It is known the $n$-sphere, the level set of $\|\cdot\|_2$, i.e.$\{x \in \mathbb{R}^{n+1}\, | \| x\|_2 =c \, \}$, has a non negative sectional curvature.
I wonder if this level set of any smooth norm $\|\cdot\|$ on $\mathbb{R}^{n+1}$ is also non negatively curved?.
Thanks in advance.
Assuming your definition of norm includes subadditivity, any norm on $\mathbb R^{n+1}$ is a convex function $f : \mathbb R^{n+1} \to [0,\infty),$ and thus the sub-level set $\Omega_c = \{ x \mid f(x) < c \}$ is a convex set.
Since norms are homogeneous and positive away from the origin, if we assume that $f$ is smooth then its derivative is non-zero in the radial direction. Thus any $c>0$ is a regular value of $f$, so the level set $\Gamma_c = \{x \mid f(x) = c \}$ is a smooth hypersurface forming the boundary of $\Omega_c.$ Convexity of the bounded region $\Omega_c$ then implies that $\Gamma_c$ has non-negative definite second fundamental form $A$.
The Gauss equation $$R(u,v,w,x) = A(u,w)A(v,x) - A(u,x)A(v,w)$$ tells us that $A \ge 0$ implies non-negative sectional curvature: to see this, take $u,v$ to be an orthonormal basis for an arbitrary plane $\Pi$, and note that we have $$\operatorname{sec}(\Pi)=R(u,v,u,v) = A(u,u)A(v,v)-A(u,v)^2 = \det(A|_\Pi) \ge 0.$$
Strict positivity is not necessarily true - for example, you could take the $\ell^1$ norm on $\mathbb R^3$ and smooth out the corners while leaving some portions of the original faces flat, where the curvature would be zero.
Likewise, if you reduce the regularity of the norm too much then you need to be careful that your question makes any sense at all - if $f$ is not at least $C^2$ then the classical sectional curvature of its level sets is ill-defined. Thus you'd need to move into the world of metric geometry, which is quickly leaving my domain of knowledge. From what I can find with a quick web search, it seems things do work out: apparently it is true that the boundary of a convex region is an Alexandrov space of non-negative curvature.
Just to illustrate the non-subadditive $p$-"norms" in $\mathbb R^2$ which have negative curvature. The numbers are $\frac{1}{p}$. Notice how aggressively non-linear they become close to 0 ( large $\frac{1}{p}$ ).
