# Can $X^*$ characterize a smooth norm on $X$?

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)