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Version 3 |
| 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 $u\in H$. |
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 |
$\mathcal F$ 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 |
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 $u\in H_1$. |
$H_1\subset H_2$ 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. |
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. |
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| 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 $(F_j,H_j)\le (F_i,H_i)$. |
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 $F(\alpha u + \beta v) = F_i( \alpha u + \beta v) = \alpha F_i(u) + \beta F_i(v) = \alpha F(u) + \beta F(v)$. |
Clearly the so constructed pair $(F,H)$ is an upper bound for the chain $(F_i,H_i)$ since $F$ is an extension of every $F_i$. |
| The so constructed pair $(F,H)$ is hence an upper bound for the chain |
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| $(F_i,H_i)$ because $F$ is an extension of every $F_i$. |
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| 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 $H=V$. |
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 $H=V$. |
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| Suppose by contradiction that there exists $v\in V\setminus H$. Then consider the vector space $H'= H+ Kv=\{ u+tv\colon |
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 $v$). |
u\in H,\quad t\in K\}$ ($H'$ is the vector space generated by $H$ and $v$). |
| Choose |
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| \[ |
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| \lambda = \sup_{x\in H}\{ F(x)-p(x-v)\}. |
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| \] |
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| We notice that given any $x,y\in H$ it holds |
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| \[ |
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| F(x)-F(y) = F(x-y) \le p(x-y) = p (x-v+v-y) \le p(x-v) + p(y-v) |
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| \] |
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| i.e. |
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| \[ |
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| F(x)-p(x-v) \le F(y) + p(y-v); |
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| \] |
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| in particular we find that $\lambda < +\infty$ and for all $y\in H$ it holds |
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| \[ |
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| F(y)-p(y-v) \le \lambda \le F(y)+p(y-v). |
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| \] |
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| Define $F'\colon H'\to K$ as follows: |
Define $F'\colon H'\to K$ as follows: |
| \[ |
F'(u+tv) = F(u) |
| F'(u+tv) = F(u) + t\lambda. |
ly $H$ is a vector subspace of $V$ and contains $U$. Define $F\colon H \to K$ by ``merging'' all $F_i$'s as follows. Given $u\in H$ there exists $i$ such that $u\in H_i$: define $F(u)=F_i(u)$. This is a good definition since if both $H_i$ and $H_j$ contain $u$ then $F_i(u)=F_j(u)$ since either $(F_i,H_i)\le (F_j,H_j)$ or $(F_j,H_j)\le (F_i,H_i)$. |
| \] |
Clearly the so constructed pair $(F,H)$ is an upper bound for the chain $(F_i,H_i)$ since $F$ is an extension of every $F_i$. |
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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 $H=V$. |
| Clearly $F'$ is a linear functional. |
Suppose by contradiction that there exists $v\in V\setminus H$. Then consider the vector space $H'= H+ Kv=\{ u+tv\colon |
| We have |
u\in H,\quad t\in K\}$ ($H'$ is the vector space generated by $H$ and $v$). |
| \[ |
Define $F'\colon H'\to K$ as follows: |
| \lvert F'(u+tv)\rvert = \lvert F(u) + t\lambda \rvert |
F'(u+tv) = F(u). |
| = \lvert t\rvert \, \lvert F(u/t) + \lambda \rvert |
Clearly $F'$ is a linear functional and it holds $\vert F'(u+tv)\vert = \vert F(u) \vert \le p(u) \le p(u+tv)$. So $(F',H')\in\mathcal F$ and $(F',H')> (F,H)$ which is a contradiction. |
| \] |
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| and by letting $y=-u/t$ by the previous estimates on $\lambda$ we obtain |
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| \[ |
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| F(u/t) + \lambda \le F(u/t) + F(-u/t) + p(-u/t-v ) = p(u/t+v) |
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| \] |
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| and |
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| \[ |
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| F(u/t) + \lambda \ge F(u/t) + F(-u/t) - p(-u/t -v) = -p(u/t+v) |
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| \] |
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| which together give |
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| \[ |
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| \lvert F(u/t) + \lambda \rvert \le p(u/t+v) |
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| \] |
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| and hence |
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| \[ |
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| \lvert F'(u+tv) \rvert \le \lvert t\rvert p(u/t+v) = p(u+tv). |
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| \] |
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| So we have proved that $(F',H')\in\mathcal F$ and $(F',H')> (F,H)$ which is a contradiction. |
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