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?