0
$\begingroup$

Suppose we have a cauchy sequence $\{a_n\}$ in a normed vector space $V$. Given a linear transformation $T:V \rightarrow V$, is the sequence $\{T(a_n)\}$ also cauchy? Or is it true only for finite dimensional normed spaces? I'd be much obliged if someone could give a proof for this, preferably an elementary one. Thanks!

$\endgroup$
  • $\begingroup$ $T$ has to be continuous. Because continuous linear transformation on a normed space is bounded: $\| Tx\| \leq M\| x\|$ for some $M>0$ and all vectors $x$. Hence if $\{ a_n\}$ is a Cauchy sequence, then $\| Ta_m-Ta_n\| \leq M\| a_m-a_n\|$ gives that $\{ Ta_n\}$ is a Cauchy sequence as well. $\endgroup$ – Janko Bracic Jan 17 '15 at 8:54
0
$\begingroup$

Given $n,m\in \mathbb N; \exists p\in \mathbb N $ such that $||a_n-a_m||<\epsilon \forall m,n\geq p$

Now $||T(a_n)-T(a_m)||=||T(a_n-a_m)||<\epsilon$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.