Does the image of the gradient of a convex function contain a ball? Let $\Phi:\mathbb{R}^n\to\mathbb{R}$ be a smooth convex function.  Suppose that $\Phi$ also satisfies the following additional properties:

*

*$\nabla \Phi(0) = 0$

*$\| \nabla \Phi(x) \| \geq 1$ for every unit vector $\|x\|=1$, where I am taking the Euclidean norm.

Is this sufficient information to conclude that the image of the unit ball $B_1$
under the map $\nabla \Phi$ also contains the unit ball (or perhaps a ball of a smaller radius?)
More precisely, letting $B_1 = \{x\in\mathbb{R}^n \mid \|x\|\leq 1\}$, I wonder if it follows that
$$
B_1 \subset \nabla \Phi(B_1) = \{\nabla \Phi(x) \mid \|x\|\leq 1\}.
$$
or if at least $B_{1/2} \subset \nabla \Phi(B_1)$ (say).
I know that this is true when $n=1$, since then the gradient is just a monotone function, and all that is needed is the intermediate value theorem.  What I am curious about is whether this holds in higher dimensions.
 A: First observation: fix any vector $\vec{x} \in \mathbb{R}^n$. Then $f: \mathbb{R} \to \mathbb{R}$ given by $f(t) = \Phi(\vec{x} t)$ must be convex; that is, $f'(t) = \vec{d} \cdot \nabla \Phi(\vec{x} t)$ is increasing. As $f'(0) = 0$, so $f'(1) = \vec{x} \cdot \nabla \Phi(\vec{x}) \geq 0$ for all $\vec{x} \in \mathbb{R}^n$.
First, we'll prove $B_1 \subseteq \nabla \Phi(B_1)$ for the case $n = 2$ (the example $\Phi(x, y) = \frac{x^2 + y^2}{2}$ shows the inclusion cannot be made strict); this should make the general case clearer. Write $S^1(r) = \{(x, y) : x^2 + y^2 = r^2\}$ (and use the shorthand $S^1$ for the unit circle $S^1(1)$), and let $\pi: \mathbb{R}^2 \setminus \{(0, 0)\} \to S^1$ be the radial projection map $\pi(\vec{x}) = \vec{x}/||\vec{x}||$. Define $a: S^1 \to S^1$ as $a(\vec{x}) = \pi(\nabla \Phi(\vec{x}))$; i.e. $a(\vec{x})$ is the angle of $\nabla \Phi(x)$. Now, $a(\vec{x})$ and $\vec{x}$ never differ by more than 90 degrees, so as $\vec{x}$ goes once around $S^1$, so does $a(\vec{x})$ (any more or fewer loops would imply that at some point, $\vec{x}$ and $a(\vec{x})$ are 180 degrees apart). So $\Phi \nabla(S_1)$ is some path that loops once around the origin and stays outside (or at most touches) the unit circle.
As $r$ shrinks from 1 to 0, $\nabla \Phi(S^1(r))$ contracts from the loop $\nabla \Phi(S^1)$ to a single point at the origin. In the process, the loop has to pass through every point in the unit disc $B_1$: the winding number of $\nabla \Phi(S^1)$ about any point in $B_1$ (besides any point on $\nabla \Phi(S^1)$ itself) is 1 and the winding number of $\nabla \Phi(S^1(0))$ about any point besides the origin is 0. Winding number is a homotopy invariant, so continuous deformations of a loop can't change the winding number of any point that the loop doesn't pass through.

Now the argument for general $n$. Again write $S^{n-1}(r)$ for the sphere of radius $r$ about the origin in $\mathbb{R}^n$, write $S^{n-1}$ as shorthand for $S^{n-1}(1)$, let $\pi: \mathbb{R}^n \setminus \{\vec{0}\} \to S^{n-1}$ be the radial projection map, and let $a = \pi \circ \nabla \Phi|_{S^{n-1}}$. As before, $a(\vec{x})$ must be in the hemisphere of $S^{n-1}$ centered on $\vec{x}$. A fortiori, $a$ does not map any point to its antipode: $(\forall \vec{x} \in S^{n-1}) a(\vec{x}) \neq -\vec{x}$.
This criterion is enough for an explicit homotopy $H: S^{n-1} \times [0, 1] \to S^{n-1}$ from $a$ to the identity map $\iota$ (cribbing from this answer here): define $H(\vec{x}, t) = \pi((1-t) \vec{x} + t a(\vec{x}))$. Then $H(\cdot, 0) = \iota$ and $H( \cdot, 1) =a$. Since $\vec{x}$ and $a(\vec{x})$ have equal magnitudes, $(1-t) \vec{x} + t a(\vec{x}) = \vec{0}$ if and only if $t = 1/2$ and $f(\vec{x}) = -\vec{x}$, which we've excluded by hypothesis, so $H$ is well-defined.
So $\nabla \Phi(S^{n-1}(1))$ is homotopic to a sphere (possibly with self-intersections) that includes no point in the interior of $B_1$; as $r$ shrinks from $1$ to $0$, $\nabla \Phi(S^{n-1}(r))$ collapses to a point at the origin, passing through every point in $B_1$ along the way. (You could formalize this into a statement about the homology group $H_{n-1}$ of $\mathbb{R}^n$ minus a point if you wanted.)
