Argthtjtr

A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?

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 $xin X$ 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 $xin X$ 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
$$
{ fin X^* mid |f| le 1 text{ and } f(x) = |x|}$$
coincides with the convex subdifferential of $F$ at $x$.
If this subdifferential is a singleton, then $F$ is differentiable.