nontrivial 'sub' Banach spaces I'm looking for a pair of Banach spaces $(X,\left\|\right\|_X)$ and $(Y,\left\|\right\|_Y)$ where $Y\subset X$, $\left\|\right\|_Y \leq \left\|\right\|_X$ and $Y$ is dense in $X$ for the $X$-norm, yet $Y\neq X$. I suppose such a pair exists, but have some trouble finding one.

The puzzling property of such a pair is of course that every $\left\|\right\|_X$-Cauchy-sequence in $Y$ is necessarily also a $\left\|\right\|_Y$-Cauchy sequence. Hence any $\left\|\right\|_X$-Cauchy sequence in $Y$ converging to some $x \in X\setminus Y$ (A sequence which exists by assumption.) must converge to some different $y \in Y$ for the $Y$-norm.

I suppose this question may be quite standard, so apologies in advance if the answer can be found in simple places (which I have not found until now.)
In order for $Y\subset X$ to be complete, $Y$ must be closed as a subset of $X$, so that $Y=\overline Y$. However, $Y$ being dense in $X$ means that $\overline Y=X$.

So, the only dense "Banach subspace" of $X$ is $X$, if the norm on $Y$ is inherited (or topologically equivalent to the inherited norm) from $X$.



On the other hand, considering $(\ell^1,\|\cdot\|_1)$ as a subset of $(\ell^2,\|\cdot\|_2)$ gives you the desired result.
In order for $Y\subset X$ to be complete, $Y$ must be closed as a subset of $X$, so that $Y=\overline Y$. However, $Y$ being dense in $X$ means that $\overline Y=X$.

So, the only dense "Banach subspace" of $X$ is $X$, if the norm on $Y$ is inherited (or topologically equivalent to the inherited norm) from $X$.



On the other hand, considering $(\ell^1,\|\cdot\|_1)$ as a subset of $(\ell^2,\|\cdot\|_2)$ gives you the desired result.