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Let $(M,d)$ be a metric space and $f\colon[0,\infty)\to[0,\infty)$ metric preseving map that is right continuous at $0$, i.e. $f$ has satisfies $$\forall x,y\in [0,\infty)\colon f(x+y)\le f(x)+f(y)\quad\text{and}\quad f(x)=0\iff x=0.$$ Furthermore, $f$ is non-decreasing and addition $$f(0^+)=\lim_{x\downarrow0}f(x)=f(0).$$

Then the composition $\Delta=f\circ d$ is again a metric on $M$. But how can I show $$(M,d) \text{ is complete}\iff (M,\Delta) \text{ is complete}?$$

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The domain of $f$ is $[0,\infty)$ but the Triangle Inequality is expressed for all $x,y\in M$. Did you mean for all $x,y \ge 0$? – Patrick Mar 23 '12 at 23:05
@bbm: Patrick was pointing out something else. There's no need to mention that $f$ is non-decreasing since this follows from the other conditions (when corrected as Patrick proposed). – joriki Mar 23 '12 at 23:41
@Brian M. Scott: I think that the rightcontinuity at $0$ is does the trick, since $f$ is non-negative, $f(0)=0$ and f is right continuous at $0$. $\exists \delta>0$ such that $x\in[0,\delta)\implies f(x)<\epsilon.$ So if $\Delta(x_m,x_n)<\epsilon$ is small, then so is $f(d(x_m,x_n))$. The previous comments then imply $d(x_m, x_n)$ is small. How can I write this down in a neat way? – mathQ Mar 24 '12 at 1:38

You simply need to show that $\langle M,d\rangle$ and $\langle M,\Delta\rangle$ have the same Cauchy sequences: it follows at once from that that one is complete iff the other is.

To get you started, suppose that $\langle p_n:n\in\Bbb N\rangle$ is a $d$-Cauchy sequence in $M$. Let $\epsilon>0$ be given. It follows from the fact that $f({0^+})=f(0)=0$ that there is a $\delta>0$ such that $f(x)<\epsilon$ whenever $0\le x<\delta$. Because $\langle p_n:n\in\Bbb N\rangle$ is $d$-Cauchy, there is an $n_\epsilon\in\Bbb N$ such that $d(p_m,p_n)<\delta$ whenever $m,n\ge n_\epsilon$. But then for $m,n\ge n_\epsilon$ we have $$\Delta(p_m,p_n)=f\left(d(p_m,p_n)\right)<\epsilon\;,$$ and it follows that $\langle p_n:n\in\Bbb N\rangle$ is $\Delta$-Cauchy as well.

Now you want to assume that $\langle p_n:n\in\Bbb N\rangle$ is $\Delta$-Cauchy and show that it’s necessarily also $d$-Cauchy. As before, start with some $\epsilon>0$. Eventually you want to come up with some $n_\epsilon\in\Bbb N$ such that $d(p_m,p_n)<\epsilon$ whenever $m,n\ge n_\epsilon$. Since you’re assuming that the sequence is $\Delta$-Cauchy, you know that you can get $\Delta(p_m,p_n)=f\left(d(p_m,p_n)\right)$ small by taking $m$ and $n$ sufficiently large; can you see how to use what you know about $f$ to translate this into making $d(p_m,p_n)$ itself small?

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