Approximate smoothness in normed linear spaces

Abstract

We introduce the notion of approximate smoothness in a normed linear space. We characterize this property and show the connections between smoothness and approximate smoothness for some spaces. As an application, we consider in particular the Birkhoff–James orthogonality and its right-additivity under the assumption of approximate smoothness.

1 Introduction

Smoothness is definitely one of the most important geometrical properties of normed linear spaces (cf. monographs [7, 13, 15] or a survey [10] for example). In particular, some natural attributes of the Birkhoff–James orthogonality relation can be derived for smooth spaces. Since in its full strength, smoothness can be sometimes a too much restrictive assumption, we are going to propose somehow relaxed, approximate version of this property. We study this concept in general, as well as for some particular linear normed spaces, polyhedral Banach spaces and direct sums of normed linear spaces.

1.1 Notations

Throughout the text, we use the symbols X, Y, Z to denote real normed linear spaces. The zero vector of a normed linear space is denoted by $\theta$, but in case of the scalar field $\mathbb {R}$, we simply use the symbol 0. By $B_X:=\{x\in X: \Vert x\Vert \le 1\}$ and $S_X:=\{x\in X: \Vert x\Vert =1\}$, we denote the unit ball and the unit sphere of X, respectively. The collection of all extreme points of $B_X$ will be denoted as $\textrm{Ext}\,B_X$. For $A\subset X$, $\textrm{diam}\,A:=\sup \nolimits _{x,y\in A}\Vert x-y\Vert$ denotes the diameter of A.

Let $X^*$ denote the dual space of X. Given $x\in X\smallsetminus \{\theta \}$, a functional $f\in S_{X^*}$ is said to be a supporting functional at x if $f(x)=\Vert x\Vert$. The collection of all supporting functionals at x will be denoted by J(x), i.e.,

$$\begin{aligned} J(x):=\{f\in S_{X^{*}}:\ f(x)=\Vert x\Vert \},\qquad x\in X\smallsetminus \{\theta \}. \end{aligned}$$

The Hahn–Banach theorem guarantees that the set J(x) is always nonempty and it is easy to see that it is also convex. It is also known that J(x) is $w^*$-compact. An element $x\in X\smallsetminus \{\theta \}$ is said to be a smooth point if J(x) is a singleton (i.e., $J(x)=\{f\}$ for a unique $f\in S_{X^*}$). A normed linear space X is called smooth if every $x\in S_X$ (hence, every $x\in X\smallsetminus \{\theta \}$) is a smooth point.

For $f\in X^*\smallsetminus \{\theta \}$, we write

$$\begin{aligned} M_{f}:=\{x\in S_{X}: |f(x)|=\Vert f\Vert \} \end{aligned}$$

and

$$\begin{aligned} M_{f}^{+}:=\{x\in S_{X}: f(x)=\Vert f\Vert \}. \end{aligned}$$

Given two elements $x,y\in X$, let

$$\begin{aligned} \overline{xy}{:=}\text{ conv }\,\{x,y\}=\{(1-t)x+ty:t\in [0,1]\} \end{aligned}$$

denote the closed line segment joining x and y. By $\mathcal {R}(X)$ we will denote the length of the “longest” line segment lying on a unit sphere (cf. [5, 20]); more precisely,

$$\begin{aligned} \mathcal {R}(X):=\sup \{\Vert x-y\Vert :\overline{xy}\subset S_X\}. \end{aligned}$$

By a hyperplane, we mean a set

$$\begin{aligned} H{:=}\{x\in X:\ f(x)=c\}, \end{aligned}$$

where $f\in X^*\smallsetminus \{\theta \}$ is a functional and $c\in \mathbb {R}$ a constant. Each hyperplane $H\subset X$ divides X into two closed half-spaces whose intersection is H itself. We call H a supporting hyperplane to the unit ball, if $B_X$ lies entirely within one of the two half-spaces and $H\cap B_X\ne \emptyset$. Equivalently, H is a supporting hyperplane for $B_X$ if and only if there exists $f\in S_{X^*}$ such that $H=\{x\in X:\ f(x)=1\}$. Notice that then we have $H\cap S_X=M_f^+$.

By $\mathfrak {H}(X)$, we denote the set of all supporting hyperplanes for the unit ball in X. Now, we introduce the notion

$$\begin{aligned} \mathcal {S}(X){:=}\sup \{\textrm{diam}\,(H\cap S_X):\ H\in \mathfrak {H}(X)\} \end{aligned}$$

and by previous observations, we have

$$\begin{aligned} \mathcal {S}(X)=\sup \{\textrm{diam}\,M_f^+:\ f\in S_{X^*}\}. \end{aligned}$$

(1.1)

Actually, using the fact that the diameter of a convex set is the supremum of lengths of segments inside the set and both $H\cap S_X$ (for $H\in \mathfrak {H}(X)$) and $M^{+}_{f}$ are convex sets, it follows that

$$\begin{aligned} \mathcal {S}(X)=\mathcal {R}(X). \end{aligned}$$

(1.2)

We define $d:X\smallsetminus \{\theta \} \longrightarrow [0,2]$ by $d(x)=\textrm{diam}\,J(x).$ We also set

$$\begin{aligned} \mathcal {E}(X){:=}\sup \{{d(x)}:\ x\in S_X\}. \end{aligned}$$

By $\Psi : X\longrightarrow X^{**}$, we denote the canonical embedding of a normed linear space X into its bidual $X^{**}$, that is

$$\begin{aligned} \Psi (x)(f){:=}f(x),\qquad x\in X, f\in X^*. \end{aligned}$$

It is known that $\Psi$ is a linear isometry and if it is surjective ($\Psi (X)=X^{**}$), then X is called a reflexive space (necessarily a Banach space).

Using the observation that $\Psi (M_{f}^{+})=J(f)\cap \Psi (X)$ and the fact that $\Psi$ is an isometry, we notice that for any normed linear space X:

$$\begin{aligned} \mathcal {E}(X)\le \mathcal {S}(X^*) \quad \text{ and }\quad \mathcal {S}(X)\le \mathcal {E}(X^*). \end{aligned}$$

(1.3)

In particular, we have

$$\begin{aligned} \mathcal {E}(X)\le \mathcal {S}(X^*)\le \mathcal {E}(X^{**}). \end{aligned}$$

(1.4)

An n-dimensional Banach space X is said to be a polyhedral Banach space if $B_X$ contains only finitely many extreme points or, equivalently, if $S_X$ is a polyhedron. For more details on polyhedral Banach spaces see [17].

We define the sign function on $\mathbb {R}$ by

$$\begin{aligned} \textrm{sgn}\,t:=\frac{t}{|t|},\quad t\in \mathbb {R}\smallsetminus \{0\}\quad \text{ and }\quad \textrm{sgn}\,0:=0. \end{aligned}$$

Let $(X,\Vert \cdot \Vert _X)$ and $(Y, \Vert \cdot \Vert _Y)$ be normed linear spaces. Then,

$$\begin{aligned} X\oplus _p Y:=\{(x,y):x\in X, y\in Y\},\qquad 1\le p \le \infty , \end{aligned}$$

is a normed linear space with respect to the following norm:

$$\begin{aligned} \Vert (x,y)\Vert _p:= {\left\{ \begin{array}{ll} (\Vert x\Vert _X^p+\Vert y\Vert _Y^p)^{\frac{1}{p}}&{} \text {if }p<\infty \\ \max \{\Vert x\Vert _X,\Vert y\Vert _Y\} &{} \text {if }p=\infty . \end{array}\right. } \end{aligned}$$

Let $1\le q\le \infty$ be conjugated to p, i.e., such that $\frac{1}{p}+\frac{1}{q}=1$ for $1<p<\infty$, $q=1$ for $p=\infty$ and $q=\infty$ for $p=1$. Then, the dual space of $X\oplus _p Y$ can be isometrically identified with $X^*\oplus _q Y^*$ in the following sense: for each $F\in (X\oplus _p Y)^*$ there exist a unique $(f,g)\in X^*\oplus _q Y^*$ such that $\Vert F\Vert =\Vert (f,g)\Vert _q$ and $F((x,y))=f(x)+g(y)$ for all $(x,y)\in X\oplus _p Y$ (compare [13, p. 5] or [8, Definition 0.18 and Proposition 0.19]).

1.2 Norm derivatives

It is well known that smoothness of $x\in S_X$ is equivalent to the Gâteaux differentiability of the norm at x. The concept of norm derivatives arises naturally from the two-sided limiting nature of the Gâteaux derivative of the norm. Let us recall the following definition and basic properties of one-sided norm derivatives.

Definition 1.1

Let X be a normed linear space and $x,y\in X$. The norm derivatives of x in the direction of y are defined as

$$\begin{aligned} \rho '_{+}(x,y):=\Vert x\Vert \lim _{\lambda \rightarrow 0^{+}}\frac{\Vert x+\lambda y\Vert -\Vert x\Vert }{\lambda }=\lim _{\lambda \rightarrow 0^{+}}\frac{\Vert x+\lambda y\Vert ^2-\Vert x\Vert ^2}{2\lambda },\\ \rho '_{-}(x,y):=\Vert x\Vert \lim _{\lambda \rightarrow 0^{-}}\frac{\Vert x+\lambda y\Vert -\Vert x\Vert }{\lambda }=\lim _{\lambda \rightarrow 0^{-}}\frac{\Vert x+\lambda y\Vert ^2-\Vert x\Vert ^2}{2\lambda }. \end{aligned}$$

The following properties of norm derivatives will be useful in this note (see [1] and [9] for proofs):

• ($\rho$-i) For all x, $y\in X$ and all $\alpha \in \mathbb {R}$,

  $$\begin{aligned} \rho '_\pm (\alpha x, y)=\rho '_\pm (x,\alpha y) = \left\{ \begin{array}{ll} \alpha \rho '_\pm (x,y) &{} \text{ if } \alpha \ge 0 \\ \alpha \rho '_\mp (x,y) &{} \text{ if } \alpha < 0. \end{array} \right. \end{aligned}$$

• ($\rho$-ii) $\rho '_{-}(x,y)\le \rho '_{+}(x,y)$. Moreover, $x\in X\smallsetminus \{\theta \}$ is smooth if and only if $\rho '_{-}(x,y)= \rho '_{+}(x,y)$ for all $y\in X$.

• ($\rho$-iii) $\rho '_{+}(x,y)=\Vert x\Vert \sup \{f(y):f\in J(x)\}$.

• ($\rho$-iv) $\rho '_{-}(x,y)=\Vert x\Vert \inf \{f(y):f\in J(x)\}$.

1.3 Birkhoff–James orthogonality

For vectors x and y in a normed space X, we say that x is Birkhoff–James orthogonal (BJ-orthogonal) to y, written as $x\perp _B y$, if

$$\begin{aligned} \Vert x+\lambda y\Vert \ge \Vert x\Vert ,\qquad \text{ for } \text{ all }\ \lambda \in \mathbb {R} \end{aligned}$$

(cf. [2, 11, 12]). James in [11, Theorem 2.1] proved that if $x \in X\smallsetminus \{\theta \}$ and $y\in X$, then $x\perp _B y$ if and only if there exists $f\in J(x)$ such that $f(y)=0$.

Chmieliński in [3] defined an approximate Birkhoff–James orthogonality as follows. Given $x,y \in X$ and $\varepsilon \in [0,1)$, x is said to be approximately orthogonal to y, written as $x\perp _B^\varepsilon y$, if

$$\begin{aligned} \Vert x+\lambda y\Vert ^2\ge \Vert x\Vert ^2-2\varepsilon \Vert x\Vert \Vert \lambda y\Vert ,\qquad \text{ for } \text{ all }\ \lambda \in \mathbb {R}. \end{aligned}$$

(1.5)

As shown in [4], condition (1.5) can be equivalently written in the form:

$$\begin{aligned} \Vert x+\lambda y\Vert \ge \Vert x\Vert -\varepsilon \Vert \lambda y\Vert ,\qquad \text{ for } \text{ all }\ \lambda \in \mathbb {R}. \end{aligned}$$

(1.6)

In [6, Theorems 2.2 and 2.3], two important characterizations of the considered approximate orthogonality were given:

$$\begin{aligned} x\perp _B^\varepsilon y\Longleftrightarrow & {} \exists \,z\in \text{ span }\{x,y\}:\ x\perp _B z,\ \Vert z-y\Vert \le \varepsilon \Vert y\Vert ; \end{aligned}$$

(1.7)

$$\begin{aligned}\Longleftrightarrow & {} \exists \,f\in J(x):\ |f(y)|\le \varepsilon \Vert y\Vert . \end{aligned}$$

(1.8)

Obviously, for any f in J(x), we have $|f(y)|\le \Vert y\Vert$. If $|f(y)|<\Vert y\Vert$ then there exists $\varepsilon \in [0,1)$ such that $|f(y)|=\varepsilon \Vert y\Vert$ whence $x\perp _B^{\varepsilon }y$. In particular, it follows from (1.8) for $x,y\in S_X$:

(1.9)

or equivalently

(1.10)

The study of approximate orthogonality has been an active area of research in recent times, particularly in the space of bounded linear operators on a Banach space (see [6, 14, 16, 18] for recent study on approximate orthogonality). Since Birkhoff–James orthogonality is closely related to the classical notion of smoothness in a normed space, the above works motivated us to introduce a suitable notion of approximate smoothness in normed spaces which will be in some sense compatible with approximate orthogonality.

In [11], James obtained a characterization of smooth points in terms of right-additivity of the Birkhoff–James orthogonality relation. Namely, $x\in X\smallsetminus \{\theta \}$ is a smooth point in X if and only if $\perp _B$ is right-additive at x, that is for any $y,z \in X$:

$$\begin{aligned} x\perp _B y\ \ \text{ and }\ \ x\perp _B z\ \implies \ x\perp _B (y+z). \end{aligned}$$

We now define the notion of an approximate right-additivity of the BJ-orthogonality.

Definition 1.2

Let X be a normed linear space and $x\in X\smallsetminus \{\theta \}$. We say that the BJ-orthogonality is approximately right-additive at x ($\varepsilon$-right-additive for some $\varepsilon \in [0,1)$) if for any $y,z \in X$:

$$\begin{aligned} x\perp _B y\ \ \text{ and }\ \ x\perp _B z\ \implies \ x\perp _B^{\varepsilon } (y+z). \end{aligned}$$

Similarly, we define the right-additivity of the approximate BJ-orthogonality.

Definition 1.3

We say that the approximate BJ-orthogonality is right-additive at x if whenever $x\perp _B^\varepsilon y$, $x\perp _B^\varepsilon z$ for some $y,z \in X$ and $\varepsilon \in [0,1)$, then there exists some $\varepsilon _1\in [0,1)$ such that $x\perp _B^{\varepsilon _1} (y+z)$.

Observe that in Definition 1.2 for a given $x\in X\smallsetminus \{\theta \}$, $\varepsilon$ is uniform in the sense that it is independent on the choice of y and z and in the Definition 1.3 for given x, y, z and $\varepsilon$, $\varepsilon _1$ may depend on all of them.

The connections between smoothness and right-additivity of the BJ-orthogonality induce a natural question on a condition, weaker than smoothness, which would characterize the approximate right-additivity of the BJ-orthogonality (or right-additivity of the approximate BJ-orthogonality). This is one of the motivations that lead to the notion of approximate smoothness which is stated in the next section.

2 Approximate smoothness and rotundity

2.1 Approximate smoothness—definition and basic properties

It is obvious that for any $x\in X\smallsetminus \{\theta \}$, $0\le {d(x)}\le 2$ and $d(x)=0$ if and only if x is a smooth point. We will be considering the case when the set J(x) is not necessarily a singleton, but its diameter is limited, in particular strictly less than 2.

Definition 2.1

Let X be a normed linear space, $x\in X\smallsetminus \{\theta \}$ and $\varepsilon \in [0,2)$. We say that x is $\varepsilon$-smooth if ${d(x)}\le \varepsilon$. When the value of $\varepsilon$ is not specified, we say that x is approximately smooth. The space X is said to be approximately smooth ($\varepsilon$-smooth) if each $x\in S_X$ is $\varepsilon _x$-smooth for some $\varepsilon _x\le \varepsilon <{2}$.

Observe that in the above definition, we intentionally excluded the possibility of $\varepsilon =2$. Otherwise, every non-zero element of a normed space would be approximately smooth and every normed space—approximately smooth one. However, our motivation is to distinguish approximately smooth points and spaces, so we restrict to $\varepsilon <2$.

The following lemma will be useful for obtaining a characterization of approximate smoothness in terms of norm derivatives.

Lemma 2.2

Let X be a normed linear space, $x\in X\smallsetminus \{\theta \}$. Then,

$$\begin{aligned} \sup \limits _{y\in S_X}\{\rho '_+(x,y)-\rho '_{-}(x,y)\}=d(x) \Vert x\Vert . \end{aligned}$$

Proof

Using the properties ($\rho$-iii) and ($\rho$-iv), for an arbitrary $y\in S_X$, we have

$$\begin{aligned} \rho '_+(x,y)-\rho '_{-}(x,y)&=\Vert x\Vert \{\sup _{f\in J(x)}f(y)-\inf _{g\in J(x)}g(y)\}\\&=\Vert x\Vert \sup _{f,g\in J(x)}\{f(y)-g(y)\}. \end{aligned}$$

Hence, $\sup \limits _{y\in S_X}\{\rho '_+(x,y)-\rho '_{-}(x,y)\}= d(x) \Vert x\Vert$. $\square$

We now prove the said characterization.

Lemma 2.3

Let X be a normed linear space, $x\in X\smallsetminus \{\theta \}$ and $\varepsilon \in [0,2)$. Then, the following conditions are equivalent:

• (i) x is $\varepsilon$-smooth.

• (ii) $\sup \limits _{y\in S_X}\{\rho '_+(x,y)-\rho '_{-}(x,y)\}\le \varepsilon \Vert x\Vert$.

Proof

(i)$\implies$ (ii) follows from Lemma 2.2.

Now, we prove (ii)$\implies$(i). Let $f,g\in J(x)$ and $y\in S_X$. Then, $\rho '_+(x,y)\ge \Vert x\Vert f(y)$, $\rho '_+(x,y)\ge \Vert x\Vert g(y)$, $\rho '_-(x,y)\le \Vert x\Vert f(y)$ and $\rho '_-(x,y)\le \Vert x\Vert g(y)$. Thus,

$$\begin{aligned} \sup \limits _{y\in S_X}\{\rho '_+(x,y)-\rho '_{-}(x,y)\}\ge \Vert x\Vert \sup _{y\in S_X}|f(y)-g(y)|=\Vert x\Vert \Vert f-g\Vert . \end{aligned}$$

Hence, (ii) yields $\Vert x\Vert \,\Vert f-g\Vert \le \varepsilon \Vert x\Vert$ and this proves (i). $\square$

As an immediate application of Lemma 2.3, we obtain a characterization of the approximate smoothness of the whole space.

Corollary 2.4

Let X be a normed linear space and $\varepsilon \in [0,2)$. Then, the following conditions are equivalent:

• (i) X is $\varepsilon$-smooth.

• (ii) $\sup \limits _{x,y\in S_X}\{\rho '_+(x,y)-\rho '_{-}(x,y)\}\le \varepsilon$.

If X is a reflexive Banach space then we notice that $\Psi (M_{f}^{+})=J(f)$ for $f\in X^*\smallsetminus \{\theta \}$. Since $\Psi$ is an isometry so it follows that $f\in X^*\smallsetminus \{\theta \}$ is $\varepsilon$-smooth if and only if $\textrm{diam}\,M_{f}^{+}\le \varepsilon$. Also, for a reflexive Banach space X using (1.4) and $\mathcal {E}(X)=\mathcal {E}(X^{**})$, we get

$$\begin{aligned} X \text{ is } \varepsilon \text{-smooth } \text{ if } \text{ and } \text{ only } \text{ if } \mathcal {S}(X^*)\le \varepsilon . \end{aligned}$$

(2.1)

Now, we consider a finite-dimensional space. We show that if each vector on a unit sphere is approximately smooth (not necessarily with the same approximation constant), then so is the whole space. Basically, this result is a consequence of Weierstrass compactness theorem.

Theorem 2.5

Let X be a finite-dimensional Banach space such that each $x\in S_X$ is $\varepsilon _x$-smooth for some $\varepsilon _x\in [0,2)$. Then, X is approximately smooth.

Proof

Let $x\in S_X$. Without loss of generality, we assume that $\varepsilon _x=d(x)$.

Let $\varepsilon :=\sup _{x\in S_X}\varepsilon _x$. Obviously, $\varepsilon \le 2$ and suppose, contrary to our claim, that $\varepsilon =2$. Then, we could find $\{x_n\}\subset S_X$ such that $\varepsilon _{x_n}\nearrow 2$. Also, for each $n\in \mathbb {N}$, we can choose $f_n,g_n\in J(x_n)$ such that $\Vert f_n-g_n\Vert >\varepsilon _{x_n}-\frac{1}{n}$.

Now, using the compactness of $S_X$, $S_{X^*}$ and the fact that $\mathcal {A}=\{(x,f,g):x\in S_X, f,g\in J(x)\}$ is a closed set in $S_{X}\times S_{X^*}\times S_{X^*}$, we can find a convergent subsequence of $\{(x_n,f_n,g_n)\}$ which we again denote by $\{(x_n,f_n,g_n)\}$. Let $(x,f,g)\in \mathcal {A}$ be the limit of $\{(x_n,f_n,g_n)\}$. Let $h:X\times X^*\times X^*\longrightarrow \mathbb {R}$ be defined by $h((x,f,g))=\Vert f-g\Vert$. Then, h is clearly a continuous function. Thus, $\lim _{n \rightarrow \infty }h((x_n,f_n,g_n))=h((x,f,g))$ and this implies that $\Vert f-g\Vert =2$. This contradicts our assumption that $\varepsilon _x\in [0,2)$, whence $\varepsilon <2$ and X is $\varepsilon$-smooth. $\square$

A normed space is smooth if and only each of its two-dimensional subspaces is smooth (cf. [15, Proposition 5.4.21]). An analogous result can be proved for approximate smoothness. However, if the approximation constant $\varepsilon$ is not fixed, we can prove it merely for finite-dimensional spaces. The authors do not know whether this assumption is essential.

Theorem 2.6

• (i) Let X be a normed linear space and $\varepsilon \in [0,2)$. Then, X is $\varepsilon$-smooth if and only if each of its two-dimensional subspaces is $\varepsilon$-smooth.

• (ii) Let X be a normed linear space. If X is approximately smooth then so is each of its two-dimensional subspaces. Moreover, if the dimension of X is finite and each of its two-dimensional subspaces is approximately smooth then X is approximately smooth.

Proof

(i). Let X be $\varepsilon$-smooth space and Y its two-dimensional subspace. For $y\in Y\smallsetminus \{\theta \}$ any two supporting functionals at y, $\tilde{f},\tilde{g}\in S_{Y^*}$ can be extended (by the Hahn-Banach theorem) to $f,g\in S_{X^*}$—supporting functionals at y in X. Now, $\Vert \tilde{f}-\tilde{g}\Vert \le \Vert f-g\Vert \le \varepsilon$ implies that Y is $\varepsilon$-smooth.

Conversely, let every two-dimensional subspace of X be $\varepsilon$-smooth. Suppose, on the contrary, that X is not $\varepsilon$-smooth. Then, there exist $x\in S_X$ and $f,g\in J(x)$ such that $\Vert f-g\Vert >\varepsilon$. Let $y\in S_X$ be such that $|(f-g)(y)|>\varepsilon$. Clearly, this choice of y implies that x and y are linearly independent. Let $Y=\text{ span }\{x,y\}$, $\tilde{f}=f|_Y$ and $\tilde{g}=g|_Y$. Then, $\tilde{f}(x)=1$, $\tilde{g}(x)=1$ implies that $\tilde{f},\tilde{g}\in S_{Y^*}$. Also, $|(f-g)(y)|>\varepsilon$ implies that $\Vert \tilde{f}-\tilde{g}\Vert >\varepsilon$. Thus, $\tilde{f},\tilde{g}\in \{h\in S_{Y^*}: h(x)=1\}$ and $\Vert \tilde{f}-\tilde{g}\Vert >\varepsilon$. This leads to the contradiction with $\varepsilon$-smoothness of Y and thus X is $\varepsilon$-smooth.

(ii). If X is approximately smooth then, as above, an application of the Hahn–Banach theorem yields approximate smoothness of any two-dimensional subspace of X.

Now, we assume that X is a finite-dimensional space and let every two-dimensional subspace of X be approximately smooth. Suppose, on the contrary, that X is not approximately smooth. Then, by Theorem 2.5, there exist $x\in S_X$ and $f,g\in J(x)$ such that $\Vert f-g\Vert =2$. Let $y\in S_X$ be such that $|(f-g)(y)|=2$. Clearly, this choice of y implies that x and y are linearly independent. Let $Y=\text{ span }\{x,y\}$, $\tilde{f}=f|_Y$ and $\tilde{g}=g|_Y$. Now, using arguments similar to the proof of (i), we arrive at the contradiction with approximate smoothness of Y and thus X is approximately smooth. $\square$

The following example shows that in some spaces the notions of smoothness and approximate smoothness can coincide.

Example 2.7

(a) Observe that if $x=(x_1,x_2,x_3,\ldots )\in \ell _1\smallsetminus \{\theta \}$ then $d(x)=2$ if $x_{i_0}=0$ for some $i_0\in \mathbb {N}$ and $d(x)=0$ if for all i. Using this observation it follows that $x\in \ell _1\smallsetminus \{\theta \}$ is $\varepsilon$-smooth for $\varepsilon \in [0,2)$ if and only if x is smooth.

(b) If $x=(x_1,x_2,x_3,\ldots )\in c_0\smallsetminus \{\theta \}$ then $d(x)=2$ if norm of x is attained at more than one coordinate and $d(x)=0$ if norm of x is attained only at one coordinate. Thus, it follows that $x\in c_0\smallsetminus \{\theta \}$ is $\varepsilon$-smooth for $\varepsilon \in [0,2)$ if and only if x is smooth.

2.2 Approximate rotundity

Rotundity (strict convexity) is another important geometrical property of normed spaces. Although rotundity and smoothness are independent properties, they are related to each other. We would like to show that it is so with their approximate counterparts.

Let X be a normed linear space and let $\varepsilon \in [0,2)$.

Definition 2.8

We say that X is $\varepsilon$-rotund (or $\varepsilon$-strictly convex) if $\mathcal {S}(X)\le \varepsilon$.

Obviously, for $\varepsilon =0$, the above condition means that each functional $f\in S_{X^*}$ supports the unit sphere in at most one point, which is equivalent to rotundity (cf. [15, Theorem 5.1.15, Corollary 5.1.16]).

Based on (1.3) and (1.4), we can easily establish connections between approximate smoothness (rotundity) of a given space and approximate rotundity (smoothness) of its dual.

Theorem 2.9

Let X be a normed linear space and let $\varepsilon \in [0,2)$.

• 1. If $X^*$ is $\varepsilon$-smooth, then X is $\varepsilon$-rotund;

• 2. If $X^*$ is $\varepsilon$-rotund, then X is $\varepsilon$-smooth;

• 3. If X is reflexive, then X is $\varepsilon$-smooth if and only if $X^*$ is $\varepsilon$-rotund and X is $\varepsilon$-rotund if and only if $X^*$ is $\varepsilon$-smooth.

3 Approximate additivity of the Birkhoff–James orthogonality in approximately smooth spaces

As we have reminded, the Birkhoff–James orthogonality is right-additive at smooth points. Although it is no longer true for non-smooth points, we will show that under a relaxed condition of approximate smoothness, right-additivity can be partially maintained.

We will start with an example showing that, in general, approximate smoothness does not imply approximate right-additivity of the Birkhoff–James orthogonality.

Example 3.1

Let $X=\mathbb {R}^2$ with a norm generated by

$$\begin{aligned} B_{\delta }:=\textrm{conv}\,\{(1,1),(0,1+\delta ),(-1,1),(-1,-1),(0,-1-\delta ),(1,-1)\} \end{aligned}$$

as a unit ball (with $\delta > 0$).

Consider the point $P=(0,1+\delta )$ and functionals

$$\begin{aligned} \begin{array}{c} f(x,y)=\frac{\delta }{1+\delta }x+\frac{1}{1+\delta } y,\\ g(x,y)=-\frac{\delta }{1+\delta }x+\frac{1}{1+\delta } y. \end{array} \end{aligned}$$

Notice that $f,g\in J(P)$ (we denote by $L_f$ and $L_g$ the respective supporting lines) and $J(P)=\textrm{conv}\,\{f,g\}$ (see Fig. 1). Then,

$$\begin{aligned} \textrm{diam}\,J(P)=\Vert f-g\Vert =\sup _{(x,y)\in B_{\delta }}\frac{2\delta }{1+\delta }|x|= \frac{2\delta }{1+\delta }. \end{aligned}$$

Thus, the considered space is $\varepsilon$-smooth at P, with arbitrarily small $\varepsilon$ (if only $\delta$ is small enough). On the other hand, for $R_1=(1,\delta )$ and $R_2=(-1,\delta )$, we have

$$\begin{aligned} P\bot _B R_1,\quad P\bot _B R_2 \end{aligned}$$

but $R_1+R_2=(0,2\delta )=\lambda P$ with $\lambda =\frac{2\delta }{1+\delta }$ hence for any $\varepsilon \in [0,1)$.

Fig. 1

Illustration to Example 3.1

Although approximate smoothness generally does not imply even approximate right-additivity (no matter how small is $\varepsilon$), the following results give some information in particular situations.

Theorem 3.2

Let X be a normed linear space and let $x\in X\smallsetminus \{\theta \}$. Let $y_1, y_2\in X\smallsetminus \{\theta \}$ be such that $x\perp _B^\varepsilon y_1$, $x\perp _B^\varepsilon y_2$, where $\varepsilon \in [0,1)$ is such that

$$\begin{aligned} 0\le \varepsilon<\frac{2\Vert y_1+y_2\Vert }{3(\Vert y_1\Vert +\Vert y_2\Vert )}<1. \end{aligned}$$

If x is $\varepsilon$-smooth, then there exists $\varepsilon _1\in [0,1)$ such that $x\perp _B^{\varepsilon _1} (y_1+y_2)$.

Proof

According to (1.8), we can find $f,g\in J(x)$ such that $|f(y_1)|\le \varepsilon \Vert y_1\Vert$ and $|g(y_2)|\le \varepsilon \Vert y_2\Vert$. Now, using $\varepsilon$-smoothness of x, we get

$$\begin{aligned} |f(y_2)|&=|(f-g)(y_2)+g(y_2)|\\&\le \Vert f-g\Vert \Vert y_2\Vert +\Vert g(y_2)\Vert \\&\le \varepsilon \Vert y_2\Vert +\varepsilon \Vert y_2\Vert \\&=2\varepsilon \Vert y_2\Vert . \end{aligned}$$

Using similar arguments, we can show that $|g(y_1)|\le 2 \varepsilon \Vert y_1\Vert$. Convexity of J(x) yields $\frac{1}{2}f+\frac{1}{2}g \in J(x)$ and

$$\begin{aligned} \left| \left( \frac{1}{2}f+\frac{1}{2}g\right) (y_1+y_2)\right|&\le \frac{1}{2}(\varepsilon \Vert y_1\Vert +2\varepsilon \Vert y_2\Vert +2\varepsilon \Vert y_1\Vert +\varepsilon \Vert y_2\Vert )\\&=\frac{3}{2}\varepsilon (\Vert y_1\Vert +\Vert y_2\Vert )\\&<\Vert y_1+y_2\Vert . \end{aligned}$$

Let ${\varepsilon _1}\in [0,1)$ be such that $|(\frac{1}{2}f+\frac{1}{2}g)(y_1+y_2)|\le \varepsilon _1 \Vert y_1+y_2\Vert$. Then, (1.8) implies that $x\perp _B^{\varepsilon _1} (y_1+y_2)$. $\square$

Theorem 3.3

Let X be a normed linear space and let $x\in X\smallsetminus \{\theta \}$. Let $y_1, y_2\in X\smallsetminus \{\theta \}$ be such that $x\perp _By_1$, $x\perp _By_2$. If x is $\varepsilon$-smooth, where $0\le \varepsilon <\frac{2\Vert y_1+y_2\Vert }{\Vert y_1\Vert +\Vert y_2\Vert }\le 2$, then there exists ${\varepsilon _1}\in [0,1)$ such that $x\perp _B^{\varepsilon _1} (y_1+y_2)$.

Proof

We find $f,g\in J(x)$ such that $f (y_1)=0$ and $g(y_2)=0$. Now, by $\varepsilon$-smoothness of x, we get

$$\begin{aligned} |f(y_2)|=|(f-g)(y_2)|\le \varepsilon \Vert y_2\Vert , \end{aligned}$$

and similarly $|g(y_1)|\le \varepsilon \Vert y_1\Vert$. Clearly, $\frac{1}{2}f+\frac{1}{2}g \in J(x)$ and

$$\begin{aligned} \left| \left( \frac{1}{2}f+\frac{1}{2}g\right) (y_1+y_2)\right| \le \frac{\varepsilon }{2}(\Vert y_1\Vert +\Vert y_2\Vert )<\Vert y_1+y_2\Vert . \end{aligned}$$

Let ${\varepsilon _1}\in [0,1)$ be such that $\left| \left( \frac{1}{2}f+\frac{1}{2}g\right) (y_1+y_2)\right| \le \varepsilon _1 \Vert y_1+y_2\Vert$. Thus, (1.8) implies that $x\perp _B^{\varepsilon _1} (y_1+y_2)$. $\square$

We now provide an example to show that in Theorems 3.2 and 3.3, approximate smoothness of x cannot be omitted. We will need the following lemma.

Lemma 3.4

Let X be a normed linear space and let $x\in S_X$. Then, the approximate Birkhoff–James orthogonality is not right-additive at x if and only if there exist $y_1,y_2\in S_X$, $f_1,f_2\in J(x)$ such that , and $\frac{y_1+y_2}{\Vert y_1+y_2\Vert }\in M_f$ for all $f\in J(x)$.

Proof

The proof relies on characterizations (1.9) and (1.10). Let $x\in S_X$ and suppose that the approximate Birkhoff–James orthogonality is not right-additive at x, i.e., there exist $y_1,y_2\in S_X$, $\varepsilon \in [0,1)$ such that $x\perp _B^\varepsilon y_1$, $x\perp _B^\varepsilon y_2$ and there does not exist any $\varepsilon _1\in [0,1)$ such that $x\perp _B^{\varepsilon _1} \frac{y_1+y_2}{\Vert y_1+y_2\Vert }$. Equivalently, we can write that (due to (1.9)) there exist $f_1,f_2\in J(x)$ such that , as well as (by (1.10)) that $\frac{y_1+y_2}{\Vert y_1+y_2\Vert }\in M_f$ for all $f\in J(x)$. $\square$

If $x\in X\smallsetminus \{\theta \}$ is $\varepsilon$-smooth and $y_1,y_2\in X$ satisfy the conditions stated in Theorem 3.2, then there exists $f\in J(x)$ such that . The following example shows that if in Theorems 3.2 or 3.3 the assumption of x being $\varepsilon$-smooth is omitted, the results are not true.

Example 3.5

Consider $X=(\mathbb {R}^2,\Vert ~\Vert _\infty )$. Let $x,y_1,y_2\in S_X$, where $x=(1,1)$, $y_1=(1,\alpha )$, $y_2=(\alpha ,1)$, $0<\alpha <\frac{1}{2}$ (see Fig. 2A). Then, we can observe, using (1.7), that $x\perp _B^\alpha y_1$ and $x\perp _B^\alpha y_2$. Moreover, $0<\alpha <\frac{1+\alpha }{3}=\frac{2\Vert y_1+y_2\Vert }{3(\Vert y_1\Vert +\Vert y_2\Vert )}$.

Now, we have $\frac{y_1+y_2}{\Vert y_1+y_2\Vert }=x\in M_f$ for all $f\in J(x)$ and, by Lemma 3.4, approximate Birkhoff–James orthogonality is not right-additive at x.

To justify essentialness of approximate smoothness in Theorem 3.3 take $x,z_1,z_2\in S_X$, where $x=(1,1)$, $z_1= (1,-\alpha )$, $z_2=(-\alpha ,1)$, $0<\alpha <\frac{1}{2}$ (see Fig. 2B). Then, $x\perp _Bz_1$, $x\perp _Bz_2$ and clearly, $\frac{z_1+z_2}{\Vert z_1+z_2\Vert }=x\in M_f$ for all $f\in J(x)$. Thus, Lemma 3.4 implies that Birkhoff–James orthogonality is not approximately right-additive at x.

Fig. 2

Illustration to Example 3.5

Finally, in this section, we give a result showing that approximate smoothness of a vector x guarantees the approximate right-additivity of the Birkhoff–James orthogonality on some restricted set of vectors.

Theorem 3.6

Let X be a real normed linear space and let $x\in X\smallsetminus \{\theta \}$ be $\varepsilon$-smooth with $\varepsilon \in [0,2)$. Suppose that $y_1,y_2\in X$ are such that $\min \{\Vert y_1\Vert ,\Vert y_2\Vert \}\le \left\| \frac{y_1+y_2}{2}\right\|$. If $x\bot _B\, y_1$ and $x\bot _B\, y_2$, then

$$\begin{aligned} x\bot _B^{\varepsilon / 2}\,(y_1+y_2). \end{aligned}$$

Proof

Since $x\bot _B\, y_1$ and $x\bot _B\, y_2$, there exist $f,g\in J(x)$ such that $f(y_1)=g(y_2)=0$. Therefore,

$$\begin{aligned} f(y_2)=(f-g)(y_2)\ \ \text{ and }\ \ g(y_1)=(g-f)(y_1) \end{aligned}$$

which, together with the assumed $\textrm{diam}\,J(x)\le \varepsilon$, gives

$$\begin{aligned} |f(y_2)|\le \varepsilon \Vert y_2\Vert \ \ \text{ and }\ \ |g(y_1)|\le \varepsilon \Vert y_1\Vert . \end{aligned}$$

Suppose that $\Vert y_1\Vert \le \Vert y_2\Vert$. It follows then from the assumption that $\Vert y_1\Vert \le \left\| \frac{y_1+y_2}{2}\right\|$ and

$$\begin{aligned} |g(y_1+y_2)|=|g(y_1)|\le \varepsilon \Vert y_1\Vert \le \frac{\varepsilon }{2}\Vert y_1+y_2\Vert . \end{aligned}$$

This means that $x\bot _B^{\varepsilon / 2}\,(y_1+y_2)$.

Similarly, if $\Vert y_2\Vert \le \Vert y_1\Vert$, we show that $|f(y_1+y_2)|\le \frac{\varepsilon }{2}\Vert y_1+y_2\Vert$, which also gives the assertion. $\square$

Note that the condition $\min \{\Vert y_1\Vert ,\Vert y_2\Vert \}\le \left\| \frac{y_1+y_2}{2}\right\|$ depends both on directions and norms of vectors $y_1,y_2$. It holds true, however, regardless of directions, if $\Vert y_1\Vert \ge 3\Vert y_2\Vert$ or $\Vert y_2\Vert \ge 3\Vert y_1\Vert$.

4 Polyhedral spaces

Now, we consider a 2-dimensional regular polyhedral Banach space X with 2n extreme points. Regularity here means that all the edges of the unit sphere are of the same length with respect to the Euclidean metric and all the interior angles are of the same measure. For such spaces, we will calculate the values of d(x) for each $x\in \textrm{Ext}\,B_X$ and the value of $\mathcal {E}(X)$.

Example 4.1

Let X be a 2-dimensional regular polyhedral Banach space with 2n ($n\ge 2$) extreme points and let $x\in \textrm{Ext}\,B_X$. Then,

$$\begin{aligned} {d(x)}=\mathcal {E}(X)= \left\{ \begin{array}{ll} 2 \tan \frac{\pi }{2n} &{} \text{ if } n \text{ is } \text{ even, } \\ 2 \tan \frac{\pi }{2n} \sin \frac{(n-1)\pi }{2n} &{} \text{ if }~n~ \text{ is } \text{ odd. } \end{array} \right. \end{aligned}$$

Fig. 3

Illustration to Example 4.1

Proof

If $x\in S_X\smallsetminus \textrm{Ext}\,B_X$ then clearly $d(x)=0$. Thus, to calculate $\mathcal {E}(X)$ it is sufficient to find d(x) for all $x\in \textrm{Ext}\,B_X$. Moreover, regularity and symmetry of $S_X$ implies that $d(x)=d(y)$ for all $x,y\in \textrm{Ext}\,B_X$.

Without loss of generality, we may assume that

$$\begin{aligned} \textrm{Ext}\,B_X=\{v_k:\ 1\le k \le 2n\}, \end{aligned}$$

where $v_k=\left( \cos \frac{(k-1)\pi }{n},\sin \frac{(k-1)\pi }{n}\right) \ \text{ for }\ 1\le k\le 2n$ (see Fig. 3).

Let $f_k$ be the unique support functional for the segment $\overline{v_kv_{k+1}}$ for $k=1,\ldots ,2n-1$ and $\overline{v_{2n}v_{1}}$ for $k=2n$ (on Fig. 3 the respective supporting lines are denoted by $L_{f_k}$ and $L_{f_{2n}}$). Some calculations, which will be omitted here, lead to the explicit formula for the value of $f_k$ at $(x,y)\in \mathbb {R}^2$:

$$\begin{aligned} f_k((x,y))=\left( x \cos \frac{(2k-1)\pi }{2n} +y \sin \frac{(2k-1)\pi }{2n}\right) \sec \frac{\pi }{2n} \end{aligned}$$

(compare with a similar result in [19, the proof of Theorem 3.1]). Moreover,

$$\begin{aligned} J(v_k)= \left\{ \begin{array}{ll} \overline{f_{k-1}f_k} &{} \text{ if } 1< k\le 2n, \\ \overline{f_{2n}f_{1}} &{} \text{ if } k=1 \end{array} \right. \end{aligned}$$

and thus

$$\begin{aligned} {d(v_k)}= \left\{ \begin{array}{ll} \Vert f_{k-1}-f_k\Vert &{} \text{ if } 1< k\le 2n, \\ \Vert f_{2n}-f_{1}\Vert &{} \text{ if } k=1. \end{array} \right. \end{aligned}$$

For $1< k\le 2n$, we have

$$\begin{aligned} (f_k-f_{k-1})((x,y))&=x\left( \cos \frac{(2k-1)\pi }{2n}-\cos \frac{(2k-3)\pi }{2n}\right) \sec \frac{\pi }{2n}\\&\quad +y\left( \sin \frac{(2k-1)\pi }{2n}-\sin \frac{(2k-3)\pi }{2n}\right) \sec \frac{\pi }{2n}\\&=\left\{ -2x\sin \frac{(k-1)\pi }{n}\sin \frac{\pi }{2n}+2y\cos \frac{(k-1)\pi }{n}\sin \frac{\pi }{2n}\right\} \sec \frac{\pi }{2n}. \end{aligned}$$

To calculate the norm of $f_k-f_{k-1}$ (supremum over the unit sphere), we use the Krein–Milman theorem and restrict ourselves to extremal points. Thus, we have for $1<k\le 2n$

$$\begin{aligned} \begin{array}{l} \Vert f_k-f_{k-1}\Vert =\\ \max _{(x,y)\in \textrm{Ext}\,B_X}\left| -2x\sin \frac{(k-1)\pi }{n}\sin \frac{\pi }{2n}+2y\cos \frac{(k-1)\pi }{n}\sin \frac{\pi }{2n}\right| \sec \frac{\pi }{2n}. \end{array} \end{aligned}$$

(4.1)

If n is even then by taking $k=\frac{n}{2}+1$ in (4.1), we get

$$\begin{aligned} \Vert f_{\frac{n}{2}+1}-f_{\frac{n}{2}}\Vert =\max _{(x,y)\in \textrm{Ext}\,B_X}2|x|\sin \frac{\pi }{2n}\sec \frac{\pi }{2n}. \end{aligned}$$

For $(x,y)\in \textrm{Ext}\,B_X$, $|x|\le 1$. Thus, $\Vert f_{\frac{n}{2}+1}-f_{\frac{n}{2}}\Vert =2 \tan \frac{\pi }{2n}$.

If n is odd, then by taking $k=n+1$ in (4.1), we get

$$\begin{aligned} \Vert f_{n+1}-f_{n}\Vert =\max _{(x,y)\in \textrm{Ext}\,B_X}2|y|\sin \frac{\pi }{2n}\sec \frac{\pi }{2n}. \end{aligned}$$

(4.2)

For $(x,y)\in \textrm{Ext}\,B_X$, $|y|\le \sin \frac{(n-1)\pi }{2n}$. Thus,

$$\begin{aligned} \Vert f_{n+1}-f_{n}\Vert =2\sin \frac{(n-1)\pi }{2n}\sin \frac{\pi }{2n}\sec \frac{\pi }{2n}=2 \tan \frac{\pi }{2n}\sin \frac{(n-1)\pi }{2n}. \end{aligned}$$

(4.3)

Now, using (4.2), (4.3), symmetry and regularity of $S_X$, we get

$$\begin{aligned} \mathcal {E}(X)={d(v_k)}= \left\{ \begin{array}{ll} 2 \tan \frac{\pi }{2n} &{} \text{ if }~n~\text{ is } \text{ even }, \\ 2 \tan \frac{\pi }{2n} \sin \frac{(n-1)\pi }{2n} &{} \text{ if }~n~\text{ is } \text{ odd. } \end{array} \right. \end{aligned}$$

$\square$

As an application of the above, we obtain the following result on approximate smoothness of a 2-dimensional regular polyhedral Banach space X with 2n extreme points.

Proposition 4.2

Let X be a 2-dimensional regular polyhedral Banach space with 2n extreme points ($n\ge 2$). Then, the following holds true.

• (i) X is $\varepsilon$-smooth for $\varepsilon \in [2 \tan \frac{\pi }{2n},2)$, if n is even.

• (ii) X is $\varepsilon$-smooth for $\varepsilon \in [2 \tan \frac{\pi }{2n} \sin \frac{(n-1)\pi }{2n},2)$, if $n\ge 3$ is odd.

In the next result, we give formulas for d(x), $x\in S_X$ and $\mathcal {E}(X)$ for a finite-dimensional polyhedral Banach space X. Note that in case of X being a finite-dimensional polyhedral Banach space, its dual $X^*$ is also a polyhedral Banach space with finitely many extreme points in $B_{X^*}$.

Theorem 4.3

Let X be a finite-dimensional polyhedral Banach space and let $x\in S_X$. Then,

• (i) $d(x)=\max \{\Vert f_i-f_j\Vert :\ f_i,f_j\in \textrm{Ext}\,B_{X^*}\ \text{ such } \text{ that }\ f_i,f_j\in J(x)\},$

• (ii)

Proof

(i) Let $x\in S_X$ and $F_{i_1},F_{i_2},\ldots ,F_{i_k}$ be the facets of $S_X$ which contain x. Let $f_{i_1},f_{i_2},\ldots , f_{i_k}\in \textrm{Ext}\,B_{X^*}$ be the unique supporting functionals for the facets $F_{i_1},F_{i_2},\ldots ,F_{i_k}$, respectively. Let $f,g\in J(x)$. Then, $f=\sum _{j=1}^{k}\alpha _j f_{i_j}$, $g=\sum _{j=1}^{k}\beta _j f_{i_j}$, where $0\le \alpha _j,\beta _j\le 1$ and $\sum _{j=1}^k\alpha _j=1$, $\sum _{j=1}^k\beta _j=1$. Now,

$$\begin{aligned} \Vert f-g\Vert= & {} \left\| \sum _{j=1}^{k}\alpha _j f_{i_j}-g\right\| =\left\| \sum _{j=1}^{k}\alpha _j f_{i_j}-\sum _{j=1}^{k}\alpha _j g\right\| \\= & {} \left\| \sum _{j=1}^{k}\alpha _j(f_{i_j}-g)\right\| \le \sum _{j=1}^{k}\alpha _j\Vert f_{i_j}-g\Vert . \end{aligned}$$

Similar arguments show that

$$\begin{aligned} \Vert f_{i_j}-g\Vert \le \sum _{\ell =1}^{k}\beta _\ell \Vert f_{i_j}-f_{i_\ell }\Vert \le \max \limits _{1\le \ell \le k}\Vert f_{i_j}-f_{i_\ell }\Vert \end{aligned}$$

for all $1\le j\le k$. Thus, $\Vert f-g\Vert \le \max \limits _{1\le j,\ell \le k}\Vert f_{i_j}-f_{i_\ell }\Vert$ and this proves the result.

(ii) Observe that if $y\in F$ for some facet F and $f,g\in J(y)$ then $f,g\in J(x)$ for some $x\in \textrm{Ext}\,B_X\cap F$. Thus, to calculate $\mathcal {E}(X)$ it is sufficient to consider $x\in \textrm{Ext}\,B_X$. Now, (ii) follows from (i) by observing that if $f,g\in J(x)$ for some $x\in \textrm{Ext}\,B_X\cap F$ and facet F then . $\square$

5 Direct sums

Given normed linear spaces X, Y, we study the space $Z=X\oplus _p Y$, $1\le p\le \infty$, and approximate smoothness of its elements. We start with a description of the set of supporting functionals.

Proposition 5.1

Let X, Y be normed linear spaces and let $Z=X\oplus _p Y$ with $1\le p\le \infty$.

• 1. If $1<p<\infty$ and q is conjugated to p, then for any $x\in X\smallsetminus \{\theta \}$ and $y\in Y\smallsetminus \{\theta \}$, we have

  • (a)
    $$\begin{aligned} \begin{array}{l} J((x,y))=\\ \left\{ \left( \frac{\Vert x\Vert ^{p-1}f}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{q}}},\frac{\Vert y\Vert ^{p-1}g}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{q}}} \right) \in S_{X^*\oplus _q Y^*}:\ f\in J(x),\ g\in J(y) \right\} , \end{array} \end{aligned}$$

    (5.1)

  • (b) $J((x,\theta ))=\{ (f,\theta )\in S_{X^*\oplus _q Y^*}:\ f\in J(x) \}$,

  • (c) $J((\theta ,y))=\{ (\theta ,g)\in S_{X^*\oplus _q Y^*}:\ g\in J(y) \}$.

• 2 If $p=1$, $x\in X\smallsetminus \{\theta \}$, $y\in Y\smallsetminus \{\theta \}$, then

  • (a) $J((x,y))=\{(f,g)\in S_{X^*\oplus _{\infty } Y^*}:\ f\in J(x),\ g\in J(y)\}$,

  • (b) $J((x,\theta ))=\{(f,g)\in S_{X^*\oplus _{\infty } Y^*}:\ f\in J(x),\ g\in B_{Y^*}\},$

  • (c) $J((\theta ,y))=\{(f,g)\in S_{X^*\oplus _{\infty } Y^*}:\ f\in B_{X^*},\ g\in J(y)\}.$

• 3 Let $p=\infty$ and $(x,y)\in Z\smallsetminus \{\theta \}$.

  • (a) If $\Vert x\Vert >\Vert y\Vert$, then

    $$\begin{aligned} J((x,y))=\{(f,\theta )\in S_{X^*\oplus _{1} Y^*}:\ f\in J(x)\}. \end{aligned}$$

  • (b) If $\Vert x\Vert <\Vert y\Vert$, then

    $$\begin{aligned} J((x,y))=\{(\theta ,g)\in S_{X^*\oplus _{1} Y^*}:\ g\in J(y)\}. \end{aligned}$$

  • (c) If $\Vert x\Vert =\Vert y\Vert$, then

    $$\begin{aligned} \{(\alpha f, (1-\alpha )g): f\in J(x), g\in J(y), 0\le \alpha \le 1\}\subseteq J((x,y)). \end{aligned}$$

Proof

(1a). Let $F\in J((x,y))$, where $F=(f,g)\in S_{X^*\oplus _q Y^*}$. Then, using Hölder’s inequality, we get

$$\begin{aligned} ( \Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{p}}= & {} \Vert (x,y)\Vert _p=F((x,y))=f(x)+g(y)\\\le & {} \Vert f\Vert \Vert x\Vert +\Vert g\Vert \Vert y\Vert \\\le & {} (\Vert f\Vert ^q+\Vert g\Vert ^q)^{\frac{1}{q}} (\Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{p}}\\= & {} (\Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{p}}. \end{aligned}$$

This shows that equality holds in Hölder’s inequality and thus

$$\begin{aligned} \Vert f\Vert =\frac{\Vert x\Vert ^{p-1}}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{q}}},~ \Vert g\Vert =\frac{\Vert y\Vert ^{p-1}}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^{\frac{1}{q}}}. \end{aligned}$$

(5.2)

Also, we get $f(x)=\Vert f\Vert \Vert x\Vert$, $g(y)=\Vert g\Vert \Vert y\Vert$ which implies that $\tilde{f}=\frac{f}{\Vert f\Vert }\in J(x)$ and $\tilde{g}=\frac{g}{\Vert g\Vert }\in J(y)$. Combining this with (5.2), we get

$$\begin{aligned} F=(f,g)=\left( \frac{\Vert x\Vert ^{p-1}\tilde{f}}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}, \frac{\Vert y\Vert ^{p-1}\tilde{g}}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}\right) \end{aligned}$$

which shows $\subseteq$ in (5.1). Now, consider

$$\begin{aligned} F=\left( \frac{\Vert x\Vert ^{p-1}f}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}, \frac{\Vert y\Vert ^{p-1}g}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}\right) \in S_{X^*\oplus _q Y^*}, \end{aligned}$$

where $f\in J(x)$ and $g\in J(y)$. Then, (we use that $(p-1)q=p$)

$$\begin{aligned} F((x,y))=\frac{\Vert x\Vert ^p+\Vert y\Vert ^p}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}=\Vert (x,y)\Vert _p. \end{aligned}$$

Thus, $F\in J((x,y))$ and this proves $\supseteq$ in (5.1) and finishes the proof of part (1a).

Parts (1b) and (1c) follow using the similar reasoning.

(2). For the proof of (2a) let $x\in X{\setminus }\{\theta \}$, $y\in Y{\setminus }\{\theta \}$, $f\in J(x)$ and $g\in J(y)$. Then, $(f,g)\in S_{ X^*\oplus _\infty Y^*}$ and $(f,g)((x,y))=f(x)+g(y)=\Vert x\Vert +\Vert y\Vert =\Vert (x,y)\Vert _1$. Thus, $(f,g)\in J((x,y))$.

For the reverse, let $(f,g)\in J((x,y))$. Then, $\Vert f\Vert \le 1$, $\Vert g\Vert \le 1$ and

$$\begin{aligned} \Vert x\Vert +\Vert y\Vert= & {} \Vert (x,y)\Vert _1=(f,g)((x,y))=f(x)+g(y)\\\le & {} \Vert f\Vert \Vert x\Vert +\Vert g\Vert \Vert y\Vert \le \Vert x\Vert +\Vert y\Vert . \end{aligned}$$

This shows that $\Vert f\Vert =1$, $\Vert g\Vert =1$, $f(x)=\Vert x\Vert$ and $g(y)=\Vert y\Vert$. Thus, $f\in J(x)$ and $g\in J(y)$.

To prove (2b) let $x\in X\setminus \{\theta \}$, $f\in J(x)$ and $g\in B_{Y^*}$. Then, $(f,g)\in S_{ X^*\oplus _\infty Y^*}$ and $(f,g)((x,\theta ))=f(x)=\Vert x\Vert =\Vert (x,\theta )\Vert _1$. Thus, $(f,g)\in J((x,\theta ))$. The reverse inclusion is clear.

The proof of (2c) is analogous.

(3). For the proof of (3a) let $x\in X{\setminus } \{\theta \}$, $y\in Y$ be such that $\Vert x\Vert >\Vert y\Vert$. Let $f\in J(x)$. Then, clearly $(f,\theta )\in S_{X^*\oplus _1Y^*}$, $(f,\theta )((x,y))=f(x)=\Vert x\Vert =\Vert (x,y)\Vert _\infty$ and thus $(f,\theta )\in J((x,y))$.

Now, let $(f,g)\in J((x,y))$, where $g\in B_{Y^*}$. If $f=\theta$, then

$$\begin{aligned} (f,g)((x,y))=g(y)\le \Vert y\Vert <\Vert (x,y)\Vert _\infty . \end{aligned}$$

Thus, . Suppose . Since $(f,g)\in S_{ X^*\oplus _1 Y^*}$, , , gives $\Vert f\Vert <1$ and $\Vert g\Vert <1$. Also, $\Vert f\Vert +\Vert g\Vert =1$ implies $\Vert g\Vert =1-\Vert f\Vert$. Thus,

$$\begin{aligned} \Vert x\Vert= & {} \Vert (x,y)\Vert _\infty =(f,g)((x,y))=f(x)+g(y)\\\le & {} \Vert f\Vert \Vert x\Vert +\Vert g\Vert \Vert y\Vert <\Vert f\Vert \Vert x\Vert +(1-\Vert f\Vert )\Vert x\Vert =\Vert x\Vert . \end{aligned}$$

The obtained contradiction proves the required form of J((x, y)).

The proof of (3b) is analogous.

Finally, we prove (3c). Let $x\in X{\setminus } \{\theta \}$, $y\in Y{\setminus } \{\theta \}$ be such that $\Vert x\Vert =\Vert y\Vert$. Let $f\in J(x)$, $g\in J(y)$ and $0\le \alpha \le 1$. Then, $(\alpha f, (1-\alpha )g)\in S_{X^*\oplus _1Y^*}$ and $(\alpha f, (1-\alpha )g)((x,y))=\alpha f(x)+(1-\alpha )g(y)=\alpha \Vert x\Vert +(1-\alpha )\Vert y\Vert =\Vert x\Vert =\Vert (x,y)\Vert _\infty$. This proves the result. $\square$

Corollary 5.2

Let X, Y be normed linear spaces and let $Z=X\oplus _p Y$ with $1\le p\le \infty$.

• 1. If $1<p<\infty$ and q is conjugated to p, then for any $x\in X\smallsetminus \{\theta \}$ and $y\in Y\smallsetminus \{\theta \}$, we have

  • (a)
    $$\begin{aligned} d((x,y))=\left( \frac{\Vert x\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p}d(x)^{q}+\frac{\Vert y\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p}d(y)^{q} \right) ^{\frac{1}{q}}, \end{aligned}$$

    (5.3)

  • (b) $d((x,\theta ))=d(x)$,

  • (c) $d((\theta ,y))=d(y)$.

• 2. If $p=1$, $x\in X\smallsetminus \{\theta \}$, $y\in Y\smallsetminus \{\theta \}$, then

  • (a) $d((x,y))=\max \{d(x),d(y)\}$,

  • (b) $d((x,\theta ))=2$,

  • (c) $d((\theta ,y))=2$.

• 3. Let $p=\infty$, $(x,y)\in Z\smallsetminus \{\theta \}$.

  • (a) If $\Vert x\Vert >\Vert y\Vert$, then $d((x,y))=d(x)$.

  • (b) If $\Vert x\Vert <\Vert y\Vert$, then $d((x,y))=d(y)$.

  • (c) If $\Vert x\Vert =\Vert y\Vert$, then $d((x,y))=2$.

Proof

We will prove (1a). Let $F,G\in J((x,y))$. On account of (5.1), there exist $f_x,g_x\in J(x)$, $f_y,g_y\in J(y)$ such that

$$\begin{aligned} F=\left( \frac{\Vert x\Vert ^{p-1}f_x}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}, \frac{\Vert y\Vert ^{p-1}f_y}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}\right) \end{aligned}$$

and

$$\begin{aligned} G=\left( \frac{\Vert x\Vert ^{p-1}g_x}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}, \frac{\Vert y\Vert ^{p-1}g_y}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}\right) . \end{aligned}$$

This gives (using $(p-1)q=p$)

$$\begin{aligned} \Vert F-G\Vert _q^q= \frac{\Vert x\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p }\Vert f_x-g_x\Vert ^q+\frac{\Vert y\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p }\Vert f_y-g_y\Vert ^q \end{aligned}$$

and thus,

$$\begin{aligned} d((x,y))^q \le \frac{\Vert x\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p }d(x)^q+\frac{\Vert y\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p }d(y)^q. \end{aligned}$$

To prove the reverse inequality fix arbitrarily $\delta >0$ and choose $f_x^{\delta },g_x^{\delta }\in J(x)$ such that $\Vert f_x^{\delta }-g_x^{\delta }\Vert ^q>d(x)^{q}-\delta$. Analogously, let $f_y^{\delta },g_y^{\delta }\in J(y)$ be chosen such that $\Vert f_y^{\delta }-g_y^{\delta }\Vert ^q>d(y)^{q}-\delta$. Define

$$\begin{aligned} F^{\delta }:=\left( \frac{\Vert x\Vert ^{p-1}f_x^{\delta }}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}, \frac{\Vert y\Vert ^{p-1}f_y^{\delta }}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}\right) \end{aligned}$$

and

$$\begin{aligned} G^{\delta }:=\left( \frac{\Vert x\Vert ^{p-1}g_x^{\delta }}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}, \frac{\Vert y\Vert ^{p-1}g_y^{\delta }}{(\Vert x\Vert ^p+\Vert y\Vert ^p)^\frac{1}{q}}\right) . \end{aligned}$$

By (5.1), $F^{\delta },G^{\delta }\in J((x,y))$ whence

$$\begin{aligned} d((x,y))^q\ge & {} \Vert F^{\delta }-G^{\delta }\Vert _q^q\\> & {} \frac{\Vert x\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p }(d(x)^q-\delta )+\frac{\Vert y\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p}(d(y)^q-\delta ) \end{aligned}$$

and since $\delta >0$ was arbitrary, we get

$$\begin{aligned} d((x,y))^q\ge \frac{\Vert x\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p }d(x)^q+\frac{\Vert y\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p}d(y)^q. \end{aligned}$$

The proofs of other cases are similar or obvious. $\square$

Now, as a straightforward consequence of Corollary 5.2, we characterize approximate smoothness of the direct sum.

Theorem 5.3

Let X, Y be normed linear spaces and $Z=X\oplus _p Y$, $1<p<\infty$. Then, the following statements hold true:

• (i) Let $x\in X\smallsetminus \{\theta \}$, $y\in Y\smallsetminus \{\theta \}$.

  • (a) If x is $\varepsilon _x$-smooth in X and y is $\varepsilon _y$-smooth in Y for $\varepsilon _x,\varepsilon _y\in [0,2)$, then (x, y) is $\varepsilon$-smooth in Z with

    $$\begin{aligned} \varepsilon :=\left( \frac{\Vert x\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p}\varepsilon _x^{q}+\frac{\Vert y\Vert ^{p}}{\Vert x\Vert ^p+\Vert y\Vert ^p}\varepsilon _y^{q}\right) ^{\frac{1}{q}}. \end{aligned}$$

  • (b) If x is $\varepsilon$-smooth in X and y is $\varepsilon$-smooth in Y for $\varepsilon \in [0,2)$, then (x, y) is $\varepsilon$-smooth in Z.

  • (c) If (x, y) is $\varepsilon$-smooth in Z for $\varepsilon \in [0,2)$, then either x is $\varepsilon$-smooth in X or y is $\varepsilon$-smooth in Y.

  • (d) (x, y) is approximately smooth if and only if either x or y is approximately smooth.

• (ii) Let $x\in X\smallsetminus \{\theta \}$. Then, $(x,\theta )\in Z$ is $\varepsilon$-smooth for $\varepsilon \in [0,2)$ if and only if x is $\varepsilon$-smooth in X.

• (iii) Let $y\in Y\smallsetminus \{\theta \}$. Then, $(\theta ,y)\in Z$ is $\varepsilon$-smooth for $\varepsilon \in [0,2)$ if and only if y is $\varepsilon$-smooth in Y.

Proof

For the proof of (i), we need to apply Corollary 5.2 (1a). In particular, for (id), we observe that $d(x,y)<2$ if and only if $d(x)<2$ or $d(y)<2$. The statements (ii) and (iii) immediately follow from Corollary 5.2 (1b) and (1c), respectively. $\square$

Theorem 5.4

Let X, Y be normed linear spaces and $Z=X\oplus _1 Y$. Then, the following hold true:

• (i)

  • (a) If x is $\varepsilon _x$-smooth in X and y is $\varepsilon _y$-smooth in Y for $\varepsilon _x,\varepsilon _y\in [0,2)$, then (x, y) is $\varepsilon$-smooth in Z with $\varepsilon :=\max \{\varepsilon _x,\varepsilon _y\}$.

  • (b) If x is $\varepsilon$-smooth in X and y is $\varepsilon$-smooth in Y for $\varepsilon \in [0,2)$, then (x, y) is $\varepsilon$-smooth in Z.

  • (c) If (x, y) is $\varepsilon$-smooth in Z for $\varepsilon \in [0,2)$, then both x and y are $\varepsilon$-smooth in X and Y, respectively.

  • (d) (x, y) is approximately smooth if and only if both x and y are approximately smooth.

• (ii) If $x\in X\smallsetminus \{\theta \}$ then $(x,\theta )\in Z$ cannot be approximately smooth.

• (iii) If $y\in Y\smallsetminus \{\theta \}$ then $(\theta ,y)\in Z$ cannot be approximately smooth.

Proof

The proof of (i) relies on Corollary 5.2 (2a). Statements (ii) and (iii) follow immediately from Corollary 5.2 (2b) and (2c), respectively. $\square$

Theorem 5.5

Let X, Y be normed linear spaces, $Z=X\oplus _\infty Y$ and $z=(x,y)\in Z\smallsetminus \{\theta \}$. Then, the following statements hold true:

• (i) Let $\Vert x\Vert >\Vert y\Vert$. Then, z is $\varepsilon$-smooth in Z for $\varepsilon \in [0,2)$ if and only if x is $\varepsilon$-smooth in X.

• (ii) Let $\Vert x\Vert <\Vert y\Vert$. Then, z is $\varepsilon$-smooth in Z for $\varepsilon \in [0,2)$ if and only if y is $\varepsilon$-smooth in Y.

• (iii) If $\Vert x\Vert =\Vert y\Vert$, then z cannot be approximately smooth.

Proof

Statements (i) and (ii) follow from Corollary 5.2 (3a) and (3b), whereas (iii) is a consequence of Corollary 5.2 (3c). $\square$

The following final example is an application of the above theorem and the fact that any $t\in \mathbb {R}\smallsetminus \{0\}$ is a smooth point of $\mathbb {R}$.

Example 5.6

Let $Z=X\oplus _\infty \mathbb {R}$ be a 3-dimensional polyhedral Banach space whose unit ball is a right prism with regular polyhedron P as its base. Then, the following hold true for $z=(x,t)\in Z\smallsetminus \{\theta \}$:

• (i) Let $\Vert x\Vert >|t|$. Then, z is $\varepsilon$-smooth in Z for $\varepsilon \in [0,2)$ if and only if x is $\varepsilon$-smooth in X.

• (ii) Let $\Vert x\Vert <|t|$. Then, z is smooth.

• (iii) If $\Vert x\Vert =|t|$, then z cannot be approximately smooth.