# Hahn-Banach Theorem/Complex Vector Space

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## 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 $x \in X_0$.

### Corollary

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\C$.

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

Let $f_0 : X_0 \to \C$ be a bounded linear functional.

Then $f_0$ can be extended to a bounded linear functional $f : X \to \C$ with:

$\norm f_{X^\ast} = \norm {f_0}_{\paren {X_0}^\ast}$

where $\norm \cdot_{X^\ast}$ and $\norm \cdot_{\paren {X_0}^\ast}$ are the norms of the normed dual spaces $X^\ast$ and $\paren {X_0}^\ast$.

## Proof

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

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

for each $x \in X_0$.

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

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

for each $x \in X$.

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

 $\ds \cmod {\map {g_0} x}$ $=$ $\ds \sqrt {\paren {\map {g_0} x}^2}$ $\ds$ $=$ $\ds \sqrt {\paren {\map {g_0} x}^2 + \paren {\map {h_0} x}^2}$ $\ds$ $=$ $\ds \cmod {\map {f_0} x}$ $\ds$ $\le$ $\ds \map p x$

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:

 $\ds \map f {\lambda x + \mu y}$ $=$ $\ds \map g {\lambda x + \mu y} - i \map g {i \paren {\lambda x + \mu y} }$ $\ds$ $=$ $\ds \lambda \map g x + \mu \map g x - i \lambda \map g {i x} - i \mu \map g {i x}$ since $g$ is $\R$-linear $\ds$ $=$ $\ds \lambda \map f x + \mu \map f y$

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:

 $\ds \map f {i x}$ $=$ $\ds \map g {i x} - i \map g {-x}$ $\ds$ $=$ $\ds \map g {i x} + i \map g x$ $\ds$ $=$ $\ds i \paren {\map g x - i \map g {i x} }$ $\ds$ $=$ $\ds i \map f x$

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 $x \in X$, and so:

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

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 we have, since $f$ is $\C$-linear:

 $\ds \cmod {\map f x}$ $=$ $\ds \lambda \map f x$ $\ds$ $=$ $\ds \map f {\lambda x}$ $\ds$ $=$ $\ds \map g {\lambda x} - i \map g {i \lambda x}$

Since:

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

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

Finally, we have:

 $\ds \cmod {\map f x}$ $=$ $\ds \map g {\lambda x}$ $\ds$ $\le$ $\ds \map p {\lambda x}$ $\ds$ $=$ $\ds \cmod \lambda \map p x$ Seminorm Axiom $\text N 2$: Positive Homogeneity $\ds$ $=$ $\ds \map p x$

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

$\blacksquare$

## Source of Name

This entry was named for Hans Hahn and Stefan Banach.