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Let $f: mathbb R^n to [0 , +infty] $ be a lower semicontinuous, convex, and positively homogenous degree-$2$ function. Prove that for all $x in mboxdom f$, we have $$ partial f(x) neq emptyset $$

Note, that the latter is equivalent to saying that there exists a $kappa > 0$ and a neighborhood $U$ of $x$ such that, for all $y in U$ we have $$- kappa | y - x| leq f(y) - f(x)$$

I was able to prove that $partial f(0) neq emptyset$, but I couldn't prove that for nonzero vectors. Actually It may not be true for those.

P.S : Positively homogenous of degree 2 means for all $t geq 0 $ we have $f(tx)= t^2 f(x).$

Ok, let's construct a counterexample, since I can't express my thoughts via comments.

Start by considering the semicircle function over $BbbR$, defined by
$$x mapsto begincases1-sqrt1 - x^2 & textif |x| le 1 \ infty & textif |x| > 1 endcases.$$
This is an example of an lsc which fails to admit subgradients at each point in the domain, as there is no subgradient at $x = pm 1$. We can see this visually by observing that the points $(pm 1, 1)$ are supported only by vertical lines.

Now, we embed this function in $BbbR^2$, then extend it to a cone. Let $C$ be the cone generated by $[-1, 1] times 1$, which is defined by the inequality $y ge |x|$. We form a function $g$ over $C$ by defining its graph to be the (non-convex) cone generated by the graph of the semicircle function over the domain $[-1, 1] times 1$. More explicitly,
$$g(x, y) = begincasesy -sqrty^2 - x^2 & textif (x, y) in C \ infty & textotherwise endcases.$$
This function is scalar positive-homogeneous (only degree $1$; we shall square it soon enough!).

I also claim that $g$ is lsc and convex. I think the most intuitive way to see this is to express $operatornameepi g$ as the closure of the cone generated by our semicircle defined over $[-1, 1]$. In particular,
$$operatornameepi g = overlineoperatornamecone S$$
where
$$S = leftx.$$
So, suppose $(a, b, c) in operatornamecone S$. We may therefore find $x, z, lambda$, the latter non-negative, such that $(a, b, c) = lambda(x, 1, z)$, $|x| le 1$, and $z ge 1 - sqrt1 - x^2$. Clearly, $lambda = b$. If $b = 0$, then $(a, b, c) = (0, 0, 0) in operatornameepi g$, so we may assume $b > 0$.

We also have $left|fracabright| = |x| le 1 implies |a| le |b| = b$, so $(a, b, c) in C$. Given the condition that $z ge 1 - sqrt1 - x^2$, we get
$$fraccb ge 1 - sqrt1 - left(fracabright)^2 implies c ge b - sqrtb^2 - a^2 = g(a, b) implies (a, b, c) in operatornameepi g.$$

On the other hand, if $(a, b, c) in operatornameepi g$, then $(a, b) in C$, hence $b ge |a|$. Suppose for the moment that $b > 0$, and consider the point
$$(x, 1, z) = left(fracab, 1, fraccbright).$$
We see that $b(x, 1, z) = (a, b, c)$, and by reverse engineering the above calculation, we get that $(x, 1, z) in S$. Hence $(a, b, c) in operatornamecone S$.

On the other hand, suppose $b = 0$. Note that this implies $a = 0$ too, by definition of $C$. We know that $c ge 0$.

Fix $varepsilon > 0$. Consider the point $(0, varepsilon, c)$. We have
$$c ge 0 = g(0, varepsilon),$$
hence this point lies in $operatornameepi g$, and by the above argument, $operatornamecone S$. Since this point is $varepsilon$-close to $(a, b, c)$, it follows $(a, b, c) in overlineoperatornamecone S$.

To finally complete the proof that $g$ is lsc and convex, note that $g$ is continuous when restricted to $C$, hence its epigraph is closed, and thus is equal to the closed convex set $overlineoperatornamecone S$.

Now, we may define $f : BbbR^2 to [0, infty]$ by $f(x) = g(x)^2$, where $infty^2 = infty$. As we are applying a monotone increasing real convex function over the positive reals to $g$, the resulting function is convex. The nice continuity properties of the squaring map also ensures that $f$ is lsc. We finally have
$$f(lambda (x, y)) = g(lambda (x, y))^2 = lambda^2 g(x, y)^2 = lambda^2 f(x, y),$$
as required. We have now completely constructed our counterexample!

The only thing left to do is prove it is a counterexample. None of the points of the form $(pm x, x)$ will have subgradients, but let's just pick on $(1, 1)$.

Suppose $(a, b) in partial f(1, 1)$. Consider points along the ray $(1, 1) + t(-1, 1)$ for $t ge 0$. Note that this ray is contained in $C$, and we have, by definition of the subgradient,
$$(a, b) cdot (-1, 1) le fracf(1 - t, 1 + t) - 1t.$$
Expanding,
beginalign*
f(1 - t, 1 + t) &= left(1 + t - sqrt(1 + t)^2 - (1 - t)^2right)^2 \
&= left(1 + t - 2sqrttright)^2 = (1 - sqrtt)^4.
endalign*
However,
$$lim_t to 0^+ frac(1 - sqrtt)^4 - 1t = -infty,$$
so for sufficiently small $t$, we have
$$fracf(1 - t, 1 + t) - 1t < (a, b) cdot (-1, 1)$$
against assumption. That is, $partial f(1, 1) = emptyset$ (finally).