Frechet differentiable implies reflexive?

Question

Note: The question has been cross-posted (and answered) on MathOverflow here.

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Let $X$ be a Banach space.

Is it true that if dual norm of $X^*$ is Frechet differentiable then $X$ is reflexive?

Answer 1

The answer is yes, as Norbert pointed out. Detailed treatment can be found in Chapter 5 of An introduction to Banach space theory by Megginson. Rough idea: if $X$ is not reflexive, then by James' theorem, there exists a unit functional $f\in X^*$ which does not attain $1$ on the unit ball $B_X$. Let $x_n$ be a sequence in $B_X$ such that $f(x_n)\to 1$. Since $x_n$ do not converge, they jump around, making the unit sphere of $X$ look pretty flat in that area. (They do converge in the weak* topology to some $z\in X^{**}$, meanwhile.) This gets in the way of the Fréchet smoothness at $f$. The actual computations are tedious: see the aforementioned sources.

Answer 2

The answer of Yes is also provided in the MO cross-post here.