Proof. (i): If $\text{dc}(x) \leq 1$, then the inequality is trivial since

\begin{equation*} \|\text{Id} + T\| \geq 1 \geq (\text{dc}(x) - 1)\|x^*\| + 1, \end{equation*}

so we may assume that $\text{dc}(x) \gt 1$. Let $x^* \in {X^*}$ with $\|x^*\| \geq 1$, δ > 0 and $0 \lt \varepsilon \lt \text{dc}(x) -1$ be given. Suppose first that $\|x^*\| = 1$. Since $\text{dc}(-x)=\text{dc}(x)$, we can find $y \in S(x^*, \delta )$ so that $\|y+x\| \gt \text{dc}(x) - \varepsilon \gt 1$. This implies that

\begin{align*} \| \text{Id} + x^* \otimes x\| &\geq \|y + x^* (y) x\| \geq \|y+x\| - \delta \gt \text{dc}(x) - \varepsilon - \delta. \end{align*}

Since δ > 0 and ɛ > 0 were chosen arbitrarily, we conclude that $\| \text{Id} + x^* \otimes x\| \geq \text{dc}(x)$.

Next, suppose that $\|x^*\| \gt 1$. Fix $y \in S\bigl(\frac{x^*}{\|x^*\|}, \delta \bigr)$ so that $\|x+y\| \gt \text{dc}(x) - \varepsilon \gt 1$. For a small enough δ > 0 such that $x^*(y) \gt 1$, we have

\begin{align*} \| \text{Id} + x^* \otimes x\| &\geq \|y + x^* (y) x\| \geq x^*(y) \|y+x\| -x^*(y) +1 \\ & \gt x^*(y) (\text{dc}(x) - \varepsilon -1) +1 \gt \|x^*\| (1-\delta) (\text{dc}(x) - \varepsilon -1) +1. \end{align*}

Again from the choice of δ > 0 and ɛ > 0, it follows that $\| \text{Id} + x^* \otimes x\| \geq (\text{dc}(x)-1) \|x^*\| +1$.

(ii): First note that in at least 2-dimensional Banach space X, the inequality is trivial if ${\delta}c(x) \leq 1$. For ${\delta}c(x) \gt 1$, fix $\varepsilon, \delta \gt 0$ small and find $y \in B_X$ such that $x^*(y) \gt 1- \delta$ and $\|x- y\| \gt {\delta}c(x) -\varepsilon \gt 1$. If $x^*(y) \leq 1$, then

\begin{equation*} \|\text{Id} - P \| \geq \|y - x^*(y) x\| \geq \|y - x\| - \delta \gt {\delta}c(x) - \varepsilon - \delta. \end{equation*}

If $x^*(y) \gt 1$, then

\begin{align*} \|\text{Id} - P \| &\geq \|y - x^*(y) x\| \\ &\geq x^*(y) \|x-y\| - x^*(y) +1 \\ & \gt x^*(y) ( {\delta}c(x) - \varepsilon - 1) + 1 \gt (1-\delta)({\delta}c(x) - \varepsilon - 1) + 1. \end{align*}

As ɛ and δ were arbitrary, this concludes the proof.

Proof. The inclusion (2.2) can be proved similarly as in [Reference Abrahamsen, Haller, Lima and Pirk3, lemma 3.5]. For the sake of completeness, we present the proof. Let ɛ > 0 be given. Let $y \in B_X$, $y_1^{**}, \ldots, y_{m}^{**} \in \Delta_{2-\alpha}^{X^{**}} (x)$, and $\lambda_1,\ldots, \lambda_m \gt 0$ be such that $\sum_{i=1}^m \lambda_i = 1$ and $\|y - \sum_{i=1}^m \lambda_i y_i^{**} \| \lt \frac{\varepsilon}{2}$. Put $E = \text{span} \{x, y, y_1^{**}, \ldots, y_m^{**} \} \subseteq X^{**}$ and let $0 \lt \eta \lt 1$ be such that $(1-\eta)\alpha \gt \alpha-\varepsilon$. By Principle of Local Reflexivity, there exists a linear operator $T : E \rightarrow X$ such that

1. (i) Te = e for every $e \in E \cap X$;

2. (ii) $(1-\eta)\|e\| \leq \|Te\| \leq (1+\eta) \| e\|$ for every $e \in E$.

It follows that $\|x - Ty_{i}^{**}\| = \|Tx - Ty_i^{**}\| \geq (1-\eta) \alpha \gt \alpha - \varepsilon$. That is, $Ty_i^{**} \in \Delta_{2-\alpha+\varepsilon}^X (x)$ for $i=1,\ldots, m$. Moreover,

\begin{align*} \left\| y - \sum_{i=1}^m \lambda_i T y_i^{**} \right\| = \left\| Ty - \sum_{i=1}^m \lambda_i T y_i^{**}\right\| \leq (1+\eta)\frac{\varepsilon}{2} \lt \varepsilon. \end{align*}

It follows that $y \in \overline{\text{co}} \,\Delta_{2-\alpha +\varepsilon}^{X} (x)$.

1. (i) By proposition 2.3, if $\text{dc}_{X^{**}} (x) \geq \alpha$, then $B_{X^{**}} \subseteq \overline{\text{co}} \Delta_{2-\alpha+\varepsilon}^{X^{**}} (x)$ for every ɛ > 0. Let us fix ɛ > 0. Note that if $y \in B_X$, then $y \in \overline{\text{co}} \Delta_{2-\alpha+\varepsilon}^{X^{**}} (x)$. By (2.2), we have that $y \in \overline{\text{co}} \Delta_{2-\alpha+\varepsilon+ \eta}^{X} (x)$ for every η > 0. In particular, $y \in \overline{\text{co}} \Delta_{2-\alpha+2\varepsilon}^{X} (x)$, which shows that $B_X \subseteq \Delta_{2-\alpha+2\varepsilon}^{X} (x)$. As this holds for every ɛ > 0, by proposition 2.3, we conclude that $\text{dc} (x) \geq \alpha$.

2. (ii) Note from proposition 2.3 that $x \in \overline{\text{co}} \Delta_{2-{\delta}c(x)+\varepsilon}^{X} (x)$ for every ɛ > 0, which is, by (2.2), equivalent to that $x \in \overline{\text{co}} \Delta_{2-{\delta}c(x)+\varepsilon}^{X^{**}} (x)$ for every ɛ > 0. It follows that ${\delta}c (x) \geq \alpha$ if and only if ${\delta}c_{X^{**}}(x) \geq \alpha$.

Proof. Suppose first that $x \in S_X$ satisfies ${\delta}c(x)=0$. Let η > 0 be given. Then, there exist ɛ > 0, δ > 0, and $y^* \in S_{X^*}$ such that $x \in S(y^*,\delta)$ and $\|y-x\| \leq \eta - \varepsilon$ for every $y \in S(y^*,\delta)$. Thus, $S(y^*,\delta)$ is a slice of B_X containing x such that $\text{diam} S(y^*,\delta) \leq 2(\eta-\varepsilon)$. Since η > 0 was arbitrary, it follows that x is a denting point of B_X.

On the other hand, let $x \in \text{dent}(B_X)$ and η > 0 be given. Fix δ > 0 and $y^* \in S_{X^*}$ such that $x \in S(y^*,\delta)$ and $\text{diam} S(y^*,\delta) \lt \eta/2$. Then $\|y-x\| \leq \eta/2$ for every $y \in S(y^*,\delta)$, so ${\delta}c(x) \leq \eta$. This finishes the proof.

Proof. Suppose that $y \in \text{dent}(B_X)$ and $\|x-y\| \lt \text{dc}(x)$. Take a slice $S = S(x^*, \delta)$ containing y such that $\text{diam} (S) \lt \eta$, where $0 \lt 2\eta \lt \text{dc}(x) - \|x-y\|$. Find $u \in S$ so that $\|x-u\| \gt \text{dc}(x) -\eta$. Then,

\begin{equation*} \|x-y\| \geq \|x-u \| - \|u-y\| \gt \text{dc}(x) - \eta -\eta \gt \|x-y\|, \end{equation*}

which is a contradiction.

Proof. It is clear that $\text{dc}(x) \geq 1- \|x\|$ from the definition. In order to show that $\text{dc}(x) \leq 1 - \|x\|$, let ɛ > 0 be given and fix $x^* \in S_{X^*}$ such that $x^*(x) = \|x\|$. We claim that there exists δ > 0 such that if $x^*(y) \gt 1-\delta$ for some $y \in B_X$, then $\bigl\|y-x/\|x\|\bigr\| \lt \varepsilon$. Otherwise, we may choose $y_n \in S(x^*, 1/n)$ for each $n \in \mathbb{N}$ such that $\bigl\| y_n - x/\|x\| \bigr\| \geq \varepsilon$ for all $n \in \mathbb{N}$. But notice that

\begin{equation*} \|y_n+x\| \geq x^*(y_n) +x^*(x) \gt 1+\|x\|-\frac{1}{n}; \end{equation*}

thus, since x is a LUR point, we can deduce that $\|y_n - x/\|x\|\| \rightarrow 0$. This contradicts the choice of $(y_n)$. Now, observe that for every $y \in S(x^*,\delta)$,

\begin{equation*} \|y-x\| \leq \left\|y - \frac{x}{\|x\|}\right\| + 1- \|x\| \lt 1-\|x\|+\varepsilon. \end{equation*}

Since ɛ > 0 was arbitrary, this shows that $\text{dc}(x) \leq 1 - \|x\|$.

Proof. Consider the slice $S=\{u \in B_H : \big\langle u, \frac{x}{\|x\|} \big\rangle \gt \|x\| - \delta\}$ with δ > 0, where $\langle \cdot, \cdot \rangle$ denotes the inner product on H. It is clear that $x \in S$. For $y \in S$, observe that

(2.3)\begin{equation} \|x-y\|^2 = \|x\|^2+\|y\|^2 - 2\langle x , y \rangle \leq \|x\|^2 + 1 - 2 \|x\| (\|x\|-\delta). \end{equation}

By letting $\delta \rightarrow 0$, the right-hand side of (2.3) tends to $1-\|x\|^2$. This implies that ${\delta}c(x) \leq \sqrt{1-\|x\|^2}$. For the reverse inequality, let $S(y, \alpha)$ be a slice containing the point x. Since $\dim(H) \geq 2$, we can find $z \in S_H$, which is orthogonal to x, i.e., $\langle z,x\rangle = 0$. By changing the sign if necessary, we may assume that $\langle z, y \rangle \geq 0$. Note that $x+ \sqrt{1-\|x\|^2 }\, z \in S_H$ and $x+ \sqrt{1-\|x\|^2 }\,z \in S(y, \alpha)$ since

\begin{equation*} \langle y, x+ \sqrt{1-\|x\|^2 } \, z \rangle \geq \langle y, x\rangle \gt 1 - \alpha. \end{equation*}

Note from $\| x -( x+ \sqrt{1-\|x\|^2 }\,z) \| = \sqrt{1-\|x\|^2 }$ that ${\delta}c(x) \geq \sqrt{1-\|x\|^2 }$.

Proof. The proof of the ‘if’ part is clear. Assume that X admits almost Daugavet points. Let ɛ > 0 and $x_0 \in B_X$ be so that $B_X \subseteq \overline{\text{co}} \,\Delta_{\varepsilon/2} (x_0)$. By proposition 2.3, it is clear that $\text{dc} (x_0) \geq 2-\frac{\varepsilon}{2}$. Note that $\|x_0\| \geq 1 - \frac{\varepsilon}{2}$. Thus, putting $z:=\frac{x_0}{\|x_0\|} \in S_X$, we have that $\|z - x_0\| \lt \frac{\varepsilon}{2}$. Since the mapping $x \mapsto \text{dc} (x)$ is 1-Lipschitz (see remark 2.2), we have

\begin{equation*} \text{dc} (z) \geq \text{dc} (x_0) - | \text{dc}(z) - \text{dc}(x_0)| \geq \text{dc} (x_0) - \|z-x_0\| \gt \left(2-\frac{\varepsilon}{2}\right) - \frac{\varepsilon}{2} = 2 - \varepsilon, \end{equation*}

which implies $\sup_{x \in S_X} \text{dc}(x) =2$.

Proof. It suffices to show that for $x \in B_X$, there exists $z \in S_X$ such that ${\delta}c(z) \geq {\delta}c(x)$. Let $x \in B_X \setminus \{0\}$ be given. We claim that ${\delta}c((x,1)) \geq {\delta}c(x)$. Let ɛ > 0, δ > 0 and $(x^*, \beta) \in S_{X^* \oplus_1 \mathbb{R}}$ with $(x,1) \in S:= S\bigl( (x^*, \beta), \delta\bigr)$ be given. Put $z:= (-\|x\|^{-1} x, 1 )$ and observe that $z \in S$. Also, $\|z-(x,1) \| = 1+\|x\| \geq {\delta}c(x) \gt {\delta}c(x) -\varepsilon$. Next, assume that $x^* \neq 0$. Note that we have

\begin{equation*} \frac{x^*}{\|x^*\|} (x) \gt \frac{1}{\|x^*\|}(1-\delta -\beta). \end{equation*}

Thus, there exists $y \in B_X$ such that $\|x^*\|^{-1} x^* (y) \gt \|x^*\|^{-1} (1-\delta -\beta)$ and $\|y-x\| \gt {\delta}c(x) - \varepsilon$. Set $z:=(y,1)$ and observe that $z \in S$ since $(x^*, \beta) (y,1) \gt 1-\delta-\beta+\beta= 1-\delta$. It is clear that

\begin{equation*} \|z- (x,1)\| = \|y-x\| \gt {\delta}c(x) -\varepsilon. \end{equation*}

This proves that ${\delta}c( (x,1)) \geq {\delta}c(x)$.

Proof. As the statements (i) and (ii) can be proved in a very similar way, we only prove (i). Suppose that $\text{dc}(x) \lt r$ for some $x \in S_X$. Then, there exist ɛ > 0 and $S=S(x^*,\delta)$ such that $\|x-y\| \leq r-\varepsilon$ for every $y \in S$. This implies that $\mathcal{T}^s (X) \leq r$. Conversely, suppose that $\mathcal{T}^s (X) \lt r$. Let ɛ > 0 be such that $\mathcal{T}^s (X) \lt r- \varepsilon \lt r$, and find $\mathcal{T}^s (X) \lt t \lt r-\varepsilon$ such that there exist $x \in S_X$ and $S=S(x^*,\delta)$ satisfying $S \subseteq B(x,t)$. That is, $\|x-y\| \leq t \lt r-\varepsilon$ for every $y \in S$. This implies that $\text{dc}(x) \lt r$.

Proof. The result [Reference Haller, Langemets, Lima, Nadel and Rueda Zoca10, proposition 2.8] guarantees that for any $n \in \mathbb{N}$, there exists a Banach space X_n satisfying that $\mathcal{T}^s (X_n) = 1/2^n$. As a matter of fact, a careful investigation of the proof shows that $\mathcal{T}^s (X_n)$ in fact coincides with $\mathcal{T}^\delta (X_n)$. Observe also that $\mathcal{T}^\delta (E_1 \oplus_1 E_2)\leq \mathcal{T}^\delta (E_1)$ for any Banach spaces E ₁ and E ₂, which is a consequence of proposition 4.11 in §4 and remark 2.15.

Now, set $X = \bigl(\oplus_n X_n\bigr)_{\ell_1}$. Then, we have that

\begin{equation*} 0\leq \mathcal{T}^s (X) \leq \mathcal{T}^\delta (X) \leq \min \{\mathcal{T}^\delta (X_n) , \mathcal{T}^\delta (Y_n)\} \leq \frac{1}{2^n}, \end{equation*}

where $Y_n = \bigl(\oplus_{k \neq n} Y_k\bigr)_{\ell_1}$ for each $n \in \mathbb{N}$. This shows that $\mathcal{T}^\delta (X) =0$.

Next, we claim that $\text{dc}(x) \gt 0$ for every $x \in S_X$. Let $x= (x_n) \in S_X$ be given. If there exists $m \in \mathbb{N}$ such that $\|x_m\| = 1$, then we have

\begin{equation*} x = (0,\ldots, 0, x_{m}, 0, \ldots ) =: (x_m, 0) \in X_m \oplus_1 Y_m. \end{equation*}

Note that $\text{dc}(x)=\text{dc}(x_m, 0) \geq \text{dc}(x_m) \geq 2^{-m}$ (see proposition 4.2 for its justification). So, suppose that $\|x_n\| \lt 1$ for every $n \in \mathbb{N}$. Let $y^*= (y_n^*) \in S_{X^*}$, ɛ > 0, δ > 0 be given. Find $k \in \mathbb{N}$ so that $\|y_k^* \| \gt 1- \delta$; so there is $y_k \in S_{X_{k}}$ such that $y_k^* (y_k) \gt 1-\delta$. Letting ι_k be the embedding from X_k into X, observe that $\iota_k (y_k) \in S(y^*, \delta)$ and

\begin{equation*} \| \iota_k (y_k) - x \| = \|y_k - x_k\| + \sum_{n\neq k } \|x_n\| \geq 1 - \|x_k\| + (1- \|x_k\|) \geq 2 -2 \sup_{n \in \mathbb{N}} \|x_n\|. \end{equation*}

This implies that $\text{dc} (x) \geq 2 - 2 \sup_{n \in \mathbb{N}} \|x_n\| \gt 0$.

Proof. Let $x = (x_n)_{n=1}^N \in B_{\ell_\infty^N}$ be given. Let ɛ > 0, δ > 0, and $y^*=(y_n^*)_{n=1}^N \in S_{\ell_1^N}$ be given. If we let $y:=(\text{sgn} (y_n^*) )_{n=1}^N \in S_{\ell_\infty^N}$, then $y^*(y)=1$; hence $y \in S(y^*,\delta)$. Notice that

\begin{equation*} |x_n - \text{sgn} (y_n^*)| \geq 1 - |x_n|. \end{equation*}

Thus, $\| x- y\| \geq \max \{1-|x_n| : n=1,\ldots, N\}$. This argument shows that $\text{dc} (x) \geq \max \{1-|x_n| : n=1,\ldots, N\}$.

Next, assume to the contrary that $\text{dc} (x) \gt \max \{1-|x_n| : n=1,\ldots, N\}$. Without loss of generality, we may assume that $x_i \geq 0$ for every $i=1,\ldots, N$. Then, we can find η > 0 such that $x_i \gt 1 - \eta \gt 1 - \text{dc}(x)$ for every $i =1,\ldots, N$. Let δ > 0 be sufficiently small so that $\delta \lt N^{-1} \eta$, and $y^* = N^{-1} (1,\ldots, 1) \in S_{\ell_1^N}$. If $y = (y_1,\ldots, y_N) \in S(y^*, \delta)$, then by a standard argument, we have

\begin{equation*} y_i \gt 1 - N \delta \gt 1- \eta \quad \text{for } i =1,\ldots, N. \end{equation*}

It follows that $|x_i - y_i| \lt \eta$ for every $i=1,\ldots, N$, that is, $\|x-y\| \lt \eta$. This implies that $\text{dc}(x) \leq \eta \lt \text{dc}(x)$, which is a contradiction.

Finally, when N = 2, it suffices to show that ${\delta}c(x) \leq \max \{1 -|x_1|, 1-|x_2|\}$. Without loss of generality, we may assume that $x_1 \geq x_2 \geq 0$. Thus, $\max \{1 -|x_1|, 1-|x_2|\} = 1-x_2$. Again, assume to the contrary that ${\delta}c(x) \gt 1-x_2$. Find η > 0 so that $x_2 \gt 1 - \eta \gt 1 - {\delta}c(x)$. If $y^* = (1/2, 1/2) \in S_{\ell_1^2}$, then

\begin{equation*} y^*(x) = \frac{x_1+x_2}{2} \gt \frac{x_1}{2} + \frac{1-\eta}{2} =: 1 - \delta. \end{equation*}

Observe that if $y=(y_1,y_2) \in S(y^*, \delta)$, then

\begin{equation*} y_1 + y_2 \gt x_1 + (1-\eta). \end{equation*}

This implies that

\begin{equation*} x_1 -y_1 \lt \eta \ \text{and } \ x_2 -y_2 \leq x_1 - y_2 \lt \eta. \end{equation*}

If $y_1 \gt x_1$, then $|x_1-y_1| = y_1 - x_1 \leq 1 -x_1 \lt \eta$. Also, if $y_1 \leq x_1$, then $|x_1 - y_1| = x_1 - y_1 \lt \eta$; hence $|x_1 - y_1| \lt \eta$ in any case. Similarly, $|x_2 - y_2| \lt \eta$, therefore $\|x-y\| \lt \eta$. Since $y \in S(y^*, \delta)$ is chosen arbitrarily, we obtain that $\|x-y\| \lt \eta$, which contradicts that ${\delta}c(x) \gt \eta$.

Proof. We first show that $\text{dc}(x) \geq 1$ for every $x \in B_{c_0}$. Observe that it suffices to prove that $\text{dc}(x) \geq 1$ for every $x \in B_{c_{00}}$. Fix $x \in B_{c_{00}}$ and we may write $x = (x_1, \ldots, x_m, 0, \ldots)$ for some $m \in \mathbb{N}$. Let ɛ > 0, δ > 0 and $y^* = (y_n^*) \in S_{\ell_1}$ be given. Pick any $y = (y_n) \in S(y^*,\delta/3) \cap S_{c_0}$ and choose k > m such that $|y_k^*| \lt \delta/3$. If we define $z = (z_n) \in S_{c_0}$ by $z_n = y_n$ for each $n \in \mathbb{N} \setminus \{k\}$ and $z_k = 1$, then

\begin{align*} y^*(z) \geq \sum_{i \neq k} y_i^*y_i - \frac{\delta}{3} \geq y^*(y) - \frac{2\delta}{3} \gt 1- \delta, \end{align*}

which yields $z \in S(y^*,\delta)$. The conclusion follows from $\|z-x\| \geq |z_k-x_k| =1$.

Now, suppose that $\text{dc}(x_0) \gt 1$ for some $x_0 = (x_n) \in B_{c_0}$. Let $A \subseteq \mathbb{N}$ be the finite subset of all indices such that $|x_i| \geq (\text{dc}(x_0) - 1)/4$ for each $i \in A$, where A is non-empty since $\|x_0\| \geq \text{dc}(x_0) - 1$. For

\begin{equation*} \varepsilon_0 = \frac{\text{dc}(x_0)-1}{2}, \,\delta_0 = \frac{\text{dc}(x_0)+1}{2|A|}\, \text{and }\, y_0^* = \frac{1}{|A|} \sum_{i \in A} \text{sgn} (x_i) e_i^* \in S_{\ell_1}, \end{equation*}

there must exist $y_0 = (y_n) \in S(y_0^*,\delta_0)$ such that

\begin{equation*} \|y_0-x_0\| \gt \text{dc}(x_0) - \varepsilon_0 = \frac{1}{2} \left( \text{dc}(x_0) + 1 \right). \end{equation*}

However, observe that, for each $i \notin A$, we have

\begin{equation*} |y_i - x_i| \leq \frac{1}{4} \left(\text{dc}(x_0) -1 \right) + 1 \lt \frac{1}{2} \left( \text{dc}(x_0) + 1 \right). \end{equation*}

It follows that there must exist $k \in A$ such that

\begin{equation*} |y_k - x_k | = \| y_0 - x_0\| \gt \frac{1}{2} \left( \text{dc}(x_0) + 1 \right) \gt 1. \end{equation*}

This, in particular, implies that $\text{sgn} (y_k) \neq \text{sgn} (x_k)$. Moreover,

\begin{equation*} |y_k| \gt \frac{1}{2} (\text{dc}( x_0) - 1). \end{equation*}

Thus, we have

\begin{align*} y_0^*(y_0) = \frac{1}{|A|} \sum_{i \in A} \text{sgn} (x_i) y_i &= \frac{1}{|A|} \sum_{i \in A \setminus \{k\}} \text{sgn} (x_i) y_i - \frac{1}{|A|} |y_k| \\ &\leq \frac{|A|-1}{|A|} - \left( \frac{\text{dc}(x_0)-1}{2|A|}\right) = 1 - \left( \frac{\text{dc}(x_0)+1}{2|A|}\right). \end{align*}

This contradicts that $y_0 \in S(y_0^*, \delta_0)$.

Proof. We may assume $x_n \geq 0$ for each $n \in \mathbb{N}$ for convenience. Let $n_0 \geq 3$ and $\{i_1, \ldots, i_{n_0}\} \subseteq \mathbb{N}$ be given. For simplicity, let us say for some $n_1 \leq n_0$ that

1. (i) $i_n = n$ for $n =1, \ldots, {n_0}$;

2. (ii) $0 \lt x_n \leq 1 -\frac{2}{n_0}$ for $n = 1,\ldots,n_1$;

3. (iii) $1 -\frac{2}{n_0} \lt x_n \leq 1$ for $n = n_1+1, \ldots, n_0$.

Define $w_1, \ldots, w_{n_0} \in c_0$ by

\begin{align*} w_1 &= \left( \phantom{....} -1,\phantom{....} \frac{n_0 x_{2} + 1}{n_0 - 1}, \cdots, \frac{n_0 x_{n_1} + 1}{n_0 - 1} , \phantom{.......}1 ,\phantom{..........} 1 \phantom{..}, \cdots, \phantom{w..........}1 ,\phantom{w.........}\! x_{n_0 + 1}, x_{n_0 + 2}, \cdots\vphantom{\frac{n_0 x_{2} + 1}{n_0 - 1}} \right), \\ w_2 &= \left( \frac{n_0 x_1 + 1}{n_0-1}, \phantom{....} -1 \phantom{....}, \cdots, \frac{n_0 x_{n_1} + 1}{n_0 - 1} , \phantom{.......}1 ,\phantom{..........} 1 \phantom{..}, \cdots, \phantom{w..........}1 ,\phantom{w.........}\! x_{n_0 + 1}, x_{n_0 + 2}, \cdots\vphantom{\frac{n_0 x_{2} + 1}{n_0 - 1}} \right), \\ &\phantom{..............}\vdots \phantom{............} \vdots \phantom{w....} \ddots \phantom{...........} \vdots \phantom{..............} \vdots \phantom{............} \vdots \phantom{....} \ddots \phantom{.............} \vdots \\ w_{n_1} &= \left( \frac{n_0 x_1 + 1}{n_0-1}, \frac{n_0 x_{2} + 1}{n_0 - 1}, \cdots, \phantom{.....} -1 , \phantom{.....} \phantom{.......}1, \quad\phantom{......} 1 \phantom{..}, \cdots, \phantom{w..........}\! 1,\phantom{w........} x_{n_0 + 1}, x_{n_0 + 2}, \cdots \right), \\ w_{n_1+1} &= \left( \frac{n_0 x_1 + 1}{n_0-1}, \frac{n_0 x_{2} + 1}{n_0 - 1}, \cdots, \frac{n_0 x_{n_1} + 1}{n_0 - 1} , n_0x_{n_1+1} - (n_0-1) , \phantom{..}1\phantom{..} , \cdots, \phantom{.}1 ,\phantom{..........} x_{n_0 + 1}, x_{n_0 + 2}, \cdots \right), \\ &\phantom{..............}\vdots \phantom{............} \vdots \phantom{w....} \ddots \phantom{...........} \vdots \phantom{..............} \vdots \phantom{............} \vdots \phantom{....} \ddots \phantom{.............} \vdots \\ w_{n_0} &= \left( \frac{n_0 x_1 + 1}{n_0-1}, \frac{n_0 x_{2} + 1}{n_0 - 1}, \cdots, \frac{n_0 x_{n_1} + 1}{n_0 - 1} , \phantom{w.....}1, \quad \phantom{.....}1\phantom{..}, \cdots, n_0x_{n_0} - (n_0-1) , x_{n_0 + 1}, x_{n_0 + 2}, \cdots \right). \end{align*}

Then, we have $\frac{1}{n_0} \sum_{n=1}^{n_0} w_n = x$ and $\|w_n-x\| \geq f_{n_0}(x_n)$ for $n = 1,\ldots,n_0$. Indeed, one can see that $\|w_n-x\| \geq 1 + x_n$ for $n=1,\ldots,n_1$ and $\|w_n-x\| \geq (n_0-1)(1-x_n)$ for $n = n_1+1,\ldots,n_0$. This shows that $x \in \operatorname{co} \Delta_{2-\min \{f_{n_0}(x_1), \ldots, f_{n_0}(x_{n_0}) \}} (x)$, and thus ${\delta}c(x) \geq \min \{f_{n_0}(x_1), \ldots, f_{n_0}(x_{n_0}) \}$.

Proof. Let $x \in B_{\ell_\infty}$ be given. First, let $S=S(y,\delta)$ be a weak$^*$-slice of $B_{\ell_\infty}$ with $y=(y_n) \in S_{\ell_1}$. Find $z=(z_n) \in S_{c_0}$ so that $z \in S(y,\delta/3)$. Choose $m \in \mathbb{N}$ such that $\sum_{n=m+1}^\infty |y_n| \lt \delta/3$. Consider $w \in B_{\ell_\infty}$ given by

\begin{equation*} w = (z_1,\ldots, z_m, -\text{sgn}(x_{m+1}), \text{sgn}(x_{m+2}), \ldots ). \end{equation*}

Then, $\langle y, w\rangle \geq \sum_{n=1}^m y_n z_n - \sum_{n=m+1}^\infty |y_n| \geq \langle y, z \rangle - 2 \sum_{n=m+1}^\infty |y_n| \gt 1 - \delta/3 - 2\delta/3$; thus $w \in S(y,\delta)$. Moreover,

\begin{equation*} \|x-w\| \geq \sup_{n \geq m+1} | x_n + \text{sgn}(x_n) | = 1+ \sup_{n \geq m+1} |x_n| \geq 1+ \limsup_n |x_n|. \end{equation*}

This proves that $\textit{w}^*dc(x) \geq 1+\limsup_n |x_n|$.

Next, assume that $\textit{w}^*dc(x) \gt 1+\limsup_n |x_n|$. Find δ > 0 sufficiently small so that

\begin{equation*} \textit{w}^*dc(x) - \delta \gt 1 + \limsup_n |x_n|. \end{equation*}

Let $N_1 \in \mathbb{N}$ and consider the vector

\begin{equation*} y^{(1)} := \left(\frac{\text{sgn}(x_1)}{N_1}, \ldots, \frac{\text{sgn}(x_{N_1})}{N_1},0,0,\ldots\right) \in S_{\ell_1}. \end{equation*}

Then, there exists $z \in S(y^{(1)}, 1/N_1)$ such that $\|x-z\| \gt \textit{w}^*dc(x) - \delta \gt 1+\limsup_n |x_n|$. Note that if there exists $i \in \{1,\ldots, N_1\}$ so that $\text{sgn}(x_i) \neq \text{sgn}(z_i)$, then

\begin{equation*} \frac{N_1-1}{N_1} \geq \frac{1}{N_1} \sum_{n \in \{1,\ldots,N_1\}\setminus\{i\}} \text{sgn}(x_n) z_n \gt \frac{1}{N_1} \sum_{n=1}^{N_1} \text{sgn}(x_n) z_n \gt 1 - \frac{1}{N_1}, \end{equation*}

which is a contradiction. Thus, $\text{sgn}(x_i)=\text{sgn}(z_i)$ for every $i=1,\ldots ,N_1$. In particular, this implies that there exists $j_1 \gt N_1$ such that $|x_{j_1} - z_{j_1}| \gt \textit{w}^*dc(x)-\delta$. Thus, $|x_{j_1}| \gt \textit{w}^*dc(x)-\delta-1$. Now, take $N_2 \gt j_1 \gt N_1$ and consider the element

\begin{equation*} y^{(2)} := \left(\frac{\text{sgn}(x_1)}{N_2}, \ldots, \frac{\text{sgn}(x_{N_2})}{N_2},0,0,\ldots\right) \in S_{\ell_1}. \end{equation*}

Arguing in the same way, we end up with finding $j_2 \gt N_2$ such that $|x_{j_2}| \gt \textit{w}^*dc(x)-\delta-1$. In this way, we find a sequence of natural numbers $\{N_1 \lt j_1 \lt N_2 \lt j_2 \lt \cdots\}$. Passing to a subsequence, we may assume that $\lim_n |x_{j_n}|$ exists, so

\begin{equation*} 1+ \lim_n |x_{j_n}| \geq \textit{w}^*dc(x) - \delta \gt 1 + \limsup_n |x_n|, \end{equation*}

which is a contradiction. This shows that $\textit{w}^*dc(x) \leq 1 + \limsup_n |x_n|$ and completes the proof.

Proof. Let $f \in L_1 (\mu)$ with $\|f\| \leq 1$ and δ > 0. Without loss of generality, let each A_n be an atom for µ. Note first that $|f\vert_{A_n}|| \mu(A_n)| \leq \|f\|$ for every $n \in \mathbb{N}$. Since µ is σ-finite, $L_1 (\mu)^* = L_\infty (\mu)$. Pick $g \in L_\infty(\mu)$ with $\|g \| = 1$. For each $n \in \mathbb{N}$, let $I_n : = \{x \in A_n : |g(x)| \gt 1-\delta/2\}$. Then, there exists $m \in \mathbb{N}$ such that I_m has a positive measure, which implies that $\chi_{I_m} = \chi_{A_m}$ almost everywhere.

Note that $h:= \frac{\text{sgn} (g |_{A_m}) }{\mu(A_m)} \chi_{A_m}$ satisfies that

\begin{equation*} g(h) = \int_{A_m} \frac{\text{sgn} (g |_{A_m}) }{\mu(A_m) } g \, d\mu = \int_{I_m} \frac{| g |_{A_m} | }{\mu(I_m) } \, d\mu \geq 1-\frac{\delta}{2} \gt 1- \delta. \end{equation*}

Observe that

\begin{align*} \|f-h\| &= \int_{A_m} \left | f- \frac{\text{sgn}(g|_{A_m})}{\mu (A_m)} \right| \, d\mu + \int_{\cup_{n\neq m} A_n} |f| \, d\mu \\ &= | \text{sgn} (g|A_m) - f|_{A_m} \mu (A_m) | + \|f \| - |f|_{A_m}| \mu(A_m) \\ &\geq 1- |f|_{A_m}| \mu (A_m) + \|f \| - |f|_{A_m}| \mu(A_m) = 1+ \|f\| - 2 | f|_{A_m}| \mu(A_m). \end{align*}

This shows that $\text{dc}(f) \geq 1+ \|f\| - 2 | f|_{A_m}| \mu(A_m)$.

Next, assume that

(3.3)\begin{equation} \text{dc}(f) \gt 1+ \|f\| - 2 \sup \left\{ |f \vert_{A_n} |\, \mu (A_n) : n \in \mathbb{N} \right\}. \end{equation}

For simplicity, we may assume that $|f|_{A_1}| \mu(A_1) = \sup\left\{|f|_{A_n}| \mu(A_n) : {n\in\mathbb{N} } \right\}$. If $|f|_{A_1}| \mu(A_1) = \|f\|$, then this would imply that

\begin{equation*} f = {f \vert_{A_1}} \chi _{A_1}. \end{equation*}

It is not difficult to check that $\text{dc} (f) = 1 - \|f\|$ in such a case, which contradicts to (3.3).

Thus, we assume that $|f|_{A_1}| \mu(A_1) \lt \|f\|$. Take $S=S(\text{sgn}(f |_{A_1} ) \chi_{A_1} , \delta)$ with δ > 0 sufficiently small so that

\begin{equation*} \delta \lt \text{dc}(f) - \bigl( 1+ \|f\| - 2 |f|_{A_1}| \mu(A_1)\bigr) \, \text{and } \, \delta \lt \|f\| - |f|_{A_1}| \mu(A_1). \end{equation*}

Let us pick $h \in S$ with $\|h\|=1$ such that $\|f-h\| \gt \text{dc}(f) - \delta$. That is,

\begin{equation*} \text{sgn} ( f|_{A_1} ) h |_{A_1} \mu(A_1) \gt 1- \delta \geq \|f \| - \delta, \end{equation*}

and

\begin{equation*} \| f-h\| \gt \text{dc}(f)-\delta \gt 1+ \|f\| - 2|f|_{A_1}| \mu(A_1). \end{equation*}

Note, in particular, that $\text{sgn} ( f|_{A_1} ) h |_{A_1} \mu(A_1) \gt |f|_{A_1}| \mu(A_1)$. Observe that

\begin{align*} \|f-h\| &= \int_{A_1} |f-h| \, d\mu + \int_{\cup_{n\geq 2} A_n} |f-h| \, d\mu \\ &= | \text{sgn} ( f|_{A_1} ) h |_{A_1} \mu(A_1) - |f|_{A_1}|\mu(A_1)| + \int_{\cup_{n\geq 2} A_n} |f-h| \, d\mu \\ &\leq \text{sgn} ( f|_{A_1} ) h |_{A_1} \mu(A_1) - |f|_{A_1}|\mu(A_1) + \|f\| - | f|_{A_1}| \mu(A_1) + 1 - | h_{A_1}| \mu(A_1) \\ &\leq 1 + \|f\| - 2 | f|_{A_1}| \mu(A_1), \end{align*}

which is a contradiction.

Proof. By considering −f instead of f when $c:= f \vert_A$ less than or equal to 0, we may assume that $c \geq 0$. Assume to the contrary that ${\delta}c (f) \gt 1+\|f\| -2c \mu(A)$. Let ɛ > 0 be such that $\varepsilon \lt {\delta}c(f) - ( 1+\|f\| - 2c \mu(A))$. Suppose that $g \in L_1 (\mu)$ satisfies that $\|g\| \leq 1$ and $\|f-g\| \geq {\delta}c (f) - \varepsilon$. Put $g \vert_A =: d$ and observe that

\begin{align*} {\delta}c(f) - \varepsilon \leq \int_\Omega |f-g| \,d\mu &= \int_A |f-g|\,d\mu + \int_{\Omega\setminus A} |f-g|\,d\mu \\ &\leq \int_A |f-g|\,d\mu + \left( \|f\| -\int_A |f| \, d\mu \right) + \left( 1 - \int_A |g| \, d\mu \right) \\ &\leq |c-d|\mu(A) + \|f\| + 1 - c \mu(A) - |d| \mu(A). \end{align*}

This implies that

(3.4)\begin{equation} c\mu(A) + d\mu(A) \leq 1+\|f\| - {\delta}c(f) +\varepsilon + |c-d| \mu(A). \end{equation}

If $c \leq d$, then (3.4) implies that

\begin{equation*} 2c \mu (A) \leq 1 + \|f\| -{\delta}c (f) + \varepsilon, \end{equation*}

which contradicts the choice of ɛ. Thus, $c \geq d$, and we have from (3.4) that

\begin{equation*} c \mu(A) + d\mu(A) \leq 1+\|f\| -{\delta}c(f) + \varepsilon + (c-d) \mu(A). \end{equation*}

It follows that

(3.5)\begin{equation} g\vert_A = d \leq \frac{1}{2\mu(A) } ( 1+ \|f\| -{\delta}c(f) + \varepsilon). \end{equation}

Now, by (iii) of proposition 2.3, we obtain that $f \in \overline{\text{co}} \,\Delta_{2-{\delta}c (f) + \eta}(f)$ for every η > 0. In particular, $f \in \overline{\text{co}} \,\Delta_{2-{\delta}c (f) + \varepsilon}(f) = \overline{\text{co}} \,\{g \in B_{L_1 (\mu)} : \|f-g\| \geq {\delta}c (f) - \varepsilon\}$. However, if $g_1, \ldots, g_m \in \Delta_{2-{\delta}c (f) + \varepsilon} (f)$ and $\lambda_1,\ldots, \lambda_m$ are positive scalars so that $\sum_{i=1}^m \lambda_i = 1$, then

\begin{align*} \left\| f - \sum_{i=1}^m \lambda_i g_i \right \| &\geq \int_A \left | \sum_{i=1}^m \lambda_i (c- d_i) \right |\,d\mu \quad (d_i := g_i \vert_{A}, \, i=1,\ldots, m) \\ &= \int_A \sum_{i=1}^m \lambda_i (c- d_i) \,d\mu \quad (\text{since } c \geq d_i, \, i=1,\ldots, m) \\ &\geq \int_A c - \frac{1}{2\mu(A) } ( 1+\|f\| -{\delta}c(f) + \varepsilon) \, d\mu \quad (\text{by } (3.5)) \\ &= \frac{1}{2} \bigl[ 2c \mu(A) - (1+\|f\| -{\delta}c(f) +\varepsilon) \bigr] \gt 0. \end{align*}

This contradicts $f \in \overline{\text{co}} \,\Delta_{2-{\delta}c (f) + \varepsilon}(f) $.

Proof. The equalities for $\text{dc}(x)$ and ${\delta}c(x)$ are immediate from propositions 3.6 and 3.7. For their weak$^*$ constants, it suffices to observe that $\textit{w}^*{\delta}c(x)\leq 1+\|x\| -2\sup \{|x_n| : n \in I \}$. Indeed, note that in the case of $\ell_1$, the slice used in remark 3.8 is nothing but a weak$^*$ slice determined by $e_k \in c_0$ for some $k \in \mathbb{N}$.

Proof. By the assumption that the set in (3.6) is finite, and by the Krein–Milman theorem, observe that $\{e^* \in \text{ext} (B_{X^*}) : e^* (x) = 1 \}$ is non-empty and finite. Let us say

\begin{equation*} \{e^* \in \text{ext} (B_{X^*}) : e^* (x) = 1 \} = \{e_1^*, \ldots, e_n^*\}. \end{equation*}

It follows that

\begin{equation*} \alpha:= \sup \{|e^*(x)| : e^* \in \text{ext} (B_{X^*}) \setminus \{\pm e_i^* :i=1,\ldots, n\} \} \lt 1. \end{equation*}

Consider the slice $S=S\bigl(g,\frac{1 - \alpha}{2n}\bigr)$, where $g = \frac{1}{n} \sum_{i=1}^n e_i^*$. If $y \in S$, then it is not difficult to check that $y \in S\bigl(e_i^*, \frac{1 - \alpha}{2}\bigr)$ for every $i =1,\ldots, n$. In particular, $|e_i^* (x -y)| \lt \frac{1 + \alpha}{2}$ for every $i =1,\ldots, n$. Thus, we observe that

\begin{align*} \|y-x\| &=\sup \{|e^*(y-x)|: e^* \in \text{ext} (B_{X^*})\} \\ &\leq \max \bigl\{1+\alpha, \max\{ |e_i^* (y-x)|: i=1,\ldots, n\} \bigr\} = 1 +\alpha \lt 2. \end{align*}

Since this holds for every $y \in S$, we conclude (3.7).

Proof. Proof of theorem 3.12

(i) $\Rightarrow$ (ii): It is a direct consequence of lemma 2.8.

(ii) $\Rightarrow$ (iii): Arguing as in the proof of (ii) $\Rightarrow$ (iii) in [Reference Veeorg26, theorem 2.1], it is enough to prove the case when M is complete. Let $u \neq v \in M$, $r,s \gt 0$ such that there exists δ > 0 satisfying

\begin{equation*} [u,v]_\delta \subseteq B(u,rd(u,v)) \cup B(v, sd(u,v)). \end{equation*}

If $r+s \geq \alpha/2$, then there is nothing to prove; so we assume that $r+s \lt \alpha/2$. By lemma 3.13(iii), there exist $x \in B(u, r d(u,v))$ and $y \in B(v, s d(u,v))$ such that $m_{x,y}$ is a denting point. By the assumption, we then have that $\|\mu - m_{x,y} \| \geq \alpha$. Note that

\begin{equation*} d(u,x) + d(v,y) \leq r d(u,v) + s d(u,v) \lt \frac{\alpha}{2} d(u,v) \leq d(u,v). \end{equation*}

Thus, we may find η > 0 such that

\begin{equation*} d(x,u) + d(v,y) = d(u,v) + d(x,y) - \eta \max \{d(x,y), d(u,v)\}. \end{equation*}

By applying lemma 3.13(ii), we obtain that

\begin{align*} \|m_{x,y} - m_{u,v}\|&=\| m_{x,y} + m_{v,u} \| \\ &= \frac{d(u,x)+d(v,y) + |d(u,v) - d(x,y)|}{\max \{d(u,v), d(x,y) \}} \\ &\leq \frac{2d(u,x) + 2 d(v,y)}{\max \{d(u,v), d(x,y) \}} \leq 2 \frac{rd(u,v)+s d(u,v)}{\max \{d(u,v),d(x,y)\}} \leq 2(r+s). \end{align*}

Consequently,

\begin{equation*} \| \mu - m_{u,v}\| \geq \| \mu - m_{x,y} \| - \| m_{x,y} - m_{u,v} \| \geq \alpha - 2(r+s). \end{equation*}

(iii) $\Rightarrow$ (i): Fix ɛ > 0 and a slice $S= S(f,\xi)$ of $B_{\mathcal{F} (M)}$. Let $u_0 \neq v_0 \in M$ be such that $f(u_0)-f(v_0) \gt (1-\xi) d(u_0,v_0)$. By lemma 3.13(i), there exists γ > 0 such that if $x\neq y \in M$ satisfy $d(x,y) \lt \gamma$, then $\| \mu - m_{x,y} \| \geq 2-\varepsilon$. Let $n \in \mathbb{N}$ and δ > 0 be such that

1. (i) $f(u_0)-f(v_0) \gt (1-\xi) (1+\delta)^n d(u_0,v_0)$;

2. (ii) $\left(1-\dfrac{\varepsilon}{4} + \delta\right)^n d(u_0,v_0) \lt \gamma$.

If $\| \mu - m_{u_0, v_0}\| \geq \alpha- \varepsilon$, then there is nothing to prove. Suppose that $\| \mu - m_{u_0,v_0} \| \lt \alpha -\varepsilon$. Then, by the assumption, there exists

\begin{equation*} p \in [u_0,v_0]_{\delta d(u_0,v_0)} \setminus \left( B\left(u_0, \frac{\varepsilon}{4}d(u_0,v_0)\right) \cup B\left(v_0, \frac{\varepsilon}{4} d(u_0,v_0)\right) \right). \end{equation*}

This allows us to choose $u_1, v_1 \in M$ such that

1. (i) $f(u_1)-f(v_1) \gt (1-\xi) (1+\delta)^{n-1} d(u_1,v_1)$;

2. (ii) $d(u_1, v_1) \lt \left(1-\dfrac{\varepsilon}{4}+\delta \right) d(u_0,v_0)$

(for its detail, see the proof of (iii) $\Rightarrow$ (i) in [Reference Veeorg26, theorem 2.1]). Assume for $k \in \{1,\ldots, n-1\}$ that

1. (i) $f(u_k)-f(v_k) \gt (1-\xi) (1+\delta)^{n-k} d(u_k,v_k)$;

2. (ii) $d(u_k, v_k) \lt \left(1-\dfrac{\varepsilon}{4}+\delta \right)^k d(u_0,v_0)$.

If $\| \mu - m_{u_k,v_k}\| \geq \alpha -\varepsilon$ for some k, then we finish the proof. If not, we end up with the case:

1. (i) $f(u_n)-f(v_n) \gt (1-\xi) d(u_n,v_n)$;

2. (ii) $d(u_n, v_n) \lt \left(1-\dfrac{\varepsilon}{4}+\delta \right)^n d(u_0,v_0) \lt \gamma$; implying that $\| \mu - m_{u_n, v_n}\| \geq 2- \varepsilon$.

Therefore, in any case, we find $u \neq v \in M$ so that $m_{u,v} \in S$ and $\| \mu - m_{u,v}\| \geq \alpha-\varepsilon$.

Proof. Let us consider

\begin{equation*} M:= ( [0,1]\times \{0\}) \cup\left\{\left(0,\frac{1}{n}\right),\left(1,\frac{1}{n}\right) : n \in \mathbb{N} \right\}\subseteq (\mathbb R^2,\Vert\cdot\Vert_2). \end{equation*}

Put $x:=(1,0)$, $y:=(0,0)$, $u_n = \bigl(1,\frac{1}{n}\bigr)$, and $v_n = \bigl(0,\frac{1}{n}\bigr)$ for each $n \in \mathbb{N}$. As there is a (1-Lipschitz) path connecting x and y, by [Reference Jung and Rueda Zoca17, proposition 4.2], the molecule $m_{x,y}$ is a Δ-point, i.e., ${\delta}c (m_{x,y})=2$.

Let us fix $n \in \mathbb{N}$, and note that $[u_n,v_n]=\{u_n,v_n\}$. Thus, it is clear to see that there exist sufficiently small $r,s,\delta \gt 0$ with $r+s \lt d(u_n,v_n)=1$ such that

\begin{equation*} [u_n,v_n]_\delta := \{p \in M : d(u_n, p) + d(v_n, p) \lt d(u_n,v_n)+\delta\} = \{u_n,v_n\}, \end{equation*}

$B(u_n,r) = \{u_n\}$, and $B(v_n, s)=\{v_n\}$. Applying (3) of lemma 3.13, we obtain that $m_{u_n, v_n}$ is a denting point.

Suppose that $\text{dc} (m_{x,y} ) \gt 0$. Take α > 0 so that $\text{dc}(m_{x,y})\geq \alpha \gt 0$. By theorem 3.12, it follows that we have $\| m_{u_n,v_n} - m_{x,y} \| \geq \alpha$ for every $n \in \mathbb{N}$. However, the distance between molecules can be estimated as follows (see [Reference Cascales, Chiclana, García-Lirola, Martín and Rueda Zoca7, lemma 1.3]):

\begin{equation*} \| m_{u_n,v_n} - m_{x,y} \| \leq 2 \frac{d(x,u_n) + d(y, v_n) }{\max \{d(x, y), d(u_n,v_n)\}} = \frac{4}{n} \rightarrow 0. \end{equation*}

This contradicts $\text{dc} (m_{x,y}) \geq \alpha \gt 0$; hence we conclude that $\text{dc}(m_{x,y})=0$.

Proof. Assume first that x ≠ 0 and y ≠ 0. Assume also $\min \left\{\text{dc} \bigl( \frac{x}{\|x\|} \bigr) , \text{dc} \bigl( \frac{y}{\|y\|}\bigr) \right\} \gt 1+\varepsilon \gt 1$ for some ɛ > 0 (otherwise, there is nothing to prove). Let $S=S\bigl((x^*,y^*),\delta\bigr)$ be a slice of $B_{X\oplus_N Y}$ with $\|(x^*,y^*)\| =1$. Take $(u,v) \in S_{X\oplus_N Y}$ so that $x^* (u) + y^* (v) \gt 1 - \delta' \gt 1-\delta$ for some $0 \lt \delta' \lt \delta$. Fix η > 0 so that $2\eta \lt \delta-\delta'$, and consider

\begin{equation*} S_1 = \begin{cases} \bigl\{z \in B_X : x^*(z) \gt x^* \bigl(\frac{u}{\|u\|}\bigr) - \eta\bigr\} &\text{when } u \neq 0 \\ B_X &\text{when } u = 0 \end{cases} \end{equation*}

and

\begin{equation*} S_2 = \begin{cases} \bigl\{w \in B_Y : y^*(w) \gt y^* \bigl(\frac{v}{\|v\|}\bigr) - \eta\bigr\} &\text{when } v \neq 0 \\ B_Y &\text{when } v = 0. \end{cases} \end{equation*}

Observe that there exist $z \in S_1$ and $w\in S_2$ such that $\bigl\|z - \frac{x}{\|x\|}\bigr\| \gt \text{dc}\bigl( \frac{x}{\|x\|}\bigr) - \frac{\varepsilon}{2}$ and $\bigl\|w - \frac{y}{\|y\|}\bigr\| \gt \text{dc}\bigl( \frac{y}{\|y\|}\bigr) - \frac{\varepsilon}{2}$. Note that

\begin{equation*} \left\|z - \frac{x}{\|x\|}\right\| \gt \text{dc}\left( \frac{x}{\|x\|}\right) - \frac{\varepsilon}{2} \gt (1+\varepsilon) - \frac{\varepsilon}{2} = 1 + \frac{\varepsilon}{2}. \end{equation*}

Then,

\begin{equation*} \bigl\|x - \|u\| z \bigr\| = \left\| \|x\| \frac{x}{\|x\|} + \|u\| (-z) \right\| \geq (\|x\| + \|u\|) \left (\text{dc}\left( \frac{x}{\|x\|}\right) - \frac{\varepsilon}{2} - 1\right), \end{equation*}

where we used the fact: if $e_1, e_2 \in B_X$ satisfy $\|e_1+e_2\|\geq 1+\alpha$ for some $\alpha\in [0,1]$, then $\|\lambda e_1 + \mu e_2\| \geq (\lambda+\mu)\alpha$ for all $\lambda,\mu\geq 0$. Similarly, we obtain

\begin{equation*} \bigl\|y-\|v\| w\bigr\| \geq (\|y\| + \|v\|) \left (\text{dc}\left( \frac{y}{\|y\|}\right) - \frac{\varepsilon}{2} - 1\right). \end{equation*}

Note that $(\|u\| z, \|v\| w) \in B_{X\oplus_N Y}$ and if $u \neq 0, v\neq 0$, then

\begin{align*} (x^*, y^*) (\|u\| z, \|v\| w) &= \|u\| x^*(z) + \|v\| y^* (w) \\ & \gt \|u\| \left( x^* \left(\frac{u}{\|u\|} \right)-\eta \right) + \|v\| \left( y^*\left(\frac{v}{\|v\|}\right) -\eta \right) \\ & \gt 1-\delta' - 2\eta \gt 1-\delta. \end{align*}

If one of u and v is 0, say v = 0, then $\|u\|=1$ and

\begin{align*} (x^*, y^*) (\|u\| z, \|v\| w) &= x^*(z) \gt x^*(u) - \eta \gt 1-\delta' -\eta \gt 1-\delta. \end{align*}

Thus, in any case, $(\|u\| z, \|v\| w) \in S$. Now, observe that

\begin{align*} \| (x,y) &- (\|u\| z, \|v\| w) \|_N \\ &= N\bigl( \|x- \|u\|z \|, \|y - \|v\|w\|\bigr) \\ &\geq N\left( (\|x\| + \|u\|) \left (\text{dc}\left( \frac{x}{\|x\|}\right) - \frac{\varepsilon}{2} - 1\right), (\|y\| + \|v\|) \left (\text{dc}\left( \frac{y}{\|y\|}\right) - \frac{\varepsilon}{2} - 1\right) \right) \\ &\geq \min \left\{\text{dc}\left( \frac{x}{\|x\|}\right) - \frac{\varepsilon}{2} - 1, \text{dc}\left( \frac{y}{\|y\|}\right) - \frac{\varepsilon}{2} - 1 \right\} N\bigl( \|x\|+\|u\|, \|y\|+\|v\|\bigr) \\ &\geq 2\gamma \min \left\{\text{dc}\left( \frac{x}{\|x\|}\right) - \frac{\varepsilon}{2} - 1, \text{dc}\left( \frac{y}{\|y\|}\right) - \frac{\varepsilon}{2} - 1 \right\}. \end{align*}

Next, assume that x = 0 or y = 0. We only prove the case when y = 0. Assume that $\text{dc}(x) \gt 1+\varepsilon$ for some ɛ > 0, otherwise there is nothing to prove. Let $S=S((x^*,y^*),\delta)$ be a slice of $B_{X\oplus_N Y}$ with $\|(x^*,y^*)\| =1$. Take $(u,v) \in S_{X\oplus_N Y}$ as above with some $0 \lt \delta' \lt \delta$. Fix η > 0 and consider $S_1, S_2$ as above too. Find $z \in S_1$ such that $\|z - x \| \gt \text{dc}(x) - \frac{\varepsilon}{2}$. Let $w \in S_2 \cap S_Y$ be arbitrarily given. Arguing as above, $(\|u\| z, \|v\| w) \in S$. Moreover,

\begin{align*} \bigl\|(x,0) - (\|u\| z, \|v\| w) \bigr\|_N &= N\bigl(\|x - \|u\|z\| , \|v\|\bigr) \\ &\geq N \left( (1 + \|u\|) \left(\text{dc} (x) - \frac{\varepsilon}{2} - 1\right), \|v\|\right) \\ &\geq \left(\text{dc} (x) - \frac{\varepsilon}{2} - 1\right) N\bigl( 1+\|u\|, \|v\| \bigr) \\ &\geq 2\gamma \left(\text{dc} (x) - \frac{\varepsilon}{2} - 1\right). \end{align*}

This finishes the proof.

Proof. (i): Let $S=S\bigl( (x^*, y^*), \delta\bigr)$ be a slice of $B_{X\oplus_1 Y}$ with $\|(x^*,y^*)\|=1$ and ɛ > 0. Without loss of generality, we may assume that $\|x^*\| = 1$. Find $u \in S(x^*, \delta)$ so that $\bigl\|u- \frac{x}{\|x\|}\bigr\| \geq \text{dc} \bigl(\frac{x}{\|x\|} \bigr) - \varepsilon$. Note that $(u,0) \in S$ and

\begin{align*} \|(u,0)-(x,y)\| = \|u-x \| + \|y\| &\geq \left\| u - \frac{x}{\|x\|}\right\| - \left\| \frac{x}{\|x\|} - x \,\right\| + \|y\| \\ &\geq \text{dc}\left(\frac{x}{\|x\|}\right) - \varepsilon - (1-\|x\|) + \|y\| \\ &= \text{dc}\left(\frac{x}{\|x\|}\right) -\varepsilon. \end{align*}

This implies that $\text{dc} ((x,y)) \geq \text{dc} \bigl(\frac{x}{\|x\|}\bigr)$. Notice that if $\|y^*\| = 1$, then, by changing the role of x and y, we observe $\text{dc} ((x,y)) \geq \text{dc} \bigl(\frac{y}{\|y\|}\bigr)$.

As the proofs of (ii) and (iii) are very similar, we only prove (ii): Let $S=S\bigl( (x^*, y^*), \delta\bigr)$ be a slice of $B_{X\oplus_1 Y}$ and ɛ > 0. Suppose that $\|x^*\| = 1$. Then, there exists $u \in S_X$ so that $x^*(u) \gt 1-\delta$. In other words, $(u,0) \in S$. In this case, $\|(0,y)-(u,0)\| = \|u\| +\|y\| = 2$. If $\|y^*\| = 1$, then we can pick $v \in S(y^*, \delta)$ so that $\| v-y\| \gt \text{dc} (y)-\varepsilon$. This implies that $(0,v) \in S$ and $\|(0,y)-(0,v)\| = \|v-y\| \gt \text{dc}(y)-\varepsilon$. In any case, we obtain a point in S, which is a distance of $\text{dc}(y)-\varepsilon$ from $(0,y)$. This proves that $\text{dc} ((0,y))\geq \text{dc} (y)$.

Proof. We only prove the case when $\|x\|=1$: Let $S=S( (x^*, y^*), \delta)$ be a slice of $B_{X\oplus_\infty, Y}$ with $\|(x^*, y^*)\| =1$ and ɛ > 0. Note that $\|x^*\| +\|y^*\| =1$. If $x^* \neq 0$, then find $u \in S\bigl(\frac{x^*}{\|x^*\|}, \delta\bigr)$ such that $\|u-x\| \gt \text{dc}(x)-\varepsilon$. Pick $v \in S\bigl(\frac{y^*}{\|y^*\|}, \delta\bigr)$ if $y^* \neq 0$ and $v \in B_Y$ arbitrarily if $y^* =0$. In any case, $(x^*,y^*) (u,v) = x^*(u) +y^*(v) \gt 1-\delta$, that is, $(u,v) \in S$. Note that

\begin{equation*} \|(x,y)-(u,v)\| = \max \bigl\{\|x-u\| ,\|y-v\|\bigr\} \gt \text{dc}(x)-\varepsilon. \end{equation*}

It remains to consider the case when $x^* = 0$. In this case, $\|y^*\| =1$. Then $(u,v) \in S$ for any $u \in B_X$ and $v \in S(y^*, \delta)$. Note that

\begin{equation*} \|(x,y)-(-x,v)\| = \max \bigl\{\|2x\| , \|y-v\| \bigr\} =2. \end{equation*}

Consequently, whether $x^*$ is 0 or not, we can always find a point in S, which is ($\text{dc}(x)-\varepsilon$)-away from (x, y). So, $\text{dc}((x,y)) \geq \text{dc}(x)$.

Proof. Let $(x^*, y^*) \in S_{(X\oplus_N Y)^*}, \alpha \gt 0$ and ɛ > 0. Choose δ > 0 so that $\delta N(1,1) \lt \varepsilon$. Find $u \in B_X$ and $v \in B_Y$ such that

\begin{equation*} x^* (u) \geq \left( 1- \frac{\alpha}{2}\right) \|x^*\| \quad \text{and} \quad y^*(v) \geq \left( 1-\frac{\alpha}{2}\right) \|y^*\|, \end{equation*}

$\|x-u\| \gt \text{dc}(x)-\delta$ and $\|y-v\| \gt \text{dc}(y)-\delta$. Find $k \geq 0, \ell \geq 0$ such that $N(k,\ell)=1$, $N\bigl((a,b)+(k,\ell)\bigr)=2$, and $k\|x^*\| + \ell \|y^*\| = 1$. Note that

\begin{equation*} (x^*,y^*)(ku,\ell v) = kx^*(u) +\ell y^*(v) \geq \left(1-\frac{\alpha}{2}\right) (k\|x^*\| +\ell \|y^*\| ) = 1-\frac{\alpha}{2}, \end{equation*}

which implies that $(ku, \ell v) \in S( (x^*,y^*), \alpha)$. By triangle inequality, we obtain $\|ax-ku\| \geq (a+k)(\text{dc}(x)-\delta-1)$ and $\|by-\ell v\| \geq (b+\ell) (\text{dc}(y) -\delta-1)$. Thus,

\begin{align*} \| (ax,by) - (ku,\ell v)\|_N &= N\bigl(\|ax-ku\|, \|by- \ell v\| \bigr) \\ &\geq N\bigl( (a+k)(\text{dc} (x)-\delta-1), (b+\ell)(\text{dc}(y)-\delta-1) \bigr) \\ &\geq N\bigl( (a+k)(\text{dc} (x)-1), (b+\ell)(\text{dc}(y)-1) \bigr) - 2 \varepsilon \\ &\geq \min \bigl\{\text{dc}(x)-1, \text{dc}(y)-1 \bigr\} N(a+k, b+\ell) - 2 \varepsilon \\ &=2 \min \bigl\{\text{dc}(x)-1, \text{dc}(y)-1 \bigr\} - 2 \varepsilon, \end{align*}

as desired.

Proof. Let ɛ > 0 and $0 \lt \eta \lt \varepsilon$. By proposition 2.3, we can find $x_1, \ldots, x_m \in \Delta_{2-{\delta}c(x) +\varepsilon} (x)$ and $y_1,\ldots, y_m \in \Delta_{2-{\delta}c(y)+\varepsilon} (y)$ such that

\begin{equation*} \left\| x - \frac{1}{m} \sum_{i=1}^m x_i \right\| \lt \eta \, \text{and } \, \left\| y - \frac{1}{m} \sum_{i=1}^m y_i \right\| \lt \eta, \end{equation*}

where we applied [Reference Abrahamsen, Haller, Lima and Pirk3, lemma 4.1] to choose the same number of vectors in X and Y. Note that

\begin{equation*} \left\| (ax,by) - \frac{1}{m} \sum_{i=1}^m (ax_i, by_i) \right\|_N = N \left( a \left\|x- \frac{1}{m} \sum_{i=1}^m x_i\right\| , b \left \| y- \frac{1}{m} \sum_{i=1}^m y_i\right\| \right) \leq \eta. \end{equation*}

Moreover, for each $i =1,\ldots, m$,

\begin{equation*} \bigl\| (ax,by)-(ax_i,by_i) \bigr\|_N = N\bigl(a \|x-x_i\|, b\|y-y_i\| \bigr) \geq \min \bigl\{{\delta}c(x)-\varepsilon, {\delta}c(y)-\varepsilon \bigr\}. \end{equation*}

Thus, by again proposition 2.3, we conclude that ${\delta}c((ax,by)) \geq \min \bigl\{{\delta}c(x), {\delta}c(y)\bigr\}$.

When a = 0 or b = 0, the above argument proves the estimates (i) and (ii).

Proof. For the assertion (i), we apply proposition 4.9 with $a = \|x\|$ and $b = \|y\|$. For (ii), use again proposition 4.9 with a = 1 and b = 1.

Proof. Thanks to corollary 4.10, it suffices to show for every $(x,0) \in S_{X \oplus_1 Y}$ that ${\delta}c((x,0)) \leq {\delta}c(x)$. Assume that ${\delta}c(x) \leq \alpha$, then there exists a slice $S(x^*,\delta)$ with $x \in S(x^*,\delta)$ such that $\|y-x\| \lt \alpha$ for every $y \in S(x^*,\delta)$. Given $0 \lt \eta \lt \delta$, by [Reference Ivakhno and Kadets14, lemma 2.1], we can find $y_1^* \in S_{X^*}$ satisfying that

\begin{equation*} x \in S(y_1^*,\eta) \subseteq S(x^*,\delta). \end{equation*}

Note that for $(u,v) \in S( (y_1^*, 0), \eta)$, we have $y_1^* (u) \gt 1-\eta$ and $\|v\| \lt \eta$. This implies that

\begin{equation*} \|(u,v) - (x,0)\| = \|u-x\| + \|v\| \lt \|u-x\| + \eta, \end{equation*}

where $u, x \in S(x^*,\delta)$. As it is clear that $(x, 0) \in S( (y_1^*, 0) , \eta)$ and η can be chosen arbitrarily small, we conclude that ${\delta}c((x,0)) \leq \alpha$.