# Reason for the term “smooth”

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 $$x\in X$$ 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 $$1 the unit ball looks smooth (its boundary is differentiable). And it is possible to show that for each $$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.

• $F'(x)=f$ is a typo? – David C. Ullrich Nov 28 '18 at 15:41
• No, the derivative of $F$ at $x$ is the linear function $f$. – gerw Nov 28 '18 at 18:01
• The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...) – David C. Ullrich Nov 28 '18 at 18:31