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!



Can $X^*$ characterize a smooth norm on $X$?
This is a known result even without any reflexivity assumption, see for instance [Smoothness and renormings in Banach spaces](https://books.google.de/books/about/Smoothness_and_renormings_in_Banach_spac.html?id=ZR_vAAAAMAAJ) 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).


