I have a continuous bounded function $f\in C^{\infty}(\mathbb{R}^k, \mathbb{R})$ with non-zero, bounded, derivative in some neighbourhood of a point $c\in\mathbb{R}^k$. I also have two sequences $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ which both tend to $c$.
For all $n$, $d(a_n,c)>d(b_n,c)$. Furthermore, for all $n$, $f(a_n)>0$ and $f(b_n)>0$ and $f(c)=0$.
I want to show that there exists some $N$ such that $f(a_n)>f(b_n)$ for all $n>N$. Formally, I want to show:
$$\exists N \in \mathbb{N}\text{ such that } \forall n > N, f(a_n)>f(b_n)$$
Is this possible? At first I thought it would be easy and could be done using the fact a $C^{\infty}$ function is locally lipschitz continuous and considering a Lipshitz continuous region about $c$ but I can't get this to work... does anyone know how to show this?
Edit: In fact the result is true as stated if $k=1$. We do that after the counterexample for $k=2$:
In $\Bbb R^2$ a counterexample is utterly trivial:
Take $c=(0,0)$, $a_n=(2/n,0)$, $b_n=(0,1/n)$, $f(x,y)=x+3y$.
(Seems to me $f$ being linear puts constraints on what sort of proof of a positive result might possibly work under what sort of additional hypotheses...)
Proof. Wlog $c=0$. Since $f'(0)\ne0$, wlog $f'(0)>0$ and then wlog $f'(0)=1$.
Now $f<0$ on $(-\delta,0)$, so wlog $a_n,b_n>0$; hence $$a_n-b_n>0.$$
And $f\in C^1$ implies $$\frac{f(a_n)-f(b_n)}{a_n-b_n}\to1;$$
hence $f(a_n)-f(b_n)>0$ for large $n$.