This question already has an answer here:

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 $$\Vert f \Vert_Z \colon= \sup \left\{ \ \frac{ \vert f(x) \vert }{\Vert x \Vert} \ \colon \ x \in Z, \ x \neq 0 \ \right\} \ \mbox{ if } \ Z \neq \{\ 0 \ \}; \ \mbox{ otherwise } \ \Vert f \Vert_Z \colon= 0.$$ 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?


marked as duplicate by user147263, user99914, user228113, Asaf Karagila axiom-of-choice Jan 24 '16 at 6:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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? $\endgroup$ – DisintegratingByParts Jul 15 '15 at 17:28
  • $\begingroup$ @TrialAndError The obvious still requires some form of the Axiom of Choice $\endgroup$ – David C. Ullrich Jul 15 '15 at 17:47
  • $\begingroup$ @TrialAndError At least the obvious version of the obvious requires AC - Andreas has just explained how to avoid it. $\endgroup$ – David C. Ullrich Jul 15 '15 at 18:34
  • $\begingroup$ @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. $\endgroup$ – DisintegratingByParts Jul 15 '15 at 18:50
  • $\begingroup$ @DavidC.Ullrich : There are skeletons buried in the assumption of separable, too: for each $x \in X$, there exists a subsequence ... . $\endgroup$ – DisintegratingByParts Jul 15 '15 at 19:00

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:


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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Andreas Blass Jul 15 '15 at 18:21
  • $\begingroup$ @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... $\endgroup$ – David C. Ullrich Jul 15 '15 at 18:33
  • 1
    $\begingroup$ @SaaqibMahmuud We do that extension exactly as in the standard proof of the Hahn-Banach theorem. $\endgroup$ – David C. Ullrich Jul 15 '15 at 19:30
  • 1
    $\begingroup$ @SaaqibMahmuud There's nothing non-constructive in the part of the proof I was referring to. For the rest of it see Andreas' comment... $\endgroup$ – David C. Ullrich Jul 15 '15 at 20:13
  • 1
    $\begingroup$ @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.) $\endgroup$ – David C. Ullrich Jul 15 '15 at 20:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.