Can $\inf \Vert Dv(x)\Vert = 0$? I am trying to prove or disprove the following: 
Assume $v\in C^1(\mathbb{R}^n)$ is real-valued, $|Dv(x)| \leq C$ for all $x\in \mathbb{R}$, $v$ is bounded below by 0 and $v$ doesn't have any local min. Prove that
$$ \inf_{x\in \mathbb{R}^n} |Dv(x)| = 0.$$
In one dimensional case it is true by a simple argument, since $v$ will be strictly decreasing at $x$ large. Can anyone help me with $n\geq 2$?
 A: For each $ n \in \Bbb N,$ let  $z_n \in R^n $  such that $v(z_n)  \leq  \inf v(x)  + \frac{1}{n}. $
Now define $f(x) := v(x) +  \frac{1}{n} \| x-z_n \|^2 .$ Since $v$ is bounded below observe that $f(x) \to + \infty$ as $\|x\| \to + \infty $ this  shows $f$ attains its infimum at some $ y \in R^n$.  I claim $\|y - z_n \| \leq 1$. This  actually follows by
$$  f(y)    \leq f(z) = v(z_n) \leq  \inf v(x)  + \frac{1}{n} \le v(y)  + \frac{1}{n} $$
Hence
$$   f(v) \le   v(y)  + \frac{1}{n}$$
which easily proves the claim.  Now to solve problem consider the sequence $\{z_n \}_{n \in N} $ constructed  in above manner. Let $y_n$ be the minimizer of the function $f$ then according to claim we have $\| y_n - z_n \|  \le 1$,  also we know $\nabla f(y_n) = \nabla v(y_n) + \frac{2}{n} (y_n - z_n)   = 0$   now let $n \to + \infty$  to get the result.

NOTE that according to my answer the assumption of boundedness of derivative is redundant ! If I am not missing something!
A: You can reduce this problem to 1d. 
Note that the 1d problem is still valid on $[0,\infty)$.
Suppose $1>\inf \vert Dv\vert = a > 0$. Define a curve $h(t):[0,\infty) \to \mathbb{R}$ as following. 
1) Start at 0. Let $u_1$ be the vector in the direction of greatest descent and that $\vert u_1 \cdot Dv(0)\vert = 1$ . Define $h_1(s)= s u_1$, $s\in [0,\infty)$. Let $s_1$ be the first point when $\vert u_1\cdot Dv(h_1(s))\vert \le a$. 
2) Now, at the point $v(\bar{h_1}(s_1))$, find the direction of greatest descent $u_2$ and repeat step 1 to get a curve ${h}_2$ defined on $[0,s_2]$, where ${h_2}(0)={h_1}(s_1)$. 
3) Repeat this to define ${h}_n$.
4) Note that $\sum s_n =\infty$, otherwise, you've got a singularity.
5) Now, define $h$ by concatenate $v(\bar{h}_1) v(\bar{h}_2)...$
By construction $\inf_{[0,\infty)}\vert h'(t)\vert \ge a$, which is a contradiction.

My initial thought on this was just pick a straight line and go with it but dot product poses a problem with that. Morally, the intuition is to just follow the path down to $-\infty$.
