The proof proceeds in two stages. First we handle the codimension-one extension (Proposition), then we use Zorn's lemma to extend to the full space. **Step 1: Codimension-one extension.** Suppose $W \subset V$ has codimension $1$, so $V = W \oplus \operatorname{span}\{v_0\}$ for some $v_0 \in V \setminus W$. Every $v \in V$ can be written uniquely as $v = w + av_0$ with $w \in W$ and $a \in \mathbb{R}$. Since $\tilde{f}$ must be linear and extend $f$, we need $\tilde{f}(v) = f(w) + a\tilde{f}(v_0)$. So it suffices to choose $\tilde{f}(v_0)$ such that $\tilde{f}(v) \le p(v)$ for all $v \in V$. For $a > 0$: the condition $f(w) + a\tilde{f}(v_0) \le p(w + av_0)$ reduces (dividing by $a$ and replacing $w/a$ by $w' \in W$) to $\tilde{f}(v_0) \le p(w' + v_0) - f(w')$ for all $w' \in W$. For $a < 0$: dividing by $-a > 0$ gives $\tilde{f}(v_0) \ge -p(w' - v_0) + f(w')$ for all $w' \in W$. Combining, we need: \begin{align*} \sup_{w_1 \in W} \left[-p(w_1 - v_0) + f(w_1)\right] \le \tilde{f}(v_0) \le \inf_{w_2 \in W} \left[p(w_2 + v_0) - f(w_2)\right]. \end{align*} This interval is non-empty because for all $w_1, w_2 \in W$: \begin{align*} f(w_1) + f(w_2) = f(w_1 + w_2) \le p(w_1 + w_2) = p((w_1 - v_0) + (w_2 + v_0)) \le p(w_1 - v_0) + p(w_2 + v_0) \end{align*} which rearranges to $-p(w_1 - v_0) + f(w_1) \le p(w_2 + v_0) - f(w_2)$. **Step 2: Zorn's lemma argument.** Define the partially ordered set \begin{align*} S := \{(\tilde{f}, \tilde{V}) : W \subset \tilde{V} \subset V \text{ subspaces}, \; \tilde{f}: \tilde{V} \to \mathbb{R} \text{ linear}, \; \tilde{f}|_W = f, \; \tilde{f}(v) \le p(v) \; \forall v \in \tilde{V}\} \end{align*} ordered by $(\tilde{f}_1, \tilde{V}_1) \le (\tilde{f}_2, \tilde{V}_2)$ if $\tilde{V}_1 \subset \tilde{V}_2$ and $\tilde{f}_2|_{\tilde{V}_1} = \tilde{f}_1$. Every totally ordered chain $\{(\tilde{f}_\alpha, \tilde{V}_\alpha)\}_{\alpha \in A}$ has an upper bound: set $\tilde{V} = \bigcup_\alpha \tilde{V}_\alpha$ and define $\tilde{f}(x) = \tilde{f}_\alpha(x)$ for any $\alpha$ with $x \in \tilde{V}_\alpha$ (well-defined by the total ordering). Then $(\tilde{f}, \tilde{V}) \in S$ is an upper bound. By Zorn's lemma, $S$ has a maximal element $(\tilde{f}, \tilde{W})$. If $\tilde{W} \ne V$, then there exists $v_0 \in V \setminus \tilde{W}$, and Step 1 provides a strict extension, contradicting maximality. Hence $\tilde{W} = V$.