If $V\subset W$ (sub) normed vector spaces, can $V^*$ be embedded in $W^*$ linearly and isometrically? If $W$ is a normed vector space and $V$ a linear subspace, Hahn-Banach tells us that $\phi\in V^*$ can be extended to $\tilde{\phi}\in W^*$. Can this mapping, $\phi\mapsto\tilde{\phi}$ be chosen to be a linear isometry?
Maybe at least in finite dimensions?
 A: Before giving an example, I'll explain why linear isometry should not be expected to  exist. For simplicity assume $W$ is reflexive. Suppose there is a linear  isometry $T:V^*\to W^*$ with the property you described. The fact that $T\phi$ is an extension of $\phi$ can be stated as: $RT=I_{V^*}$ where $R:W^*\to V^*$ is the restriction operator. 
Taking adjoints yields $T^*R^*=I_V$ where $R^*: V\to W$ is simply the inclusion map. Thus, $T^*$ is a norm-$1$ projection from $W$ onto $V$. (One says that $V$ is $1$-complemented in $W$.) This is rare. Being $1$-complemented means there is a linear projection of $W$ onto $V$ such that the image of the unit ball is contained in its intersection with $V$.
And here is a concrete example of a subspace $V$ that is not $1$-complemented. Let $W=\ell_\infty^3$, the   three-dimensional space with norm $\max(|x|,|y|,|z|)$. Let $V$ be the plane $x+y+z=0$. As you can see on the picture below, the intersection of $V$ with the unit ball of $W$ is a regular hexagon. Each vertex of this hexagon is a midpoint of an edge of the cube. The only way to keep the image of this edge within the hexagon is to project it into a point. But this can't happen, because these edges span $W$; the kernel of projection would be the entire space.
 
