4
$\begingroup$

I need to prove that there is an isomorphic isometry from $c_0$ to some subspace of $C[0,1]$. Researching a bit, it looks like it follows from Banach-Mazur theorem, but we haven't studied it, at least looks that way from my notes. Is there a "simple", direct way to prove this given that our domain is specifically $c_0$?

$\endgroup$
6
$\begingroup$

Let $x$ be in $c_0$. Define the function $f$ in $C[0,1]$ as follows: $f(1/n)=x_n$ for every $n$ and linear in between. The image of $c_0$ by this functional is clearly a subspace, and the norm is conserved.

$\endgroup$
  • $\begingroup$ I was thinking something similar but didn't think of extending it linearly, which got me stuck - thanks! $\endgroup$ – ctlaltdefeat Mar 5 '13 at 18:02

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.