Existence of Support Functional

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Theorem

Let $\mathbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$.

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$.

Let $x \in X$.


Then there exists $f \in X^\ast$ such that:

$(1): \quad$ $\norm f_{X^\ast} = 1$
$(2): \quad$ $\map f x = \norm x_X$.

That is:

there exists a support functional at $x$.


Proof

Let:

$U = \span {\set x}$

Then $U$ consists precisely of the $u \in X$ of the form:

$u = \alpha x$

for $\alpha \in \mathbb F$.

From Linear Span is Linear Subspace, we have:

$U$ is a linear subspace of $X$.

Let $\struct {U^\ast, \norm \cdot_{U^\ast} }$ be the normed dual space of $U$.

Define $f_0 : U \to \R$ by:

$\map {f_0} {\alpha x} = \alpha \norm x_X$

for each $\alpha \in \mathbb F$.

In particular, we have:

$\map {f_0} x = \norm x$

We show that this is a linear functional.

Let $u, v \in U$ and $k, l \in \mathbb F$.

Then there exists $\alpha, \beta \in \mathbb F$ such that:

$u = \alpha x$

and:

$v = \beta x$

We then have:

\(\ds \map {f_0} {k u + l v}\) \(=\) \(\ds \map {f_0} {k \alpha x + l \beta x}\)
\(\ds \) \(=\) \(\ds \map {f_0} {\paren {k \alpha + l \beta} x}\)
\(\ds \) \(=\) \(\ds \paren {k \alpha + l \beta} \norm x_X\)
\(\ds \) \(=\) \(\ds k \alpha \norm x_X + l \beta \norm x_X\)
\(\ds \) \(=\) \(\ds k \map {f_0} u + l \map {f_0} v\)

so $f_0$ is a linear functional.

Now we show that $f_0 \in X^\ast$ and:

$\norm {f_0}_{U^\ast} = 1$

Let $u \in U$ and write:

$u = \alpha x$

for $\alpha \in \mathbb F$.

We then have:

\(\ds \size {\map {f_0} u}\) \(=\) \(\ds \size {\map {f_0} {\alpha x} }\)
\(\ds \) \(=\) \(\ds \size {\alpha \norm x_X}\)
\(\ds \) \(=\) \(\ds \size \alpha \norm x_X\)
\(\ds \) \(=\) \(\ds \norm {\alpha x}_X\) Definition of Norm
\(\ds \) \(=\) \(\ds \norm u_X\)

so we have that $f_0$ is bounded.

That is, $f \in U^\ast$.

We then have:

$\ds \sup_{\norm u_X = 1} \size {\map {f_0} u} = 1$

That is, from the definition of dual norm, we have:

$\norm {f_0}_{U^\ast} = 1$

We apply:

Hahn-Banach Theorem: Real Vector Space: Corollary 2 if $\mathbb F = \R$
Hahn-Banach Theorem: Complex Vector Space: Corollary if $\mathbb F = \C$

to find that there exists $f \in X^\ast$ such that:

$f$ extends $f_0$ to $X$

and:

$\ds \norm f_{X^\ast} = \norm {f_0}_{U^\ast} = 1$

Since $f$ extends $f_0$, we have:

$\map f x = \map {f_0} x = \norm x$

So $f$ is the required linear functional.

$\blacksquare$


Sources

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