Frechet differentiable implies reflexive?

Question

Let $X$ be a Banach space. Is it true that if dual norm of $X^*$ is Frechet differentiable then $X$ is reflexive? Can any one help me? thanks

I think that finite dimensional $l^p$ spaces are already a counter-example. The norm is not differentiable at zero. Or am I missing something? — Igor Khavkine, Commented Jun 12, 2014 at 21:50
@ Dear Benjamin Dickman , thanks, your answer is very perfect — user62498, Commented Aug 29, 2014 at 7:41

Martin Sleziak · Accepted Answer · 2014-06-14 08:47:31Z

Yes, this is true. See Theorem 8.6 in:

Fabian, M. J. (Ed.). (2001). Functional analysis and infinite-dimensional geometry (Vol. 8). Springer.

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(Lemma 8.4 not visible in Google Books.) The citations for the above are:

[Jam2] James, R. C. (1964). Weak compactness and reflexivity. Israel Journal of Mathematics, 2(2), 101-119.

and

[Dis2] Diestel, J. (1984). Sequences and series in Banach spaces (Vol. 13). Berlin Heidelberg New York: Springer.

Alternatively, see Corollary 2.3 in:

Haghshenas, H. (2009). A Compilation of some Well-Known Results in Renorming Theory. arXiv preprint arXiv:0901.3029.

(Here the author proves the stated fact as a consequence of Smulian’s theorem, which is mentioned in the comments above.)

Other classical results can also be found in:

Diestel, J. (1975). Geometry of Banach spaces: selected topics (Vol. 485). Berlin: Springer-Verlag.

(I think this particular fact is again proved on p. 33.)

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