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I have an exercise to show the following:

Every separable Banach space is isometrically isomorphic to a subspace of $\ell_{\infty}$.

My question is that whether one really needs the space to be complete?

Here's my attempt at proving the statement for general normed spaces:

Let $X$ be a separable normed space. Then there exists $D=\{x_1,x_2,\ldots,\}\subseteq X$ such that $\overline{D}=X$.

Without loss of generality, suppose $0\notin D$. As a consequence of the Hahn-Banach theorem (see Corollary 2), for each $n\in\mathbb N$, there exists $f_n\in X^*$ such that $f_n(x_n)=\| x_n\|$ and $\|f_n\|=1$.

Now define, $T:X\to \ell_{\infty}$ as $T(x)= (f_n(x))$. As $\|f_n\|=1$, so $T$ is bounded and $\|T\|\le 1$.

Finally, let $x\in X$ such that $\|x\|=1$. Then for each $\epsilon>0$, there exists $x_n\in D$ such that $\|x_n-x\|<\epsilon$. Thus $|1-f_n(x)|=|f_n(x_n)-f_n(x)|<\epsilon$. Thus for each $0<\epsilon<1$, there exists $n\in\mathbb N$ such that $|f_n(x)|>1-\epsilon$. So, $\|T(x)\|\ge 1=\|x\|$. Thus $\|T\|\ge 1$.

It follows that $T$ is isometric.

Edit: My question is not a duplicate of the suggested question, as completeness is not highlighted in any of the answers.

Also, as @AnneBauval pointed out, the above set $D$ needs to be replaced with $\left\{ \frac{x}{\|x\|} : x\in D\right\}$.

Guest
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  • Do you rather mean $T(x)= (f_n(x))$? 2) Why $|1-f_n(x)|=|f_n(x_n)-f_n(x)|$? 3) "It follows that $T$ is isometric" not from $|T|\ge1$ but from the (yet imho unproved) previous sentence. 4) Of course if the theorem is proved for a separable Banach, you can apply it to the completion of any separable normed space.
    • – Anne Bauval Nov 16 '22 at 14:13
  • Question: When you defined T, we must have $(f_n(x_n)) \in l^\infty$. But why is $\sup_{n \in \mathbb{N}}||x_n|| < \infty$? – user57 Nov 16 '22 at 14:14
  • @AnneBauval 1) Yes, edited. 2) Because $f_n(x_n)=1$ by choice of $f_n$. 3) Isometry follows from $|T|=1$. – Guest Nov 16 '22 at 14:17
  • No, $f_n(x_n)=|x_n|.$ You should take a dense sequence in the unit sphere. 3) By no means.
    • – Anne Bauval Nov 16 '22 at 14:19
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    Does this answer your question? Isometric Embedding of a separable Banach Space into $\ell^{\infty}$ – Anne Bauval Nov 16 '22 at 14:23
  • @AnneBauval No, it doesn't because it doesn't highlight where completeness of the space is used. – Guest Nov 16 '22 at 14:25
  • Doesn’t the Hahn-Banach theorem rely on completeness? – FShrike Nov 16 '22 at 14:26
  • @FShrike The reason I linked the corollary of Hahn-Banach theorem I used is to highlight that it doesn't require completeness. – Guest Nov 16 '22 at 14:27
  • @Guest The duplicate is a corrected version of your proof (cf. points 2 and 3 in my first comment), and point 4 answered your question. – Anne Bauval Nov 16 '22 at 14:30
  • @AnneBauval Ah, so every separable normed space sits inside its completion which sits inside $\ell^{\infty}$? But anyway, the proof of the duplicate holds true for normed spaces and there is no need to go via completion. – Guest Nov 16 '22 at 14:33
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    On that, I agree (Silentmovie already pointed it as a comment to David Mitra's answer in the duplicate). – Anne Bauval Nov 16 '22 at 14:35
  • @AnneBauval Thanks, the fact that the statement was stated everywhere only for Banach spaces was the point of confusion for me (since the proofs didn't rely on completion). You can add a short answer and I'll accept it to close the question. – Guest Nov 16 '22 at 14:36
  • I think the question can be just closed as a duplicate. May be we should append your remark to the answers over there. But do you now agree with my points 2 and 3? – Anne Bauval Nov 16 '22 at 14:37
  • I agree without your point 2. For 3, I don't see why $|T|=1$ doesn't imply $T$ is an isometry. – Guest Nov 16 '22 at 14:50
  • Even an automorphism $f$ of $\Bbb R^2$ may have norm $1$ without being an isometry. E.g. $f(x,y)=(x,y/2).$ – Anne Bauval Nov 16 '22 at 22:53
  • @AnneBauval But $|Tx|=|x|$, so $T$ is injective, hence an isometry. – Guest Nov 16 '22 at 23:01
  • Yes, that is why I wrote "It follows that T is isometric" not from ∥T∥≥1 but from the previous sentence" (which showed precisely that $|Tx|=|x|$). Here again, $T$ is an isometry follows from $|Tx|=|x|$ and not from injectivity (my $f$ was injective). – Anne Bauval Nov 16 '22 at 23:09