2
$\begingroup$

Does there exist such an operator $T$ (bounded and 1-1 on a normed space $X$, it's range $R_{T}$ is dense in $X$ and $T^{-1}: R_{T}\to X$ is bounded) which is not surjective? In other words, does there exist a normed space isometrically isomorphic to a proper dense subspace of it?

I know that if $X$ is a Banach space, then $R_{T}$ must equal to $X$ because $R_{T}$ is closed in $X$. But what if $X$ is just a normed space?

Thank you for your help.

$\endgroup$
5
  • $\begingroup$ So a special case would be a normed space $X$ isometric to a dense proper subspace of itself. $\endgroup$ Jan 19, 2022 at 8:27
  • $\begingroup$ Yes, maybe I should edit my question. Thank you for your comment. $\endgroup$
    – shepherd
    Jan 19, 2022 at 8:33
  • $\begingroup$ Take a surjective isometry $T$ on a Banach space $Y$, a dense (non-closed ) subspace $Y_0\subset Y$ such that $T(Y_0)=Y_0$, and $y\in Y\setminus Y_0$. Put $X\subset Y$ to be the linear span of $\{ T^n y: n\geq0 \}\cup Y_0$. It's easy to arrange $y\notin T(X)$. $\endgroup$ Jan 19, 2022 at 8:44
  • $\begingroup$ Maybe that's a right way to give an exmple. Let me think... $\endgroup$
    – shepherd
    Jan 19, 2022 at 8:57
  • $\begingroup$ @NarutakaOZAWA: Is it that easy? For instance, if $T$ is the identity map your argument does not work. $\endgroup$
    – Alex M.
    Jan 20, 2022 at 21:48

1 Answer 1

2
$\begingroup$

Take a pre-Hilbert space with countable Hamel (algebraic) dimension (i.e. the standard scalar product on the space of eventually zero sequences)

There is a non-continuos linear functional, i.e. a non-closed (hence dense) hyperplane. It has again algebraic countable dimension, hence again countable Hilbert dimension, so it is isometrically isomorphic to the given space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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