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?

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.

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.

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

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.

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?

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.