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Am reading Kreyszig's proof of the Hahn-Banach Theorem on page 214:

Let $X$ be a real vector space and $p$ a sublinear functional on $X$. Let $f$ be a linear functional defined on a subspace $Z$ of $X$ which satisfies

$$f(x)\leq p(x) \hspace{1cm} \text{for all } x\in Z$$

Then $f$ has a linear extension $\tilde{f}$ on $X$ satisfying

$$\tilde{f}(x)\leq p(x) \hspace{1cm} \text{for all } x\in X$$

The proof reveals that, given $y_1\in X\setminus Z$, it is possible to extend $f$ to the subspace spanned by $Z\cup \{y_1\}$ while satisfying the required majorization on this subspace. At the end of the proof, Kreyszig explains that we can possibly avoid using Zorn's lemma with the following argument:

First extend $f$ to $Z_1=span \{ Z\cup \{y_1\}\}$ as above and call the extension $\tilde{f}_1$. Inductively suppose $\tilde{f}_n$ is a linear extension of $\tilde{f}_{n-1}$ satisfying the required majorization on its domain $Z_n$. If $X=Z_n$ then let $\tilde{f}_{n+1}:=\tilde{f}_n$. Otherwise let $y_{n+1}\in X\setminus Z_{n}$ and let $\tilde{f}_{n+1}$ be the extension of $\tilde{f}_n$ to $Z_{n+1}=span \{ Z_n\cup \{y_{n+1}\}\}$ (i.e. the extension mentionned above and it satisfies the required majorization on $Z_{n+1}$). By the principle of recursive definition $\tilde{f}_n$ is defined for all $n\geq 1$. Now let $\tilde{f}:=\cup_{n=1}^\infty \tilde{f}_n$. We check that $\tilde{f}$ is a well-defined linear functional on the subspace $Z_{\infty}:=\cup_{n=1}^\infty Z_n$ and that it satisfies the required majoriation on $Z_{\infty}$.

If it so happens that $Z_{\infty}=X$ then we are done. But how can we know ex-ante whether this procedure will be sufficient? For example, if $X\setminus Z$ is finite then we must have $Z_{\infty}=X$, for otherwise $\{y_n\}$ would be an infinite subset of $X\setminus Z$.

Are there other cases?

Alphie
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  • Is there an assumption that $X$ is separable? – Asaf Karagila Mar 30 '21 at 16:17
  • @AsafKaragila Ok I looked at the post you linked for the case $X$ separable. The answer there uses the axiom of choice. But isn't just a recursive definition of functions sufficient? I mean I don't pick an extension but construct a definite one. – Alphie Mar 30 '21 at 16:33
  • Did you follow on the duplicate's duplicate? https://math.stackexchange.com/questions/1623210/hahn-banach-theorem-for-separable-spaces-without-zorns-lemma the point is that you can choose a canonical point from a closed subset of $\Bbb R$. And once you defined it on a dense subset you can take the unique extension to the space. – Asaf Karagila Mar 30 '21 at 16:35
  • @AsafKaragila Ok I see thank you. And what about the case of $X$ finite dimensional? The procedure above should work right? – Alphie Mar 30 '21 at 16:44
  • No choice is needed for finite dimensional spaces since ZF proves AC for finite families of sets. – Asaf Karagila Mar 30 '21 at 16:49
  • @AsafKaragila In this case I would extend the basis ${v_1,\dots,v_k}$ of $Z=:Z_k$ to a basis ${v_1,\dots,v_n}$ for $X$ and extend recursively $f$ to $Z_i=span { Z_{i-1}\cup {v_{i}}}$ for $i=k+1,\dots, n$? – Alphie Mar 30 '21 at 17:10
  • Yes, that's fine. Not to mention that finite dimensional spaces (over $\Bbb R$) are separable. – Asaf Karagila Mar 30 '21 at 17:11

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