**Step 1: Define the functional on a one-dimensional subspace.** Fix $v \in V \setminus \{0\}$ and set $W = \operatorname{span}\{v\}$. Define $f: W \to \mathbb{R}$ by $f(\lambda v) := \lambda \|v\|_V$. Then $f$ is linear, and $\|f\|_{W^*} = 1$ since $|f(\lambda v)| = |\lambda| \cdot \|v\|_V = \|\lambda v\|_V$ for all $\lambda \in \mathbb{R}$. **Step 2: Extend to all of $V$.** By the [Norm-Preserving Extension](/theorems/880), there exists $f_v \in V^*$ with $f_v|_W = f$ and $\|f_v\|_{V^*} = \|f\|_{W^*} = 1$. **Step 3: Verify.** We have $f_v(v) = f(v) = \|v\|_V$ and $\|f_v\|_{V^*} = 1$.