**Step 1: Apply the Hahn-Banach theorem.** Define $p: V \to \mathbb{R}$ by $p(x) := \|f\|_{W^*} \cdot \|x\|_V$. This [function](/page/Function) is positively homogeneous (from the homogeneity of the norm) and subadditive (from the triangle inequality). For all $w \in W$: $f(w) \le |f(w)| \le \|f\|_{W^*} \cdot \|w\|_V = p(w)$. By the [Hahn-Banach Theorem](/theorems/879), there exists a linear extension $\tilde{f}: V \to \mathbb{R}$ with $\tilde{f}|_W = f$ and $\tilde{f}(v) \le p(v) = \|f\|_{W^*} \cdot \|v\|_V$ for all $v \in V$. **Step 2: Obtain the absolute-value bound.** Applying the inequality to both $v$ and $-v$: \begin{align*} \tilde{f}(v) \le \|f\|_{W^*} \cdot \|v\|_V \quad \text{and} \quad \tilde{f}(-v) \le \|f\|_{W^*} \cdot \|v\|_V. \end{align*} Since $\tilde{f}(-v) = -\tilde{f}(v)$, the second inequality gives $-\tilde{f}(v) \le \|f\|_{W^*} \cdot \|v\|_V$. Combining: $|\tilde{f}(v)| \le \|f\|_{W^*} \cdot \|v\|_V$ for all $v \in V$. **Step 3: Verify the norm equality.** The bound $|\tilde{f}(v)| \le \|f\|_{W^*} \cdot \|v\|_V$ gives $\|\tilde{f}\|_{V^*} \le \|f\|_{W^*}$. Conversely, since $\tilde{f}|_W = f$ and $W \subset V$: \begin{align*} \|\tilde{f}\|_{V^*} = \sup_{\substack{v \in V \setminus \{0\}}} \frac{|\tilde{f}(v)|}{\|v\|_V} \ge \sup_{\substack{w \in W \setminus \{0\}}} \frac{|f(w)|}{\|w\|_V} = \|f\|_{W^*}. \end{align*} Combining gives $\|\tilde{f}\|_{V^*} = \|f\|_{W^*}$.