Dense subspace of a normed linear space

Question

If $V \subset U$ both normed linear spaces, and $V$ is not dense in $U$, then is it possible to construct a linear functional $l$ on $U$ so that $l = 0$ on $V$ but $l$ is not identically $0$? Do we require the Hahn-Banach Theorem to extend this functional to all of $U$?

Answer 1

Take the quotient space $X=U/\bar{V}$. Take a non-zero functional $f$ on $X$. Let $\pi : U \to X$ be the projection map. Then

$$ f \circ \pi $$ is what you want.

Answer 2

Take $x\in U$ \ $\bar V.$ Let $d=\inf \{\|x-y\|:y\in \bar V\}.$

Let $W$ be the vector subspace generated by $\{x\}\cup \bar V$.

$W$ is closed in $U.$

For scalar $a$ and for $y\in \bar V$, let $f(ax+y)=ad.$

Then $f$ is a well-defined linear functional on $W,$ with norm $1$.

Hahn-Banach extends $f$ to a functional on $U$ with norm $1$.