# Bounded linear operator on a normed space with bounded inverse and dense range

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?

• So a special case would be a normed space $X$ isometric to a dense proper subspace of itself. Jan 19 at 8:27
• 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 $y\notin T(X)$. Jan 19 at 8:44
• @NarutakaOZAWA: Is it that easy? For instance, if $T$ is the identity map your argument does not work. Jan 20 at 21:48