# There exists an isometric embedding

Let $W$ be a closed linear subspace of a normed vector space $V$. Let $i_V: V \to V^{**}$. and $i_W: W \to W^{**}$ be the canonical embeddings of V and W into their second duals. Prove that there exists an isometric embedding $\Phi: W^{**} \to V^{**}$. Show that $\Phi(W^{**}) = (W^{\perp})^{\perp}$. Can you help me to prove this?

$(W^{\perp})^{\perp}=\{\Gamma \in V^* | F(f) = 0 \quad \text{for all} \quad f \in V^* s.t. f(W)=0\}$ I think I have to use Hahn Banach theorem, but I don't know how.

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There is a natural inclusion map $i:W\to V$ that is obviously distance-preserving. It induces a map $i^{\ast\ast}:W^{\ast\ast}\to V^{\ast\ast}$ that might be worth looking at. –  Brian M. Scott Apr 4 '12 at 7:06

The only possible answer to this question is that $\Phi = i^{\ast\ast}: W^{\ast\ast} \to V^{\ast\ast}$ where $i: W \to V$ is the inclusion (note that $\Phi$ is not just any embedding—it is forced upon us). We only have to check that it has the desired properties.
So, let us check that $i^{\ast\ast}$ is an isometric embedding, that is, we want to check that for $w^{\ast\ast} \in W^{\ast\ast}$ we have $$\tag{1} \sup_{\|w^\ast\|_{W^\ast} \leq 1}{ \left| \langle w^{\ast\ast}, w^\ast\rangle_{W^{\ast\ast},W^\ast} \right| } = \|w^{\ast\ast}\|_{W^{\ast\ast}} \stackrel{\color{red}{(!!)}}{=} \|i^{\ast\ast}w^{\ast\ast}\|_{V^{\ast\ast}} = \sup_{\|v^{\ast}\|_{V^{\ast}} \leq 1}{ \left| \langle i^{\ast\ast}w^{\ast\ast},v^{\ast}\rangle_{V^{\ast\ast},V^{\ast}} \right| }. \quad$$ Notice that $\|i\| =1$ implies that $\|i^\ast\| = 1$ and thus $\|i^{\ast\ast}\| = 1$, whence $$\tag{2} \|i^{\ast\ast}w^{\ast\ast}\|_{V^{\ast\ast}} \leq \|i^{\ast\ast}\|\,\|w^{\ast\ast}\|_{W^{\ast\ast}} = \|w^{\ast\ast}\|_{W^{\ast\ast}}.$$ To prove the other inequality $\|w^{\ast\ast}\|_{W^{\ast\ast}} \leq \|i^{\ast\ast}w^{\ast\ast}\|_{V^{\ast\ast}}$, let $v^\ast \in V^\ast$ and compute $$\tag{3} \langle i^{\ast\ast}w^{\ast\ast}, v^\ast \rangle_{V^{\ast\ast},V^{\ast}} = \langle w^{\ast\ast}, i^{\ast}v^{\ast}\rangle_{W^{\ast\ast},W^\ast} = \langle w^{\ast\ast}, v^\ast|_{W}\rangle_{W^{\ast\ast}, W^{\ast}}$$ since $(i^\ast v^{\ast})(w) = v^\ast(i(w)) = v^\ast|_{W}(w)$ for all $w \in W$. For every $\varepsilon \gt 0$ we can find $w^{\ast} \in W^{\ast}$ with $\|w^{\ast}\|_{W^\ast} = 1$ such that $\left|\langle w^{\ast\ast}, w^{\ast} \rangle\right|_{W^{\ast\ast},W^\ast} \geq \|w^{\ast\ast}\|_{W^{\ast\ast}} - \varepsilon$. By Hahn-Banach we can extend $w^{\ast} \in W^\ast$ to a linear functional $v^\ast \in V^\ast$ of norm $1$. Combining the right hand side of $(1)$ with $(3)$ and our choice of $w^\ast$, we get $$\tag{4} \|i^{\ast\ast}w^{\ast\ast}\|_{V^{\ast\ast}} \geq \|w^{\ast\ast}\|_{W^{\ast\ast}} - \varepsilon,$$ and, as $\varepsilon \gt 0$ was arbitrary, $(4)$ together with $(2)$ establishes the desired equality $\color{red}{(!!)}$ of $(1)$.