For what kind of $f$ is $(\mathbb{R}, |f(.)-f(.)|)$ complete?

Question

Let $f$ an increasing function on $\mathbb{R}$. We note $d(x,y) := |x-y|$ the usual distance.

We set $\delta(x,y):=|f(x)-f(y)|$. It is a distance on the real line.

My question : which condition on $f$ should hold to have $(\mathbb{R},\delta)$ complete metric space ?

Answer 1

It is a complete space iff the range of $f$ is closed in the standard topology.

First of all note that a metric space is complete iff every Cauchy sequence admits a convergent subsequence (this is an easy exercise).

Now if $x_n$ is a Cauchy sequence wrt. your new metric, the sequence $f(x_n)$ is bounded. But this means it contains a convergent subsequence and since the range of $f$ is closed, there is a $y$ so that $f(y)$ is the limit of our subsequence in the standard topology. But this means that $x_n$ converges to $y$ in the new topology.

To see the other direction, assume the range is not closed and choose a sequence $f(x_n)$ in the range whose limit point does not lie in the range of $f$. Now $x_n$ is a Cauchy sequence in the new topology without a limit.