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Theorem: Let $q:X\rightarrow\mathbb{R}$ be a sublinear functional on a real linear space $X$. Let $M$ be a linear subspace of $X$ and suppose that $f:M\rightarrow\mathbb{R}$ is a linear functional such that $f(x)\leqslant q(x)$ for all $x\in M$. Then there exists a linear functional $F:X\rightarrow\mathbb{R}$ such that $F(x)\leqslant q(x)$ for all $x\in X$ and $F(x)=f(x)$ for all $x\in M$.

Proof: let $S$ be the collection of all pairs $(M_1,f_1)$, where $M_1$ is a linear subspace of $X$ such that $M\subseteq M_1$ and $f_1:M_1\rightarrow\mathbb{R}$ is a linear extension of $f$ satisfying $f_1(x)\leqslant q(x)$ for all $x\in M_1$. By the claim above, $S\neq\emptyset$. For any $(M_1,f_1),(M_2,f_2)\in S$, define $(M_1,f_1)\preccurlyeq(M_2,f_2)$ if $M_1\subseteq M_2$ and $f_2$ is a linear extension of $f_1$. Then $\preccurlyeq$ is a partial order on $S$. Let $C=\{(M_i,f_i):i\in I\}$ be a chain in $S$. Let $\tilde{M}=\bigcup\limits_{i\in I}M_i$. Then $\tilde{M}$ is a linear subspace of $X$. Define $\tilde{f}:\tilde{M}\rightarrow\mathbb{R}$ by $\tilde{f}(x)=f_i(x)$ if $x\in M_i$ for some $i\in I$.

My questions are:

  1. In Conways's book, $\tilde{M}$ is a linear subspace of $X$ since $C$ is a chain in $S$. I don't get it. $\tilde{M}$ is a linear subspace of $X$ isn't because each $M_i$ is a linear subspace of $X$? Why do we need $C$ is a chain in $S$?

  2. How to show $\tilde{f}$ is linear? My attempt is :

For all $x,y\in\tilde{M}$, there exists $j\in I$ such that $x,y\in M_j$. For all $\alpha\in\mathbb{R}$, we have $$\tilde{f}(x+\alpha y)=f_j(x+\alpha y)=f_j(x)+\alpha f_j(y)=\tilde{f}(x)+\alpha\tilde{f}(y).$$ Hence, $\tilde{f}:\tilde{M}\rightarrow\mathbb{R}$ is linear. But my professor said it is not clear at all!! Can someone help me point out which part is missing?

Thank you in advance!

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For 1, in general a union of subspaces is not a subspace. But it is, if the union is increasing.

Your argument for the linearity of $\bar f$ is fine. What you are not doing, and maybe that's what your prof expect, is showing that $\bar f$ is well-defined. That again requires that $C$ is a chain and that the $f_j$ are extensions.

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  • $\begingroup$ Thanks for your reply. Can you be more specific about a union of subspaces is not a subspace. But it is, if the union is increasing? And I have shown $\tilde{f}$ is well defined but I didn't write it down on this page. My proof about the linearity is fine right? $\endgroup$ – Answer Lee Apr 2 '18 at 3:39
  • $\begingroup$ As I said in the answer, your argument for linearity is fine. As for the union of subspaces, give it a try in $\mathbb R^2$. $\endgroup$ – Martin Argerami Apr 2 '18 at 4:44

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