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\author[]{Samir Hamad}
\address{Mathematisches Seminar, Christian-Albrechts-Universit\"at zu Kiel, Kiel, Germany}
\email{hamad@math.uni-kiel.de}
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\title[The Daugavet property of bidual spaces]
{The Daugavet property of bidual spaces}
\date{November 2026}
\subjclass[2020]{46B04, 46B20}
\keywords{Daugavet property, bidual space, finite-dimensional extension}
\thanks{Working draft.}

\begin{document}
\begin{abstract}
We construct a separable Banach space whose bidual has the Daugavet property. The construction proceeds by successively extending finite-dimensional spaces so that prescribed vectors are approximated by averages of newly added unit vectors. A simultaneous extension argument produces finitely many fixed functionals whose norm-attaining faces approximate the unit ball of the bidual.
\end{abstract}
\maketitle

\section{Introduction}

Throughout the paper all spaces, scalars and operators are real. We say that a normed space $X$ has the \emph{Daugavet property}, shortly $X \in \mathrm{DPr}$, if
\[
\|\mathrm{Id} + T\| = 1 + \|T\|,
\]
for every rank-one operator $T \colon X \to X$. The following characterization was obtained in \cite[Lemma~2.2]{kadets}.

\begin{theorem}
Let $X$ be a normed space. The following assertions are equivalent:
\begin{enumerate}
\item $X$ has the Daugavet property.
\item For every $y \in S_X$, $x^* \in S_{X^*}$, and every $\varepsilon > 0$, there exists $x \in S_X$ such that
\[
x^*(x) \ge 1 - \varepsilon
\quad \text{and} \quad
\|x + y\| \ge 2 - \varepsilon.
\]
\item For every $\varepsilon > 0$ and every $y \in S_X$, the closed convex hull of
\[
\{\, u \in (1+\varepsilon) B_X \mid \|y + u\| \ge 2 - \varepsilon \,\}
\]
contains $S_X$.
\end{enumerate}
\end{theorem}

We give a positive answer to the question of V.~M.~Kadets and
D.~Werner recorded in \cite[Problem~2.1, p.~204]{plichko-popov},
which asks whether there exists a Banach space whose bidual
has the Daugavet property. The bidual problem is motivated in part by the fact that classical Daugavet spaces such as \(C[0,1]\) and \(L^1[0,1]\) lose the Daugavet property upon passing to their biduals.

\begin{theorem}\label{thm:main}
There exists a separable Banach space $X$ such that $X^{**}$
has the Daugavet property.
\end{theorem}

We use successive finite-dimensional extensions, as in
\cite[Sections 2--3]{hamad}. The additional ingredient is a contractive
extension into an auxiliary space $V_{C,M}$, with one prescribed
coordinate. It yields a family of functionals chosen independently of
the point of $B_X$ being decomposed.

\section{The finite-dimensional extension}\label{sec:extension}

Let $E$ be a finite-dimensional normed space, $p\in B_E$, $\delta>0$
and $n\in\N$. Put $q_n=n^{-1}(1,\ldots,1)$ and define
$F=F(E,p,\delta,n)=E\oplus\R^n$ by
\begin{equation}\label{eq:infnorm}
 \norm{(x,b)}_F=
 \inf_{t\in\R}\bigl(
   \norm{x+tp}_E+\norm{b-tq_n}_1+\delta|t|
 \bigr).
\end{equation}
This is the quotient norm induced by the surjection
\[
 Q:E\oplus_1\ell_1^n\oplus_1(\R,\delta|\cdot|)\longrightarrow F,
 \qquad Q(y,b,t)=(y-tp,b+tq_n).
\]
We identify $E$ with $E\oplus\{0\}$ and put $h_j=(0,e_j)\in F$.

\begin{lemma}\label{lem:extension}
The inclusion $E\hookrightarrow F$ is isometric, and
\begin{equation}\label{eq:meanerror}
 h_j\in B_F\quad(1\le j\le n),\qquad
 \norm{\frac1n\sum_{j=1}^n h_j-p}_F\le\delta.
\end{equation}
Under the pairing
$(f,\alpha)(x,b)=f(x)+\sum_{j=1}^n\alpha_jb_j$, one has
\begin{equation}\label{eq:dualnorm}
 \norm{(f,\alpha)}_{F^*}
 =\max\left\{
   \norm f_{E^*},\ \norm\alpha_\infty,\
   \frac1\delta\abs{\frac1n\sum_{j=1}^n\alpha_j-f(p)}
 \right\}.
\end{equation}
If $n\delta\ge2$, then
\begin{equation}\label{eq:localdaugavet}
 \norm{h_j}_F=1,\qquad
 \norm{x+h_j}_F=2\quad(x\in S_E,\ 1\le j\le n).
\end{equation}
\end{lemma}

\begin{proof}
For $x\in E$,
\[
 \norm{x+tp}_E+(1+\delta)|t|
 \ge\norm x_E+(1+\delta-\norm p_E)|t|
 \ge\norm x_E.
\]
Together with $t=0$, this proves isometry. Taking $t=0$ at $h_j$
and $t=1$ at $(-p,q_n)$ proves \eqref{eq:meanerror}.
The adjoint of the quotient map is
\[
 Q^*(f,\alpha)=
 \left(f,\alpha,\frac1n\sum_{j=1}^n\alpha_j-f(p)\right),
\]
which gives \eqref{eq:dualnorm}.
If $n\delta\ge2$, taking $f=0$ and $\alpha=e_j$ in
\eqref{eq:dualnorm} gives $\norm{h_j}_F=1$.
Finally, fix $x\in S_E$ and $j$, and choose $f\in S_{E^*}$ with
$f(x)=1$. Set $\alpha_j=1$ and $\alpha_k=f(p)$ for $k\ne j$.
Then
\[
 \abs{\frac1n\sum_{k=1}^n\alpha_k-f(p)}
 =\frac{1-f(p)}n\le\frac2n\le\delta.
\]
Thus $(f,\alpha)\in B_{F^*}$ and $(f,\alpha)(x+h_j)=2$.
The triangle inequality gives \eqref{eq:localdaugavet}.
\end{proof}

\begin{lemma}\label{lem:operator}
Let $Y$ be a normed space, $T:E\to Y$ linear and $b_1,\ldots,b_n\in Y$.
The unique linear extension $\widetilde T:F\to Y$ satisfying
$\widetilde T|_E=T$ and $\widetilde T h_j=b_j$ has norm
\begin{equation}\label{eq:operatornorm}
 \opnorm{\widetilde T}{F,Y}
 =\max\left\{
   \opnorm T{E,Y},\ \max_{1\le j\le n}\norm{b_j}_Y,\
   \frac1\delta\norm{\frac1n\sum_{j=1}^n b_j-Tp}_Y
 \right\}.
\end{equation}
\end{lemma}

\begin{proof}
Apply \eqref{eq:dualnorm} to
$y^*\circ\widetilde T=(y^*\circ T,(y^*(b_j))_{j=1}^n)$
and take the supremum over $y^*\in B_{Y^*}$.
\end{proof}

\section{The inductive construction}\label{sec:induction}

Put $E_1=\R$ and $P_0=\varnothing$. Given $E_s$, choose a finite
$2^{-s}$-net $N_s\subset B_{E_s}$ and set
\[
 P_s=P_{s-1}\cup N_s,\qquad n_s=(s+2)!,\qquad m_s=s|P_s|.
\]
Enumerate
\[
 P_s\times\{1,\ldots,s\}
 =\{(p_{s,r},k_{s,r}):1\le r\le m_s\}.
\]
Starting with $F_{s,0}=E_s$, define
\begin{equation}\label{eq:schedule}
 F_{s,r}=F\left(F_{s,r-1},p_{s,r},\frac1{k_{s,r}},n_s\right)
 \quad(1\le r\le m_s),\qquad E_{s+1}=F_{s,m_s}.
\end{equation}
Here every $p_{s,r}$ is chosen in $E_s$ before the extensions in
\eqref{eq:schedule}. Lemma~\ref{lem:extension} gives isometric inclusions
\[
 E_s=F_{s,0}\subset F_{s,1}\subset\cdots\subset F_{s,m_s}=E_{s+1}.
\]
Let $X$ be the completion of $X_0=\bigcup_{s\ge1}E_s$, and put
$P=\bigcup_{s\ge1}P_s$. Then $X$ is separable, and
\begin{equation}\label{eq:density}
 \overline P^{\,\norm{\cdot}_X}=B_X.
\end{equation}
Indeed, $\bigcup_s B_{E_s}$ is dense in $B_X$, and
$\operatorname{dist}(x,P_s)\le2^{-s}$ for $x\in B_{E_r}$ and $s\ge r$.

For each fixed $p\in P$ and $k,M\in\N$, the construction contains
arbitrarily late elementary extensions
\begin{equation}\label{eq:recurrence}
 F(E,p,1/k,n)\qquad\text{with }M\mid n.
\end{equation}
In fact, $p\in P_s$, $k\le s$ and $M\mid(s+2)!$ for all sufficiently
large $s$. 

\section{Extensions with a prescribed coordinate}\label{sec:lifting}

Fix $C\ge1$ and $M\in\N$. On $V=V_{C,M}=\R\oplus\R^M$ put
\begin{equation}\label{eq:Vnorm}
 \norm{(a,c)}_V=
 \max\left\{
 |a|,\ \max_{1\le i\le M}|a+c_i|,\
 \max_{A\subseteq\{1,\ldots,M\}}
 \abs{a+\frac1C\sum_{i\in A}c_i}
 \right\}.
\end{equation}
Write $K=B_V$, $L(a,c)=a$, $\pi_i(a,c)=c_i$ and
$c_i^\pm=\max\{\pm c_i,0\}$. Then
\begin{equation}\label{eq:K}
 (a,c)\in K\quad\Longleftrightarrow\quad
 \begin{cases}
 |a|\le1,\quad -1-a\le c_i\le1-a &(1\le i\le M),\\
 \norm{c^+}_1\le C(1-a),\quad \norm{c^-}_1\le C(1+a).
 \end{cases}
\end{equation}
In particular,
\begin{equation}\label{eq:cbounds}
 \norm c_\infty\le2,\qquad \norm c_1\le2C
 \quad((a,c)\in K).
\end{equation}
If $(a,c)\in K$ and $|a|<1$, then
\begin{equation}\label{eq:faces}
 v_+=\left(1,-\frac{2c^-}{1+a}\right)\in K,\qquad
 v_-=\left(-1,\frac{2c^+}{1-a}\right)\in K,\qquad
 (a,c)=\frac{1+a}{2}v_++\frac{1-a}{2}v_-.
\end{equation}
All assertions follow directly from \eqref{eq:K}.

\begin{lemma}\label{lem:lifting}
Let $F=F(E,p,\delta,n)$, $f_F\in B_{F^*}$ and $0<\theta<1$.
Every contraction $T_E:E\to V_{C,M}$ with
$L\circ T_E=\theta f_F|_E$ admits a contraction $T_F:F\to V_{C,M}$
such that
\begin{equation}\label{eq:preservedbase}
 T_F|_E=T_E,\qquad L\circ T_F=\theta f_F.
\end{equation}
If, in addition,
\begin{equation}\label{eq:refreshcondition}
 M\mid n,\qquad M\ge C,\qquad \frac4C\le(1-\theta)\delta,
\end{equation}
then, for any partition $\{1,\ldots,n\}=J_1\sqcup\cdots\sqcup J_M$
with $|J_i|=n/M$, the extension may be chosen to satisfy
\begin{equation}\label{eq:saturation}
 (L+\pi_i)(T_Fh_j)=1\qquad(j\in J_i,\ 1\le i\le M).
\end{equation}
\end{lemma}

\begin{proof}
Put $a_0=\theta f_F(p)$ and $a_j=\theta f_F(h_j)$.
Since $|a_0|\le\theta<1$, apply \eqref{eq:faces} to $T_Ep$ and write
\[
 T_Ep=\frac{1+a_0}{2}v_++\frac{1-a_0}{2}v_-,\qquad
 b_j=\frac{1+a_j}{2}v_++\frac{1-a_j}{2}v_-.
\]
Then $b_j\in K$, $Lb_j=a_j$, and \eqref{eq:meanerror} gives
\begin{equation}\label{eq:regularslack}
 \norm{\frac1n\sum_{j=1}^n b_j-T_Ep}_V
 =\abs{\frac1n\sum_{j=1}^n a_j-a_0}
   \norm{\frac{v_+-v_-}{2}}_V
 \le\theta\delta.
\end{equation}
Thus $T_F|_E=T_E$, $T_Fh_j=b_j$ defines a contraction by
\eqref{eq:operatornorm}, and satisfies \eqref{eq:preservedbase}.

Assume \eqref{eq:refreshcondition}. For $(a,c)\in K$, define
\[
 R_i(a,c)=\left(a,\left(1-\frac1C\right)c+\beta_i e_i\right),
 \qquad \beta_i=1-a-\left(1-\frac1C\right)c_i\in[0,2].
\]
If $R_i(a,c)=(a,c')$, then $c'_i=1-a$, the coordinate bounds in
\eqref{eq:K} hold, and
\[
 \norm{(c')^+}_1
 \le\left(1-\frac1C\right)C(1-a)+(1-a)=C(1-a),\qquad
 \norm{(c')^-}_1\le\left(1-\frac1C\right)C(1+a).
\]
Hence $R_i(K)\subset K$.
Write $b_j=(a_j,c^{(j)})$, let $\ell(j)=i$ for $j\in J_i$, and put
\[
 \beta_i^{(j)}=1-a_j-\left(1-\frac1C\right)c_i^{(j)},
 \qquad b'_j=R_{\ell(j)}b_j.
\]
The mean change has the form
\[
 \frac1n\sum_{j=1}^n(b'_j-b_j)=(0,d),\qquad
 d=-\frac1{nC}\sum_{j=1}^n c^{(j)}
   +\frac1n\sum_{j=1}^n\beta_{\ell(j)}^{(j)}e_{\ell(j)}.
\]
Since $|J_i|=n/M$, \eqref{eq:cbounds} yields
\[
 \norm d_\infty\le\frac2C+\frac2M,\qquad
 \norm d_1\le4,\qquad
 \norm{(0,d)}_V
 \le\max\left\{\norm d_\infty,\frac{\norm d_1}{C}\right\}
 \le\frac4C.
\]
Together with \eqref{eq:regularslack}, this gives
\[
 \norm{\frac1n\sum_{j=1}^n b'_j-T_Ep}_V
 \le\theta\delta+\frac4C\le\delta.
\]
Now prescribe $T_F|_E=T_E$ and $T_Fh_j=b'_j$.
Formula~\eqref{eq:operatornorm} proves contractivity;
$Lb'_j=a_j$ and $(L+\pi_i)b'_j=1$ for $j\in J_i$ prove the rest.
\end{proof}

\section{Fixed functionals and bidual decompositions}\label{sec:fixed}

The space $X$ and its chain of finite-dimensional subspaces are now fixed.

\begin{proposition}\label{prop:fixedfaces}
Let $f\in B_{X^*}$, $0<\theta<1$, $k_0\in\N$, $\delta_0=1/k_0$,
$C\ge1$ and $M\in\N$ satisfy
\begin{equation}\label{eq:globalparams}
 M\ge C,\qquad \frac4C\le(1-\theta)\delta_0.
\end{equation}
There exist $g_1,\ldots,g_M\in X^*$ such that, with
$\phi_i=\theta f+g_i$, one has
\begin{equation}\label{eq:signbound}
 \phi_i\in S_{X^*},\qquad
 \norm{\sum_{i=1}^M\sigma_i g_i}_{X^*}\le2C
 \quad(\sigma_i\in\{-1,1\}).
\end{equation}
For these same functionals, put
$\mathcal F_i=\{z\in B_{X^{**}}:z(\phi_i)=1\}$. Then
\begin{equation}\label{eq:bidualfaces}
 B_{X^{**}}\subset
 \frac1M(\mathcal F_1+\cdots+\mathcal F_M)+\delta_0B_{X^{**}}.
\end{equation}
The functionals depend on $f,\theta,\delta_0,C,M$, but are chosen
independently of the vector in $B_{X^{**}}$ to be decomposed.
\end{proposition}

\begin{proof}
Set $V=V_{C,M}$ and $T_{E_1}x=(\theta f(x),0)$.
Along the fixed chain \eqref{eq:schedule}, apply Lemma~\ref{lem:lifting}
to each inclusion $E\subset F=F(E,p,\delta,n)$, always with
$f_F=f|_F$. Use the additional conclusion whenever
$\delta=\delta_0$ and $M\mid n$. The resulting contractions satisfy
\[
 T_F|_E=T_E,\qquad L\circ T_F=\theta f|_F.
\]
They define a contraction on $X_0$ and hence a contraction $T:X\to V$
with $L\circ T=\theta f$. Put
\[
 g_i=\pi_i\circ T,\qquad \phi_i=(L+\pi_i)\circ T.
\]
By \eqref{eq:Vnorm} and \eqref{eq:cbounds},
\[
 \norm{\phi_i}_{X^*}\le1,\qquad
 \sup_{x\in B_X}\sum_{i=1}^M|g_i(x)|\le2C,
\]
which implies the sign estimate in \eqref{eq:signbound}.

Fix $p\in P$. By \eqref{eq:recurrence}, choose an elementary extension
at $p$ with $\delta=\delta_0$ and $M\mid n$.
For its partition $J_1,\ldots,J_M$, set
\[
 z_i(p)=\frac Mn\sum_{j\in J_i}h_j\in B_X.
\]
Equations~\eqref{eq:meanerror} and \eqref{eq:saturation} give
\begin{equation}\label{eq:pfaces}
 \phi_i(z_i(p))=1,\qquad
 \norm{p-\frac1M\sum_{i=1}^M z_i(p)}_X\le\delta_0.
\end{equation}
In particular, $\norm{\phi_i}_{X^*}=1$ and each $\mathcal F_i$ is nonempty.

Identify $X$ with its canonical image in $X^{**}$.
The sets $\mathcal F_i$ are weak-star compact. Therefore
\[
 Q=\frac1M(\mathcal F_1+\cdots+\mathcal F_M)+\delta_0B_{X^{**}}
\]
is weak-star compact, being the image of a finite product of weak-star
compact sets under addition and scalar multiplication.
By \eqref{eq:pfaces} and the fact that $\phi_i$ are chosen before the selection of $p$, $P\subset Q$. Goldstine's theorem and
\eqref{eq:density} now give
\[
 B_{X^{**}}=\overline P^{\,w^*}\subset Q,
\]
as required.
\end{proof}

\section{Proof of the main theorem}\label{sec:conclusion}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Fix $u\in S_{X^{**}}$ and $0<\varepsilon,\tau<1$.
Choose, in this order,
\begin{align*}
 &\eta=\frac\varepsilon8,\qquad \theta=1-\frac\varepsilon4,
 \qquad f\in S_{X^*}\text{ with }u(f)>1-\eta,\\
 &k_0\in\N\text{ with }\delta_0=\frac1{k_0}<\frac\tau2,\\
 &C\ge\max\left\{1,\frac4{(1-\theta)\delta_0}\right\},
 \qquad M\in\N,\quad
 M\ge\max\left\{C,\frac{8C}{\eta\tau}\right\}.
\end{align*}
Thus $\eta,\theta$ depend only on $\varepsilon$;
$\delta_0$ depends only on $\tau$; and $C,M$ may be chosen as
functions of $(\varepsilon,\tau)$ alone.
Apply Proposition~\ref{prop:fixedfaces} to obtain $g_i$ and $\phi_i$.
Choosing $\sigma_i$ according to the sign of $u(g_i)$ gives
\begin{equation}\label{eq:badcount}
 \sum_{i=1}^M|u(g_i)|
 =u\left(\sum_{i=1}^M\sigma_i g_i\right)\le2C.
\end{equation}
Put
\[
 I=\{i:|u(g_i)|\le\eta\},\qquad b=M-|I|.
\]
Then $b\le2C/\eta$ and $b/M\le\tau/4<1$.
For every $i\in I$,
\begin{equation}\label{eq:goodwitness}
 u(\phi_i)=\theta u(f)+u(g_i)>\theta(1-\eta)-\eta.
\end{equation}

Let $v\in B_{X^{**}}$. By \eqref{eq:bidualfaces}, there are
$z_i\in\mathcal F_i$ with
\[
 \norm{v-\frac1M\sum_{i=1}^M z_i}_{X^{**}}\le\delta_0.
\]
For $i\in I$, \eqref{eq:goodwitness} yields
\[
 \norm{u+z_i}_{X^{**}}
 \ge(u+z_i)(\phi_i)
 >1+\theta(1-\eta)-\eta
 =2-\bigl((1-\theta)+(1+\theta)\eta\bigr)
 >2-\varepsilon,
\]
since $(1-\theta)+(1+\theta)\eta
=\varepsilon/2-\varepsilon^2/32<\varepsilon$.
Consequently
\[
w=\frac{1}{\lvert I\rvert}\sum_{i\in I}z_i
\in \operatorname{conv}\left\{
z\in B_{X^{**}} : \|u+z\|\ge 2-\varepsilon
\right\}.
\]
Splitting the original average into the retained and discarded
summands, we obtain
\[
\frac1M\sum_{i=1}^M z_i
=\frac1M\sum_{i\in I}z_i+\frac1M\sum_{i\notin I}z_i
=\frac{M-b}{M}\,w+\frac1M\sum_{i\notin I}z_i.
\]
Subtracting \(w\) therefore yields
\[
\frac1M\sum_{i=1}^M z_i-w
=\frac1M\sum_{i\notin I}z_i-\frac bM\,w.
\]
There are exactly \(b\) discarded indices. Hence
\[
\begin{aligned}
\left\|\frac1M\sum_{i=1}^M z_i-w\right\|_{X^{**}}
&\le \frac1M\sum_{i\notin I}\|z_i\|_{X^{**}}
     +\frac bM\|w\|_{X^{**}}\\
&\le \frac bM+\frac bM
=\frac{2b}{M}.
\end{aligned}
\]
Combining this estimate with
\[
\left\|v-\frac1M\sum_{i=1}^M z_i\right\|_{X^{**}}\le\delta_0,
\]
we conclude that
\[
\begin{aligned}
\|v-w\|_{X^{**}}
&\le
\left\|v-\frac1M\sum_{i=1}^M z_i\right\|_{X^{**}}
+
\left\|\frac1M\sum_{i=1}^M z_i-w\right\|_{X^{**}}\\
&\le \delta_0+\frac{2b}{M}\\
&\le \delta_0+\frac{4C}{\eta M}\\
&<\tau.
\end{aligned}
\]
Here the penultimate inequality uses \(b\le 2C/\eta\), and the
last inequality follows from the choice of the parameters.
Since $v$ and $\tau$ were arbitrary,
$B_{X^{**}}=\overline{\operatorname{conv}}\left\{
z\in B_{X^{**}} : \|u+z\|\ge 2-\varepsilon
\right\}$ with respect to the norm topology in $X^{**}$.
The characterization Theorem 1.1 completes the proof.
\end{proof}


\section*{Acknowledgments}
The author used OpenAI's GPT-6 Astra to assist with mathematical exploration,
the development and checking of proofs, and the presentation of arguments. The author takes full responsibility for the mathematical content and any
remaining errors.


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\bibitem{hamad}
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\bibitem{kadets}
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\end{thebibliography}
\end{document}