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 $1 \leq \dfrac{|x_0 - x|}{\delta}$, $\forall x \in X_0$.

Define $Y = \text{Span}{x_0, X_0} = X_0 \oplus \mathbb{K} \cdot x_0$. Then for each $y \in Y$ we can find a unique $\lambda \in \mathbb{K}$ such that $u = \lambda x_0 + x$, $x \in X_0$. Define $u: Y \to \mathbb{K}$ by $u(y) = u(\lambda x_0 + x) = \lambda$. It is well defined and linear.

Furthermore, we have: $$|u(y)| = |\lambda| \leq |\lambda| \frac{|x _0 + x|}{\delta} = \frac{1}{\delta} |y| \quad \text{for} \lambda \neq 0$$ If $\lambda = 0$, then $y \in X_0$ and $u(y) = 0 \leq \frac{1}{\delta} |y|$.

Therefore, we obtain
$$ u(y) \leq \frac{1}{\delta} |y| \quad\forall y \in Y $$ By Hahn-Banach’s Theorem, we can extend $u$ to $f: X \to \mathbb{K}$ such that $f|_Y = u$ and $|f(x)| \leq \dfrac{1}{\delta} |x|$, $\forall x \in X$. Therefore $f(x_0) = u(x_0) = 1$ and $x \in X_0 \Rightarrow f(x) = 0$.