If $|\nabla F| > 1$ and $|F| \le 1$, is there a zero nearby? I saw this claim, stated without much explanation, in an article I'm reading:

Let $F:\mathbb{R}^n\to\mathbb{R}$ be a $C^1$ function which satisfies $|\nabla F|>1$ everywhere. 
We know that $|F(0)| \le 1$.
Then there is a point $x$ which is at most distance $1$ away from $0$, such that $F(x) = 0$.

I'm trying to justify this claim for my own understanding. Here is how I proved it:

According to (???), we can find a (unique?) "flow curve" $\gamma:\mathbb{R}\to\mathbb{R}^n$ satisfying:
  $$\dot \gamma(t) = \nabla F(\gamma(t))$$
  $$\gamma(0) = 0$$
  Look at $f:\mathbb{R}\to\mathbb{R}:t \mapsto F(\gamma(t))$. It satisfies $\dot f(t) = \nabla F(\gamma(t)) \cdot \dot \gamma(t) = ||\dot \gamma(t)||^2 > 1$, and $|f(0)| \le 1$. 
By a simple 1-dimensional calculus argument, there is a $t_0 \in \left[-1, 1\right]$ such that $f(t_0) = 0$.
The length of the arc $\gamma$ between the origin and our zero is $$\left|\int_0^{t_0} ||\dot \gamma(t)||\,dt\right| \stackrel{\text{Cauchy-Schwarz}}{\le}\sqrt{|t_0| \cdot \left|\int_0^{t_0} ||\dot \gamma(t)||^2\,dt\right|} = \sqrt{|t_0| \cdot \left|\int_0^{t_0} \dot f(t)\,dt\right|} \le \sqrt{|t_0 f(0)|} \le 1$$
  so the distance between $x = \gamma(t_0)$ and $0$ is at most $1$.

My questions:


*

*Is the proof correct?

*What is the justification for the existence of $\gamma$?

*Is there a simpler proof, particularly one that doesn't use the theory of ODEs?
 A: I tend to think you need $\nabla F$ to be locally Lipschitz, which would be implied by assuming $F$ is $C^{2}$. An existence proof requires finding a fixed point of the map
$$
       M(\vec{\gamma})=\vec{v}_{0}+\int_{0}^{t}\nabla F(\vec{\gamma}(t))\,dt,
$$
where $\vec{v}_{0}=0$ in this case. The simplest contractive mapping arguments will require $\vec{G}=\nabla F$ to satisfy some local Lipschitz condition $|\vec{G}(\vec{x})-\vec{G}(\vec{y})|\le K|x-y|$ for all $\vec{x}$, $\vec{y}$ near $\vec{\gamma}_{0}$. A more sophisticated fixed point method might allow you to relax the standard Lipschitz condition, but I cannot see how to show that $M$ maps some set of curves into itself from continuity of $\nabla F$ alone, and such a mapping requirement for $M$ would be a starting point for any fixed point method.
The argument given doesn't refer to the value of $F(0)$, and it must. If you choose $\gamma$ so that $F(\gamma(t))$ is increasing (as it has been chosen), and $F(0) = 1$, then that's wrong. So, you need to at least say "without loss of generality, assume $-1 \le F(0) < 0$".
The idea is a good one: at each point of the curve $\gamma$, you are headed in the direction of steepest ascent, i.e., along the gradient. How else would you choose a shortest curve (or something close) to reach a zero of $F$? I think that's okay, and I would doubt that there's a good substitute when you think about how that curve was chosen. It does make more sense, though, that $F$ should be $C^{2}$ or something similar; otherwise, getting a curve guaranteed to lead to a zero of $F$ doesn't seem intuitive.
