### Archive

Posts Tagged ‘Hahn-Banach’

## Separable space 2

Denote $c_0=\{(\alpha_n) \subset \Bbb{C} : \alpha_n \to 0\}$, the set of complex sequences which converge to $0$. Furthermore, consider the sequences $x_n=((n+1)^{-j})_j$. Prove that the closed linear span of $\{x_n : n=1,2,..\}$ is in fact $c_0$.

## non-Separable space Example 1

Prove that the space $C=\{f: [1,\infty) \to \Bbb{C},\ f \text{ continuous and bounded }\}$ is not separable.

## Hahn-Banach application.

Suppose $X$ is a normed space and $X_0$ is a closed subspace of $X$ and $x_0 \in X \setminus X_0$. Then we can find $f \in X^\prime$ such that $f(x_0)=1$ and $f(x)=0,\ \forall x \in X_0$.

Categories: Functional Analysis

## Hahn-Banach application. The dual is not trivial

Denote by $X^\prime=\{ f: X \to \mathbb{K} : f$ is linear and continuous $\}$ where $X$ is a Banach space over $\mathbb{K}$. Prove that $X^\prime \neq \{0\}$, in fact, for every $x \neq 0 \in X$, we can find $f \in X^\prime$ such that $f(x)=\|x\|$ and $\|f\|=1$. Read more…

## 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 $|F(x)| \leq p(x),\ \forall x \in X$. Read more…

Categories: Functional Analysis Tags:

## Hahn-Banach Theorem (real version)

Suppose $X$ is a vector space over $\mathbb{R}$, $p:\mathbb{X} \to \mathbb{R}$ has the following properties: $p(\lambda x)=\lambda p(x),\ \forall x \in X, \lambda \in\mathbb{R}_+$ and $p(x+y)\leq p(x)+p(y),\ \forall x,y \in X$.
Let $X_0$ be a subspace of $X$ and $u: X_0\to \mathbb{R}$ a linear functional such that $u(x) \leq p(x),\ \forall x \in X_0$.
Then we can find $f:X \to \mathbb{R}$ a linear functional such that $f| _{X_0} =u$ and $f(x) \leq u(x),\ \forall x \in X$. Read more…