Hahn-Banach Theorem/Complex Vector Space

Theorem
Let $X$ be a vector space over $\C$.

Let $p : X \to \R$ be a seminorm on $X$.

Let $X_0$ be a linear subspace of $X$.

Let $f_0 : X_0 \to \C$ be a linear functional such that:


 * $\cmod {\map {f_0} x} \le \map p x$ for each $x \in X_0$.

Then there exists a linear functional $f$ defined on the whole space $X$ which extends $f_0$ and satisfies:


 * $\cmod {\map f x} \le \map p x$ for each $x \in X$.

That is, there exists a linear functional $f : X \to \C$ such that:


 * $\cmod {\map f x} \le \map p x$ for each $x \in X$

and:


 * $\map f x = \map {f_0} x$ for each $x \in X_0$.

Proof
Define $g_0 : X_0 \to \R$ by:


 * $\map {g_0} x = \map \Re {\map {f_0} x}$

for each $x \in X_0$.

By Real Part of Linear Functional is Linear Functional, $g_0$ is an $\R$-linear functional.

Also, define $h_0 : X_0 \to \R$ by:


 * $\map {h_0} x = \map \Im {\map {f_0} x}$

for each $x \in X$.

By Imaginary Part of Linear Functional is Linear Functional, $h_0$ is an $\R$-linear functional.

Now, for each $x \in X$, we have:

Let $X_\R$ be the realification of $X$.

By Hahn-Banach Theorem: Real Vector Space on $X_\R$, there exists a $\R$-linear functional $g : X \to \R$ extending $g_0$ and satisfying:


 * $\cmod {\map g x} \le \map p x$

for each $x \in X$.

Define $f : X \to \C$ by:


 * $\map f x = \map g x - i \map g {i x}$

for each $x \in X$.

Then for $\lambda, \mu \in \R$ and $x, y \in X$ we have:

so $f$ is $\R$-linear.

To show that $f$ is $\C$-linear, we need to show that:
 * $\map f {i x} = i \map f x$

for each $x \in X$.

We have:

We now want to show that $f$ extends $f_0$.

Let $F_0$ be the restriction of $f$ to $X_0$.

We want to show that:
 * $f_0 = F_0$

We have:


 * $\map \Re {\map f x} = \map g x$

for each $x \in X$.

Hence:


 * $\map \Re {\map f x} = \map {g_0} x = \map \Re {\map {f_0} x}$

for each $x \in X_0$.

So:


 * $\map \Re {\map {F_0} x} = \map \Re {\map {f_0} x}$

for each $x \in X_0$.

From Linear Functional on Complex Vector Space is Uniquely Determined by Real Part, it follows that:


 * $\map {F_0} x = \map {f_0} x$

for each $x \in X_0$.

So $f$ indeed extends $f_0$.

Now take $x \in X$.

Pick $\lambda \in \C$ such that:
 * $\cmod \lambda = 1$

and:
 * $\lambda \map f x = \cmod {\map f x}$

Then, since $f$ is $\C$-linear:

Since:


 * $\cmod {\map f x} \in \R$

we have that:
 * $\map g {i \lambda x} = 0$

Finally, we have:

So $f$ is a linear functional satisfying our requirements.