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It's more known that normed space can be isometrically embedded to its bidual, which is Banach.

natural embedding of normed linear space is an isometry

However, what does one need to isometrically embed normed space to a dense subset of its bidual?

mavavilj
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  • The first problem I see is that a dense subset cannot necessarily contains a vectorial subspace. For example $\mathbb Q^2\setminus(0,0)$ is dense in $\mathbb R^2$. – Piquito May 12 '18 at 18:41

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For Banach spaces, the condition you're asking for is reflexivity. The isometric image of a complete space is complete, because every Cauchy sequence converges to a point in the space. In particular it is closed. So if it were dense in the bidual, it would be the entire thing.

So now assume that some incomplete normed space $Y$ has the property that it is dense in its bidual. Denote the embedding $i:Y\to Y^{**}$. If $X$ is the metric completion of $Y$, then $X^*$ is isomorphic to $Y^*$ by the density of $Y$ in $X$, so every functional on $Y$ uniquely determines a functional on $X$, and therefore $X^{**}$ is naturally isomorphic to $Y^{**}$. In particular, $i(Y)$ being dense implies that $i(X)=Y^{**}$ by the universal property of the completion. But this means that $X^{**}=i(X)$, so $X$ is reflexive. So the condition you're looking for is the completion being reflexive.

Ashwin Trisal
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