Dtyjlui

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
$$
nabla^2 f[x] (z,z) ge s vert z - (zcdot x)x vert^2
$$
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?

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...

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: $lambdanabla f(lambda x) = lambda nabla f(x)$. 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.

I think you need to assume $fin C^{2}(mathbb{R}^{n}setminus{0})$ and not
$fin C^{2}(mathbb{R}^{n})$. Otherwise, if $t>0$,
$$
frac{f(te_{i})-f(0)}{t}=frac{tfleft( e_{i}right) }{t}=f(e_{i}),
$$
while if $t<0$,
$$
frac{f(te_{i})-f(0)}{t}=-frac{tfleft( -e_{i}right) }{t}=-f(-e_{i}),
$$
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
$fgeq0$. But then by convexity, writing
$$
x=frac{1}{n}sum_{i=1}^{n}nx_{i}e_{i}%
$$
you get
$$
0leq f(x)leqfrac{1}{n}sum_{i=1}^{n}f(nx_{i}e_{i})=frac{1}{n}sum
_{i=1}^{n}n|x_{i}|f((operatorname*{sgn}x_{i})e_{i})=0,
$$
which implies that $f=0$.

So LinAlg's conjecture is true.

Note also that if a function $fin C^{1}(mathbb{R}^{n}setminus{0})$ 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 $t>0$.