Let $T: X \rightarrow Y$ be a continuous linear operator. $(X,\rho), (Y,\xi)$ are linear metric spaces and $\{x_n\} \subset X$ is a Cauchy sequence.

I need to show that $\{Tx_n\}$ is a Cauchy sequence.

For $T$ to be continuous means:

$\forall \varepsilon>0\,\, \exists \delta>0$ such that $\rho(x,y)<\delta \implies \xi(Tx,Ty)<\varepsilon.$

And for sequence to be Cauchy means:

$\rho(x_n,x_m)\rightarrow 0$ for $m,n \rightarrow \infty.$

How do I show that $\xi(Tx_n,Tx_m)\rightarrow 0$ for $m,n \rightarrow \infty?$ So far I suspect that the operator needs to be linear, because there are counterexamples for nonlinear continuous mappings (Namely $\{\frac{1}{n}\}$ and $T(x)=\frac{1}{x}$.).

I'm sorry if I wasn't thorough enough with my search but, I've only seen a similar question here with normed vector spaces instead of metric or the mapping was uniformly continuous and I wasn't able to translate it to my problem.

EDIT: Forgot to include this attribute of metrics: $\rho(x_1+x,x_2+x)=\rho(x_1,x_2) \forall x_1,x_2 \in X$

  • $\begingroup$ What are $X,Y$? Vector spaces, groups, ...? (So that we can use the attribute linear for a map between them.) Are the metrics compatible with the algebraic operations? $\endgroup$
    – dan_fulea
    Nov 27, 2018 at 17:17
  • $\begingroup$ @YadatiKiran I know a definiton for a bounded operator in normed vector space, is there a metric vector space equivalent? $\endgroup$ Nov 27, 2018 at 17:26
  • $\begingroup$ @adamkyjovsky: My argument is absurd. Sorry. $\endgroup$ Nov 27, 2018 at 17:31
  • $\begingroup$ @dan_fulea X,Y are vector spaces, I called them linear because one of my professors does that. $\endgroup$ Nov 27, 2018 at 17:39
  • $\begingroup$ @adam kyjovsky ... and are the metrics compatible with the addition and scalat multiplication on the vector fields $X$, $Y$ (which are vector fields over the same field $\Bbb R$)? Do we really need the linearity of $T$? $\endgroup$
    – dan_fulea
    Nov 27, 2018 at 17:58

1 Answer 1


The central theme here doesn't need linearity:

If $(X, \rho), (Y, \xi)$ are metric spaces, $f : X \to Y$ is uniformly continuous, and $\{x_i\}$ is a Cauchy sequence in $X$, then $\{f(x_i)\}$ is a Cauchy sequence in $Y$.

  • That $\{x_i\}$ is Cauchy means that $\forall \epsilon > 0, \exists N$ such that $\forall m,n > N, \rho(x_n, x_m) < \epsilon$
  • That $f$ is uniformly continuous means that $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall x, y\in X,\rho(x,y) <\delta\implies\xi(f(x), f(y)) < \epsilon$. Note the difference between this and ordinary continuity:

$f$ is continuous if $\forall y \in X$ and $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall x\in X,\rho(x,y) <\delta\implies \xi(f(x), f(y)) < \epsilon$.

Regular continuity allows you to choose different $\delta$ values for each $y$, while uniform continuity requires a single $\delta$ to work for every $y$.

To prove the theorem above, we need to show that $\forall \epsilon > 0, \exists N$ such that $\forall m,n > N, \xi(f(x_n), f(x_m)) < \epsilon$. This is straight-forward: Given $\epsilon$, pick the corresponding $\delta$ that satisfies the uniform continuity condition on $f$. Use this $\delta$ as the "epsilon" for the Cauchy condition on $\{x_i\}$ and consider the corresponding $N$. For $m,n > N$, we have $\rho(x_n, x_m) < \delta$. Now doesn't that look a lot like the uniform continuity condition on $f$?

So why does your problem have the linearity condition? Because you are only given that $T$ is continuous, not uniformly continuous. What you still need to do is to show that because of the linearity, continuity of $T$ implies uniform continuity. This is a simple calculation, as continuity of $T$ at $y = 0$ is sufficient to prove uniform continuity everywhere (by the translation invariance of the two metrics that you added late to your post).

  • $\begingroup$ Thank you, this clears it up. $\endgroup$ Nov 29, 2018 at 11:01

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