The finite Jung constant in Banach spaces

Abstract

We study in this paper the finite Jung constant, its interplay with Kottman’s constant and its meaning regarding the geometry of Banach spaces.

1 Introduction: the Jung constants

Given a bounded subset $A\subset X$ the diameter of A is defined as $\delta (A) = \sup \{ \Vert a-b\Vert : a, b\in A\}$, while the radius of A in X is defined by $r_X(A) = \inf _{b\in X} \sup _{a\in A} \Vert a-b\Vert$. If the infimum is attached at a point b then this point is called a center for A; if only $\sup _{a\in A} \Vert a-b\Vert \le r_X(A)+\varepsilon$ then b will be called an $\varepsilon$-center. The Jung constant [16] of A is defined as

$$\begin{aligned} J(A) = \frac{2r_X(A)}{\delta (A)} \end{aligned}$$

while the Jung constant of X is the supremum $J(X) = \sup J(A)$ taken over all closed bounded sets A with $\delta (A)>0$. A combination of results by Davis [11], Franchetti in [14] and Lindenstrauss [19] show that a Banach space is 1-injective if and only if $J(X)=1$. Recall that a Banach space X is $\lambda$-injective if for every Banach space F and every subspace E of F every operator $t: E\rightarrow X$ has an extension $T: F\rightarrow X$ with $\Vert T\Vert \le \lambda \Vert t\Vert$.

Two important variations of this notion [2, 4] are $\lambda$-separable injectivity, when the property above holds when F is separable; and local $\lambda$-injectivity, when the preceding property holds when F is finite dimensional. We can consider the corresponding variation of Jung’s constant for separability and obtain the separable Jung constant $J_s(\cdot )$, introduced in [7] as

$$\begin{aligned} J_s(X) = \sup J(A) \end{aligned}$$

where the supremum is taken over all separable closed bounded sets A with $\delta (A)>0$. In this paper we will consider the finite Jung constant introduced by Amir [1] (see also [5]) and defined as

$$\begin{aligned} J_f(X) = \sup J(A) \end{aligned}$$

where the supremum is taken over all finite sets A with $\delta (A)>0$.

The first type of characterization we are interested in this paper was obtained by Davis [11]: a Banach space X is 1-injective if and only if $J(X)=1$. We obtained in [7] the corresponding characterization for $J_s$: a Banach space X is separably 1-injective if and only if $J_s(X)=1$. Our first set of results in this paper provided in Sect. 2 deal with the characterization of the spaces X for which $J_f(X)=1$. It was (implicitly) proved by Bayod and Masa [3] that $J_f (X) = 1$ if and only if X is a Lindenstrauss space. This fact was reproved in [25], while in [13] it was observed that a careful reading of [19] yields the same characterization. Moreover, they show [13, Theorem 2.7] that $J_f (X) = 1$ if and only if every four-point set of diameter 1 has radius 1/2 and a center.

2 Banach spaces with finite Jung constant 1

In this Section, we will prove the Lindenstrauss–Bayod–Masa characterization of Lindenstrauss spaces through a new equality $J_f(X) = J_s(X_\mathcal {U})$ for some ultrapower of X. This characterization will have a few interesting consequences.

Semenov and Franchetti [26, Lemma 2.4] show that if Y, X are Banach spaces such that for each $\varepsilon >0$ the space X contains $(1+\varepsilon )$-isomorphic $(1+\varepsilon )$-complemented copy of Y then $J(Y)\le J(X)$. In particular, if Y is a $1^+$-complemented subspace of X then $J(Y)\le J(X)$. We generalize this:

Lemma 2.1

Let $Y\subset X$ and Z be Banach spaces.

• (1) If Y is $\lambda ^+$-complemented in X then $J(Y)\le \lambda J(X)$, $J_s(Y)\le \lambda J_s(X)$ and $J_f(Y)\le \lambda J_f(X)$.

• (2) If Y is locally $\lambda ^+$-complemented in X then $J_f(Y)\le \lambda J_f(X)$

• (3) $J(X\oplus _\infty Z)= \max \{J(X), J(Z)\}$; $J_s(X\oplus _\infty Z)= \max \{J_s(X), J_s(Z)\}$; $J_f(X\oplus _\infty Z)= \max \{J_f(X), J_f(Z)\}$

Proof

(1) To avoid further confusion, given $Y\subset X$ and $A\subset Y$ let $r_Y(A)$ (resp. $r_X(A))$ denote the radius of A in Y (resp. in X). Thus, we need to show that $r_Y(A)\le \lambda r_X(A)$ for every $A\subset Y$ with $0<\delta (A)\le 1$. Let P be a projection on X onto Y with $\Vert P\Vert \le \lambda$. If $x\in X$ and $\Vert a-x\Vert \le r$ for each $a\in A$, taking $y=Px\in Y$ we have $\Vert a-y\Vert =\Vert P(a-x)\Vert \le \lambda \Vert a-x\Vert \le r$ and the result follows.

(2) Assume now that $A\subset Y$ is finite and Y is locally $\lambda ^+$-complemented in X. If $x\in X$ and $\Vert a-x\Vert \le r$ for each $a\in A$, pick $P: A+ [x]\longrightarrow Y$ a projection with $\Vert P\Vert \le \lambda +\varepsilon$ so that $\Vert a-Py\Vert =\Vert P(a-x)\Vert \le (\lambda +\varepsilon )\Vert a-x\Vert \le (\lambda +\varepsilon )r$ and the result follows as well.

(3) We make the proof for J, but the proofs for $J_s$ and $J_f$ are analogous. By (1), $J(X\oplus _\infty Y)\ge \max \{J(X), J(Y)\}$. Conversely, let $A\subset X\oplus _\infty Y$ with $0<\delta (A)\le 1$. Let $\pi _X$ the canonical projection onto X. The sets $B=\pi _X(A)\subset X$ and $C=(I-\pi _X)(A)\subset Y$ satisfy $\delta (B), \delta (C)\le 1$. If $\delta (B)=0$ then $\delta (C)=\delta (A)$ and $r_Y(C)=r_{X\oplus _\infty Y}(A)$, and analogously when $\delta (C)=0$. Assume then that $\delta (B)>0$ and $\delta (C)>0$. If we fix $\varepsilon >0$ and pick $x_0\in X$ and $y_0\in Y$ such that $\Vert x_0-b\Vert <r_X(B)+\varepsilon$ for each $b\in B$ and $\Vert y_0-c\Vert <r_Y(C)+\varepsilon$ for each $c\in C$ then $\Vert (x_0,y_0)-(b,c)\Vert = \max \{\Vert x_0-b\Vert ,\Vert y_0-c\Vert \}<\max \{r_X(B),r_Y(C)\}+\varepsilon$ for each $(b,c)\in A$. On the other hand, $\delta (A) = \max \{\delta _X(B), \delta _Y(C)\}$. Therefore

$$\begin{aligned} J(X\oplus _\infty Y) = \sup \frac{2r(A)}{\delta (A)}\le \sup \frac{2\max \{r_X(B),r_Y(C)\}}{\delta (A)} \le \max \{J(X), J(Y)\}. \end{aligned}$$

$\square$

It is clear that this result says nothing for $\lambda \ge 2$. One could suspect that the parameter $\lambda$ plays no role in either (1) or (2). Let us show it is not so. We discuss (2) first, and recall from [4, Chapter 9] the existence of an exact sequence $0\rightarrow C[0,1] \rightarrow \Omega \rightarrow c_0 \rightarrow 0$ in which $\Omega$ cannot be renormed to be a Lindenstrauss space. This sequence can be placed in a commutative diagram

which therefore (see again [4]) yields a commutative diagram

in which the lower sequence locally $1^+$-splits and the space P is isomorphic to $\ell _\infty \oplus _\infty c_0$. Thus, after renorming, we have a locally $\lambda ^+$-split sequence

in which $J_f(\ell _\infty \oplus _\infty )=1$ but $J_f(\Omega )>1$. The bidual sequence

provides a counterexample for (1): it splits by [4], namely, $\Omega ^{**}$ is complemented in $\ell _\infty ^{**} \oplus _\infty \ell _\infty$; it cannot be 1-complemented because otherwise, by the Principle of Local Reflexivity, $\Omega$ would be locally $1^+$-complemented in $\ell _\infty \oplus _\infty c_0$, which is not the case. Hence $\Omega ^{**}$ is not 1-injective, and therefore $J(\Omega ^{**})>1$, while $J(\ell _\infty ^{**} \oplus _\infty \ell _\infty )=1$. The same works, under the Continuum Hypothesis, regarding $J_s$: if $\Omega$ were 1-separably injective then it would be universally 1-separably injective (see [2]), hence 1-complemented in $\ell _\infty \oplus _\infty c_0$, that we know it is not.

As a consequence of the results in [7, 9], J(X) (resp. $J_s(X)$) and $J(X^{**})$ (resp. $J_s(X^{**})$) can be different since $J(c_0)=2=J_s(c_0)$ while $J(\ell _\infty )=J_s(\ell _\infty )=1$. In general, given a countably incomplete ultrafilter $\mathcal {U}$ on ${\mathbb {N}}$, one has $J_s(C[0,1])=2$ but $J_s(C[0,1]_\mathcal {U})=1$ since according to [2] the ultrapower of a Lindenstrauss space is 1-separably injective. On the other hand, $J(\ell _\infty )=1$ but $J((\ell _\infty )_\mathcal {U})>1$ since, again according to [2] no infinite-dimensional ultrapower is injective. The finite Jung constant behaves, however, differently:

Proposition 2.2

$J_f(X)= J_f(X^{**})$.

Proof

The inequality $J_f(X) \le J_f(X^{**})$ follows from (2) in the previous Lemma. Next observe that for a certain ultrafilter $\mathcal {U}$ the space $X^{**}$ is 1-complemented in $X_\mathcal {U}$, and therefore $J_f(X^{**})^\le J_f(X_\mathcal {U})$ by part (1) of the Lemma above. It remains to show that $J_f(X_\mathcal {U})\le J_f(X)$. Let $\varepsilon >0$ be fixed and pick a finite set $A = \{a^1, \dots , a^m\} \subset X_\mathcal {U}$ with $\delta (A)=\Vert a^u - a^v\Vert = 1$ such that $J_f(X_\mathcal {U})\le 2r(A) + \varepsilon$ and let a be an $\varepsilon$-approximate center for A; namely $\Vert a^j - a\Vert \le r(A) + \varepsilon$ for $1\le j\le m$. Assume that $a^j = [a_1^j, \dots , a^j_n, \dots ]$ and $a = [a_1, \dots , a_n, \dots ]$. Since the sets $U_j =\{n: \Vert a^j_n - a_n\Vert \le r(A) + 2\varepsilon \}$ belong to $\mathcal {U}$ for $j=1,..., m$ as well as $U=\{n: \Vert a^u_n - a_n^v\Vert \ge 1 - \varepsilon \}$ so does $\bigcap _j U_j \cap U$. Thus, picking n in this set, $B=\{a^1_n, \dots , a^m_n\} \subset X$ has diameter at least $1-\varepsilon$ and the point $a_n$ is a $2\varepsilon$-approximate center for B. All this yields

$$\begin{aligned} J_f(X)\ge \frac{J_f(X_\mathcal {U}) + 2\varepsilon }{1-\varepsilon }. \end{aligned}$$

$\square$

It is then immediate from the previous argument that also $J_f(X) = J_f(X_{\mathcal {U}})$ for every ultrafilter $\mathcal {U}$. When $\mathcal {U}$ is countably incomplete, the argument can be improved to

Theorem 2.3

If $\mathcal {U}$ is a countably incomplete ultrafilter then $J_f(X)= J_s(X_\mathcal {U})$.

Proof

We just need to prove the inequality $J_s(X_\mathcal {U})\le J_f(X)$. Let $\varepsilon >0$ be fixed and pick a countable set $A = \{a^m: m\in {\mathbb {N}}\} \subset X_\mathcal {U}$ with $\delta (A) = 1$ such that $J_s(X_\mathcal {U})\le 2r(A) + \varepsilon$. There is no loss of generality in assuming that $\Vert a^i-a^j\Vert =1$ for all $i,j\in {\mathbb {N}}$ just to simplify future choices. Let a be an $\varepsilon$-approximate center for A; namely $\Vert a^m - a\Vert \le r(A) + \varepsilon$ for all m. Set as before $a^m = [a_1^m, \dots , a^m_n, \dots ]$ and $a = [a_1, \dots , a_n, \dots ]$. The sets $U_m =\{n: \Vert a^m_n - a_n\Vert \le r(A) + 2\varepsilon \}$ belong to $\mathcal {U}$ for all m as well as $U_{u,v} = \{n: \Vert a^u_n - a_n^v\Vert \ge 1 - \varepsilon \}$ for all $u,v\in {\mathbb {N}}$. Now proceed orderly: pick $k\in \bigcap _{m=1}^k U_m \cap \bigcap _{1\le u,v\le k} U_{u, v}\in \mathcal {U}$ and form the set $B_k=\{a^1_k, \dots , a^k_k\} \subset X$, who has diameter at least $1-\varepsilon$ and the point $a_k$ is a $2\varepsilon$-approximate center for $B_k$. The only problem that could appear is if some $b_k\in X$ yields a “better” center for $B_k$, namely $\Vert b - b_k\Vert \le \alpha < J_s(X_\mathcal {U})$ for some $\alpha$ and all $b\in B_k$. But if this happens for an infinite set $M\subset {\mathbb {N}}$ then the element $b1_M$ having $b_k$ at the corresponding place of $1_M$ is a “better” center for A in $X_\mathcal {U}$, namely $\Vert a - b1_M\Vert \le \alpha$, which is a contradiction as $\varepsilon \rightarrow 0$. Therefore

$$\begin{aligned} J_f(X)\ge \frac{J_s(X_\mathcal {U}) + 2\varepsilon }{1-\varepsilon }. \end{aligned}$$

$\square$

We draw now some consequences. The first of them is a new proof for the Lindenstrauss–Bayod–Masa characterization of Lindenstraus spaces:

Proposition 2.4

A Banach space X is a Lindenstrauss space if and only if $J_f(X)=1$

Proof

If X is a Lindenstrauss space, $X_\mathcal {U}$ is 1-separably injective [2] and therefore $J_s(X_\mathcal {U}) =1$ according to [7], which proves the necessity. On the other hand, if $J_f(X)=1$ then also $J_s(X_\mathcal {U})=1$ and thus $X_\mathcal {U}$ is 1-separably injective [7]. It must therefore be Lindenstrauss space [2], as well as X by the principle of local reflexivity. $\square$

Amir [1, p.5] shows that $J_f(X^*)=J(X^*)$ for every dual space. Moreover:

Corollary 2.5

If X is 1-complemented in $X^{**}$ then $J_f(X)=J_s(X)$.

Proof

$J_s(X)\le J_s(X^{**})\le J_s(X_\mathcal {U}) = J_f(X)$. $\square$

3 The interplay between the finite Jung and Kottman constants

In [21, Theorem 6] it is shown that $2\le J_s(X)K(X)$. If X is an infinite-dimensional Banach space with unit ball B(X), the finite Kottman constant of X is defined as

$$\begin{aligned} K_f(X)=\; \sup \{r>0: \forall n \in {\mathbb {N}}\quad \exists \; A: |A|=n\; \textrm{and}\; \inf _{i\ne j} \Vert x_i -x_j\Vert \ge r \}. \end{aligned}$$

Since $J_f(X)\le J_s(X)$ but $K(X)\le K_f(X)$ it is worth checking the finite analog. A combinatorial argumentation could be: if X is not reflexive then $K_f(X)=2$ [18]; while if X is reflexive then [1] $J_f(X)=J(X)$ and therefore $K_f(X)J_f(X) = K_f(X)J(X) \ge K(X)J(X)\ge 2$. Let us present a straight proof.

Lemma 3.1

$2\le J_f(X) K_f(X)$.

Proof

Given $\varepsilon >0$, pick a set $A= \{x_1, \dots , x_N\}$ such that $K_f(X)+\varepsilon \ge \Vert x_i - x_j\Vert \ge K_f(X)-\varepsilon$ (use [7, Lemma 5]). Since $2r_A/\delta (A) \le J_f(X)$ one has $r_X(A) \le \frac{1}{2} J_f(X)\delta (A)\le \frac{1}{2} J_f(X)(K_f(X) + \varepsilon )$. Pick p such that $\Vert x_i -p\Vert \le \frac{1}{2} (J_f(X)+ \varepsilon ) (K_f(X) + \varepsilon )$ and therefore the ball centered at p with radius $\frac{1}{2} (J_f(X)+ \varepsilon ) (K_f(X) + \varepsilon )$ contains a finite set $(K_f(X)-\varepsilon )$-separated, and therefore the unit ball contains a finite set $\frac{K_f(X)-\varepsilon }{\frac{1}{2} (J_f(X)+ \varepsilon ) (K_f(X) + \varepsilon )}$-separated; hence

$$\begin{aligned} \frac{K_f(X)-\varepsilon }{\frac{1}{2} (J_f(X)+ \varepsilon ) (K_f(X) + \varepsilon )} \le K_f(X) \end{aligned}$$

and, therefore,

$$\begin{aligned} K_f(X) \le \frac{1}{2} J_f(X) K_f(X)^2 \Longrightarrow 2 \le J_f(X) K_f(X). \end{aligned}$$

. $\square$

One deduces from here and Proposition 2.4 that $K_f(X)=2$ for every Lindenstrauss space. It had been however shown in [6, Proposition 3.4] that $K(X)=2$ for every $\mathcal L_\infty$-space. We continue our study recalling the following result from Pichugov [23, Assertion]. We present it in its original formulation even if some terms appear unexplained in our context. The consequence we seek, namely, Pichugov’s inequality (3.1) is however clear:

Lemma 3.2

Let a closed convex set M in $X^n$ have Chebyshev radius r. Then the point y is its Chebyshev center if and only if there is a natural number $N \le n + l$, such that

• (a) there are points $x_i$ in M $(i = 1\dots N)$ such that $\Vert y_i - y\Vert =r$.

• (b) there are functionals $f_i$ in $(X^n)^*$ $(i = 1\dots N)$ such that $\Vert f_i\Vert =1$ and $\langle f_i, x_i -y\rangle =\Vert x_i-y\Vert$.

• (c) there are numbers $a_i$ $(i = 1\dots N)$, $\sum _i^N a_i=1$, $a_i\ge 0$ such that $\sum _i^N a_if_i=0$.

From there one deduces the following version of Pichugov’s inequality (see [15]):

$$\begin{aligned} J(A) \le \sup \sum _{i, j =1}^n \alpha _i\alpha _j\Vert f_i - f_j\Vert _{X^*} \end{aligned}$$

(3.1)

for $A\subset X$ a finite set of cardinality n, and the supremum is taken over finite families $f_1, \dots , f_n$ of elements of $X^*$ and scalars $\alpha _1, \dots , \alpha _n$ with $\Vert f_i\Vert _{X^*}\le 1$, $\alpha _i\ge 0$, $\sum \alpha _i=1$ and $(\sum \alpha _if_i)|_A =0$.

Proof

Pichugov’s inequality [15, (5)] yields $2r_X(A) = \sum _{i, j =1}^n \alpha _i\alpha _j \langle x_i-x_j, f_i - f_j\rangle$ from where

$$\begin{aligned} 2r_X(A) \le \sup \sum _{i, j =1}^n \alpha _i\alpha _j \Vert x_i-x_j\Vert \Vert f_i - f_j\Vert \le \delta (A)\sup \sum _{i, j =1}^n \alpha _i\alpha _j\Vert f_i - f_j\Vert \end{aligned}$$

$\square$

One gets

Proposition 3.3

If X is infinite-dimensional, $J_f(X)\le K_f(X^*)$.

Proof

We begin recalling from [6] that $K_f(X)=K(X_\mathcal {U})$ for every countably incomplete ultrafilter $\mathcal {U}$ on ${\mathbb {N}}$. Set for each $u\in {\mathbb {N}}$ a finite set $A_u\subset X$ such that $J(X)\le J(A_u)+\varepsilon u^{-1}$ and $J(A_u)\le J(A_v)$ when $u\le v$. Pick for each u elements $f^u_1,..., f_n^u$ as in Pichugov’s inequality and form the elements $F_n = [f^1_n, f^2_n, f_n^3, \dots , ]\in X^*_\mathcal {U}$ where we understand that $f_n^k=0$ for $n> n(k)$. Since $\Vert F_i - F_j\Vert _{X^*_\mathcal {U}}=\lim _\mathcal {U}\Vert f_i^u - f_j^u\Vert$ means that for every $\varepsilon >0$, $\{u: |\Vert f_i^u - f_j^u\Vert _{X^*} - \Vert F_i - F_j\Vert _{X^*_\mathcal {U}}|\le \varepsilon \}\in \mathcal {U}$, we get that $\Vert f_i^u - f_j^u\Vert _{X^*} \le \Vert F_i - F_j\Vert _{X^*_\mathcal {U}}+ \varepsilon$ for all u in a set of $\mathcal {U}$. Thus

$$\begin{aligned} J(X)= & {} \sup J(A_u)\\= & {} \lim _\mathcal {U}J(A_u)\\\le & {} \lim _\mathcal {U}\sum _{i, j} \alpha _i^u\alpha _j^u\Vert f_i^u - f_j^u\Vert _{X^*}\\\le & {} \lim _\mathcal {U}\sum _{i, j} \alpha _i^u\alpha _j^u\left( \Vert F_i - F_j\Vert _{X^*_\mathcal {U}} + \varepsilon \right) \end{aligned}$$

and since no infinite subset $M\subset {\mathbb {N}}$ exists such that $\Vert F_m - F_n\Vert > K(X^*_\mathcal {U})$ for $m,n\in M$ we get

$$\begin{aligned}\le & {} \lim _\mathcal {U}\sum _{i, j} \alpha _i^u\alpha _j^u \left( K(X^*_\mathcal {U}) + \varepsilon \right) \\\le & {} K(X^*_\mathcal {U})+\varepsilon .\end{aligned}$$

$\square$

The converse is obviously false since $J_f(c_0)=1$ and $K_f(\ell _1)=2$. The inequality above belongs to the world of finite constants since

• it is not true that $J_f(X)\le K(X^*)$ as the example of $\ell _p$-spaces, $1<p<2$, shows.

• Consequently it is not true that $J(X)\le K(X^*)$ either.

In [9] we showed that K(X) and $K(X^{**})$ are not necessarily equal. However

Proposition 3.4

$K_f(X)=K_f(X^{**})$.

Proof

We need the following version of the Principle of Local Reflexivity (see [20]): for each finite-dimensional subspace $E\subset X^{**}$ and each $\varepsilon >0$ there is a $(1+\varepsilon )$-isometry $T: E \rightarrow X$ such that $T|_{X} = id_{E\cap X}$. Pick now a finite set $\{x^{**}_1, \dots , x^{**}_N\}$ such that $\Vert x^{**}_n - x^{**}_m\Vert \ge K_f(X^{**})-\varepsilon$; hence

$$\begin{aligned} \Vert Tx^{**}_n - Tx^{**}_m\Vert \ge (1-\varepsilon )\Vert x^{**}_n - x^{**}_m\Vert \ge (1-\varepsilon )( K_f(X^{**})-\varepsilon ), \end{aligned}$$

for $\Vert Tx^{**}_n\Vert \le (1+\varepsilon )$, which is enough to conclude. $\square$

A combination of the inequality above with Proposition 3.3 provides a remarkable symmetry:

Corollary 3.5

If X is infinite-dimensional, $J_f(X)J_f(X^*)\le K_f(X)K_f(X^*)$.

A combination of Proposition 3.3 with the estimates in [7] yields

• (1) $K_f(Y)J_f(X) \le 2e_1^f(Y,X).$

• (2) $K_f(Y)J_s(X) \le 2e_1^s(Y,X^{**}).$

Here $e_1^f(Y, X)$ (resp. $e_1^s(Y,X)$) is the infimum of all $\lambda >0$ such that for finite (resp. separable) subset M of Y and every $y\in Y$, every Lipschitz map $f:M\rightarrow Z$ admits a Lipschitz extension $F: M\cup {\{y\}}\rightarrow Z$ with $Lip(F)\le \lambda Lip(f)$.

Corollary 3.6

$J_f(Y)J_f(X) \le 2e_1^f(Y^*,X).$ In particular, $J_f(X) \le \sqrt{2e_1^f(X^{**},X)}.$

4 Jung constants and interpolation

In [10] we studied the behavior of Kottman’s constants regarding complex interpolation obtaining the continuity of $K(\cdot )$ with respect to the interpolation parameter and the interpolation inequality: if $(X_0, X_1)$ is an interpolation pair and $X_\theta = (X_0, X_1)_\theta$ is the complex interpolation space obtained at $\theta$ then $K(X_\theta ) \le K(X_0)^{1-\theta } K(X_1)^\theta$. The behavior of the Jung constants regarding interpolation is necessarily quite different since an inequality $J(X_\theta ) \le J(X_0)^{1-\theta } J(X_1)^\theta$ does not hold since $J(L_\infty )=1$, $J(L_p)=2^{1-1/p}$ for $2\le p<\infty$ and $L_3 =(L_2,L_\infty )_\theta$ for $\theta =1/3$. Moreover, the characterizations of spaces X with $J_f(X), J_s(X)$ or J(X) equal to 1 makes an interpolation inequality such as $J(X_\theta ) \le J(X_0)^{1-\theta } J(X_1)^\theta$ impossible since one can obtain reflexive spaces as interpolation between injective spaces. The following explicit example was provided to us by Manuel González.

Example 4.1

Pick the space $\ell _\infty (1/n)= \{x: \sup _n \frac{1}{n} x_n<\infty \}$ endowed with the $\sup$ norm and consider the pair $(\ell _\infty ,\ell _\infty (1/n))$. The canonical inclusion $\ell _\infty \longrightarrow \ell _\infty (1(n))$ is compact, hence there are reflexive interpolation spaces, whose Jung constants must be greater than 1. However $J(\ell _\infty )= J(\ell _\infty (1/n))=1$.

Observe that this shows that $J(\cdot )$ does not satisfy an interpolation inequality for either the real or complex methods. However, if we denote the complex interpolation space as $X_\theta = (X_0, X_1)_\theta$, we have

Proposition 4.2

The Jung and finite Jung constants are continuous with respect to the interpolation parameter on (0, 1), but not on [0, 1].

Proof

The lack of continuity at the extremes has already been shown. In fact, [26, p. 870] already observed that $J(X_\theta )$ can be discontinuous at the boundary points. To show the continuity at the interior, let us recall the definition of the Kadets metric. Let M, N be closed subspaces of a Banach space Z, and let $B_M$ denote the unit ball of M. The gap g(M, N) between M and N is defined by

$$\begin{aligned} g(M,N) = \max \left\{ \sup _{x\in B_M}\text {dist}(x,B_N), \sup _{y\in B_N}\text {dist}(y,B_M)\right\} , \end{aligned}$$

The Kadets metric $d_K(X,Y)$ between two Banach spaces X and Y is the infimum of the gap g(i(X), j(Y)) taken over all the isometric embeddings of i, j of X, Y into a common Banach space. It turns out that $J(\cdot )$, $J_s(\cdot )$ and $J_f(\cdot )$ are continuous with respect to the Banach–Mazur metric: if $T:X\longrightarrow Y$ is an isomorphism with $\max \Vert T\Vert \Vert T^{-1}\Vert \le \alpha$ then $|J(X) - J(Y)|\le (\alpha ^2 -1)\min \{J(X), J(Y)\}$. We show now the continuity of the Jung constants with respect to the Kadets metric. We will need a few general facts that will be useful. Given a bounded set $A\subset X$ with approximate center a, a translation $x\rightarrow x-a$ allows us to work with the set $A'= A- a$ contained in the ball of radius $r_M(A)$ and center 0. We can change now $A'$ by $A''= r_X(A)^{-1}A'$ and still $J(A'')=J(A)$. In other words, with regard to the calculus of the Jung constants of Z there is no loss of generality in assuming that A has radius (or diameter) 1 and is contained in the unit ball has has approximate center 0. What is not true, as simple examples show, is that a subset of the ball must have its center inside the ball. Let us show now:

Claim. $J_f, J_s$ and J are continuous with respect to the gap.

Let us make the proof for $J_f$. Let $M, N \subset X$ two closed subspaces of a Banach space X. Fix M and $\varepsilon >0$. Let us call $g:B_M\rightarrow B_N$ (resp. $g': B_N\rightarrow B_M$) a function such that $\Vert x - g(x)\Vert \le g(M,N)+\varepsilon$ (resp. $\Vert x - g'(x)\Vert \le g(M,N)+\varepsilon$). There is no loss of generality in assuming that $g, g'$ are homogeneous since $\Vert \frac{x}{2} - \frac{1}{2}g(x)\Vert \le g(M,N)+\varepsilon$. Pick a finite set $A=\{a_1, \dots , a_n\}$ in $B_M$ with $\varepsilon$-center 0 and such that $J_f(M) < J(A) + \varepsilon$. Form the set $B=\{g(a_1),\dots , g(a_n)\}$ and let b be an $\varepsilon$-center for it. It is easy to check that $\Vert b\Vert \le 2+2\varepsilon$ (see the comment after the proof).

$$\begin{aligned} r(A)\le & {} \sup \Vert a_i - g'\left( \frac{b}{2}\right) \Vert \\= & {} \sup \left\| \frac{a_i}{2} + \frac{a_i}{2} - g\left( \frac{a_i}{2}\right) + g\left( \frac{a_i}{2}\right) - \frac{b}{2} + \frac{b}{2} -g'\left( \frac{b}{2}\right) \right\| \\\le & {} \sup \left( \left\| \frac{a_i}{2} \right\| + \left\| \frac{a_i}{2} - g\left( \frac{a_i}{2}\right) \right\| + \left\| g\left( \frac{a_i}{2}\right) - \frac{b}{2}\right\| + \left\| \frac{b}{2} - g'\left( \frac{b}{2}\right) \right\| \right) \\\le & {} \frac{1}{2}r(A) + \sup \left\| g\left( \frac{a_i}{2}\right) - \frac{b}{2}\right\| + 2g(M,N) + \varepsilon \\\le & {} \frac{1}{2}r(A) + \frac{1}{2} r(B) + 2g(M,N) + 2\varepsilon \end{aligned}$$

which yields

$$\begin{aligned} r(A) \le r(B) + 4g(M,N) + 4\varepsilon . \end{aligned}$$

On the other hand

$$\begin{aligned} \delta (A)= & {} \sup \Vert a_i - a_j\Vert \\ {}= & {} \Vert a_i - g(a_i) + g(a_i) - g(a_j) + g(a_j) - a_j\Vert \\ {}\ge & {} \delta (B) - 2g(M,N) - 2\varepsilon . \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{2r_N(A)}{\delta (A)} \le \frac{ 2r(B) + 8g(M,N) + 8\varepsilon }{\delta (B) - 2g(M,N) - 2\varepsilon } \end{aligned}$$

and thus

$$\begin{aligned} J_f(M) \le J_f(N) + F \end{aligned}$$

for some positive and continuous function F such that $f(0)=0$. Doing the same replacing M by N we get the other inequality, and thus

$$\begin{aligned} \lim _{g(M,N)\rightarrow 0} \left| J_f(M) - J_f(N)\right| =0 \end{aligned}$$

which is the continuity (not the uniform continuity, as it is the case of the Kottman’s constants) of $J_f$ with respect to the gap.

The continuity with respect to the Kadets metric is now immediate taking into account that if i, j are isometric embeddings, $J(X)=J(iX)$ and $J(Y)=J(jY)$. The continuity with respect to the interpolation parameter follows as we proved in [10] for the Kottman’s constants: Kalton and Ostrovskii [17] proved that the Kadets metric is continuous with respect to the interpolation parameter; precisely,

$$\begin{aligned} d_K(X_\theta , X_\eta )\le 2 \left| \frac{\sin \left( \pi (\theta -\eta )/2\right) }{\sin \left( \pi (\theta +\eta )/2\right) }\right| . \end{aligned}$$

Thus, the Jung constants are continuous with respect to the interpolation parameter. $\square$

The following is an example of a set in the unit ball of $c_0$ with a center having norm 2: pick $A=\{ \sum _i^n e_i: n\in \mathbb N\}$ and set $2e_1$. The continuity with respect to the Kadets metric in combination with the fact that given an exact sequence $0\rightarrow Y \rightarrow Z \rightarrow X \rightarrow 0$ one has $d_K(Z, Y\oplus _\infty X)=0$ (see [10]) yields that Z can be renormed to have (finite, separable) Jung constant $\max \{J(Y), J(X)\}$. However, the example $0\rightarrow C[0,1] \rightarrow \Omega \rightarrow c_0 \rightarrow 0$ presented earlier shows that, however, no renorming of Z such that $J(Z)=\max \{J(Y), J(X)\}$ can, in general, be achieved.

5 Open questions

Most of our open questions wheel around the validity of the interpolation inequality

$$\begin{aligned} J_f(X_\theta )\le J_f(X_0)^{1-\theta }J_f(X_1)^{\theta } \end{aligned}$$

which is false. Could it be true on a restricted context? Say, when also $X_0, X_1$ are superreflexive interpolation spaces, or infinite-dimensional spaces with a common unconditional basis?

• Does the inequality $2\le J_f(X)J_f(X^*)$ hold for infinite-dimensional spaces? The inequality fails for finite-dimensional spaces since $J(\ell _1^n)= 2n/n+1$ (see [12, 15]) and $J(\ell _\infty ^n)=1$. Observe that when $X, X^*$ have a common unconditional basis then $(X, X^*)_{1/2} = \ell _2$ and thus the interpolation inequality would yield $\sqrt{2}\le J_f(X)^{1/2} J_f(X^*)^{1/2}$, which is the inequality above.

• Does the interpolation formula hold for pairs $(E_0, E_1)$ of rearrangement invariant Banach lattices with $E_0, E_1 \ne L_\infty$? R.i. Banach lattices can be seen as a generalized form of Banach spaces with unconditional or symmetric basis.

• A Banach lattice E is said to be a $\theta$-Hilbert space ($0<\theta <1$) if $E=(F,L_2)_\theta$ for some r.i. space F. Each $\theta$-Hilbert space is a r.i. space (see [24]). Is it true that $J(E)\le J(L_2)^\theta J(F)^{1-\theta }= J(F)^{1-\theta } 2^{\theta /2}$ for E a $\theta$-Hilbert space? This would generalize the inequality $J(E)\le 2\cdot 2^{\theta /2}$ in [26, Theorem 2.2.].

• Can a reflexive space X be renormed in such a way that $J(X) = J(X^*)$? Recall that there are reflexive spaces for which $J(X)\ne J(X^*)$, say $J\left( \ell _2(\ell _1^n)\right) =2$ and $J\left( \ell _2(\ell _\infty ^n)\right) =\sqrt{2}$. This example is from Amir [1, 2.15.b], although he erroneously writes that this space has Jung constant 1.