Continuous linear operator preserving Cauchy sequence in metric vector spaces

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: $$\rho(x_1+x,x_2+x)=\rho(x_1,x_2) \forall x_1,x_2 \in X$$

• 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? – dan_fulea Nov 27 '18 at 17:17
• @YadatiKiran I know a definiton for a bounded operator in normed vector space, is there a metric vector space equivalent? – adam kyjovsky Nov 27 '18 at 17:26
• @adamkyjovsky: My argument is absurd. Sorry. – Yadati Kiran Nov 27 '18 at 17:31
• @dan_fulea X,Y are vector spaces, I called them linear because one of my professors does that. – adam kyjovsky Nov 27 '18 at 17:39
• @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 $T$? – dan_fulea Nov 27 '18 at 17:58

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).

• Thank you, this clears it up. – adam kyjovsky Nov 29 '18 at 11:01