Let us call a norm on $\mathbb{R}^n$ smooth if its restriction $\| \cdot \|:\mathbb{R}^n\setminus \{ 0 \} \to \mathbb{R}$ is a smooth map.

Suppose the unit sphere of a norm $\| \cdot \|$ is an embedded submanifold of $\mathbb{R}^n$. Is $\| \cdot \|$ necessarily smooth?


The converse statement is easy:

Assume $\| \cdot \|$ is smooth. Then it's a submersion (considered as a map $\mathbb{R}^n\setminus \{ 0 \} \to \mathbb{R}$). To see it's differential is nonzero just note that $df_x(x)=\|x\|$. So, the unit sphere is an inverse image of a regular value.

  • 1
    $\begingroup$ See answer to 1) here: mathoverflow.net/questions/223933/… $\endgroup$ – Deane Yang Dec 13 '15 at 23:34
  • 1
    $\begingroup$ Thanks. If I understood correctly your answer, it also implies that if the unit sphere is a $C^k$ submanifold (hypersurface) , the norm is also $C^k$. (and similarly for $C^{\infty}$). $\endgroup$ – Asaf Shachar Dec 14 '15 at 11:01
  • $\begingroup$ Yes, that's right. $\endgroup$ – Deane Yang Dec 14 '15 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.