Let $Y$ be a dense subspace of a normed linear space $X$. Show that $Y^*$ is isometrically isomorphic to $X^*$.
Define $F : Y^* \to X^*$ as $F(f\mid_Y) = f$.
Clearly the map $F$ is an isometry. And the extension is unique due to the fact that if $f$ is a continuous linear functional and $Y \subseteq X$ is a dense subspace such that $f|_Y = 0$, then $f = 0$. Thus the map is well defined.
Clearly $F$ is bijective and $F^{-1}$ also exists clearly. Thus we can say that $F$ is an isomorphism and isometry.
Is the above argument correct?
Note: Given two normed vector spaces $V$ and $W$, a isometry is a map $f : V \to W$ that preserves the norms i.e. $\mid \mid f(v) \mid \mid = \mid \mid u \mid\mid$.
For every $f: Y \to \mathbb{R}$ by Hahn-Banach there is some extension $\overline{f}:X \to \mathbb{R}$, with $\|f\| = \|\overline{f}\|$. This extension is unique as $\mathbb{R}$ is Hausdorff and $Y$ is dense. So there is a well-defined isometry $F: Y^\ast \to X^ast$ defined by $F(f) = \overline{f}$. As usual, the unicity implies that $F$ is linear as well. And the map that sends $f: X \to \mathbb{R}$ to $f|Y : Y \to \mathbb{R}$ is its obvious inverse, so $F$ is onto, as well.