1
$\begingroup$

Suppose, that we have $||.||_1$ and $||.||_2$ norms defined on an arbitrary $X$ vector space. $X$ is a complete space with both of them.

For all $x_n, n \in \mathbb{N} \subset X$ series, if $\lim_{n \to \infty} ||x_n||_1 = 0$, then $\lim_{n \to \infty} ||x_n||_2 = 0$ is also true.

Prove, that if this is true, then the two norms are equivalent!

The definition of equivalent norms is, that there exists such $c_1, c_2$ finite constants, so that $$\forall x \in X, c_1 ||x||_1 \le ||x||_2 \le c_2||x||_1$$

I simply have no idea, how to prove such statement. If $X$ is complete with this norm, then the limit of this series must be in $X$. The second part, that if the norm of the first is $0$, then the norm of the second is $0$ too, then any $c_1,c_2$ will do. I have no idea how to go forward.

Any help appreciated.

$\endgroup$
  • $\begingroup$ Is this true? If $x_n \to 0$ in $l^1$ then $x_n \to 0$ in $l^2$ (sequence spaces), but the norms on these spaces aren't equivalent. $\endgroup$ – πr8 May 1 '16 at 8:24
  • $\begingroup$ Don't you know the open mapping theorem (or closed graph or Banach's isomorphy theorem)? $\endgroup$ – Jochen May 1 '16 at 10:55

Your Answer

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

Browse other questions tagged or ask your own question.