A subspace $Y$ of a Banach space is said to be Hahn-Banach smooth if every $f\in Y^*$ has unique norm preserving extension to whole $X$. This notion is related to many other geometric properties of Banach space, viz. rotundity of the dual space, intersection properties of balls in the space etc. It is well known that if $X^*$ is strictly convex then any subspace of $X$ is Hahn-Banach smooth and vice versa-A well-known result by R. R. Phelps. The list also includes the subspaces of type $\{f\in C(K):f|_D=0\}$ in $C(K)$, where $K$ is compact $T_2$ and $D\subseteq K$ is closed. These are so-called M-ideals or the two-sided ideals of a commutative $C^*$ algebra but of course much more stronger than Hahn-Banach smoothness.

My question is if $Y$ is a subspace of a Banach space $X$ which is Hahn-Banach smooth then is it necessary that $Y^{\perp\perp}$ has also similar property in $X^{**}$? One can assume that $X=C(K)$ for some compact Hausdorff $K$. More generally if $X$ is a $L_1$ predual space.


$Y^{\bot\bot}$ need not be what you call Hahn-Banach smooth [see below] in $X^{\bot\bot}$: Take $X=L_1[0,1]$ and a smooth point of the unit sphere, e.g. $x=1$. Let $Y$ be the linear span of $x$, which is reflexive; hence $Y=Y^{\bot\bot}$. Since $x$ is smooth there is unique HB-extension from $Y$ to $X$, but not to $X^{**}$. To see this note that $X$ is an $L$-summand in $X^{**}$; i.e., $X^{**}=X\oplus_1 X_s$ and consequently $X^{***}=X_s^\bot \oplus_\infty X^\bot \cong X^* \oplus_\infty X^\bot$. Hence one can add elements of $X^\bot$ of norm $1$ to the extension in $X^*$ and still have an HB-extension in $X^{***}$.

A word about notation: What you describe is known as ''$Y$ has property (U) in $X$'' and was studied thoroughly by Phelps. The special case ``$X$ has property (U) in $X^{**}$'' was termed Hahn-Banach smoothness by Sullivan. The result you attribute to Phelps is actually due to Taylor (Duke J. 1939) with the converse due to Foguel (PAMS 1958).

  • $\begingroup$ Can we say anything about the space of type $C(K)$? If we follow the above argument, if $f\in C(K)$ is a smooth point then is it necessary that $f$ remains smooth in $C(K)^{**}$? As we know bi-dual of $C(K)$ is of the form $C(\Omega)$ for some compact $T_2$ $\Omega$. $\endgroup$ – Tanmoy Paul Feb 10 '19 at 16:12
  • $\begingroup$ @Tanmoy: Let $K=\alpha(\mathbf{N})=\mathbf{N}\cup\{\infty\}$. Let $f(n)=1-\frac1n$ and $f(\infty)=1$. Then $f\in C(K)$ is a smooth point of the unit ball, normed by the limit functional. However, in the bidual ($= \ell_\infty( \mathbf{N}\cup\{\infty\} )$) this function is normed by every Banach limit, so $f$ is not smooth in the bidual. $\endgroup$ – Dirk Werner Feb 11 '19 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.