4

Here's Theorem 4.3-2 (i.e. the Hahn Banach theorem for normed spaces):

Let $f$ be a bounded linear functional defined on a subspace $Z$ of a normed space $X$. Then there exists a bounbed linear functional $\tilde{f}$ on $X$ such that $$\tilde{f}(x) = f(x) \ \mbox{ for all } \ x\in Z,$$ and $$\Vert \tilde{f} \Vert_X = \Vert f \Vert_Z, $$ where $$\lVert f \rVert_Z := \sup \left\{ \frac{ \lvert f(x) \rvert }{\lVert x \rVert} \colon x \in Z, x \neq 0 \right\} \ \mbox{ if } Z \neq \{ 0 \}; \\ \mbox{ otherwise } \lVert f \rVert_Z := 0.$$ And, $$\Vert \tilde{f} \Vert_X \colon= \sup \left\{ \ \frac{ \vert \tilde{f}(x) \vert }{\Vert x \Vert} \ \colon \ x \in X, \ x \neq 0 \ \right\}.$$

I think I'm clear about the proof of this beautiful result. It uses the Hahn Banach Theorem for Complex Vector Spaces, which uses the Hahn Banach Theorem for Real Vector spaces, and the latter uses the Zorn's lemma.

Now if $X$ is a separable normed space, then is there a proof of the above result that doesn't involve the use of the Zorn's lemma?

Saaqib Mahmood
  • 26,762
  • Have you tried the obvious where you start with a countable dense subet ${ x_n }{n=1}^{\infty}$ and extend one vector at a time to the subspace $\mathcal{M}{k}$ spanned by ${ x_1,x_2,\cdots,x_k}$? Or did you try that and get stuck? Any thoughts? – Disintegrating By Parts Jul 15 '15 at 17:28
  • @TrialAndError The obvious still requires some form of the Axiom of Choice – David C. Ullrich Jul 15 '15 at 17:47
  • @TrialAndError At least the obvious version of the obvious requires AC - Andreas has just explained how to avoid it. – David C. Ullrich Jul 15 '15 at 18:34
  • @DavidC.Ullrich : But not Zorn's lemma, which was the focus of the question. You're basically dealing with countable in the separable case, and the use of AC doesn't require Zorn's lemma. – Disintegrating By Parts Jul 15 '15 at 18:50
  • @DavidC.Ullrich : There are skeletons buried in the assumption of separable, too: for each $x \in X$, there exists a subsequence ... . – Disintegrating By Parts Jul 15 '15 at 19:00
  • @TrialAndError Asking whether something requires AC is meaningful mathematically - the something is presumably a theorem of ZFC and the question is whether it's a theorem of ZF. Given that AC is equivalent to Zorn, I don't see how to make anything precise of the notion "requires AC but not Zorn". In some sense it's true what you say, the use of AC does not require Zorn. But in that sense nothing requires Zorn, or anything else. – David C. Ullrich Jul 15 '15 at 19:06
  • @TrialAndError If by "skeletons" you mean that the last step, extending the locally uniformly continuous function from a dense subspace to the whole space, requires AC, no it really doesn't. Say $S$ is a dense subset of the metric space $X$, $g:S\to\mathbb R$ is uniformly continuous, and we want to extend $g$ to $X$. We can define $f(x)$ by "If $(x_n)$ is a sequence in $S$ and $x_n\to x$ then...". No need to "choose" $(x_n)$, because we get the same $f(x)$ regardless of which sequence we use. – David C. Ullrich Jul 15 '15 at 19:12
  • @DavidC.Ullrich : You didn't read the question, did you? He asked for a proof that didn't use Zorn's lemma. I outlined one. The assumption of separability means you don't have to choose a sequence, and I wanted to point that out. – Disintegrating By Parts Jul 15 '15 at 19:14
  • @TrialAndError, how do we extend $f$ from $Z$ to $M_1$ without using the Hahn Banach theorems so that the extension has the same norm as $f$? Could you please elaborate? I've tried, but haven't been able to keep the norm unchanged. – Saaqib Mahmood Jul 15 '15 at 19:30
  • You use the process of the Hahn-Banach theorem to extend one vector at a time. The proof for that level of extension is basic and does not require Zorn's lemma. – Disintegrating By Parts Jul 15 '15 at 20:02
  • @TrialAndError, I would appreciate if you could take time posting an answer. As far as I can see it, the Hahn Banach theorem is a non-constructive result, while in this particuar case we require a constructive approach to extend $f$ from $Z$ to $M_1$. – Saaqib Mahmood Jul 15 '15 at 20:15
  • @SaaqibMahmuud : If you have not read a proof of Hahn-Banach, then you need to do that. In the proof, there is a standard argument for extending by one vector. – Disintegrating By Parts Jul 15 '15 at 20:57

1 Answers1

3

Edit: This is all totally wrong. Well, the proof I had in mind uses AC although it might not be obvious that it does so, so it seems worth explaining that. But it's been pointed out that AC can be avoided. Thanks to Andreas Blass (some time ago I said he should post an answer to replace mine, he never did.) First the original, explaining where AC is used, then how to avoid it:

Original:

The separable case still requires some form of the Axiom of Choice, although it's less clear why.

Say $(x_n)$ is a dense sequence of elements of $X$. Say $Z_n$ is the span of $Z$ and $x_1,\dots,x_n$.

Now you simply extend your functional to $Z_1,$ then to $Z_2$, etc. You find you've extended it to the union of the $Z_n$, which is dense in $X$, and now since the extension is uniformly continuous on bounded sets you're done.

Where does AC come in? Well, first you "choose" an extension to $Z_1$, of the possibly infinite number of possible extensions. Then you "choose" an extension to $Z_2$... you have infinitely many choices to make.

EDIT: But you don't really need AC. Start by enumerating the rationals $r_1,\dots$. At each stage in the construction the proof gives you an interval $[a,b]$ such that if you define the value of the functional at $x_n$ to be anything in $[a,b]$ then the norm does not increase. If $a=b$ there is no choice to be made. If $a<b$ choose the $r_j\in[a,b]$ with $j$ minimal.

David C. Ullrich
  • 89,985
  • 2
    I think you can avoid choice here. Start by fixing an enumeration of the rationals. When extending your functional $f$ from $Z_{n-1}$ to $Z_n$, essentailly you're just picking a value for $f(x_n)$, and the usual proof gives an interval of possible values to choose from. If the interval is degenerate, i.e., if it contains just a single number, then of course set $f(x_n)$ equal to that number. Otherwise, the interval contains some rational numbers; set $f(x_n)$ equal to the first of those rational numbers, where "first" refers to the enumeration of the rationals that you fixed at the beginning. – Andreas Blass Jul 15 '15 at 18:21
  • @AndreasBlass Ah, very good. So I learned something today... I want to delete my answer. But that would make your comment disappear, and your comment is the only correct answer so far. My vote would be you post an answer, then remind me so I can delete mine. Up to you of course... – David C. Ullrich Jul 15 '15 at 18:33
  • @DavidC.Ullrich, how do we extend $f$ from $Z$ to $Z_1$ without using the Hahn Banach theorems so that the extension has the same norm as $f$? Could you please elaborate? – Saaqib Mahmood Jul 15 '15 at 19:22
  • 1
    @SaaqibMahmuud We do that extension exactly as in the standard proof of the Hahn-Banach theorem. – David C. Ullrich Jul 15 '15 at 19:30
  • @DavidC.Ullrich, the Hahn Banach theorem is a non-constructive result, isn't it? I suppose here we need a constructive approach. – Saaqib Mahmood Jul 15 '15 at 20:11
  • 1
    @SaaqibMahmuud There's nothing non-constructive in the part of the proof I was referring to. For the rest of it see Andreas' comment... – David C. Ullrich Jul 15 '15 at 20:13
  • @DavidC.Ullrich, Can you please edit your answer to explicitly show how we can extend $f$ from $Z$ to $Z_1$ so that the norm remains unchanged? – Saaqib Mahmood Jul 15 '15 at 20:17
  • 1
    @SaaqibMahmuud Don't you have a proof of the Hahn-Banach theorem in some book? (That part of the argument doesn't use Zorn's lemma.) – David C. Ullrich Jul 15 '15 at 20:22
  • @DavidC.Ullrich I have edited your post - feel free to revert the edit if you disagree with it. I thought that it is good to make comments more obvious for readers of this post. And since Andreas Blass did not post his comments as a separate answer, this seemed as the cleanest solution to me. – Martin Sleziak Jan 24 '16 at 03:14
  • Other possibility how to deal with posts answered in comments is to make a separate CW answer. See the discussions on meta here and here. – Martin Sleziak Jan 24 '16 at 03:15
  • @MartinSleziak Thanks. See my comment - long ago I suggested he make a post to replace mine, he didn't and I forgot about it. Much cleaner now. – David C. Ullrich Jan 24 '16 at 04:55