I was reading about the Hahn-Banach Theorem, its many versions and their proofs. It's known that in the proofs we need Zorn's Lemma. But in the book that I'm reading, the author said if $X$ is a separable space then it's possible to prove the Hahn-Banach Theorem without the Zorn's Lemma.

How can we show there is a suitable extension of a linear functional without Zorn's lemma?


2 Answers 2


Note: I completely rewrote my previous attempt - posted in another (now deleted) answer, since the previous version was incorrect. Thanks to Asaf Karagila for pointing out the problem with my previous proof. And also to Eric Wofsey for simplifying some steps in the proof.

Let us hope that this time I have avoided mistakes. The formulation of the version of HBT which I am trying to prove in ZF is adapted from the claims mentioned in this paper. Some remarks explaining what this paper proves about the relation between ZF, AC and HBT for separable spaces can be found below.

As the OP wrote, there are several formulations of Hahn-Banach Theorem. So maybe it is good to start by clearly stating this result.

Hahn-Banach Theorem. Let $X$ be a vector space and let $p:X\to{\mathbb R}$ be any sublinear function. Let $M$ be a vector subspace of $X$ and let $f:M\to{\mathbb R}$ be a linear functional dominated by $p$ on $M$. Then there is a linear extension $\widehat{f}$ of $f$ to $X$ that is dominated by $p$ on $X$.

The formulation that $f$ is dominated by $p$ on $M$ means that $(\forall x\in M) f(x)\le p(x)$.

This is basically the usual formulation of Hahn-Banach Theorem; although you can find many slight variations.

We will try to prove this in ZF:

Hahn-Banach Theorem in ZF. Let $X$ be a separable topological vector space and let $p:X\to{\mathbb R}$ be a continuous sublinear function. Let $M$ be a vector subspace of $X$ and let $f:M\to{\mathbb R}$ be a linear functional dominated by $p$ on $M$. Then there is a linear extension $\widehat{f}$ of $f$ to $X$ that is dominated by $p$ on $X$.

You may notice that this implies that every bounded functional on a subspace $M$ of a separable normed space $X$ can be extended to a functional on $X$ with the same norm. This is is the version of HBT stated here.

Notice a few changes in the formulation of the theorem. Since we want to speak about separable spaces, we need some kind of topology. So it is not that surprising that we work now with topological vector spaces. Maybe it is a little bit surprising that we require $p$ to be continuous. (On the other hand, if we want to use separability, then it is perhaps natural that we will probably use somewhere in the proof that $p$ behaves reasonable w.r.t. the topological structure.) But let us try to prove this version first. We will return to the question, whether continuity of $p$ is needed, later.

The standard proof of HBT uses as one of the steps the following fact:

Lemma. Let $X$ be a vector space and let $p:X\to{\mathbb R}$ be a sublinear function. Let $M$ be a vector subspace of $X$ and let $f:M\to{\mathbb R}$ be a linear functional dominated by $p$ on $M$. Let $x\in X$ and let $c\in\mathbb R$ be a number such that $$\sup_{y\in M} [f(y)-p(y-{x})] \le c \le \inf_{y\in M} [p(y+{x})-f(y)].$$

Then there exists a linear function $\overline f \colon [M\cup\{x\}]\to \mathbb R$ which extends $f$, it is dominated by $p$ on $[M\cup\{x\}]$ and $$f(x)=c.$$

Now we can prove this version of HBT in the following steps:

  • Prove the above lemma. And also check that under the assumptions of the lemma $$\sup_{y\in M} [f(y)-p(y-x)] \le \inf_{y\in M} [p(y+{x})-f(y)].$$ which means that we have at least one possible choice of $c$ in such situation.
  • If $X$ is separable and $\{x_n; n\in\mathbb N\}$ is a countable dense subset of $X$, then we can prove using induction and the above lemma that there exists a linear functional $f_n$ defined on $A_n=[M\cup\{x_1,\dots,x_n\}]$ which agrees with $f$ on $M$ and is dominated by $p$ on $A_n$. Moreover, each $f_n$ extends $f_{n-1}$.
  • Notice that in the induction step of this proof we choose some value of $c$ from a non-empty interval. Since the interval is closed, we can simply take the left endpoint at each step of the induction. So we do not need Axiom of Choice here.
  • The above gives us a new linear functional $g$ defined on a dense subspace $A=\bigcup A_n$. (Given simply by $g=\bigcup f_n$, i.e., $g(x)=f_n(x)$ if $x\in A_n$.) Moreover, $g$ is dominated by $p$ on $A$ and $g$ extends $f$.
  • Now it only remains to get from the function defined on the set $A$ the extension to the $\overline A=X$. I.e., we want to show get a function $\widehat f \colon X \to \mathbb R$ which is also linear, dominated by $p$ and fulfills $\widehat f|_M=f$. If we have $x\in X$ then there exists a sequence $(a_n)$ of points of $A$ such that $a_n\to x$. We define $$\widehat f(x):=\lim\limits_{n\to\infty} g(a_n).$$ If we can show that this indeed defines a function (i.e., that the above limit exists and that the value of $\widehat f(x)$ does not depend on the choice of the sequence $(a_n)$), then proving that $\widehat f$ is linear, extends $f$ and is dominated by $p$ is more-or-less straightforward. (Notice that here we will have to use continuity of $p$.)

Let us have a closer look at a more detailed proof of existence and uniqueness of the above limit. (Since this is precisely the place where my previous attempt failed.)

It is worth pointing out that we repeatedly use continuity of $p$. To be more precise, we use $p(a_n)\to p(x)$.

We will use rather simple observation that $g(x)=-g(-x)\ge -p(-x)$, so we have $$-p(-x)\le g(x) \le p(x)$$ for any $x\in A$. Now if we have a sequence $a_n\to x$, we can use the inequality $$-p(a_n-a_m) \le g(a_m-a_n) \le p(a_m-a_n)$$ to show that the sequence $(g(a_n))$ is Cauchy sequence in $\mathbb R$, and therefore it has a limit. Similarly if we have $a_n\to a$ and $b_n\to a$, we can use $$-p(b_n-a_n)\le g(a_n-b_n) \le p(a_n-b_n)$$ to show that $\lim g(a_n)=\lim g(b_n)$.

Putting all things mentioned above together we get the proof of the above result.

Would it be possible to prove stronger result - to omit condition that $p$ is continuous and just leave the sublinearity? The answer is no, which shows that these problems might be a bit more subtle than they could appear on the first glance. It is shown in the paper Juliette Dodu and Marianne Morillon: The Hahn-Banach Property and the Axiom of Choice (Mathematical Logic Quarterly, Volume 45, Issue 3, pages 299–314, 1999, DOI: 10.1002/malq.19990450303) that if every separable normed space satisfies the Hahn-Banach property (i.e., if the above theorem holds without the assumption that the sublinear function $p$ is continuous), then there are non-trivial finitely-additive measures on the set of positive integers. Therefore this result cannot be shown in ZF. (See Theorem 6 and Corollary 4 in Section 9 of this paper for details.)

The authors also say the following:

If a topological vector space $E$ has a dense subset which is well-orderable, then $E$ satisfies the continuous Hahn-Banach property, and the proof relies on some transfinite recursion and the following classical lemma... In particular, separable norme spaces satisfy the continuous Hahn-Banach property.

The "classical lemma" mentioned there is the lemma I formulated above. And if the dense subset is countable, then this is the above claim (for which I tried at least to sketch a proof).

If $X$ is not separable, then the argument above cannot be used to get a function $\widehat f$ defined on the whole space $X$. But the usual proof of Hahn-Banach Theorem is along the same lines, we just use Zorn's Lemma or transfinite induction instead of mathematical induction. (Zorn's lemma is known to be equivalent to the Axiom of Choice. The proof based on transfinite induction uses AC too, since we start by choosing some well-ordering of $X$. Well-ordering theorem is equivalent to AC.)

The only step in which the above proof and the usual proof of HBT differ substantially is the extensions from a dense subspace to the whole space. This is the point of the proof where we used continuity of the sublinear function $p$. (And this condition is not needed if AC is available.)

  • $\begingroup$ What is $A$? Is $f $ defined in it? (I think you was supposed to put $\lim f_n (a_n) $ in place of $\lim f (a_n) $ ...) $\endgroup$
    – Filburt
    Jan 11, 2018 at 13:42
  • $\begingroup$ @Filburt I have edited the post to clarify and correct this. I meant $A=\bigcup A_n$. $\endgroup$ Jan 11, 2018 at 15:50
  • $\begingroup$ Since $X$ is separable, then it contains a dense subset, denoted as $\hat{X}$. By stating that $\hat{X}=\{x_n:n\in \mathbb{N}\}$, does one need axiom of countable choice, which is beyond ZF? $\endgroup$
    – lychtalent
    Apr 26, 2022 at 8:50
  • $\begingroup$ @lychtalent I do not see how this uses AC. This is simply the definition of a countable set - it is a set which is in bijection with $\mathbb N$. $\endgroup$ Apr 26, 2022 at 8:54
  • $\begingroup$ I was referring to this one: en.m.wikipedia.org/wiki/Axiom_of_countable_choice. I misunderstood axiom of countable choice in my question. Clear now $\endgroup$
    – lychtalent
    Apr 26, 2022 at 9:00

Just use the same method as in the usual proof of Hahn-Banach to extend your functional to each point of a countable dense subset one at a time by induction. You then get that the functional is defined on a dense subspace of $X$, and so then you can extend it to all of $X$ by just taking limits (and can check that this is well-defined because the functional is bounded).

  • $\begingroup$ Thanks for your atention. I'm understand the idea of your proof. But how can I to extend my functional to each point of a countable dense subset one at a time by induction? If $f:Z\subset X\to \Bbb{K}$, and $M\subset X$ is countable and dense, given $x\in M-Z$, how to extend $f$ to $\hat{f}:Z+[x]\to \Bbb{K}$? $\endgroup$
    – Irddo
    Jan 23, 2016 at 6:03
  • $\begingroup$ This is exactly what the usual proof of Hahn-Banach does (when proving that a maximal extension must be defined on all of $X$). $\endgroup$ Jan 23, 2016 at 6:04
  • $\begingroup$ Then I should to use the Zorn Lemma? $\endgroup$
    – Irddo
    Jan 23, 2016 at 6:06
  • 1
    $\begingroup$ Ok, I will try again. Thank you! $\endgroup$
    – Irddo
    Jan 23, 2016 at 6:08
  • 4
    $\begingroup$ Eric, it's a bit misleading to say that we copy the argument from Zorn's lemma. For most people who are untrained in the Way of the Choice, Zorn's lemma is just this magical box where you have a partial order and you conclude there is a maximal under some condition. What we do here is either appeal to something like Koenig's lemma, or essentially reprove Zorn's lemma for countable partial orders. But a lot of people are not really aware of that because they don't have a trained mind for these sort of arguments. So "copying the argument from Zorn's lemma" can be quite confusing. $\endgroup$
    – Asaf Karagila
    Jan 28, 2016 at 7:01

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