### Advanced Analysis, Notes 12: Banach spaces (application: existence of Haar measure)

This post is dedicated to my number one follower for her fourteenth birthday,which I spoiled…

Let $G$ be a compact abelian group. By this we mean that $G$ is at once both an abelian group and a compact Hausdorff topological space, and that the group operations are continuous, meaning that $g \mapsto g^{-1}$ is continuous on $G$ and $(g,h) \mapsto g+h$ is continuous as a map from $G \times G$ to $G$. It is known that there exists a regular Borel measure $\mu$ on $G$, called the Haar measure, which is non-negative, satisfies $\mu(G) = 1$, and is translation invariant:

$\forall g \in G . \mu(g+ E) = \mu(E) ,$

for every Borel set $E \subseteq G$. In fact, the Haar measure is known to exist in greater generality ($G$ does not have to be commutative and if one allows $\mu$ to be infinite then $G$ can also be merely locally compact). The Haar measure is an indispensable tool in representation theory and in ergodic theory. In this post we will use the weak* compactness of the unit ball of the dual to give a slick proof of the existence of the Haar measure in the abelian compact case.

#### 1. The Kakutani–Markov fixed point theorem

The following theorem can be stated and proved in greater generality with more–or–less the same proof as we present.

Theorem 1 (Kakutani–Markov fixed point theorem): Let $X$ be a Banach space, and let $\mathcal{F}$ be a commuting family of weak-$*$ continuous linear maps on $X^*$. Suppose that $C$ is a non-empty, weak*–compact and convex subset of $X^*$ such that $T(C) \subseteq C$ for all $T \in \mathcal{F}$. Then $\mathcal{F}$ has a common fixed point in $C$, i.e., there is an $x \in C$ such that $Tx = x$ for all $T \in \mathcal{F}$

Proof: Let us first prove the theorem in the case where $\mathcal{F} = \{T\}$. Choose some $y_0 \in C$. Construct the averages

$y_N = \frac{1}{N+1} \sum_{n=0}^{N} T^n y_0 .$

Since $C$ is convex and $T(C) \subseteq C$, we have that $y_N \in C$ for all $C$. Since $C$ is compact, the sequence $\{y_N\}_{N=1}^\infty$ contains a subnet $\{y_{N(\alpha)}\}$ that converges to some $y \in C$. To show that $Ty = y$, we will prove that $\langle x, Ty-y \rangle = 0$ for all $x \in X$.

Fix $x \in X$. Then $x$ is continuous on $X^*$ in the weak* topology, hence bounded on the weak*–compact set $C$, say $|\langle x, c \rangle| \leq M_x$ for all $c \in C$. Now for any $N$,

$|\langle x , Ty_N - y_N \rangle | = \frac{1}{N+1} |\langle x, T^{N+1} y_0 - y_0 \rangle| \leq \frac{2M_x}{N+1} .$

Thus $|\langle x, T y - y \rangle| = \lim_\alpha |\langle x, T y_{N(\alpha)} - y_{N(\alpha)} \rangle | = 0$ (recall that part of the definition of subnet is that $N(\alpha)$ goes to infinity with $\alpha$). That completes the proof for the case where $\mathcal{F}$ is a singleton.

Now let $\mathcal{F}$ be arbitrary, and for every finite $F \in \mathcal{F}$, denote $A_F = \{c \in C : \forall T \in F . Tc = c\}$. $A_F$ is evidently closed. We will show that the family $\{A_F : F \subseteq \mathcal{F} \textrm{ finite}\}$ has the finite intersection property. It will follow from compactness that there is some $x \in \cap_F A_F$, which will be the sought after fixed point. Now, $A_{F_1} \cap A_{F_2} = A_{F_1 \cup F_2}$, so it follows that we only have to show that every $A_F$ is non-empty. This is done by induction on the cardinality $|F|$ of $F$. If $|F| = 1$, then $A_F \neq \emptyset$ by the first half of the proof. Suppose that $A_F$ is not empty, and let $T \in \mathcal{F}$. Then for every $y \in A_F$, we have for all $S \in F$

$STy = TSy = Ty .$

Therefore $T(A_F) \subseteq A_F$, so by the first half of the proof there is some $z \in A_{F}$ fixed under $T$. In other words, $z \in A_{F \cup \{T\}}$, so this set is not empty.

#### 2. The existence of Haar measure for abelian compact groups

We can now prove the existence of Haar measure for compact abelian groups.

Theorem 2: Let $G$ be a compact Hausdorff abelian group. Then there exists a Haar measure for $G$. That is, there is a regular Borel probability measure $\mu$ on $G$ that is translation invariant.

Proof: For every $g \in G$, let $L_g : C(G) \rightarrow C(G)$ be the translation operator given by  $(L_g f)(h) = f(g-h)$. We will find a regular Borel probability measure $\mu$ on $G$ such that for all $g \in G$,

(*) $\int f d \mu = \int L_g f d \mu .$

The regularity of the measure together with Urysohn’s Lemma then implies that $\mu$ satisfies $\mu(g+E) = \mu(E)$ for all Borel $E$ and all $g \in G$ (this might be trickier than it first seems).

Consider the family $\mathcal{F} = \{T_g : g \in G\}$ of operators on $M(X) = C(X)^*$ given by $T_g = L_g^*$. Then by Exercise G in Notes 11 $T_g$ is weak* continuous for all $g$. Moreover, for all $f \in C(X), \nu \in M(X)$ and $g,h \in G$,

$\langle T_g T_h \nu, f \rangle = \langle \nu, L_h L_g f \rangle = \langle \nu, L_g L_h f \rangle = \langle T_h T_g \nu , f \rangle .$

Therefore $\mathcal{F}$ is a commuting family. Now let $C$ be the subset of $M(X)$ consisting of all probability measures. Then it is easy to see that $C$ is weak* closed and convex, and that $\mathcal{F}$ leaves $C$ invariant. By the Kakutani–Markov Theorem, $\mathcal{F}$ has a fixed point $\mu \in C$. By definition of $C$, $\mu$ is a regular Borel probability measure on $G$. By the fact that $\mu$ is a fixed point for $\mathcal{F}$ we have $\langle T_g \mu , f \rangle = \langle \mu , f \rangle$, which is just another way of writing (*). That completes the proof.

Exercise A: It may seem as if the same argument would give the existence of a translation invariant regular Borel probability measure on a locally compact Hausdorff space.

1. Prove why the theorem fails for non–compact spaces.
2. What part of the argument breaks down?
3. Make sure you know why that part doesn’t break down in the compact case.