If $\{fx_n\}$ is a Cauchy sequence with $f$, a continuous self map on a complete metric space, we know that $\{x_n\}$ need not be Cauchy. Is it true for a uniformly continuous $f$?

Here's my take: Suppose $\{x_n\}$ is not Cauchy, then for some $\alpha>0$, $d(x_n,x_m)\geq\alpha$, and since $f$ is uniformly continuous, $\forall\varepsilon>0$ there exists $\delta>0$ with $d(fx_n,fx_m)<\varepsilon$ whenever $d(x_n,x_m)<\delta$.

Choosing $\delta=\alpha$, we get $d(fx_n,fx_m)\geq\varepsilon^\prime$ for some $\varepsilon^\prime>0$, a contradiction.

I believe there might be some flaws in this argument.

And one more question, if not, then under what minimum additional condition on $f$, like injective?


On $[-1,1]$ let $f(x)=x^{2}$ and $x_n=(-1)^{n}$. Then $f(x_n)$ is Cauchy and $f$ is uniformly continuous but $x_n$ is not Cauchy.

If $f$ is injective and $X$ is compact then the conclusion holds.

Proof: $f(x_n)$ converges to some point $y$. If $x$ is any limit point of $(x_n)$ then there is a subsequence converging to $x$ and continuity of $f$ gives $f(x)=y$. Since $f$ is injective this shows that you cannot have more then one limit point. Hence $(x_n)$ is convergent, therefore Cauchy.

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  • $\begingroup$ @markhaokip I have edited the answer.. $\endgroup$ – Kavi Rama Murthy May 2 at 7:43
  • $\begingroup$ Thanks, you have been a God sent for my Cauchy problems. Another querry: If $f$ is bijective and continuous in the original question, then $\{x_n\}$ is Cauchy. Am I right? $\endgroup$ – mark haokip May 2 at 10:38

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