Consider the family of all possible extensions of $f$ , i.e. the set $\mathcal F$ of all pairings $(F,H)$ where $H$ is a vector subspace of $X$ containing $U$ and $F$ is a linear map $F\colon H \to K$ such that $F(u)=f(u)$ for all $u\in U$ and $\vert F(u)\vert \le p(u)$ for all $u\in H$ . $\mathcal F$ is naturally endowed with an partial order relation: given $(F_1,H_1),(F_2,H_2)\in \mathcal F$ we say that $(F_1,H_1)\le (F_2,H_2)$ iff $F_2$ is an extension of $F_1$ that is $H_1\subset H_2$ and $F_2(u)=F_1(u)$ for all $u\in H_1$ . We want to apply Zorn's Lemma to $\mathcal F$ so we are going to prove that every chain in $\mathcal F$ has an upper bound.

Let $(F_i,H_i)$ be the elements of a chain in $\mathcal F$ . Define $H=\bigcup_i H_i$ . Clearly $H$ is a vector subspace of $V$ and contains $U$ . Define $F\colon H \to K$ by ``merging'' all $F_i$ 's as follows. Given $u\in H$ there exists $i$ such that $u\in H_i$ : define $F(u)=F_i(u)$ . This is a good definition since if both $H_i$ and $H_j$ contain $u$ then $F_i(u)=F_j(u)$ in fact either $(F_i,H_i)\le (F_j,H_j)$ or $(F_j,H_j)\le (F_i,H_i)$ . Notice that the map $F$ is linear, in fact given any two vectors $u,v\in H$ there exists $i$ such that $u,v\in H_i$ and hence $F(\alpha u + \beta v) = F_i( \alpha u + \beta v) = \alpha F_i(u) + \beta F_i(v) = \alpha F(u) + \beta F(v)$ . The so constructed pair $(F,H)$ is hence an upper bound for the chain $(F_i,H_i)$ because $F$ is an extension of every $F_i$ .

Zorn's Lemma then assures that there exists a maximal element $(F,H)\in \mathcal F$ . To complete the proof we will only need to prove that $H=V$ .

Suppose by contradiction that there exists $v\in V\setminus H$ . Then consider the vector space $H'= H+ Kv=\{ u+tv\colon u\in H,\quad t\in K\}$ ($H'$ is the vector space generated by $H$ and $v$ ). Choose$$ \lambda = \sup_{x\in H}\{ F(x)-p(x-v)\}.$$ We notice that given any $x,y\in H$ it holds$$ F(x)-F(y) = F(x-y) \le p(x-y) = p (x-v+v-y) \le p(x-v) + p(y-v)$$ i.e.$$ F(x)-p(x-v) \le F(y) + p(y-v);$$ in particular we find that $\lambda < +\infty$ and for all $y\in H$ it holds$$ F(y)-p(y-v) \le \lambda \le F(y)+p(y-v).$$

Define $F'\colon H'\to K$ as follows:$$ F'(u+tv) = F(u) + t\lambda.$$

Clearly $F'$ is a linear functional. We have$$ \lvert F'(u+tv)\rvert = \lvert F(u) + t\lambda \rvert = \lvert t\rvert \, \lvert F(u/t) + \lambda \rvert$$ and by letting $y=-u/t$ by the previous estimates on $\lambda$ we obtain$$ F(u/t) + \lambda \le F(u/t) + F(-u/t) + p(-u/t-v ) = p(u/t+v)$$ and$$ F(u/t) + \lambda \ge F(u/t) + F(-u/t) - p(-u/t -v) = -p(u/t+v)$$ which together give$$ \lvert F(u/t) + \lambda \rvert \le p(u/t+v)$$ and hence$$ \lvert F'(u+tv) \rvert \le \lvert t\rvert p(u/t+v) = p(u+tv).$$

So we have proved that $(F',H')\in\mathcal F$ and $(F',H')> (F,H)$ which is a contradiction.