Reason for the term "smooth"

Let us consider the space $X=\mathbb R^2$ with the $\ell^p$-norm.

Then for $p=1$ or $p=\infty$ we can see that the unit ball has kinks, and does not look smooth. It can be shown, that there are points $x\in X$ such that there is more than one functional $f$ with $\|f\|=1$ and $f(x)=\|x\|=1$. (For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)

For $p$ with $1<p<\infty$ the unit ball looks smooth (its boundary is differentiable). And it is possible to show that for each $x\in X$ there is exactly one functional $f$ with $\|f\|=f(x)=\|x\|=1$.

I hope this is sufficient motivation for the term "smooth" for a normed space.

This condition implies that the function $F \colon x \mapsto \|x\|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$. Indeed, the set $$ \{ f\in X^* \mid \|f\| \le 1 \text{ and } f(x) = \|x\|\}$$ coincides with the convex subdifferential of $F$ at $x$. If this subdifferential is a singleton, then $F$ is differentiable.