Proving the bijection of a continuous function $f$ such that $|f(x)-f(y)| \geq \log(1+|x-y|) \forall x,y$ in $\mathbb{R}.$ In this question for proving the bijection, I can easily check that this is an injection. But how can one show that this function takes all the real values? by taking $y=0$ I can show that $|f(x)-f(0)| $ can be made as large as required by using the fact that it is larger than the log term. But how can I show that f assumes the negative values as well as the positive ones?
Any help is highly appreciated.
 A: Any injective continuous function on $\mathbb R$ is monotonic. Suppose $f$ is increasing. Hence $f(x)-f(y) \geq \log (1+|x-y|)$ for $x \geq y$. Letting $x \to \infty $ we see that the supremum of $f$ is $\infty$ and letting $y \to -\infty$ we see that the infimum of $f$ is $-\infty$.
A: An injective continuous function on any interval is strictly monotone, so we can assume $f$ is strictly increasing on $\mathbb R$.
To show that $f$ is surjective, it suffices to show that $$\lim_{x\to \infty}f(x)=\infty, \lim_{x\to -\infty}f(x)=-\infty$$
and let the Intermediate Value Theorem complete the remaining works.
We will use contradiction proof here: suppose $\lim_{x\to \infty}f(x)<\infty$, then since $f$ is increasing, there is a positive number $M$ such that $\lim_{x\to \infty}f(x)=M$. Choose a real number $R$ such that $$f(x)>M-1\quad\text{ whenever }\quad x\geq R,$$
so
$$1\geq |f(R+10)-f(R)|\geq \log(1+10),$$
a contradiction. We get $f(x)\to \infty$ when $x\to \infty$, so the same goes for $-\infty$.
