Is there always a linear isometry from a normed vector space into its dual?

Question

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 $\tilde E'$ ...

Answer 1

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 $p>2$.

Answer 2

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).