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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.)

5th decile
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1 Answers1

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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.

Ben Grossmann
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  • Well, drop that final assumption. – 5th decile Sep 08 '15 at 11:07
  • In most contexts, the norm on a subspace is necessarily the inherited norm. – Ben Grossmann Sep 08 '15 at 11:12
  • About your edit: isn't it l_1 that is a subset l_2 and not the other way around? – 5th decile Sep 08 '15 at 11:27
  • If something has a finite $1$-norm, then it necessarily has a finite $2$-norm. So yes, you're right. – Ben Grossmann Sep 08 '15 at 11:31
  • Again about your (edited) edit: the 1-norm is greater than the 2-norm. Your example will BTW fail also if you try any other l_p l_q pairs. – 5th decile Sep 08 '15 at 11:59
  • The $1$-norm is not necessarily greater, unless each $\xi_j$ in the sequence satisfies $|\xi_j| \leq 1$. I'm not so sure about other $p,q$ pairs. I think $\ell^p \subset \ell^q$ when $p < q$. However, things go wrong when one of the spaces is $\ell^\infty$ (that is, density fails). – Ben Grossmann Sep 08 '15 at 12:10