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

In this post we will see how the Hahn-Banach key lemma from Part II allows us to prove the Hahn-Banach theorem for extensions of bounded linear functionals on separable normed spaces, without the need to use Zorn's Lemma (or equivalent).

First, we recall an easy result about extensions of bounded linear functionals to the closure of a linear subspace.

Let $\F$ be $\R$ or $\C$, and let $(E,\jnorm)$ be a normed space over $\F.$ Now let $F$ be a linear subspace of $E$. Then the closure of $F$ in $E$, $\overline{F}$, is also a linear subspace of $E$. As usual, we give $F$ and $\overline{F}$ the norm obtained by restricting $E$'s norm, $\jnorm$.

Proposition 1 (extensions of linear functionals to the closure of a subspace)

Let $\psi \in F^*$. Then $\psi$ has a unique extension $\tilde \psi \in {\overline{F}\,}^*$ (the dual space of $\overline{F}$), and we have

$$\|\tilde\psi\|_\op = \|\psi\|_\op\,.$$

Comments

• This is a special case of a more general result about extensions of continuous linear maps from $F$ into a Banach space $Y$. (Of course, $(\F,|\cdot|)$ is a Banach space.)
• There is only one possible definition we can try for $\tilde\psi$, and it works. For every $x$ in $\overline{F}$ and every sequence $(x_n)$ in $F$ which converges to $x$, we have to have 
  $$\tilde\psi(x) = \lim_{n\to\infty} \psi(x_n)\,.$$
  Fortunately you can show that this limit exists, and is independent of which such sequence $(x_n)$ in $F$ we take. The rest is fairly routine.
• We could also note that $\psi$ is Lipschitz continuous and hence uniformly continuous on $F$. Thus $\psi$ has a unique continuous extension to $\overline{F}$, which gives us $\tilde \psi$. Again, the rest is fairly routine.

We are now ready to prove the Hahn-Banach theorem for separable normed spaces over $\R$.

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

Let $(E,\jnorm)$ be a separable 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 $\psi$.

Comments

• As usual, in this theorem and its proof, every linear subspace of $E$ is regarded as a normed space by restricting $E$'s norm, $\jnorm.$
• We denote the linear span of a subset $A$ of $E$ by $\lin A$.
• We start by applying the Hahn-Banach key lemma repeatedly to obtain a sequence of successive extensions of $\psi$ from $F$ to a nested increasing sequence of linear subspaces of $E$ whose union is dense in $E$. 
• From this we obtain an extension of $\psi$ to the relevant union of subspaces, and then apply Proposition 1 above to obtain the extension $\phi$ in $E^*$ that we want.
• In applying the Hahn-Banach key lemma we might appear to be using some form of sequential choice. However, inspection of our proof of the key lemma shows that, once we have specified $y$ there, we can make explicit which constant $c$ we prefer. (For example it could be the minimum constant that works.)

Proof

Since $E$ is separable, there is a countable dense subset $S$ of $E$. Say 

$$S=\{x_1,x_2,x_3,\dots\}\,.$$

Set $F_0=F$ and, for each $n \in \N$, set

$$F_n=\lin(F\cup\{x_1,x_2,\dots,x_n\})= F+\lin\{x_1,x_2,\dots,x_n\}\,.$$

Note that, for all $n \in \N$, we have

$$F_n = F_{n-1} + \lin\{x_n\}\,,$$

and so $F_{n-1}$ has codimension $1$ or $0$ in $F_n$.

We now choose, inductively, $\psi_n \in F_n^*$ as follows (for $n \in \N\cup\{0\}$).

We set $\psi_0=\psi \in F^*=F_0^*$.

Now let $n \in \N$ and suppose that we have chosen $\psi_{n-1} \in F_{n-1}^*$.

• If $x_n \in F_{n-1}$, then $F_n=F_{n-1}$ and we take $\psi_n = \psi_{n-1} \in F_n^*$.
• Otherwise $x_n \in F_n \setminus F_{n-1}$, and $F_n = F_{n-1} \oplus \lin\{x_n\}\,.$ In this case we can apply the Hahn-Banach key lemma to obtain $\psi_n \in F_n^*$ which extends $\psi_{n-1},$ and such that $\|\psi_n\|_\op = \|\psi_{n-1}\|_\op\,.$

The induction may now proceed. Note:

• For all $n \in \N$, we have $\|\psi_n\|_\op = \|\psi\|_\op$.
• For all $m,n \in \N\cup\{0\}$ with $m \leq n$, $\psi_n$ is an extension of $\psi_m$.

Set 

$$F_\omega = \bigcup_{n=0}^\infty F_n = \lin(F \cup \{x_1,x_2,\dots\})=F+\lin\{x_1,x_2,\dots\}\,.$$

We may define $\psi_\omega \in F_\omega^*$ as follows: for each $n \in \N$ and each $x \in F_n$, set

$$\psi_\omega(x) = \psi_n(x)\,.$$

• We need to check that this is well-defined, but fortunately the "consistency" of our extensions above ensures this. If $m,n \in \N\cup\{0\}$ and $x$ is in both $F_m$ and $F_n$, then (as noted above), $\psi_m(x)=\psi_n(x)$.
•  It is easy to check that $\psi_\omega$ is both linear and continuous, with $\|\psi_\omega\|_\op =\|\psi\|_\op\,.$
• Clearly $F_\omega$ is dense in $E$.

Finally, we apply Proposition 1 above, and let $\phi \in E^*$ be the unique continuous linear extension to $E$ of $\psi_\omega$. Then $\phi$ is an extension of $\psi$, and 

$$\|\phi\|_\op = \|\psi_\omega\|_\op =\|\psi\|_\op\,,$$ as required.$\quad\square$

• In the proof above we only really needed $F+\lin\{x_1,x_2,\dots\}$ to be dense in $E$. For example, this is guaranteed if $\{x_1,x_2,\dots\}$ is a Schauder basis of $E$.
• In the next post, we see how the general/separable complex Hahn-Banach theorem (for extensions of bounded linear functionals on normed spaces) can be deduced from the corresponding real Hahn-Banach theorem.
• We will then give a proof of the general (real) Hahn-Banach theorem .
• In fact we will give two proofs: one the usual proof by Zorn's lemma, another a proof by transfinite induction.
• You may already be able to see how the transfinite induction argument will go based on the inductive construction above, where we stopped at the first infinite ordinal $\omega$.