An introduction to the Hahn-Banach extension theorem: Part VI

In this post we sketch a second proof of the Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces, this time using transfinite induction. We omit some of the details, which the reader can fill in. In particular, some of the details are similar to those involved in proving the claim made in Part V (when we obtained an upper bound for a chain).

For a non-zero ordinal $\gamma$, recall that, with ordinal interval notation, we have $\gamma=[0,\gamma)$. That is, $\gamma$ is equal to the set of all ordinals which are strictly less than $\gamma$. However, we use ordinal interval notation throughout. If you prefer to use $\gamma$ instead of $[0,\gamma)$ that is fine, of course!

Looking at the Wikipedia page on the Hahn-Banach Theorem, apparently both Hahn and Banach (independently) used transfinite induction in their proofs, rather than Zorn's Lemma.

Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$) 

Let $(E,\jnorm)$ be a normed space over $\R,$ and let $F$ be a linear subspace of $E$. Let $\psi \in F^*$. Then there exists $\phi \in E^*$ with $\|\phi\|_\op=\|\psi\|_\op$ and such that $\phi$ is an extension of $\psi$.

Sketch of a proof using transfinite induction

[The reader should fill in any details of claims which appear unobvious.]

We may assume that $F\neq E$. Let $B$ be a non-empty subset of $E$ with the property that $F+\lin B$ is dense in $E$.

[For example, we could take $B$ to be $E$, or $E\setminus F$, or a Hamel basis for $E$, etc.]

Then there exists a non-zero ordinal $\gamma$ such that there is a surjection $\alpha \mapsto v_\alpha$ from the ordinal interval $[0,\gamma)$ to $B$. 

[Here we use the Well Ordering Principle if (as is generally the case) $B$ doesn't come with a natural well order.]

We then have

\[B= \{v_\alpha: \alpha \in [0,\gamma)\}\,.\]

We set $F_0=F$ and, more generally, we define linear subspaces $F_\alpha$ $(\alpha \in [0,\gamma])$ of $E$ by

\[F_\alpha=\lin(F \cup \{v_\beta: \beta \in [0,\alpha)\})=F+\lin\{v_\beta: \beta \in [0,\alpha)\}\,.\]

Note that, for all $\alpha \in [0,\gamma)$, we have $F_{\alpha+1}=F_\alpha+\lin\{v_\alpha\}.$ Also, for every non-zero limit ordinal $\delta\leq \gamma$, we have

\[F_\delta=\bigcup_{\alpha \in [0,\delta)} F_\alpha\,.\]

Note that $F_\gamma$ is dense in $E$.

We now choose/define inductively linear functionals $\psi_\alpha \in E_\alpha^*$ $(\alpha \in [0,\gamma])$ satisfying the following two conditions:

$(i)_\alpha\qquad\qquad \|\psi_\alpha\|_\op=\|\psi\|_\op$

and

$(ii)_\alpha\qquad\qquad$for all $\beta \in [0,\alpha]$, we have $\psi_\alpha|_{F_\beta} = \psi_\beta\,.$

Here is a sketch of how the induction works.

We set $\psi_0=\psi \in F^*=F_0^*$. Clearly $(i)_0$ and $(ii)_0$ hold.

Now let $\delta \in (0,\gamma]$ and suppose that, for all $\alpha \in [0,\delta)$, we have chosen $\psi_\alpha$ satisfying  $(i)_\alpha$ and $(ii)_\alpha$.

If $\delta=\alpha+1$ for some ordinal $\alpha$, then we have $F_\delta=F_\alpha+\lin\{v_\alpha\}$. We can use the Hahn-Banach key lemma to extend $\psi_\alpha$ to $\psi_\delta \in F_\delta^*$ with $\|\psi_\delta\|_\op = \|\psi_\alpha\|_\op$. Then it is clear that $(i)_\delta$ and $(ii)_\delta$ hold.

If, instead, $\delta$ is a non-zero limit ordinal, then we have 

\[F_\delta = \bigcup_{\alpha\in [0,\delta)} F_\alpha\,.\]

There is then a unique function $\psi_\delta:F_\delta \to \R$  such that, for all $\alpha \in [0,\delta)$, we have $\psi_\delta|_{F_\alpha}=\psi_\alpha$. 

[Here we use the "consistency" provided by $(ii)_\alpha$ for all $\alpha\in [0,\delta)$.]

Moreover, $\psi_\delta \in F_\delta^*$ and $\|\psi_\delta\|_\op = \|\psi\|_\op$.

[Here the details are similar to the proof of the claim in Part V, this time using both $(i)_\alpha$ and $(ii)_\alpha$ for $\alpha\in[0,\delta)$.]

It is now clear that $(i)_\delta$ and $(ii)_\delta$ are satisfied.

The induction may proceed.

This now gives us $\psi_\gamma \in F_\gamma^*$ with $\psi_\gamma|_F = \psi$ and $\|\psi_\gamma\|_\op = \|\psi_\op\|$.

Depending on how we chose $B$ at the start, we might have $F_\gamma=E$, in which case we can just take $\phi=\psi_\gamma$.

Otherwise, we still have that $F_\gamma$ is dense in $E$. Thus $\psi_\gamma$ has a unique extension to $\phi \in E^*$, and we have $\|\phi\|_\op=\|\psi_\gamma\|_\op = \|\psi_\op\|$, as required. $\quad\square$

This is the last post in this series. However, I may return to the Hahn-Banach topic later in order to discuss a more general version in terms of  sublinear functions. See the Wikipedia page on the Hahn-Banach Theorem. In particular, see the section on the Geometric Hahn-Banach separation theorems.