Suppose that $X$ is a normed vector space then the dual space $X'$ of $X$ is a Banach space equipped with the dual norm https://en.wikipedia.org/wiki/Dual_norm, then we can consider $X''$, the double dual space of $X$, then we can consider the canonical embedding $i:X\rightarrow X'' \quad v\rightarrow i(v)$ here $i(v)\in X''$, $i(v)$ map $f\in X'$ to $f(v)$, I have proved that $i(v)$ is linear but how to show that $i(v)$ is an isometry?
2026-03-31 10:09:26.1774951766
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Prove that the canonical embedding from $X$ to $X''$ is isometric
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We have for any $v \in X$ \begin{align} \lVert i(v)\rVert_{X''}=& \sup_{f \in X', \lVert f \rVert_{X'}\leq 1} \langle i(v), f \rangle_{X'',X'} \\ =& \sup_{f \in X', \lVert f \rVert_{X'}\leq 1} \langle f,v\rangle_{X',X} \\ =& \lVert v \rVert_{X} \, , \end{align} where the last equality is a consequence of the Hahn-Banach theorem.
The fact that is an injective map it’s a consequence of the Hahn-Banach theorem
In fact if you take $v\neq 0$, then $v$ is not contained in the closure of $U:=\{0\}$, that is zero, so that there exists a functional $f\in X’$ such that $g(v)=1$ and $||g||=||v||^{-1}$.
This means $i(v)(g)=1$ and so $i(v)$ is different from zero.
Regarding the surjectivity, you can’t prove it because it’s not true anymore in general for infinite dimensional vector spaces.
A space that satisfies this condition is called Reflexive space
In the finite dimensional case you have always that is surjective because $X$ and $X’$ have the same dimension, so that $X$ and $X’’$ have to be the same dimension and every linear injective map between two spaces of the same dimension is an isomorphism.
Now we can try to prove that is an isometry:
It is sufficient to prove that $||i(v)||=||v||$ for each $v$
$||i(v)||=sup_{f}\frac{|f(v)|}{||f||}\leq sup_f \frac{||f||\cdot||v||}{||f||}=||v||$
Thus $||i(v)||\leq ||v||$
But now by the same consequence of before of Hahn-Banach theorem there exists $g$ such that $g(v)=1$ and $||g||=||v||^{-1}$, so that
$||i(v)||=sup_{f}\frac{|f(v)|}{||f||}\geq \frac{1}{||v||^{-1}} =||v||$
and you have done.