Banach-Alaoglu Theorem/Lemma 3

Lemma for Banach-Alaoglu Theorem
Let $X$ be a normed vector space.

Denote by $B$ the closed unit ball in $X$.

Let $X^*$ be the dual of $X$.

Denote by $B^*$ the closed unit ball in $X^*$.

Let $\map \FF B = \closedint {-1} 1^B$ be the topological space of functions from $B$ to $\closedint {-1} 1$.

By Tychonoff's Theorem, $\map \FF B$ is compact with respect to the product topology.

We define the restriction map:
 * $R: B^* \to \map \FF B$

by:
 * $\map R \psi = \psi \restriction_B$

$R \sqbrk {B^*}$ is a closed subset of $\map \FF B$.

Proof
Below, we will show that $f \in R \sqbrk {B^*}$ :

First, we see how the result follows from the above claim.

That is, we show that $\map \FF B \setminus R \sqbrk {B^*}$ is open.

For $u \in \closedint {-1} 1$ and $r \in \R_{>0}$, let $\map {I_r} u$ denote the open $r$-ball of $u$ in $\closedint {-1} 1$, i.e.:
 * $\map {I_r} u := \openint {u - r} {u + r} \cap \closedint {-1} 1$

Let $f \in \map \FF B \setminus R \sqbrk {B^*}$

It suffices to find an open set $U_f$ in $\map \FF B$ such that:
 * $f \in U_f \subseteq \map \FF B \setminus R \sqbrk {B^*}$

In view of the claim above, either $(1)$ or $(2)$ must fail for $f$.

Suppose $(1)$ fails.

That is, there exist $x, y \in B$ such that $x + y \in B$ satisfying:
 * $3 \epsilon := \size {\map f x + \map f y - \map f {x + y} } > 0$

Thus we can choose:
 * $U_f := \pi_x^{-1} \sqbrk { \map {I_\epsilon} {\map f x} } \cap \pi_y^{-1} \sqbrk { \map {I_\epsilon} {\map f y} } \cap \pi_{x+y}^{-1} \sqbrk { \map {I_\epsilon} {\map f {x+y} } }$

Suppose $(2)$ fails.

That is, there exist $x \in B$ and $\alpha \in \R_{>0}$ such that $\alpha x \in B$ satisfying:
 * $2 \epsilon := \size {\map f {\alpha x} - \alpha \map f x} > 0$

Thus we can choose:
 * $U_f := \pi_{\alpha x}^{-1} \sqbrk { \map {I_\epsilon} { \map f {\alpha x} } } \cap \pi_x^{-1} \sqbrk { \map {I_{\epsilon / \alpha } } {\map f x} }$

Now, we show the above claim.

Let $f \in R \sqbrk {B^*}$.

That is, there is a $\psi \in B^\ast$ such that $f = \map R \psi$.

$\psi$ satisfies both $(1)$ and $(2)$, as $\psi \in X^\ast$, especially.

Thus $f$ also satisfies them, because it is a restriction of $\psi$.

Conversely, let $f \in \map \FF B$ be satisfying $(1)$ and $(2)$.

Then, define a mapping $\phi : X \to \R$ by:
 * $\forall x \in X \setminus \set 0 : \map \psi x := \norm x \map f {\dfrac x {\norm x} }$

and $\map \psi 0 := 0$.

We show that $\psi \in B^\ast$ such that $\map R \psi = f$.

Observe that $\map f 0 = 0$, since $\map f 0 = \map f {0 + 0} = \map f 0 + \map f 0$ by $(1)$.

Thus, for $x \in X$ such that $x \ne 0$:

Furthermore, for all $x,y \in X$ such that $x + y \ne 0$:

Thus for $x,y \in X$:
 * $\map \psi {x+y} = \map \psi x + \map \psi y$

In particular, for all $x \in X$:
 * $\map \psi {-x} = - \map \psi x$

For $x \in X$ and $\alpha \in \R$ such that $\alpha x \ne 0$:

We also have by Definition of $\psi$:
 * $\forall x \in X \setminus \set 0 : \size {\map \psi x} = \norm x \size {\map f {\frac x {\norm x} } } \le \norm x$

Thus $\psi \in B^\ast$.

Finally, we have $\map R \psi = f$, since for all $x \in B \setminus \set 0$:

and $\map \psi 0 = 0 = \map f 0$.