# Strict convexity and equivalent conditions

I have asked on Stackexchange but so far no feedback so I am reposting here hoping for some help.

Let $$f \colon \mathbb R^n \to [0,+\infty)$$ be a convex, positively 1-homogeneous function, i.e. it holds $$f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda) f(y), \qquad \forall \lambda \in [0,1], \, \forall x,y \in \mathbb R^n$$ and $$f(\lambda x) = \lambda f(x), \qquad \forall \lambda >0, \, x \in \mathbb R^n.$$

Let me denote by $$E_c := \{ f \le c\}$$ the sub-level set at height $$c$$. I have been asked to show that the sub-level sets are homothetic and indeed I have proved that $$E_c = cE_1$$ for every $$c>0$$. Now, assuming that $$f$$ is also of class $$C^2(\mathbb R^n)$$, I have to investigate the validity of the following equivalence:

(A) The set $$E_1$$ is strictly convex;

(B) There exists $$s>0$$ such that $$\nabla^2 f[x] (z,z) \ge s \vert z – (z\cdot x)x \vert^2$$ for every $$x,z \in \mathbb R^n$$.

Q. Is it true that (A) iff (B)?

I have no idea on how to face the question, and I am looking for some intuition behind condition (B). What is it actually saying? How could it be related to sub-level sets?

ADDENDUM: I think I got some intuition behind condition (B). If I am not wrong that should be a uniform bound (from below) on the Gaussian curvature of the level set $$\{f=1\}$$ (assuming that all points are regular, i.e. $$\vert \nabla f \vert \ne 0$$). This is an easy consequence of the formula contained in this page. Do you agree on this consideration? Now the question becomes: how is it possible to relate a bound on the Gaussian curvature of the level set $$\{f=1\}$$ with its convexity?