Hahn-Banach Theorem/Real Vector Space/Corollary 2

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

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

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

Then $f_0$ can be extended to a bounded linear functional $f : X \to \R$ 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
First suppose that:


 * $\norm {f_0}_{\paren {X_0}^\ast} = 0$

Then, from positive definiteness of $\norm \cdot_{\paren {X_0}^\ast}$, we have:


 * $f_0 = 0$

That is:


 * $\map {f_0} x = 0$

for each $x \in X_0$.

Then the bounded linear functional $f : X \to \R$ with:


 * $\map f x = 0$

clearly extends $f_0$ with:


 * $\norm f_{X^\ast} = 0 = \norm {f_0}_{\paren {X_0}^\ast}$.

Now take:


 * $\norm {f_0}_{\paren {X_0}^\ast} \ne 0$

From Fundamental Property of Norm on Bounded Linear Functionals, we have:


 * $\size {\map {f_0} x} \le \norm {f_0}_{\paren {X_0}^\ast} \norm x$

for each $x \in X_0$.

Now define $p : X \to \R$ by:


 * $\map p x = \norm {f_0}_{\paren {X_0}^\ast} \norm x$

for each $x \in X$, so that:


 * $\size {\map {f_0} x} \le \map p x$

for each $x \in X_0$.

From Positive Scalar Multiple of Norm on Vector Space is Norm, we have:


 * $p$ is a norm.

In particular, from Norm on Vector Space is Seminorm, we have:


 * $p$ is a seminorm.

So, applying Hahn-Banach Theorem for Real Vector Spaces: Corollary 1, we have that:


 * there exists an extension $f : X \to \R$ of $f_0$ such that:


 * $\size {\map f x} \le \map p x$ for each $x \in X$.

We now only need to check that:


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

Since:


 * $\size {\map f x} \le \norm {f_0}_{\paren {X_0}^\ast} \norm x$

From the definition of the norm on bounded linear functionals, we then have:


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

We have:


 * $\set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X} \subseteq \set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X_0}$

Since $f$ extends $f_0$ we have:


 * $\set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X} = \set {C > 0 : \size {\map {f_0} x} \le C \norm x \text { for all } x \in X_0}$

So, from Infimum of Subset, we have:


 * $\inf \set {C > 0 : \size {\map {f_0} x} \le C \norm x \text { for all } x \in X_0} \le \inf \set {C > 0 : \size {\map f x} \le C \norm x \text { for all } x \in X}$

From the definition of the norm on bounded linear functionals, we then have:


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

So we obtain the desired:


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