Strict convexity and equivalent conditions

Question

Let $f \colon \mathbb R^n \to [0,+\infty)$ be a convex, positively 1-homogeneous function, i.e. it holds $$\tag{1} f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda) f(y), \qquad \forall \lambda \in [0,1], \, \forall x,y \in \mathbb R^n $$ and $$\tag{2} f(\lambda x) = \lambda f(x), \qquad \forall \lambda >0, \, x \in \mathbb R^n. $$

Let me denote by $E_c := \{ f \le c\}$ the sub-level set at height $c$. I have been asked to show that Assumption (2) implies that the sub-level sets are homothetic. 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 $x,z \in \mathbb R^n$, being $\nabla^2 f[x](z,z) = \langle (\nabla^2 f)(x) \cdot z, z \rangle$ the quadratic form associated to the Hessian matrix of $f$ (evaluated at $x$).

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?

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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. $\vert \nabla f(x) \vert \ne 0$ for every $x \in \{f=1\}$: is this assumption necessary?). 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 (uniform) bound on the Gaussian curvature of the level set $\{f=1\}$ with the (strict) convexity of the set $\{f \le 1\}$?

To me this makes sense, because the strict convexity of $\{f \le 1\}$ roughly means that its boundary has no flat parts, and its boundary should be related to the level set - which has curvature bounded from below, i.e. it is well-round...

Answer 1

The only function I can think of that satisfies all your conditions (convex, range $[0,\infty)$, positively 1-homogeneous, $C^2$) is the function that always takes the value $0$.

You cannot have a positively 1-homogeneous function for which $E_1$ is strictly convex: the function is linear on the line segment $\{ \lambda x : f(\lambda x) \leq 1 \} \subseteq E_1$ (the line segment cannot be a single point or an empty set).

Now let's look at condition B. Since (2) holds for all $x$, the derivatives have to be equal as well: $\lambda\nabla f(\lambda x) = \lambda \nabla f(x)$. It follows that $\nabla f(\lambda x)$ is the same for all $\lambda$, so the second derivative at $0$ in the direction $z$ is $0$: $\nabla^2f[0](z,z)=0$.

As neither condition (A) nor condition (B) can ever hold, they are equivalent.

Answer 2

I think you need to assume $f\in C^{2}(\mathbb{R}^{n}\setminus\{0\})$ and not $f\in C^{2}(\mathbb{R}^{n})$. Otherwise, if $t>0$, $$ \frac{f(te_{i})-f(0)}{t}=\frac{tf\left( e_{i}\right) }{t}=f(e_{i}), $$ while if $t<0$, $$ \frac{f(te_{i})-f(0)}{t}=-\frac{tf\left( -e_{i}\right) }{t}=-f(-e_{i}), $$ and so if $\frac{\partial f}{\partial x_{i}}(0)$ exists, then necessarily, $f(e_{i})=-f(-e_{i})$, which is only possible if $f(e_{i})=f(-e_{i})=0$, since $f\geq0$. But then by convexity, writing $$ x=\frac{1}{n}\sum_{i=1}^{n}nx_{i}e_{i}% $$ you get $$ 0\leq f(x)\leq\frac{1}{n}\sum_{i=1}^{n}f(nx_{i}e_{i})=\frac{1}{n}\sum _{i=1}^{n}n|x_{i}|f((\operatorname*{sgn}x_{i})e_{i})=0, $$ which implies that $f=0$.

So LinAlg's conjecture is true.

Note also that if a function $f\in C^{1}(\mathbb{R}^{n}\setminus\{0\})$ is positively homogeneous of degree one, then its partial derivatives $\frac{\partial f}{\partial x_{i}}$ are positively homogeneous of degree zero, so $\frac{\partial f}{\partial x_{i}}(tx)=\frac{\partial f}{\partial x_{i}% }(x)$ for all $t>0$.