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

In this post we look at how every complex normed space can also be regarded as a real normed space, and how this fits with bounded linear functionals and the Hahn-Banach theorem.

Let $(E,\jnorm)$ be a normed space over $\C$. Then $E$ is also a normed space over $\R$, with the same addition, but only considering multiplication by real scalars. This has no effect on the metric induced by $\jnorm$ or the topology. In particular, separability is not affected by this. Still, let us denote the real version by $E_\R$, so that $(E_\R,\jnorm)$ is the corresponding real normed space.

If $E$ is finite-dimensional (over $\C$), then $E_\R$  is finite-dimensional (over $\R$), but the dimension doubles. Indeed, if $E$ is $n$-dimensional over $\C$ with basis $x_1,x_2,\dots,x_n$ (over $\C$), then it is easy to see that $x_1, \i x_1, x_2, \i x_2, \dots, x_n, \i x_n$ is a basis for $E_\R$ (over $\R$).

Now let $F$ be a (complex) linear subspace of $E$. Then $F_\R$ is a real linear subspace of $E_R$. However if $F$ has complex codimension $1$ in $E$, then $F_\R$ has real codimension $2$ in $E_\R$.

As usual we restrict $\jnorm$ to $F$ to make $F$ into a complex normed space, and $F_\R$ into a real normed space.

We now have two different dual spaces for each of $E$ and $F$: the complex dual space (the set of continuous complex-linear maps from these spaces to $\C$) and the real dual space (continuous real-linear maps from these spaces to $\R$). Here we use the notation $E^*$ and $F^*$ for the complex dual spaces, and $E_\R^*$and $F_\R^*$ for the real dual spaces.

As it turns out, there is a very close connection between these dual spaces.

Let $\phi \in E^*$ (so that $\phi$ is a continuous, complex-linear map from $E$ to $\C$).

Then we can define a continuous, real-linear map $\phi_\R$ from $E_\R$ to $\R$ by taking the real part: for all $x\in E$, we set 

$$\phi_\R(x) = \Re(\phi(x))\,.$$

It is easy to see that $\phi_\R \in E_\R^*.$ Perhaps less obvious is the fact that the map $\phi \mapsto \phi_\R$ is a surjective (real-linear) isometry from $E^*$ (or $(E^*)_\R$) to $E_\R^*$.

It is clear that $\|\phi_\R\|_\op \leq \|\phi\|_\op$, because $|\Re(\phi(x))| \leq |\phi(x)|$.

For the reverse inequality, let $x \in E$ with $\|x\|\leq 1$. We can then choose a complex number $\alpha$ with $|\alpha|=1$ and such that $\alpha \phi(x)= |\phi(x)|$.

Set $y=\alpha x$. Then $\|y\|\leq 1$, and $\phi_\R(y) = \Re(\phi(y)) = |\phi(x)|.$ Thus $|\phi(x)| \leq \|\phi_\R\|_\op.$

Taking $\sup$ over all such $x$ gives us $\|\phi\|_\op \leq \|\phi_\R\|_\op,$ and so we have equality. So the map $\phi \mapsto \phi_\R$ preserves operator norms.

That isn't exactly the same as being an isometry in general, if you don't have linearity! But here the map is clearly real-linear (from $(E^*)_\R$ to $E_\R^*$), and hence it really is an isometry.

Perhaps the least obvious fact is that our map is surjective. How do we know that we can get everything in $E_\R^*$ this way?

Let $\theta \in E_\R^*$. We would like to find a $\phi \in E^*$ such that $\theta = \phi_\R$. 

Let $x \in E$. We need to have 

$$\theta(x)= \Re(\phi(x))\,.$$

 However, if $\phi$ is to be complex-linear, we will also need to have

$$\theta(\i x) = \Re(\phi(\i x)) = \Re(\i \phi(x)) = -\Im(\phi(x))\,.$$

As a result, the only hope is to define $\phi(x) = \theta(x) - \i\theta(\i x).$

This map $\phi$ is clearly a continuous, real-linear map from $E_\R$ to $\C_\R,$ and, for all $x \in E$, we have $\theta(x)=\Re(\phi(x)).$ But is $\phi$ complex-linear from $E$ to $\C$? Given that we already have real-linearity, it is sufficient to show that, for all $x \in E$, we have

$$\phi(\i x) = \i \phi(x)\,.$$

But this is easy to check: we have 

$$\phi(\i x) = \theta(\i x) - \i\theta(\i (\i x)) = \theta(\i x) -\i \theta (-x) =\theta(\i x) + \i \theta (x) =  \i(\theta(x)-\i\theta(\i x)) = \i\phi(x)\,.$$

In case that looks like a fluke, remember that we originally defined $\phi$ to ensure that, for all $x \in E$, we had $\theta(x)= \Re(\phi(x))$ and $\theta(\i x) =  -\Im(\phi(x)) = \Re(\i \phi(x)).$ And of course we also have $\theta(\i x) =\Re(\phi(\i x))$. So, for all $x \in E$, we have

$$\Re (\phi(\i x))=\Re(\i \phi(x)),$$

As you can easily check, this equality of the real parts is sufficient to imply that a real-linear map $\phi$ from a complex vector space to $\C$ is actually complex linear.

I have a feeling there is probably an even better reason why this was bound to work. Anyway, for whatever reason, it works! We have $\phi \in E^*$, and $\phi_\R = \theta,$ as required. From above, we also know that

$$\|\phi\|_\op=\|\phi_\R|_\op=\|\theta\|_\op\,.$$

Armed with this information, we can now see how to deduce the complex Hahn-Banach theorem from the real Hahn-Banach theorem. Let $\psi \in F^*$. Then $\psi_\R \in F_\R^*$, and $\psi_\R$ has the same operator norm as $\psi.$ The real Hahn-Banach theorem allows us to extend $\psi_\R$ to some $\theta \in E_\R^*$, still with the same operator norm. Then there is a unique $\phi \in E^*$ with $\phi_\R=\theta$, and this $\phi$ again has the same operator norm. Finally, $\phi$ must be an extension of $\psi$: for example, for all $x \in F$, we have 

$$\phi(x)= \theta(x) -\i \theta(\i x) = \psi_\R(x) - \i \psi_\R(\i x) = \psi(x)\,.$$

Of course, if $E$ is separable, then we only need the separable version of the real Hahn-Banach theorem to deduce the separable version of the complex Hahn-Banach theorem.

In the next post, we return to prove the general case of the real Hahn-Banach theorem, where $E$ is not necessarily separable. But for this we need to use the axiom of choice somewhere, unless we are presented with something like a well-ordered Hamel basis for a dense linear subspace of $E$, in which case we can just use transfinite induction and imitate the proof we gave in the separable case.