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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.

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  • $\begingroup$ So a special case would be a normed space $X$ isometric to a dense proper subspace of itself. $\endgroup$ Jan 19 at 8:27
  • $\begingroup$ Yes, maybe I should edit my question. Thank you for your comment. $\endgroup$
    – shepherd
    Jan 19 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 at 8:44
  • $\begingroup$ Maybe that's a right way to give an exmple. Let me think... $\endgroup$
    – shepherd
    Jan 19 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 at 21:48

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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.

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