If we have that $X \subset V$ is dense linear subspace. Where $V$ is normed space. I can show that for any $f \in X^{*}$, there exists a unique extension $\bar{f}$. I want to know if it can be shown that the mapping $\phi(f) = \bar{f}$ is an isometry?
My idea is to use the following two results. The first is a Corollary of the Hahn-Banach Theorem.
Corollary:
Let $X \subset V$ be a linear subspace. If $f \in X^{*}$ then there exists $\bar{f} \in V^{*}$ that extends $f$ such that $$\Vert \bar{f} \Vert_{V^{*}} = \Vert f \Vert_{X^{*}}$$
and the
Theorem:
Let $X$ be a subspace of some metric space $(V, \rho_{1})$ and $(Y,\rho_{2})$ a complete metric space. Then, if $f:X \rightarrow Y$ is uniformly continuous there exists a unique extension $\bar{f}: \bar{X} \rightarrow Y$ which is uniformly continuous.
Together I think this gives the result I'm looking for.
Best Answer
This was sorted out in comments, so I'll add a slightly different point of view. Every bounded linear operator $T:X\to V$ has the adjoint operator $T^*:V^*\to X^*$ defined by $T^*(f)=f\circ T$. The definition shows that $\|T^*\|\le \|T\|$. (In fact, equality holds by Hahn-Banach, but we do not need this here).
In the special case when $T$ is an isometric embedding with dense image, the operator $T^*$ is surjective, because every $f\in X^*$ extends to an element $\bar f\in V^*$ via $\bar f(x) = \lim_{y\to x} f(y)$. Also, since $$|\bar f(x)|\le \lim_{y\to x}\|f\|\, \|y\| = \|f\|\,\|x\|$$ we have $\|\bar f\|\le \|f\|$. Hence, $T^*$ is a surjective isometry in this case.