In Pugh’s Real Mathematical Analysis, in order to bring an example that norms do not necessarily come from inner products it is stated that the unit sphere for every norm induced by an inner product is smooth but for the maximum norm $||.||_{max}$ the unit sphere is not smooth.
I know that intuitively the author by smooth means having no corners, but what is the mathematical definition of smooth in this context?
Best Answer
Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($\star$) There is a smooth function $F:\mathbb{R}^n \rightarrow \mathbb{R}$ with $S=\{x\vert F(x) = 1\}$ and $\nabla F(x)\neq 0$ for all $x\in S$.
Now if $S$ is the unit sphere with respect to an inner product $\langle \cdot , \cdot \rangle$, then you can take $F(x) := \langle x , x \rangle$. It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $\nabla F(x) = 2x$, hence ($\star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($\star$) is not satisfied. First note that $F(x) = \Vert x \Vert_{\max}^2$ is not a smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_n\in S$ which approach $x_0$ from different faces. Since $\nabla F(x) \perp S$ (with respect to the usual dot-product) for all $x\in S$, we get $$\nabla F(x_0)= \lim \nabla F(a_n) \perp \lim \nabla F(b_n) = \nabla F(x_0)$$ and hence $\nabla F(x_0)= 0$, a contradiction.