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Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$ and $E$ be a normed $\mathbb K$-vector space.

Can we show that there is a linear isometry from $E$ into $E'$ or is there a counterexample?

I think this should be true: My idea is that we should be able to consider the linear isometry $\iota_1$ from $E$ into its completion $\tilde E$ and the linear isometry $\iota_2$ from $\tilde E$ into its dual $\tilde E'$ ...

0xbadf00d
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    I would say that this is so false, that it is hard to even find a single nontrivial example for which this holds. By this I mean that I cannot think of any non-Hilbert, infinite dimensional Banach space $E$ which admits a linear isometry onto $E’$. I kind of remember a book, called “The James forest”, which dealt with such examples, but it could be a trick of my memory. – Giuseppe Negro Feb 04 '21 at 18:28
  • In any case, years ago I posed vaguely similar questions, here and especially here (the second link is IMHO better than the first). There, the isometries that are considered are non-linear. The second link has a very interesting answer concerning the existence and non-existence of such non-linear isometries. – Giuseppe Negro Feb 04 '21 at 18:32

2 Answers2

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This is not true. Take $E=c_0$ and $E'=\ell _1$. Assume there is any linear isometry from $E$ to $E'$, i.e., $T : E\to E'$ linear and continuous, and such that $\|Tx\|_{\ell^1} = \|x\|_{c_0}$. Let $x_n \rightharpoonup x$ in $E$, then $Tx_n \rightharpoonup Tx$ in $\ell^1$. By the Schur property of $l^1$, $\|Tx_n -Tx\|_{\ell^1}\to 0 $. Due to the isometry property, $x_n \to x$ in $c_0$. Hence every weakly converging sequence in $c_0$ is strongly converging, which is absurd (take the sequence of unit vectors).

Falcon
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daw
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Pitt's theorem asserts that every bounded linear operator from $\ell_p$ into $\ell_q$ is compact whenever $1 ≤ q<p< ∞$. Thus the answer is also negative for $\ell_p$, for every finite $p>2$.

Ruy
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  • When we consider the bidual $E''$ instead, we know that the map $\iota:E\ni x\mapsto\langle x,;\cdot;\rangle$ is a linear isometry. So, the completion of $E$ is given by $F:=\overline{\iota X}$. This should literally mean that there is an isometric isomorphism from $E$ onto a dense subset (namely $\iota X$) of a Banach space. However, I've often read that every normed space is isomorphic to a Banach space; but isn't that wrong? It should be only isomorphic to that dense subset. Or am I missing something? – 0xbadf00d Feb 22 '21 at 20:18
  • A non complete space may be isomorphic to a Banach space only if by isomorphism you mean algebraic isomorphism, i.e. not preserving norm or topology. – Ruy Feb 22 '21 at 20:26
  • What does that mean in the situation above? Is $E$ algebraically isomorphic to $F$? I still would think that it is only algebraically isomorphic to the dense subspace $\iota X$. Or am I wrong? – 0xbadf00d Feb 22 '21 at 20:28
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    The map you constructed into the double dual is an algebraic isomorphism, as well as an isometry, onto $\iota E$. It is not an isomorphism onto $F$, unless $E$ is already complete. However it is possible that there is another map which is an algebraic isomorphism from $E$ to $F$. However, such a map cannot be isometric unless $E$ is already complete. – Ruy Feb 22 '21 at 20:38
  • Thank you for the explanation. – 0xbadf00d Feb 22 '21 at 20:40