“There are some things which cannot be learned quickly, and time, which is all we have, must be paid heavily for their acquiring. They are the very simplest things, and because it takes a man’s life to know them the little new that each man gets from life is very costly and the only heritage he has to leave.” - Ernest Hemingway (More…)
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Real Hahn-Banach Theorem
Suppose $X$ is a vector space over $\mathbb{R}$, $p: X \to \mathbb{R}$ has the following properties: $p(X) = \lambda p(x)$, $\forall x \in X$, $\lambda \in \mathbb{R}_+$ 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 $f(x) \leq u(x)$, $\forall x \in X$.
Riesz Representation Theorem
1. Riesz Representation Theorem # Let $H$ be a Hilbert space over $\mathbb{R}$ or $\mathbb{C}$, and $T$ be a bounded linear functional on $H$ (a bounded operator from $H$ to the field $\mathbb{R}$ or $\mathbb{C}$, where $H$ is defined over that field). The following is known as the Riesz Representation Theorem: Theorem 1: If $T$ is a bounded linear functional on the Hilbert space $H$, then there exists $g \in H$ such that for every $f \in H$, we have: $$ T(f) = \langle f, g \rangle. $$
The application of Hahn-Banach Theorem 01
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 $f \in X’$ such that $f(x_0) = 1$ and $f(x) = 0$, $\forall x \in X_0$. Proof: Since $x_0 \notin X_0$, we can find $\delta > 0$ such that $|x_0 - x| \geq \delta$, $\forall x \in X_0$, which is equivalent to $1 \leq \dfrac{|x_0 - x|}{\delta}$, $\forall x \in X_0$.
The application of Hahn-Banach Theorem 02
$X'$ = $\{ f: X \to \mathbb{K} \}$ where $f$ is is linear and continuous and $X$ is a Banach space over $\mathbb{K}$. Prove that $X' \neq {0}$, in fact, for every $x \neq 0 \in X$, we can find $f \in X’$ such that $f(x) = |x|$ and $|f| = 1$. Proof: Pick $x_0 \in X$. Define $X_0 = x_0 \cdot \mathbb{K}$, a subspace of $X$, and $g: X_0 \to \mathbb{K}$, $g(x) = x$, which is linear. Since $g$ and $|\cdot|$ satisfy the conditions of the Hahn-Banach theorem, we can find $f: X \to \mathbb{K}$ such that $f|_{X_0} = g$, $f$ is linear and $f(x) \leq |x|$, $\forall x \in X$. Therefore $f(x_0) = g(x_0) = |x_0|$ and $|f| \leq 1$. The equality $f(x_0) = |x_0|$ guarantees that $|f| = 1$.