I am wondering the following question: Let $X$ be a separable Banach space which is linearly isomorphic to a dual Banach space $Y^*$. Is there a Banach space $Z$ such that $X$ is lineraly isometric to the dual of $Z$: $X=Z^*$.

I think that the answer is no, but I do not have a counterexample. Since $L_1$ is not isometric to any dual Banach space, maybe one can find a dual Banach space which is isomorphic to $L_1$...

To finish, do that change anythink if I suppose $X$ to be almost linearly isometric to $Y^*$ ? By almost linearly isometric I mean that for every $\varepsilon >0$ there exist $T$: $X \to Y$ a linear isomorphism satisfying $\|T\| \|T^{-1}\| \leq 1+\varepsilon$.


For every non-reflexive Banach space $X$, the dual space $X^*$ can be renormed in such a way that the renorming is not isometric to any dual space. See Corollary 2.8 here (beware the typo in the statement).

This norm can be introduced quite explicitly. Take a norm-one functional $g\in X^{*}$ such that for some norm-one $x^{**}\in X^{**}\setminus X$ one has $\langle x^{**}, g\rangle = 1$. Then the norm $$\|f\|^2 = |\langle x^{**}, f\rangle|^2 + \|f - \langle x^{**}, f\rangle g\|^2 $$ is an equivalent norm on $X^*$ that does not arise from any duality.

More generally, every non-reflexive Banach space can be renormed so as not to be isometrically a dual space.

W. J. Davis, W. B. Johnson, A renorming of nonreflexive Banach spaces. Proc. Amer. Math. Soc. 37 (1973), 486–488.

There are even more far reaching extensions of this result based on the observation that dual spaces are 1-complemented their second duals. (See p. 4 here for explanation and references.)

For more concrete, non-separable examples let us take $X=\ell_\infty$ and let us regard it as $C(\beta \mathbb{N})$. Pick $p\in \beta \mathbb{N} \setminus \mathbb{N}$. Then $\delta_p\in C(\beta\mathbb{N})^*$ given by $$\langle \delta_p, f\rangle = f(p)\quad (f\in C(\beta \mathbb{N}))$$ is a norm-one functional hence its kernel is a hyperplane so it is isomorphic to $X$. On the other hand, there are no extreme points in the unit ball of $\ker \delta_p$ so it cannot be isometically a dual space.

  • $\begingroup$ Thank you very much for your answer, I was not aware of that renorming result. You have perfectly answered my question, and I appreciate your concrete example. $\endgroup$ – C. Petitjean Nov 23 '16 at 11:52

I am not sure, whether the following is true, but it sounds promising.

Let $T : X \to Y^*$ be your isomorphism. We define a new norm on $Y^*$ by $\|y^*\|'_{Y^*} := \|T^{-1}y^*\|_X$ and on $Y$ via $\|y\|'_Y := \sup_{\|y^*\|'_{Y^*}\le1} y^*(y)$.

Then, $\|\cdot\|'_{Y^*}$ should be the dual norm to $\|\cdot\|'_Y$ and $T$ becomes an isometry.

  • $\begingroup$ What do you mean with "[you're] not sure, whether the following is true"? About which part(s) have you doubts? $\endgroup$ – Daniel Fischer Nov 23 '16 at 10:19
  • $\begingroup$ I am not sure whether the norm on $Y^*$ is really the dual norm. But this should be easy to check. $\endgroup$ – gerw Nov 23 '16 at 10:30
  • $\begingroup$ Yes, it is, by the bipolar theorem. $\endgroup$ – Daniel Fischer Nov 23 '16 at 10:36
  • $\begingroup$ Unfortunately this will not work, please see my answer. $\endgroup$ – Tomek Kania Nov 23 '16 at 11:08
  • 2
    $\begingroup$ Ah, screw it, the bipolar theorem only gives us that it's the dual norm if $Y$ is reflexive. If $Y$ is non-reflexive, the weak$^\ast$-closure of the (norm-)closed $\lVert\cdot\rVert_{Y^\ast}'$-unit ball $B$ can be larger than $B$. $\endgroup$ – Daniel Fischer Nov 23 '16 at 11:23

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