Let $K$ be a compact operator between two normed spaces. If $K(X)$ is closed, does this necessarily imply that $K(B)$ is closed?

where $B$ is the closed unit ball?

  • 1
    $\begingroup$ Take the linear functional $x=(1/2,1/4,1/8,\ldots)\in\ell_1$ on $X=c_0$. Its range is all of $\Bbb R$, but the image of $B(c_0)$ under $x$ is $(-1,1)$. $\endgroup$ – David Mitra Jul 23 '16 at 18:46

Consider $f:c_0\rightarrow R$ defined by $f(e_i)={1\over 2^i}$, it is compact and the image of the unit ball is $(-2,2)$. where $c_0$ is the set of sequences with finitely non zero terms endowed with $\|\|_{\infty}$.

  • $\begingroup$ So $(-2,2)$ is closed but not compact, right? $\endgroup$ – MorganeMaPh Jul 23 '16 at 18:48
  • $\begingroup$ it is neither closed nor compact. $\endgroup$ – Tsemo Aristide Jul 23 '16 at 18:48
  • $\begingroup$ All right! But know if in addition I assume K to be an open operator... then it would hold right? $\endgroup$ – MorganeMaPh Jul 23 '16 at 18:52
  • $\begingroup$ yes, it is also open $\endgroup$ – Tsemo Aristide Jul 23 '16 at 18:53
  • $\begingroup$ But in your example, is $f(e_i)$ closed? $\endgroup$ – MorganeMaPh Jul 23 '16 at 18:53

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