat{\otimes}_{\varepsilon}Z$ ($X\widehat{\otimes}_{\pi}Z$) (see
[R] for details). We denote the injective (projective) norm of $\varphi\in
X\widehat{\otimes}_{\varepsilon}Z$ ($\varphi\in X\widehat{\otimes}_{\pi}Z$) by
$\varepsilon(\varphi)$ ($\pi(\varphi)$). It is well known that if one of the
spaces $X^{*},Z^{*}$ has the RNP and one of them has the approximation
property, $\left(X\widehat{\otimes}_{\varepsilon}Z\right)^{*}\cong
X^{*}\widehat{\otimes}_{\pi}Z^{*}$ (see [R, Pg. 114]). In the case of the
projective tensor product, we have
$(X\widehat{\otimes}_{\pi}Z)^{*}\cong\mathcal{L}(X,Z^{*})$: the space of all
bounded linear operators from $X$ to $Z^{*}$ (see [R, Pg.24]). Let us also
recall that if $Y$ is a subspace of $Z$, then
$X\widehat{\otimes}_{\varepsilon}Y$ is a subspace of
$X\widehat{\otimes}_{\varepsilon}Z$; by contrast, $X\widehat{\otimes}_{\pi}Y$
is not necessarily a subspace of $X\widehat{\otimes}_{\pi}Z$ in general.
$X\widehat{\otimes}_{\pi}Y$ is a subspace of $X\widehat{\otimes}_{\pi}Z$ if
$Y$ is an ideal in $Z$ (see [RA, Theorem 1.(i)]).

A routine generalization for property-$(HB)$ of [DPR, Theorem 3.6] leads to
the following.

######  Theorem 3.2.2.

Let $X$ be a Banach space such that $X^{*}$ has the RNP. Let $E$ be an
$L_{1}$-predual space and let $Y$ be a subspace of $X$. Then,
$E\widehat{\otimes}_{\varepsilon}Y$ has property-$(HB)$ in
$E\widehat{\otimes}_{\varepsilon}X$ if and only if $Y$ has property-$(HB)$ in
$X$.

It is well known that $C(K,X)\cong C(K)\widehat{\otimes}_{\varepsilon}X$ for
any Banach space $X$.

######  Corollary 3.2.3.

Let $Y$ be a subspace of $X$, and let $X^{*}$ has the RNP. Then, $Y$ has
property-$(HB)$ in $X$ if and only if $C(K,Y)$ has property-$(HB)$ in
$C(K,X)$.

######  Remark 3.2.4.

As stated earlier, if $Y$ is an ideal of a Banach space $Z$, then the
projective tensor project $Y\widehat{\otimes}_{\pi}X$ is a subspace of
$Z\widehat{\otimes}_{\pi}X$. In addition, if $X$ is a Banach space, then
$L_{1}(\mu)\widehat{\otimes}_{\pi}X\cong L_{1}(\mu,X)$ (see [R, Example
2.19]). Along with [BR, Example 2.3], one can conclude that property-$(U)$ may
not be stable under a projective tensor product.

However, we can identify a class of functionals in
$(X\widehat{\otimes}_{\pi}Y)^{*}$ that still have unique norm-preserving
extensions over $X\widehat{\otimes}_{\pi}Z$.

######  Theorem 3.2.5.

Let $X$ and $Z$ be Banach spaces, and $Y$ be a subspace of $Z$. If $Y$ has
property-$(SU)$ in $Z$, then every isometry in
$\mathcal{L}(X,Y^{*})\cong(X\widehat{\otimes}_{\pi}Y)^{*}$ has a unique norm-
preserving extension to $(X\widehat{\otimes}_{\pi}Z)$.

###### Proof.

Let $S\in\mathcal{L}(X,Y^{*})$ be an isometry and $T_{1},T_{2}$ be two
distinct norm-preserving extensions of $S$ to $(X\widehat{\otimes}_{\pi}Z)$.
Let $x_{0}\in S_{X}$ such that $T_{1}x_{0}\neq T_{2}x_{0}$. As
$T_{1}|_{(X\widehat{\otimes}_{\pi}Y)}=S=T_{1}|_{(X\widehat{\otimes}_{\pi}Y)}$,
we have $T_{1}(x_{0}\otimes y)=S(x_{0}\otimes y)=T_{2}(x_{0}\otimes y)$, that
is, $T_{1}x_{0}(y)=Sx_{0}(y)=T_{2}x_{0}(y)$ for all $y\in Y$. Hence,
$T_{1}x_{0}$ and $T_{2}x_{0}$ are two linear extensions of $Sx_{0}$. Because
$S$ is an isometry, we have $\|Sx_{0}\|=\|x_{0}\|=1$. Now,
$\|S\|=\|T_{i}\|\geq\|T_{i}x_{0}\|\geq\|Sx_{0}\|=\|x_{0}\|=1=\|S\|$, $i=1,2$.
Consequently, $\|T_{1}x_{0}\|=\|T_{2}x_{0}\|=\|Sx_{0}\|$. Hence, $T_{1}x_{0}$
and $T_{2}x_{0}$ are two distinct norm-preserving extensions of $Sx_{0}$. ∎

We have an affirmative answer to the converse of Theorem 3.2.5.

######  Theorem 3.2.6.

Let $X$ and $Z$ be Banach spaces, and $Y$ be a subspace of $Z$ such that
$(X\widehat{\otimes}_{\pi}Y)$ is a subspace of $(X\widehat{\otimes}_{\pi}Z)$.
If $(X\widehat{\otimes}_{\pi}Y)$ has property-$(U)$ in
$(X\widehat{\otimes}_{\pi}Z)$, then $Y$ has property-$(U)$ in $Z$.

###### Proof.

Let $y^{*}\in S_{Y^{*}}$ and distinct $z_{1}^{*},z_{2}^{*}\in HB(y^{*})$.
Choose $x^{*}\in S_{X^{*}}$ and define the following maps:

$S(x)=x^{*}(x)y^{*}\text{ for }x\in X\mbox{~{}and}$
$T_{1}(x)=x^{*}(x)z_{1}^{*},~{}T_{2}(x)=x^{*}(x)z_{2}^{*}\text{ for }x\in X.$

One can check that
$S\in\mathcal{L}(X,Y^{*}),T_{1},T_{2}\in\mathcal{L}(X,Z^{*})$ and
$\|S\|=\|T_{1}\|=\|T_{2}\|$. We consider $S\in(X\widehat{\otimes}_{\pi}Y)^{*}$
and $T_{1},T_{2}\in(X\widehat{\otimes}_{\pi}Z)^{*}$. For any $x\otimes y\in
X\otimes_{\pi}Y$, we have $S(x\otimes
y)=Sx(y)=x^{*}(x)y^{*}(y)=x^{*}(x)z_{1}^{*}(y)=T_{1}x(y)=T_{1}(x\otimes y)$,
and we obtain $S(x\otimes y)=T_{2}(x\otimes y)$. If $D$ is the linear span of
all simple tensors in $X\otimes_{\pi}Y$, then $T_{1}|_{D}=T_{2}|_{D}=S$. Based
on the density of $D$ in $(X\widehat{\otimes}_{\pi}Y)$, we have
$T_{1}|_{(X\widehat{\otimes}_{\pi}Y)}=T_{2}|_{(X\widehat{\otimes}_{\pi}Y)}=S$.
Hence, distinct $T_{1},T_{2}\in HB(S)$. ∎

###  3.3. On quotient spaces and higher duals of Banach spaces

Let $W,Y$ be subspaces of $X$ such that $W\subseteq Y\subseteq X$, where $Y$
has property-$(SU)$ in $X$; then, it is shown in [DPR] that $Y/W$ has
property-$(SU)$ in $X/W$. One can check $(Y/W)^{\\#}=\\{w^{\perp}\in
W^{\perp}_{X^{*}}:\|w^{\perp}|_{Y}\|=\|w^{\perp}\|\\}$.

######  Theorem 3.3.1.

Let $W,Y$ be subspaces of $X$ where $W\subseteq Y\subseteq X$. If $Y$ has
property-$(HB)$ in $X$, then $Y/W$ has property-$(HB)$ in $X/W$.

###### Proof.

We have that $Y/W$ has property-$(SU)$ in $X/W$. Let $x^{*}\in
W^{\perp}_{X^{*}}(=(X/W)^{*})$ and $x^{*}=y^{\\#}+y^{\perp}$, where
$y^{\\#}\in(Y/W)^{\\#}$ and $y^{\perp}\in
Y^{\perp}_{W^{\perp}}(=(Y/W)^{\perp})$. It remains to show
$\|x^{*}\|\geq\|y^{\perp}\|$. Since $(Y/W)^{\\#}=\\{w^{\perp}\in
W^{\perp}_{X^{*}}:\|w^{\perp}|_{Y}\|=\|w^{\perp}\|\\}$ and
$y^{\\#}\in(Y/W)^{\\#}$, we have $y^{\\#}\in Y^{\\#}$. Moreover, $y^{\perp}\in
Y^{\perp}_{W^{\perp}}\subseteq Y^{\perp}_{X^{*}}$; we can view
$x^{*}=y^{\\#}+y^{\perp}$, where $y^{\\#}$ and $y^{\perp}$ are obtained from
$Y^{\\#}$ and $Y^{\perp}$, respectively. Hence, $\|x^{*}\|\geq\|y^{\perp}\|$
as $Y$ has property-$(HB)$ in $X$. ∎

The converse of Theorem 3.3.1 is true under the assumption that $W$ is an
$M$-ideal in $X$.

######  Theorem 3.3.2.

Let $J$ and $Y$ be subspaces of $X$ with $J\subseteq Y\subseteq X$, and let
$J$ be an $M$-ideal in $X$. If $Y/J$ has property-$(HB)$ in $X/J$, then $Y$
has property-$(HB)$ in $X$.

###### Proof.

First, we have that $Y$ has property-$(SU)$ in $X$ (see [DPR, Theorem 4.5]).
Consider $x^{*}\in X^{*}$ and $x^{*}=y^{\\#}+y^{\perp}$, where $y^{\\#}\in
Y^{\\#}$, $y^{\perp}\in Y^{\perp}$. Our aim is to show that
$\|x^{*}\|\geq\|y^{\perp}\|$. Let $\widetilde{x^{*}}\in X^{*}$ be the unique
norm-preserving extension of $x^{*}|_{J}$. Clearly,
$x^{*}-\widetilde{x^{*}},y^{\\#}-\widetilde{x^{*}},y^{\perp}\in J^{\perp}$.
Moreover, as $\widetilde{x^{*}}\in J^{\\#}$, we have $\widetilde{x^{*}}\in
Y^{\\#}$; hence, $y^{\\#}-\widetilde{x^{*}}\in Y^{\\#}$. Thus,
$x^{*}-\widetilde{x^{*}}=y^{\\#}-\widetilde{x^{*}}+y^{\perp}$, where
$x^{*}-\widetilde{x^{*}}\in J^{\perp}(=(X/J)^{*})$,
$y^{\\#}-\widetilde{x^{*}}\in(Y/J)^{\\#}(=Y^{\\#}_{J^{\perp}})$ and
$y^{\perp}\in Y^{\perp}_{J^{\perp}}(=(Y/J)^{\perp})$. Consequently,
$\|x^{*}-\widetilde{x^{*}}\|\geq\|y^{\perp}\|$ as $Y/J$ has property-$(HB)$ in
$X/J$. By contrast, $x^{*}=x^{*}-\widetilde{x^{*}}+\widetilde{x^{*}}$, where
$x^{*}-\widetilde{x^{*}}\in J^{\perp}$, $\widetilde{x^{*}}\in J^{\\#}$, and
$J$ is an $M$-ideal in $X$, implying that
$\|x^{*}\|\geq\|x^{*}-\widetilde{x^{*}}\|$ and finally
$\|x^{*}\|\geq\|y^{\perp}\|$. ∎

Combining the cases for an $L_{1}$-predual or an $M$-embedded space, the
following can be obtained:

######  Theorem 3.3.3.

Let $X$ be an $L_{1}$-predual (or an $M$-embedded) space, and let $Y$ be a
finite co-dimensional subspace of $X$. Then, $Y^{\perp\perp}$ has
property-$(HB)$ in $X^{**}$ if $Y$ has property-$(HB)$ in $X$.

###### Proof.

Let $Y$ have property-$(HB)$ in $X$ and $P$ be the corresponding bi-
contractive projection on $X^{*}$ with $\ker(P)=Y^{\perp}$. Thus $P^{**}$ is a
bi-contractive projection with $\ker(P^{**})=Y^{\perp\perp\perp}$. By Theorem
3.1.4, it is sufficient to prove that $Y^{\perp\perp}$ has property-$(U)$ in
$X^{**}$.

In both cases, $X^{*}$ is the $L$-summand in $X^{***}$. Since
$Y^{\perp}(=\textrm{span}\\{x_{i}^{*}:1\leq i\leq n\\})$ is Chebyshev in
$X^{*}$, $Y^{\perp\perp\perp}=\textrm{span}\\{\widehat{x_{i}^{*}}:1\leq i\leq
n\\}$ is Chebyshev in $X^{***}$, where $\widehat{x_{i}^{*}}$’s are the
canonical images of $x_{i}^{*}$’s in $X^{***}$. Hence $Y^{\perp\perp}$ has
property-$(U)$ in $X^{**}$. ∎

Our next observation follows from the definition of property-$(U)$ ($(SU)$).

######  Proposition 3.3.4.

Let $Z$ and $Y$ be the subspaces of $X$ with $Z\subseteq Y\subseteq X$.

  * $(a)$

If $Z$ has property-$(SU)$ in $X$, then $Z$ has property-$(SU)$ in $Y$.

  * $(b)$

If $Z$ (and $Y$) has property-$(SU)$ in $Y$ (and $X$), then $Z$ has
property-$(SU)$ in $X$.

######  Theorem 3.3.5.

Let $Z$ and $Y$ be subspaces of $X$ with $Z\subseteq Y\subseteq X$. If $Z$ has
property-$(HB)$ in $X$, then $Z$ has property-$(HB)$ in Y.

###### Proof.

Suppose that $Z$ has property-$(HB)$ in $X$. From Proposition 3.3.4, $Z$ has
property-$(SU)$ in $Y$. Let $y^{*}\in Y^{*}$ and $x^{*}\in HB(y^{*})$. Let
$y^{*}=y^{\\#}+y^{\perp}$, where $y^{\\#}\in Z^{\\#}_{Y^{*}}$ and
$y^{\perp}(\neq 0)\in Z^{\perp}_{Y^{*}}$. It remains to be shown that
$\|y^{*}\|\geq\|y^{\perp}\|$. Because $Z$ has property-$(HB)$ in $X$, we have
$\|x^{*}\|>\|x^{\\#}\|$ and $\|x^{*}\|\geq\|x^{\perp}\|$, where
$x^{*}=x^{\\#}+x^{\perp}$, $x^{\\#}\in Z^{\\#}_{X^{*}}$, $x^{\perp}(\neq 0)\in
Z^{\perp}_{X^{*}}$. Now, $y^{*}=x^{*}|_{Y}=x^{\\#}|_{Y}+x^{\perp}|_{Y}$. Since
$x^{\perp}|_{Y}\in Z^{\perp}_{Y^{*}}$ and $x^{\\#}|_{Y}\in Z^{\\#}_{Y^{*}}$,
because of the uniqueness of the decomposition of $y^{*}$, we have
$y^{\\#}=x^{\\#}|_{Y}$ and $y^{\perp}=x^{\perp}|_{Y}$. Hence,
$\|y^{*}\|=\|x^{*}\|\geq\|x^{\perp}\|\geq\|x^{\perp}|_{Y}\|=\|y^{\perp}\|$. ∎

######  Remark 3.3.6.

Example 3.4.5 shows that even if $Y$ has property-$(SU)$ in $X$, the converse
of Theorem 3.3.5 may not be true.

Suppose that $X$ has property-$(U)$ in $X^{**}$. Our next theorem provides a
sufficient condition for subspaces to have property-$(HB)$ in their biduals.
Note that $Y$ has property-$(HB)$ in $X$ if $Y$ has property-$(HB)$ in
$X^{**}$, following from Theorem 3.3.5.

######  Theorem 3.3.7.

Let $Y$ be a subspace of $X$, and let $X$ have property-$(U)$ in $X^{**}$. If
$Y$ has property-$(HB)$ in $X$, then $Y$ has property-$(HB)$ in $Y^{**}$.

###### Proof.

Clearly, $Y$ has property-$(SU)$ in $Y^{**}$. Let $y^{***}\in Y^{***}$ and
suppose $y^{***}=y^{\\#}+y^{\perp}$ where $y^{\\#}\in Y^{\\#}_{Y^{***}}$ and
$y^{\perp}\in Y^{\perp}_{Y^{***}}$. It remains to be shown that
$\|y^{***}\|\geq\|y^{\perp}\|$. Let $P:X^{*}\rightarrow Y^{\\#}$ be a bi-
contractive projection with $\ker(P)=Y^{\perp}$. Hence,
$P^{**}:X^{***}\rightarrow Y^{\\#\perp\perp}$ is a bi-contractive projection
with $\ker(P^{**})=Y^{\perp\perp\perp}$. As $Y$ has property-$(HB)$ in $X$,
$Y^{*}$ and $Y^{***}$ are subspaces of $X^{*}$ and $X^{***}$, respectively.
Thus, $y^{***}\in X^{***}$, $y^{\\#}\in Y^{\\#}_{Y^{***}}\cong Y^{*}\cong
Y^{\\#}_{X^{*}}\subseteq Y^{\\#\perp\perp}_{X^{***}}$ and $y^{\perp}\in
Y^{\perp}_{Y^{***}}\subseteq Y^{\perp\perp\perp}_{X^{***}}$. Hence,
$\|y^{\perp}\|=\|(I-P^{**})y^{***}\|\leq\|y^{***}\|$. ∎

###  3.4. Examples

We begin with the following example, which ensures that the property-$(wU)$ is
strictly weaker than the property-$(U)$.

######  Example 3.4.1.

Let $X$ be a non-reflexive Banach space such that $X^{*}$ is separable. Let
$\mathcal{N}(X)$ be the set of all equivalent norms on $X$ , equipped with the
topology of uniform convergence on bounded subsets of X . Suppose $M$ be a
Polish space and $A$ and $B$ be two analytic subsets of $M$ with $A\subseteq
B$. By [GYK, Theorem 3], there exists a continuous map
$\Phi:M\rightarrow\mathcal{N}(X)$ such that $(X,\Phi(t))$ is smooth Banach
space but it dual is not a strictly convex Banach space, for $t\in B\setminus
A$, where $(X,\Phi(t))$ denotes the Banach $X$ endowed with the norm
$\Phi(t)$.

Since the dual space $(X^{*},\Phi(t)^{*})$ is not strictly convex there exists
$x^{*}\in X^{*}$ such that $\\{tx^{*}:t\in\mathbb{R}\\}$ is not a Chebyshev
subspace. Hence $ker(x^{*})$ can not have property-$(U)$ in $(X,\Phi(t))$ but
any subspace has property-$(wU)$ in $(X,\Phi(t))$ (see [LA, Theorem 2.4]).

We now give a few examples that have property-$(SU)$ but not $(HB)$.

######  Theorem 3.4.2.

Let $(a_{n})\in S_{\ell_{1}}$ such that
$1>\sup_{n\in\mathbb{N}}|a_{n}|>\frac{1}{2}$. Then, $\ker((a_{n}))\subseteq
c_{0}$ has property-$(SU)$ but does not have property-$(HB)$ in $c_{0}$.

###### Proof.

Let $X=c_{0}$, $Y=\ker((a_{n}))$. Because
$\sup_{n\in\mathbb{N}}|a_{n}|>\frac{1}{2}$, there exists a unique
$N\in\mathbb{N}$ such that $|a_{N}|>\frac{1}{2}$. By [DPR, Theorem 6.6], the
metric projection $P_{Y^{\perp}}:X^{*}\rightarrow Y^{\perp}$ is defined as

$P_{Y^{\perp}}((x_{n}))=\\{\frac{x_{N}}{a_{N}}(a_{1},a_{2},a_{3},\ldots)\\}\text{
for }(x_{n})\in\ell_{1}.$

Clearly, $P_{Y^{\perp}}$ is linear and $\|I-P_{Y^{\perp}}\|=1$. If we choose
$(x_{n})\in S_{\ell_{1}}$ such that $|x_{N}|>|a_{N}|$, then
$\|P_{Y^{\perp}}(x_{n})\|>1=\|(x_{n})\|_{1}$, i.e., $\|P_{Y^{\perp}}\|>1$.
Hence, by Theorem 3.1.4.$(c)$, $Y$ does not have property-$(HB)$ in $c_{0}$. ∎

From Theorem 3.4.2 it is clear that if $\ker((a_{n}))$ has property-$(HB)$ in
$c_{0}$ then $\sup_{n\in\mathbb{N}}|a_{n}|=1$ which in turn $(a_{n})=e_{m}$
for some $m$. Hence we have the following.

######  Corollary 3.4.3.

Let $Y$ be a hyperplane in $c_{0}$. $Y$ has property-$(HB)$ in $c_{0}$ if and
only if $Y$ is an $M$-summand in $c_{0}$.

######  Example 3.4.4.

Let $m,n\in\mathbb{N}$, and $m<n$. Then, the subspace
$\textrm{span}~{}\\{e_{1},\dots,e_{m}\\}$ has property-$(SU)$ but does not
have property-$(HB)$ in $(\mathbb{R}^{n},\|.\|_{*})$, where
$\|(x_{i})_{i=1}^{n}\|_{*}=\max\\{|x_{1}|,\dots,|x_{n-1}|,\frac{|x_{1}+\dots+x_{n}|}{n}\\}$
for $(x_{i})_{i=1}^{n}\in\mathbb{R}^{n}$.

###### Proof.

Let $Y=\textrm{span}~{}\\{e_{1},\dots,e_{m}\\}$ and
$X=(\mathbb{R}^{n},\|.\|_{*})$. It is easy to observe that
$X^{*}=(\mathbb{R}^{n},\|.\|^{*})$, where
$\|(a_{i})_{i=1}^{n}\|^{*}=\sum_{i=1}^{n-1}|a_{i}-a_{n}|+n|a_{n}|$ for
$(a_{i})_{i=1}^{n}\in\mathbb{R}^{n}$.

Let $G=\textrm{span}~{}\\{e_{1},\dots,e_{m}\\}$ be a subspace of $X^{*}$. It
is evident that $X^{*}=G\oplus Y^{\perp}$. In fact,
$(a_{i})_{i=1}^{n}=(a_{1},\dots,a_{m},0,\dots,0)+(0,\dots,0,a_{m+1},\dots,a_{n})$
for an element $(a)_{i=1}^{n}\in X^{*}$, where
$(a_{1},\dots,a_{m},0,\dots,0)\in G$ and $(0,\dots,0,a_{m+1},\dots,a_{n})\in
Y^{\perp}$. Using the triangle inequality, we have
$\|(a_{1},\dots,a_{m},0,\dots,0)\|^{*}\leq\|(a_{1},\ldots,a_{n})\|^{*}$ and
obtain a strict inequality if $a_{j}$’s are not all zeros for $m<j\leq n$.
Thus, $Y$ has property-$(SU)$ in $X$.

Consider the decomposition
$(b_{i})_{i=1}^{n}=(b_{1},\dots,b_{m},0,\dots,0)+(0,\dots,0,b_{m+1},\dots,b_{n})$,
where $b_{i}=1$, $1\leq i\leq n$, but
$\|(b_{i})_{i=1}^{n}\|^{*}=n<n+m=\|(0,\dots,0,b_{m+1},\dots,b_{n})\|^{*}$.
Hence, $Y$ does not have property-$(HB)$. ∎

We conclude this section by showing that property-$(HB)$ cannot be transferred
via property-$(SU)$ in general.

######  Example 3.4.5.

Let $X=(\mathbb{R}^{3},\|.\|_{\infty})$ and
$Y=\\{(x,y,z)\in\mathbb{R}^{3}:x+y+6z=0\\}$, then
$Y^{\\#}=\\{(x,y,0):x,y\in\mathbb{R}\\}\subseteq(\mathbb{R}^{3},\|.\|_{1})$.
Because $Y^{\\#}$ is linear, $Y$ has property-$(SU)$ in $X$. Let
$Z=\ker((1,0,0))\subseteq Y$, i.e., $Z=\\{(0,y,z):y+6z=0\\}$. Clearly,
$Z_{Y^{*}}^{\perp}=\emph{span}\\{(1,0,0)\\}$ is an $L$-summand in $Y^{*}$.
Hence, $Z$ is an M-ideal in $Y$, and $Y$ has property-$(SU)$ in $X$. It is
clear that plane $Z$ does not intersect the upper and bottom surfaces of the
unit ball $B_{(\mathbb{R}^{3},\|.\|_{\infty})}$. Thus,
$\emph{ext}(B_{Z})=\\{\pm(0,1,-\frac{1}{6})\\}$, and we can show that
$Z^{\\#}=\\{(0,y,0):y\in\mathbb{R}\\}$. An easy calculation indicates that
$Z^{\perp}_{X^{*}}=\emph{span}\\{(1,0,0),(0,1,6)\\}$. Let $(0,0,1)\in X^{*}$.
Then, the unique decomposition of $(0,0,1)$ is

$(0,0,1)=(0,-\frac{1}{6},0)+(0,\frac{1}{6},1),$

and $\|(0,\frac{1}{6},1)\|_{1}=\frac{7}{6}>1=\|(0,0,1)\|_{1}$. Hence, $Z$ does
not have property-$(HB)$ in $X$.

##  4\. On property-$(k-U)$ in Banach spaces

###  4.1. Characterization

We derive a characterization for reflexive subspaces with property-$(k-U)$ in
Theorem 4.1.5. Let $x\in X\setminus\\{0\\}$ and $x^{*}\in
X^{*}\setminus\\{0\\}$. Let us recall from Section 1 that
$S(X^{*},x)=\\{x^{*}\in S_{X^{*}}:x^{*}(x)=\|x\|\\}$; we also define
$S(X,x^{*})=\\{x\in S_{X}:x^{*}(x)=\|x^{*}\|\\}$.

######  Proposition 4.1.1.

Let $Y$ be a subspace of $X$. If $Y$ is a $k$-Chebyshev subspace of $X$, then
$\dim(\\{S(X,y^{\perp})-x\\}\cap Y)<k$ for all $x\in S(X,y^{\perp})$, for all
$y^{\perp}\in S_{Y^{\perp}}$.

###### Proof.

If possible, let $y^{\perp}\in S_{Y^{\perp}}$ such that
$\dim(\\{S(X,y^{\perp})-x\\}\cap Y)\geq k$ for some $x\in S(X,y^{\perp})$. Let
$y_{1},y_{2},\dots,y_{k}\in\\{S(X,y^{\perp})-x\\}\cap Y$ be $k$ linearly
independent elements. Since $x,y_{i}+x\in S(X,y^{\perp})$, $1\leq i\leq k$, we
have $\|y_{i}+x\|=1=\|x\|$. Hence, for $y\in Y$, $\|x-y\|\geq
y^{\perp}(x-y)=y^{\perp}(x)=\|y^{\perp}\|=1=\|x\|$, which is true for all
$y\in Y$. Consequently, $d(x,Y)=\|x\|=\|x+y_{i}\|$, that is, $0,-y_{i}\in
P_{Y}(x)$ for $1\leq i\leq k$. Therefore, $Y$ is not a $k$-Chebyshev subspace
of $X$. ∎

The converse of the aforementioned theorem is true for the proximal subspace
of $X$. We first prove a similar result for the hyperplanes in $X$.

######  Proposition 4.1.2.

Let $Y=\ker(x^{*})$ be a proximinal hyperplane of $X$ for some $x^{*}\in
S_{X^{*}}$. If $Y$ is not a $k$-Chebyshev subspace of $X$, then
$\dim(\\{S(X,x^{*})-x\\}\cap Y)\geq k$ for some $x\in S(X,x^{*})$.

###### Proof.

Let $Y$ not be a $k$-Chebyshev subspace of $X$. Let $x\in X$ be such that
$d(x,Y)=1=\|x\|=x^{*}(x)$ and $\dim(P_{Y}(x))\geq k$.

Let $y_{1},\dots,y_{k}\in Y$ be $k$ linearly independent elements such that
$\|y_{i}-x\|=d(x,Y)=1$, $i=1,\dots,k$. Then, we have $-y_{i}\in S(X,x^{*})-x$,
where $x\in S(X,x^{*})$. Hence, the result follows. ∎

We now generalize the aforementioned proposition for proximinal subspaces of
$X$.

######  Proposition 4.1.3.

Let $Y$ be a proximinal subspace of $X$. If $Y$ is not a $k$-Chebyshev
subspace of $X$, then $\dim(\\{S(X,y^{\perp})-x\\}\cap Y)\geq k$ for some
$x\in S(X,y^{\perp})$ and for some $y^{\perp}\in S_{Y^{\perp}}$.

###### Proof.

If $Y$ is not a $k$-Chebyshev subspace of $X$, then there exists $x_{0}\in
X\setminus Y$ such that $Y$ is not a $k$-Chebyshev subspace of
$Z=\textrm{span}\\{Y\cup x_{0}\\}$. Clearly, $Y$ is proximinal in $Z$, and
there exists $z^{*}\in S_{Z^{*}}$ such that $Y=\ker(z^{*})$. From Proposition
4.1.2, we have $\dim(\\{S(Z,z^{*})-z\\}\cap Y)\geq k$ for some $z\in
S(Z,z^{*})$. If $x^{*}$ is a Hahn–Banach extension of $z^{*}$ over $X$, then
$\dim(\\{S(X,x^{*})-z\\}\cap Y)\geq k$, and the result follows. ∎

Hence, we have the following.

######  Theorem 4.1.4.

Let $Y$ be a proximinal subspace of $X$. $Y$ is a $k$-Chebyshev subspace of
$X$ if and only if $\dim(\\{S(X,y^{\perp})-x\\}\cap Y)<k$ for all $x\in
S(X,y^{\perp})$ and for all $y^{\perp}\in S_{Y^{\perp}}$.

We now state Theorem 1.2.1 in terms of the characterization derived in Theorem
4.1.4.

######  Theorem 4.1.5.

Let $Y$ be a subspace of $X$. Consider the following statements:

  * $(a)$

$Y$ has property-$(k-U)$ in $X$.

  * $(b)$

$Y^{\perp}$ is a $k$-Chebyshev subspace of $X^{*}$.

  * $(c)$

$\dim(\\{S(X^{*},x)-x^{*}\\}\cap Y^{\perp})<k$ for all $x^{*}\in S(X^{*},x)$
and for all $x\in S_{Y}$.

Then, $(a)\Longleftrightarrow(b)$ and $(b)\Rightarrow(c)$. If $Y$ is
reflexive, then $(c)\Rightarrow(b)$.

###### Proof.

$(a)\Longleftrightarrow(b)$ follows from [X].

For $(b)\Rightarrow(c)$, assume that there exists $x\in S_{Y}$ such that
$\dim(\\{S(X^{*},x)-x^{*}\\}\cap Y^{\perp})\geq k$ for some $x^{*}\in
S(X^{*},x)$. Identifying $x$ in $Y^{**}$ with its canonical image, say
$\widehat{x}$, the aforementioned inequality can be stated as
$\dim(\\{S(X^{*},\widehat{x})-x^{*}\\}\cap Y^{\perp})\geq k$. By Theorem
4.1.1, $Y^{\perp}$ is not a $k$-Chebyshev subspace.

$(c)\Rightarrow(b)$ follows from Theorem 4.1.4 as $Y=Y^{\perp\perp}$ when $Y$
is reflexive. ∎

###  4.2. Applications to subspaces of Bochner Integrable functions

We assume $(\Omega,\Sigma,\mu)$ to be a measure space.

######  Theorem 4.2.1.

Let $\mu$ be non-atomic and $Y=\emph{span}\\{\chi_{A_{i}}:1\leq i\leq n\\}$ be
an $n$-dimensional subspace of $L_{1}(\mu)$, where
$A_{1},A_{2},\dots,A_{n}\in\Sigma$ are disjoint, and
$\mu(\Omega\setminus\cup_{i=1}^{n}A_{i})>0$. Then, $Y$ does not have
property-$(k-U)$ for any $k\in\mathbb{N}$.

###### Proof.

Let $B=\Omega\setminus\cup_{i=1}^{n}A_{i}$. Because $\mu(B)>0$ and $\mu$ is a
non-atomic measure, there exists $B_{1}\subseteq B$ such that
$\mu(B_{1})<\mu(B)$. Hence, $\mu(B\setminus B_{1})>0$, and we find
$B_{2}\subseteq B\setminus B_{1}$ such that $0<\mu(B_{2})<\mu(B\setminus
B_{1})$. Continuing this process, we obtain $k$-disjoint sets
$(B_{j})_{j=1}^{k}$ of the positive measure. Let $f\in Y$; it is clear that
$f=0$ a.e. on $B$. Now, $g_{j}:=\chi_{B_{j}}$ are $k$ linearly independent
members of $L_{\infty}(\mu)$. Setting $g=\textrm{sgn}(f)$ on
$\cup_{i=1}^{n}A_{i}$, $g=0$ a.e. on $B$. One can verify that $g_{j}+g,g\in
S(L_{1}(\mu)^{*},f)$ for $1\leq j\leq k$. Hence, $g_{j}\in
S(L_{1}(\mu)^{*},f)-g$ and $g_{j}\in Y^{\perp}$ as $B_{j}\cap(\cup
A_{i})=\phi$, $1\leq j\leq k$. From Theorem 4.1.5, it is clear that $Y$ does
not have property-$(k-U)$. ∎

######  Proposition 4.2.2.

Let $\mu$ be non-atomic and $Y$ be an $n$-dimensional subspace of
$L_{1}(\mu)$. Then, $Y$ is an $(n+1)$-Chebyshev subspace, but not a
$k$-Chebyshev subspace, for $1\leq k\leq n$.

###### Proof.

An extreme point $g\in B_{L_{\infty}(\mu)}$ exists such that $g|_{Y}=0$ (see
[P, Theorem 2.5]). Hence, we have $|g|=1$ a.e. and $\int_{\Omega}gfd\mu=0$ for
all $f\in Y$. Let $\\{f_{1},\dots,f_{n}\\}$ be a basis for $Y$ and
$h=g\sum_{i=1}^{n}|f_{i}|$. One can check that $d(h,Y)=\|h\|$. It is evident
that $\textrm{sgn}(g\sum_{i=1}^{n}|f_{i}|-f_{i})=\textrm{sgn}(g)=g$, as
$|g|=1$. Thus, $\|h-f_{i}\|=\int g(g\sum_{i=1}^{n}|f_{i}|-f_{i})d\mu=\|h\|$.
Hence, $0,f_{1},\dots,f_{n}\in P_{Y}(h)$. Therefore, $Y$ is not a
$k$-Chebyshev subspace for $1\leq k\leq n$.

Being an $n$-dimensional subspace, $Y$ is always an $(n+1)$-Chebyshev
subspace. ∎

######  Theorem 4.2.3.

Let $Z\subseteq Y\subseteq X$ be subspaces of $X$. If $Y$ has property-$(k-U)$
in $X$, then $Y/Z$ has property-$(k-U)$ in $X/Z$.

###### Proof.

Let $Y/Z$ does not have property-$(k-U)$ in $X/Z$. Let $y^{*}\in
Z^{\perp}_{Y^{*}}$ and $x^{*}_{i}\in Z^{\perp}_{X^{*}}$, and $1\leq i\leq k+1$
be an affinely independent norm, preserving the extensions of $y^{*}$. As
$x^{*}_{i}|_{Y}=y^{*}$ and $x_{i}^{*}$, $i=1,2,\dots,k+1$ are affinely
independent norm-preserving extensions of $y^{*}\in Y^{*}$; $Y$ does not have
property-$(k-U)$ in $X$. ∎

The converse part of this result is true under the assumption that $Z$ is an
$M$-ideal in $X$.

######  Theorem 4.2.4.

Let $Y,Z$ be the subspace of $X$ such that $Z\subseteq Y\subseteq X$. Suppose
$Z$ is an $M$-ideal in $X$ and $Y/Z$ has property-$(k-U)$ in $X/Z$. Then, $Y$
has property-$(k-U)$ in $X$.

###### Proof.

If possible, let $Y$ not have property-$(k-U)$ in $X$, that is, $Y^{\perp}$ is
not a $k$-Chebyshev subspace of $X^{*}$. Let $x^{*}\in X^{*}$ and
$y_{i}^{\perp},y_{0}^{\perp}\in P_{Y^{\perp}}(x^{*})$, $i=1,2,\dots,k$, such
that the vectors in $\\{y_{i}^{\perp}-y_{0}^{\perp}:1\leq i\leq k\\}$ are
linearly independent. As $Z$ is a $M$-ideal in $X$, there exists a subspace
$G$ of $X^{*}$ such that $X^{*}=Z^{\perp}\oplus_{\ell_{1}}G$. Suppose
$x^{*}=z^{\perp}+z^{\\#}$, where $z^{\perp}\in Z^{\perp}$, $z^{\\#}\in G$.

$\displaystyle\mbox{Now}~{}d(x^{*},Y^{\perp})$ $\displaystyle=$
$\displaystyle\inf_{y^{\perp}\in Y^{\perp}}\|x^{*}-y^{\perp}\|$
$\displaystyle=$ $\displaystyle\inf_{y^{\perp}\in
Y^{\perp}}(\|z^{\perp}-y^{\perp}\|+\|z^{\\#}\|)$ $\displaystyle=$
$\displaystyle d(z^{\perp},Y^{\perp})+\|z^{\\#}\|$ $\displaystyle\leq$
$\displaystyle\|z^{\perp}-y_{i}^{\perp}\|+\|z^{\\#}\|$ $\displaystyle=$
$\displaystyle\|x^{*}-y_{i}^{\perp}\|=d(x^{*},Y^{\perp}).$

Hence, $d(z^{\perp},Y^{\perp})=\|z^{\perp}-y_{i}^{\perp}\|$, $i=0,1,\dots,k$.
Therefore, $y_{i}^{\perp},y_{0}^{\perp}\in P_{Y^{\perp}}(z^{\perp})$,
$i=1,2,\dots,k$. As $z^{\perp}\in(X/Z)^{*}=Z^{\perp}$ and
$y_{i}^{\perp}\in(Y/Z)^{\perp}=Y^{\perp}_{Z^{\perp}}=Y^{\perp}$, $(Y/Z)$ does
not have property-$(k-U)$ in $(X/Z)$. ∎

######  Theorem 4.2.5.

Let $X$ be an $L_{1}$-predual (or an $M$-embedded) space, and let $Y$ be a
finite co-dimensional subspace of $X$. Then, $Y$ has property-$(k-U)$ in $X$
if and only if $Y^{\perp\perp}$ has property-$(k-U)$ in $X^{**}$.

###### Proof.

Let $Y$ not have property-$(k-U)$ in $X$. Let $y^{*}\in S_{Y^{*}}$ and
$x_{1}^{*},\dots,x_{k+1}^{*}\in HB(y^{*})$ be affinely independent. Then, the
canonical images $\widehat{x_{1}^{*}},\dots,\widehat{x_{k+1}^{*}}\in
HB(\widehat{y^{*}})$. It is known that $Y^{\perp\perp*}\cong Y^{***}$,
$Y^{\perp\perp}$ does not have property-$(k-U)$ in $X^{**}$.

Suppose that $Y^{\perp\perp}$ does not have property-$(k-U)$ in $X^{**}$, that
is, $Y^{\perp\perp\perp}$ is not a $k$-Chebyshev subspace of $X^{***}$.
Because $Y$ is a finite co-dimensional subspace, we have $Y^{\perp}\cong
Y^{\perp\perp\perp}$. It is known that $X^{*}$ is an $L$-summand in $X^{***}$
when $X$ is an $L_{1}$-predual or an $M$-embedded space. Thus, $Y^{\perp}$ is
not a $k$-Chebyshev subspace of $X^{***}$, and $X^{*}$ is an $L$-summand in
$X^{***}$. Using similar arguments in the proof of Theorem 4.2.4, we find that
$Y^{\perp}$ is not a $k$-Chebyshev subspace of $X^{*}$. ∎

The next theorem establishes that property-$(k-U)$ remains hereditary for the
class of subspaces that are ideals in a Banach space. Let us recall that if
$Y$ is an ideal in $X$ and $P:X^{*}\rightarrow X^{*}$ is a corresponding
norm-$1$ projection, then $R(P)\cong Y^{*}$ (see [OP, Pg.297]).

######  Theorem 4.2.6.

Let $Y$ be an ideal in $X$. If $X$ has property-$(k-U)$ in $X^{**}$, then $Y$
has property-$(k-U)$ in $Y^{**}$.

###### Proof.

Let $Y$ not have property-$(k-U)$ in $Y^{**}$. Let $y^{*}\in S_{Y^{*}}$ and
$y^{***}_{1},\dots,y^{***}_{k+1}\in HB(y^{*})$ be affinely independent. Now,
$Y^{*}$ and $Y^{***}$ are subspaces (up to isometric) of $X^{*}$ and
$X^{***}$, respectively. Thus, $X$ does not have property-$(k-U)$ in $X^{**}$.
∎

###  4.3. Property-$(k-U)$ and multismoothness in Banach spaces

######  Definition 4.3.1.

[LR] Let $X$ be a Banach space and $x\in S_{X}$.

  * $(a)$

$x$ is said to be a $k$-smooth point if $\dim(S(X^{*},x))\leq k-1$.

  * $(b)$

A Banach space $X$ is said to be $k$-smooth if $\dim(S(X^{*},x))\leq k-1$ for
all $x\in S_{X}$.

######  Theorem 4.3.2.

Let $Y$ be a subspace of $X$. Suppose every $y\in S_{Y}$ is a $k$-smooth point
in $X$. Then, any reflexive subspace $Z$ of $Y$ has property-$(k-U)$ in $X$.

###### Proof.

Suppose every $y\in S_{Y}$ is a $k$-smooth point. Let $Z$ be a reflexive
subspace of $X$ that does not have property-$(k-U)$ in $X$. Let $z^{*}\in
S_{Z^{*}}$ and $x^{*}_{1},\dots,x_{k+1}^{*}\in HB(y^{*})$ be affinely
independent. Because $Z$ is reflexive, there exists $z_{0}\in S_{Z}$ such that
$z^{*}(z_{0})=1$ and $x_{i}^{*}(z_{0})=1$, $1\leq i\leq k+1$. Hence,
$\dim(S(X^{*},z_{0}))\geq k$, that is, $z_{0}$ is not a $k$-smooth point. ∎

In particular, we have the following.

######  Proposition 4.3.3.

Let $F$ be a finite-dimensional subspace of $X$. Suppose that every
$x\in\emph{ext}(B_{F})$ is a $k$-smooth point in $X$. Then, any subspace $Z$
of $F$ has property-$(k-U)$ in $X$.

###### Proof.

Let $x\in S_{F}$, then $(x_{i})_{i=1}^{n}\subseteq\textrm{ext}(B_{F})$ such
that $x=\sum_{i=1}^{n}\lambda_{i}x_{i}$ for some $(\lambda_{i})_{i=1}^{n}$
with $\lambda_{i}\geq 0$ and $\sum_{i}\lambda_{i}=1$. Clearly, we have
$S(X^{*},x)\subseteq\cap_{i=1}^{n}S(X^{*},x_{i})$. Hence, every $x\in S_{F}$
is a $k$-smooth point in $X$. The result follows from Theorem 4.3.2. ∎

We now identify some subspaces of $c_{0}$ that have property-$(k-U)$.

######  Theorem 4.3.4.

Let $F$ be a finite-dimensional subspace of $c_{0}$. Then, $F$ has
property-$(k-U)$ in $c_{0}$ for some $k\in\mathbb{N}$.

###### Proof.

Because $F$ is a finite-dimensional subspace of $c_{0}$, we have that
$\textrm{ext}(B_{F})$ is finite set and
$B_{F}=\textrm{conv}(\textrm{ext}(B_{F}))$. Let
$\textrm{ext}(B_{F})=\\{x_{1},\dots,x_{n}\\}$. As $x_{i}\in c_{0}$ and
$\|x_{i}\|_{\infty}=1$, there exists $k_{i}\in\mathbb{N}$ such that
$|x_{i}(j)|=1$, where $j\in\\{n_{1},\dots,n_{k_{i}}\\}$. It follows that there
exists a large $k$ such that all $x_{i}^{\prime}$s are $k$-smooth. Proposition
4.3.3 leads to the conclusion. ∎

######  Theorem 4.3.5.

Let $F$ be a finite dimensional subspace of $c_{0}$, which is also an ideal in
$c_{0}$, then $F^{*}$ is a $k$-strictly convex subspace of $\ell_{1}$ for some
natural $k$.

###### Proof.

First, $F^{*}$ is a subspace of $\ell_{1}$. We have already observed that $F$
has property-$(k-U)$ for some $k$. It is proven that any point in $S_{F}$ is a
$k$-smooth point. Hence, $F$ is actually a $k$-smooth subspace of $c_{0}$.
Because $F=(F^{*})^{*}$, we have that $F^{*}$ is k-strictly convex. ∎

## References

  * [BR] S. Basu, T. S. S. R. K. Rao, Some stability results for asymptotic norming properties of Banach spaces, Colloq. Math. 75 (2) (1998) 271–284.
  * [DPR] S. Daptari, T. Paul, T. S. S. R. K. Rao, Stability of unique Hahn-Banach extensions and associated linear projections. To appear in linear multilinear algebra http://dx.doi.org/10.1080/03081087.2021.1945526.
  * [DU] J. Diestel, J. J. Uhl, Vector measures, Mathematical Surveys, 15. American Mathematical Society, Providence, R.I. (1977).
  * [F] S. R. Foguel, On a theorem by A. E. Taylor, Proc. Amer. Math. Soc. 9 (325), (1958).
  * [GYK] G. Godefroy, M. Yahdi, R Kaufman, The topological complexity of a natural class of norms of Banach spaces, Proceedings of the International Conference ”Analyse $\&$ Logique” (Mons, 1997). Ann. Pure Appl. Logic, 111 (1-2) (2001), 3–13.
  * [HWW] P. Harmand, D. Warner, W. Warner, M-ideals in Banach spaces and Banach algebras, Lecture Notes in Mathematics, 1547, Springer-Verlag, Berlin, viii+387 (1993).
  * [H] J. Hennefeld, $M$-ideals, HB-subspaces, and compact operators, Indiana Univ. Math. J. 28 (6) (1979) 927–934.
  * [LH] H. E. Lacey, The isometric theory of classical Banach spaces, Die Grundlehren der mathematischen Wissenschaften, 208. Springer-Verlag, New York, Heidelberg (1974).
  * [L] Å. Lima, Intersection properties of balls and subspaces in Banach spaces, Trans. Amer. Math. Soc. 227 (1977) 1–62.
  * [LA] Å. Lima, Uniqueness of Hahn-Banach extensions and liftings of linear dependencies , Math. Scand. 53 (1983) 97–113.
  * [LR] B. L. Lin, T. S. S. R. K. Rao, Multismoothness in Banach spaces, Int. J. Math. Math. Sci., Art. ID 52382 (2007) 12 .
  * [O] $\grave{E}$. F. Oya, Strong uniqueness of the extension of linear continuous functionals according to the Hahn-Banach theorem, Math. Notes 43, 1-2 (1988) 134–139.
  * [OP] E. Oja, M. Poldvere, On subspaces of Banach spaces where every functional has unique norm-preserving extension, Studia Math., 117 (3) (1996) 289–306.
  * [P] R. R. Phelps, Uniqueness of Hahn-Banach extensions and unique best approximation, Trans. Amer. Math. Soc. 95 (1960) 238–255.
  * [RA] T. S. S. R. K. Rao, On ideals in Banach spaces, Rocky Mountain J. Math. 31 (2001) 595–609.
  * [R] R. A. Ryan, Introduction to tensor products of Banach spaces, Springer Monographs in Mathematics. Springer-Verlag London, Ltd., London, xiv+225 (2002).
  * [MO] D. Serre, Why is the Hahn-Banach theorem so important?, https://mathoverflow.net/q/26568
  * [S] I. Singer, Best approximation in normed linear spaces by elements of linear subspaces, Die Grundlehren der mathematischen Wissenschaften, 171, Springer-Verlag, New York-Berlin (1970).
  * [T] A. E. Taylor, The extension of linear functionals, Duke Math. J. 5, 538–547 (1939).
  * [X] Ji Hong Xu, Norm-preserving extensions and best approximations, J. Math. Anal. Appl. 183 (3) (1994) 631–638.


Typography-MNIST (TMNIST): an MNIST-Style Image Dataset to Categorize Glyphs
and Font-Styles

Nimish Magre | Nicholas Brown
---|---
Northeastern University | Northeastern University
<EMAIL_ADDRESS>|<EMAIL_ADDRESS>

## Abstract

We present Typography-MNIST (TMNIST), a dataset comprising of 565,292 MNIST-
style grayscale images representing 1,812 unique glyphs in varied styles of
1,355 Google-fonts111https://fonts.google.com/.The glyph-list contains common
characters from over 150 of the modern and historical language scripts with
symbol sets, and each font-style represents varying subsets of the total
unique glyphs. The dataset has been developed as part of the Cognitive Type
project [1] which aims to develop eye-tracking tools for real-time mapping of
type to cognition and to create computational tools that allow for the easy
design of typefaces with cognitive properties such as readability. The dataset
and scripts to generate MNIST-style images for glyphs in different font styles
are freely available at https://github.com/aiskunks/CognitiveType.

##  1 Introduction

Since its introduction in 1998 by LeCun et al. [3], the MNIST dataset,
comprising of handwritten digit images for classes 0-9, has become
increasingly popular with deep learning researchers. Researchers are able to
test their algorithms on this real-world data relatively quicker due to its
small sized samples and minimal pre processing requirements. This is precisely
the reason behind producing the TMNIST dataset with pre processing steps that
mirror the MNIST dataset. Therefore, each grayscale image produced in the
TMNIST dataset has a standard $28\times 28$ size with the glyph image centered
and sized to $20\times 20$.

With this work, we hope to provide researchers with a practical dataset for
performing font and glyph classification tasks. In comparison to both MNIST
and EMNIST [2] (an extended version of MNIST with lower and uppercase Latin
alphabets), TMNIST could prove to be a more challenging classification dataset
due to the introduction of a higher number of classes in terms of both glyph-
labels and font-styles.

##  2 Typography-MNIST Dataset

The Typography-MNIST dataset relies on the freely accessible binary Google-
font files 222https://github.com/google/fonts and a combination of all unique
sample-glyphs represented by each of the 1,355 Google fonts1\. Since Google
forbids web-scraping from the Google-fonts subsite, and a comprehensive list
of all possible glyphs that can be configured by each of the Google-fonts is
unavailable, a list of concatenated sample glyphs configured by each of the
1,355 Google-fonts was manually attained from the Google-fonts site 1. Using
this list, a unique set of 1,819 glyphs was compiled and is available along
with the dataset.
The following conversion pipeline was then used to generate MNIST style images
from the list of unique glyphs and binary font files:

  1. 1.

import the .ttf/.otf binary font file

  2. 2.

verify if the glyph can be configured by the particular font file

  3. 3.

display the glyph in the particular font-style on a blank white canvas with
font-size 28 and font-color ’black’

  4. 4.

convert the image to gray-scale and invert image pixel values

  5. 5.

crop the glyph portion of the image and resize to $20\times 20$ without losing
aspect ratio

  6. 6.

add zero padding to resize the image to $28\times 28$

  7. 7.

compute the weighted average (center of mass) of the pixel intensities and
recenter this point to the center of the $28\times 28$ field

Fig. 1: Visualization of the process used to generate TMNIST data. The two
examples depicted display the Devanagari Letter Pha in the Rajdhani-Bold font-
style and the Latin Capital Letter C in the Aladin-Regular font-style
respectively.

The dataset is stored in a csv file where the first column represents the
Font-style, the second column represents the Glyph-label and the remaining 786
columns represent the $28\times 28$ gray-scale image pixel values. The Font-
style (ex: aladin-regular, abel-Bold) is obtained from the binary font-file
name whereas the Glyph-label (ex: LATIN CAPITAL LETTER C, DEVANAGARI LETTER
PHA) is obtained through the Unicode Character
Database333https://www.unicode.org/reports/tr44/ for the specific glyph.
Certain glyphs such as a are represented as a combination of two Unicode data-
names and therefore the glyph label is represented as a combination of the two
Unicode data-names separated by the ’+’ symbol. Hence, the glyph-label for the
character a is (DEVANAGARI LETTER A + DEVANAGARI SIGN ANUSVARA).

Two subsets of the dataset that contain images of digits from 0-9 only
(TMNIST-Digit)444https://www.kaggle.com/nimishmagre/tmnist-typeface-mnist and
images of upper and lower-case Latin alphabets (TMNIST-
Alphabet)555https://www.kaggle.com/nikbearbrown/tmnist-alphabet-94-characters
respectively, along with the original TMNIST
dataset666https://www.kaggle.com/nimishmagre/tmnist-glyphs-1812-characters
have also been released. The promising classification results obtained on
these images are available with the datasets and validate their utility. A
more informal method of randomly displaying 100 images from the dataset and
manually verifying the labels was also successfully applied to validate the
data.

The following table summarizes the TMNIST dataset files released so far:

Table 1: Files contained in the Typography-MNIST dataset

##  3 Conclusion

This paper introduces Typography-MNIST (TMNIST), an MNIST-style image dataset
of unique glyphs presented in varying font-styles. The dataset is intended to
provide researchers the advantages of the MNIST data whilst performing glyph
and font-style classification tasks. An extended version of the dataset with
elastic distortions is also available for use and the users are free to
augment the dataset further if required.

## References

  * [1] Nicholas Brown.  Cognitive Type mapping typography to cognition, 2021.
  * [2] Gregory Cohen, Saeed Afshar, Jonathan Tapson, and André van Schaik.  Emnist: Extending mnist to handwritten letters.  In 2017 International Joint Conference on Neural Networks (IJCNN), pages 2921–2926, 2017.
  * [3] Yann LeCun, Léon Bottou, Yoshua Bengio, and Patrick Haffner.  Gradient-based learning applied to document recognition.  Proceedings of the IEEE, 86(11):2278–2324, 1998.