Home > Functional Analysis > Hahn-Banach (complex version)

Hahn-Banach (complex version)

Let X be a complex vector space, X_0 one of its subspaces, p: X \to \mathbb{R}_+ such that p(\lambda x)=|\lambda| p(x),\ \forall \lambda \in \mathbb{C}, x \in X and p(x+y) \leq p(x)+p(y),\ \forall x,y \in X, satisfying |f(x)| \leq p(x),\ \forall x \in X_0, where f:X_0 \to \mathbb{C} is linear.
Under these conditions, there exists a linear functional F :X \to \mathbb{C} such that F| _{X_0}=f and |F(x)| \leq p(x),\ \forall x \in X.
Solution: Since f is linear, it follows that Re\ f :X_0 \to \mathbb{R} is linear and Re\ f(x) \leq |f(x)| \leq p(x),\ \forall x \in X_0. By the Hahn-Banach Theorem (real version) there exists g: X \to \mathbb{R} a linear functional such that g is an extension for Re\ f and g(x) \leq p(x),\ \forall x \in X. We also have g(x)=-g(-x)\geq -p(x) so |g(x)| \leq p(x),\ \forall x \in X
Define now F(x)=g(x)-ig(ix),\ \forall x \in X. This is obviously linear and if x \in X_0 we have F(x)=g(x)-ig(ix)=Re\ f(x)- i Re\ i f(x)= Re\ f(x)+i Im\ f(x)=f(x),\ \forall x \in X_0.
For the last part we have |F(x)|=e^{i\theta} F(x)=F(e^{i\theta} x)=g(e^{i\theta}x), because this is a real number. Further more, we have g(e^{i\theta}x) \leq p(e^{i\theta} x)=p(x). Combinig the two above, we get |F(x)|\leq p(x),\ \forall x \in X, which solves the theorem. \Box

Categories: Functional Analysis Tags:
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: