Given a sequence $(x_n)$ of real numbers satisfying
$$|x_n – x_m| \leq \frac 12 |x_{n-1} – x_{m-1}|$$
does it follow that $(x_n)$ is Cauchy?
This question arose in an attempted answer to this question. I feel like this should be false. All I can think of is the double sequence $a_{n,m} = 2^{n-2m}$ which satisfies the analogous inequality above and does not go to $0$ for the product topology. But it does not satisfy the triangle inequality $a_{n,m} \leq a_{n,k} + a_{k,m}$ so it's quite far from a counterexample.
Best Answer
Yes.
Set $a=|x_1-x_0|$. By induction, applying your assumption with $n=m+1$, $$ |x_{m+1}-x_m| \le 2^{-m}a $$
This makes the sequence Cauchy: If you're given $\varepsilon$, choose $N$ large enough that $2^{-N}<\varepsilon/2a$. Then for every $n,m>N$ you have $|x_n-x_m|<2\cdot 2^{-N}a$.