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I conjecture following statement in the Banach space $X$ with norm $\|.\|$.

Let $B \subseteq X$ be the unit closed ball in $X$ then,

$\|.\|$ is smooth, in sense $\|.\|$ is differentiable at any nonzero point in $X$, if and only if for any point, say $ x_0 \in \text{bd}(B)$ there is a unique $f \in X^*$ such that $\|f\|=f(x_0).$

My thought: I think proving Left $\rightarrow$ Right is easy It is just a separation, and using smoothness of norm we can show that $f\in X^*$ is unique, but I am interested see your rigorous proof. I don't have any idea about the other way! Even I am not sure it is true! However I am quite positive it is true when $X$ is finite dimension, still any proof in this particular case would be appreciated . (I first assumed $X$ is a reflexive,(I don't know why I did that! maybe because my unexplainable intuition) anyway you may assume reflexivity)

Thanks for nice comments and answers inadvance!

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This is a known result even without any reflexivity assumption, see for instance Smoothness and renormings in Banach spaces by Deville, Godefroy and Zizler, Chapter I, Corollary 1.5. The proof there is based on a characterization of differentiability for the norm function on a Banach space by Smulyan (Theorem 1.4 in the same book).

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  • $\begingroup$ Thank you, I saw that, since my definition of smoothness was based on Ferchet differentiability, Right to Left is not true in any Banach space.... $\endgroup$
    – Red shoes
    Jun 7 '17 at 9:05

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