Is it true that the set $\{x\colon \partial\phi(x)\subset B_r(0)\}$ is convex when $\phi$ is convex?

Question

Let $\phi\colon \mathbb{R}^n\to \mathbb{R}$ be convex. Is it true that the sets $$ A_r = \{x\in \mathbb{R}^n\;\colon\; \partial \phi(x) \subset B_r(0)\},\quad r>0, $$ are convex?

For $n=1$ this follows from monotonicity. Indeed, for any pair of points such that $x_0<x_1$ and any intermediate point $x_t=(1-t)x_0+tx_1$, $t\in [0,1]$, it follows that $$ p_{x_0}\leq p_{x_t}\leq p_{x_1}\quad \forall p_{x_0}\in \partial\phi(x_0),\quad \forall p_{x_1}\in \partial\phi(x_1),\quad \forall p_{x_t}\in \partial\phi(x_t). $$ Hence, $$ -r \le \min\{\partial\phi(x_0)\}\le p_{x_t} \le \max\{\partial\phi(x_1)\}\le r. $$ But what about $n\ge 2$?

Answer 1

The answer is no. Indeed, consider the case when $n=2$ and $$f(x,y)=\max(| x|,2 | y|)$$ for all $(x,y)\in\mathbb R^2$.

Then for $(x,y)$ near $(\pm1,0)$ we have $f(x,y)=|x|$ and hence $\partial f(\pm1,0)\subseteq B_1(0)$. On the other hand, $f(0,y)=2|y|$, so that $\partial f(0,0)\not\subseteq B_1(0)$.