[proofplan] We invoke [Zorn's Lemma](/theorems/???) on the poset of dominated linear extensions of $g$ to subspaces containing $Y$, ordered by extension. The pair $(g, Y)$ shows the poset is non-empty, and the union along any chain produces an upper bound, so a maximal element $(f, Z)$ exists. The crux is to show that the domain of any maximal element must be all of $X$: if $Z \ne X$ we extend $f$ to a strictly larger subspace $Z_1 = \operatorname{span}(Z \cup \{x_0\})$ by selecting a value $\alpha = f_1(x_0)$ inside an interval whose non-emptiness comes from subadditivity of $p$ and linearity of $f$. This contradicts maximality, forcing $Z = X$. [/proofplan]

[step:Set up the poset of dominated linear extensions and verify it is non-empty] Define the set \begin{align*} \mathcal{P} := \{(h, Z) : Y \subset Z \subset X \text{ is a subspace},\ h: Z \to \mathbb{R} \text{ linear},\ h|_Y = g,\ h(z) \le p(z)\ \forall z \in Z\}. \end{align*} Order $\mathcal{P}$ by declaring $(h, Z) \preceq (h', Z')$ iff $Z \subseteq Z'$ and $h'|_Z = h$. This is reflexive, antisymmetric (if $Z = Z'$ and the restriction agrees, then $h = h'$), and transitive, so $(\mathcal{P}, \preceq)$ is a partially ordered set. The hypothesis $g(y) \le p(y)$ for $y \in Y$ shows that $(g, Y) \in \mathcal{P}$, hence $\mathcal{P} \ne \varnothing$. [/step]

[step:Verify every chain in $\mathcal{P}$ has an upper bound] Let $\mathcal{C} = \{(h_i, Z_i)\}_{i \in I}$ be a chain in $(\mathcal{P}, \preceq)$. Define \begin{align*} Z &:= \bigcup_{i \in I} Z_i, \\ h: Z &\to \mathbb{R}, \qquad h(z) := h_i(z) \text{ if } z \in Z_i. \end{align*} We check that $(h, Z) \in \mathcal{P}$ and dominates every $(h_i, Z_i)$. **$Z$ is a subspace.** Given $z_1, z_2 \in Z$ and scalars $\lambda_1, \lambda_2 \in \mathbb{R}$, there are $i_1, i_2 \in I$ with $z_k \in Z_{i_k}$. Since $\mathcal{C}$ is a chain, either $Z_{i_1} \subseteq Z_{i_2}$ or $Z_{i_2} \subseteq Z_{i_1}$; without loss of generality the former, so both $z_1, z_2 \in Z_{i_2}$ and hence $\lambda_1 z_1 + \lambda_2 z_2 \in Z_{i_2} \subseteq Z$. **$h$ is well-defined.** If $z \in Z_i \cap Z_j$, by the chain condition one of $Z_i, Z_j$ is contained in the other, say $Z_i \subseteq Z_j$. Since $(h_i, Z_i) \preceq (h_j, Z_j)$, we have $h_j|_{Z_i} = h_i$, so $h_j(z) = h_i(z)$. **$h$ is linear.** For $z_1, z_2 \in Z$, find $i \in I$ with $z_1, z_2 \in Z_i$ as above; linearity follows from linearity of $h_i$. **$h|_Y = g$.** For $y \in Y \subseteq Z_i$ for any $i$, $h(y) = h_i(y) = g(y)$. **$h \le p$ on $Z$.** For $z \in Z$, choose $i$ with $z \in Z_i$; then $h(z) = h_i(z) \le p(z)$ since $(h_i, Z_i) \in \mathcal{P}$. Hence $(h, Z) \in \mathcal{P}$, and by construction $(h_i, Z_i) \preceq (h, Z)$ for every $i$. [/step]

[step:Apply Zorn's Lemma to obtain a maximal dominated extension $(f, Z)$] By the previous step every chain in $(\mathcal{P}, \preceq)$ has an upper bound, so [Zorn's Lemma](/theorems/???) applies and yields a maximal element $(f, Z) \in \mathcal{P}$. [/step]

[step:Set up the one-step extension to a strictly larger subspace]Suppose for contradiction that $Z \ne X$. Pick $x_0 \in X \setminus Z$ and set \begin{align*} Z_1 := \operatorname{span}(Z \cup \{x_0\}) = \{z + \lambda x_0 : z \in Z,\ \lambda \in \mathbb{R}\}. \end{align*} Since $x_0 \notin Z$, the decomposition $z + \lambda x_0$ of any element of $Z_1$ is unique. Therefore any linear $f_1: Z_1 \to \mathbb{R}$ extending $f$ is determined by the single scalar \begin{align*} \alpha := f_1(x_0), \end{align*} via $f_1(z + \lambda x_0) = f(z) + \lambda \alpha$.[/step]

[guided]The construction so far has produced a maximal dominated extension $(f, Z)$. To reach a contradiction with maximality, we need to show that whenever $Z \ne X$ we can produce a strictly larger dominated extension. The natural way to enlarge $Z$ by one dimension is to adjoin a single vector $x_0 \in X \setminus Z$, forming the span $Z_1 = \operatorname{span}(Z \cup \{x_0\})$. Why is the decomposition $z + \lambda x_0$ unique? If $z + \lambda x_0 = z' + \lambda' x_0$ with $z, z' \in Z$, then $(\lambda - \lambda') x_0 = z' - z \in Z$. If $\lambda \ne \lambda'$, this would give $x_0 = (z' - z)/(\lambda - \lambda') \in Z$, contradicting $x_0 \notin Z$. So $\lambda = \lambda'$ and hence $z = z'$. Consequently, **a linear extension $f_1$ of $f$ to $Z_1$ is parametrised by a single real number $\alpha = f_1(x_0)$**, with the formula $f_1(z + \lambda x_0) = f(z) + \lambda \alpha$. The remaining task is to choose $\alpha$ so that $f_1(z + \lambda x_0) \le p(z + \lambda x_0)$ for all $z \in Z$, $\lambda \in \mathbb{R}$.[/guided]

[step:Reduce the domination condition to an interval constraint on $\alpha$]We seek $\alpha \in \mathbb{R}$ such that \begin{align*} f(z) + \lambda \alpha \le p(z + \lambda x_0) \qquad \text{for all } z \in Z,\ \lambda \in \mathbb{R}. \end{align*} *Case $\lambda = 0$:* the inequality becomes $f(z) \le p(z)$, which holds since $(f, Z) \in \mathcal{P}$. *Case $\lambda > 0$:* divide by $\lambda$ and substitute $z_2 := z/\lambda \in Z$. Using positive homogeneity $p(\lambda u) = \lambda p(u)$ for $\lambda > 0$ and linearity of $f$, the inequality becomes \begin{align*} \alpha \le p(z_2 + x_0) - f(z_2). \end{align*} *Case $\lambda < 0$:* set $\mu := -\lambda > 0$ and $z_1 := z/\mu \in Z$. Then $z + \lambda x_0 = \mu(z_1 - x_0)$, so by positive homogeneity $p(z + \lambda x_0) = \mu\, p(z_1 - x_0)$, while linearity of $f$ gives $f(z) + \lambda\alpha = \mu f(z_1) - \mu \alpha$. Dividing by $\mu > 0$ and rearranging yields \begin{align*} f(z_1) - p(z_1 - x_0) \le \alpha. \end{align*} Combining the two non-trivial cases, the existence of $\alpha$ amounts to the existence of a real number satisfying \begin{align*} A := \sup_{z_1 \in Z}\bigl(f(z_1) - p(z_1 - x_0)\bigr) \;\le\; \alpha \;\le\; \inf_{z_2 \in Z}\bigl(p(z_2 + x_0) - f(z_2)\bigr) =: B. \end{align*}[/step]

[guided]The condition $f_1(z + \lambda x_0) \le p(z + \lambda x_0)$ must hold for *every* pair $(z, \lambda) \in Z \times \mathbb{R}$. The strategy is to use positive homogeneity of $p$ to **reduce all $\lambda \ne 0$ cases to $\lambda = \pm 1$**, then read off a single interval constraint on $\alpha$. For $\lambda > 0$: divide by $\lambda$. Setting $z_2 = z/\lambda \in Z$, \begin{align*} \frac{f(z) + \lambda \alpha}{\lambda} = f(z_2) + \alpha, \qquad \frac{p(z + \lambda x_0)}{\lambda} = p\!\left(\tfrac{z}{\lambda} + x_0\right) = p(z_2 + x_0), \end{align*} where we used positive homogeneity $p(\lambda u) = \lambda p(u)$ for $\lambda > 0$ and linearity of $f$. The inequality becomes $\alpha \le p(z_2 + x_0) - f(z_2)$. For $\lambda < 0$: set $\mu = -\lambda > 0$ and $z_1 = z/\mu \in Z$. Then $z + \lambda x_0 = \mu z_1 - \mu x_0 = \mu(z_1 - x_0)$, and positive homogeneity gives $p(z + \lambda x_0) = \mu\, p(z_1 - x_0)$. Linearity of $f$ gives $f(z) + \lambda \alpha = \mu f(z_1) - \mu\alpha$. Dividing by $\mu > 0$ and rearranging: $f(z_1) - p(z_1 - x_0) \le \alpha$. (Note that we cannot use "$p(-u) = p(u)$"; subadditive positively homogeneous functions need *not* be even — that holds only for semi-norms.) Taking the supremum over all $z_1 \in Z$ in the first inequality and the infimum over all $z_2 \in Z$ in the second yields \begin{align*} A := \sup_{z_1 \in Z}\bigl(f(z_1) - p(z_1 - x_0)\bigr) \;\le\; \alpha \;\le\; \inf_{z_2 \in Z}\bigl(p(z_2 + x_0) - f(z_2)\bigr) =: B. \end{align*} The existence of $\alpha \in [A, B]$ is equivalent to $A \le B$, which is what we must verify next.[/guided]

[step:Show the supremum is at most the infimum via subadditivity of $p$] We claim that for all $z_1, z_2 \in Z$, \begin{align*} f(z_1) - p(z_1 - x_0) \le p(z_2 + x_0) - f(z_2). \end{align*} By linearity of $f$: \begin{align*} f(z_1) + f(z_2) = f(z_1 + z_2). \end{align*} Since $(f, Z) \in \mathcal{P}$, we have $f \le p$ on $Z$, so \begin{align*} f(z_1 + z_2) \le p(z_1 + z_2) = p\bigl((z_1 - x_0) + (z_2 + x_0)\bigr) \le p(z_1 - x_0) + p(z_2 + x_0), \end{align*} where the last inequality is subadditivity of $p$. Rearranging: \begin{align*} f(z_1) - p(z_1 - x_0) \le p(z_2 + x_0) - f(z_2), \end{align*} as claimed. Taking the supremum over $z_1 \in Z$ on the left and the infimum over $z_2 \in Z$ on the right gives $A \le B$. [/step]

[step:Pick $\alpha$ in the interval to obtain a strict extension and contradict maximality] By the previous step the interval $[A, B]$ is non-empty, so we may select any $\alpha \in [A, B]$. Define \begin{align*} f_1: Z_1 &\to \mathbb{R}, \\ z + \lambda x_0 &\mapsto f(z) + \lambda \alpha. \end{align*} Then $f_1$ is linear (immediate from uniqueness of the decomposition $z + \lambda x_0$), $f_1|_Z = f$ (set $\lambda = 0$), and the derivation in the previous step shows $f_1(z + \lambda x_0) \le p(z + \lambda x_0)$ for all $\lambda \in \mathbb{R}$, $z \in Z$. Hence $(f_1, Z_1) \in \mathcal{P}$. Moreover $Z \subsetneq Z_1$ since $x_0 \in Z_1 \setminus Z$, so $(f, Z) \prec (f_1, Z_1)$, contradicting the maximality of $(f, Z)$. [/step]

[step:Conclude $Z = X$ and assemble the dominated extension] The contradiction in the previous step shows that the assumption $Z \ne X$ was false; hence $Z = X$. Therefore $f: X \to \mathbb{R}$ is a linear functional with $f|_Y = g$ and $f(x) \le p(x)$ for all $x \in X$. This is the dominated extension claimed by the theorem. Non-uniqueness of the extension is visible from the construction: at each one-step extension above, $\alpha$ may be chosen to be any value in the (potentially non-degenerate) interval $[A, B]$. [/step]