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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?

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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.

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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.

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

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