Inequality $||DG(t)||\leq c||G(t)||$ implies $G\equiv 0$? Continuous and differentiable function $G:[0,\infty)\to \mathbb{R}^n$ satisfies $G(0) = 0\in\mathbb{R}^n$ and there exists $c>0$ such that for all $t>0$ we have
$$||DG(t)||\leq c||G(t)||$$
Does it mean that $G\equiv 0$?
Remark:
$G(t) = (g_1(t),g_2(t),...,g_n(t))$, $DG(t)$ is the derivative of $G$ and $||\circ||$ is euclidean norm.
If $n=1$ then this statement is true, but every proof I've seen relied on Mean Value Theorem. As of yet I was unable to generalize it to higher dimensions or provide counter examples.
 A: Without loss of generality, $c=1.$ First suppose that $0\le x\le 1$ and let $M=\sup \{G(t):t\in [0,1]\}.$ Now then
$G(x)=\int_0^xG'(t)dt\Rightarrow |G(x)|\le \int_0^x|G'(t)|dt \le \int_0^x|G(t)|dt.$
By the mean-value inequality$^1$, there is an $0<\alpha<t$ such that
$|G(t)|\le t|G'(\alpha)|$ so $|G(x)|\le \frac{x^2}{2}|G'(\alpha)|\le \frac{x^2}{2}|G(\alpha)|\le M\frac{x^2}{2}.$
But this means that $|G(x)|\le M\frac{x^2}{2}$ and so inductively, that $|G(x)|\le M\frac{x^n}{n!}$ for each integer $n.$
We conclude that $G=0$ in $[0,1].$ Now define $G_1(t)=G(t+1).$ Then, $G_1$ satisfies the same hypothesis as $G$ and so $G_1=0$ on $[0,1]$ or what is the same thing, $G=0$ on $[1,2].$ Now, another induction proves the claim.
$^1$ This is the mean-value inequality for vector-valued functions. Rudin proves it this way: define $z:=G(b)-G(a)$ and $\varphi(t)=\langle z,G(t)\rangle$ and apply the single-variable mean-value theorem to find a $\alpha\in (a,b)$ such that $\varphi(b)-\varphi(a)=\varphi'(\alpha)=\langle z,G'(\alpha)\rangle.$ But we also have $\varphi(b)-\varphi(a)=|z|^2$ so the result follows by the Schwarz inequality.
A: Let $M= \max_{x\in [0,1]} \|G(x\|$. Then  $\|G'(x)\| \le c M$ for all $x\in [0,1]$. Now, for all $x\in [0,1]$ we have
$$G(x) = G(x) - G(0) =\int_{0}^x G'(t) dt$$
and so
$$\|G(x) \|\le c M x$$
for all $x\in [0,1]$.  Let's show that in general
$$\|G(x) \le \frac{c^n M x^n}{n!}$$
for all $x\in [0,1]$. By induction on $n$. For $n=1$, checked. Assume true for $n$. Then we have $\|G'(x)\|\le c\cdot \frac{c^n M x^n}{n!}$, and so
$$\|G(x) \| \le \int_{0}^x \|G'(t)\| dt \le c^{n+1} M \int_{0}^x \frac{t^n}{n!} = \frac{c^{n+1} M x^{n+1}}{(n+1)!}$$
Now, observe that $ \frac{c^n M x^n}{n!}\to 0$ as $n\to \infty$.
Note: We assumed that we can use the Newton-Leibniz formula.
