Compact groups have property (T) by Haar averaging

Fix a strongly continuous unitary representation $\pi$ and a unit vector $\xi$, and put $f(x)=\pi (x)\xi$. For each integer $n\ge 1$, let ${\mathcal{U}}_n$ be the set of open subsets $U\subseteq K$ on which $\parallel f(y)-f(z)\parallel <1\mathrm{/}n$ for all $y,z\in U$. Continuity makes ${\mathcal{U}}_n$ an open cover. Compactness gives a finite subcover; discard empty members, and finite choice supplies one sample point $x_j$ in each remaining member $U_j$. AC chooses such a finite subcover and its sample points for every $n$. Set $E_1=U_1$ and $E_j=U_j\setminus {\bigcup }_{r<j}{U}_r$; these are a finite Borel partition of $K$. The simple function $s_n={\sum }_{j}f(x_j){\mathbf{1}}_{E_j}$ satisfies ${\sup }_{x\in K}\parallel f(x)-s_n(x)\parallel \le 1\mathrm{/}n$. Thus $f$ is strongly measurable; since $\parallel f(x)\parallel =1$ and $\mu (K)=1$, [F4] makes it Bochner integrable.

Define $\eta ={\int }_{K}fd\mu$. The displacement $d(x)=\parallel f(x)-\xi \parallel$ is continuous, so [F3] gives a maximum $M={\sup }_{x\in K}d(x)$. For each simple approximant $s_n={\sum }_{j}f(x_j){\mathbf{1}}_{E_j}$ from step 1.1, [F5] and $\mu (K)=1$ give $\parallel {\int }_K{s}_{n}d\mu -\xi \parallel =\parallel {\sum }_j\mu (E_j)(f(x_j)-\xi )\parallel \le {\sum }_j\mu (E_j)d(x_j)\le M{\sum }_j\mu (E_j)=M$. Passing to the Bochner-integral limit yields $\parallel \eta -\xi \parallel \le M$. If $\xi$ is $(K,\epsilon )$-invariant, then $d(x)<\epsilon$ for every $x$; since the maximum is attained, $M<\epsilon$. If instead the explicit hypothesis ${\sup }_{K}d<1$ holds, the same bound gives $\parallel \eta -\xi \parallel <1$, hence $\eta \mathrm{\ne }0$.

For each $h\in K$, bounded linearity of $\pi (h)$ and [F5] give $\pi (h)\eta ={\int }_K\pi (h)f(x)d\mu (x)={\int }_{K}f(hx)d\mu (x)$. To see the last integral equals $\eta$, use the simple approximants from step 1.1: if $s_n={\sum }_j{v}_j{\mathbf{1}}_{E_j}$, then $s_n(hx)={\sum }_j{v}_j{\mathbf{1}}_{h^{-1}{E}_j}(x)$, and [F5]–[F6] give ${\int }_K{s}_n(hx)d\mu ={\sum }_j\mu (h^{-1}{E}_j)v_j={\sum }_j\mu (E_j)v_j={\int }_K{s}_n(x)d\mu$. The functions $s_n(h\cdot )$ are simple and converge uniformly to $f(h\cdot )$, whose norm is constantly one, so [F4] makes $f(h\cdot )$ Bochner integrable and [F5] makes these integrals converge to ${\int }_{K}f(hx)d\mu (x)$. The original simple integrals converge to $\eta$. Therefore $\pi (h)\eta =\eta$ for every $h\in K$.

If $0<\epsilon \le 1$ and $\xi$ is a $(K,\epsilon )$-invariant unit vector, step 2.1 gives $\parallel \eta -\xi \parallel \le M<\epsilon \le 1$, so $\eta \mathrm{\ne }0$; step 2.2 makes it invariant. Thus $(K,\epsilon )$ is a Kazhdan pair for every such $\epsilon$. A representation on the zero Hilbert space has no unit vector and satisfies the pair implication vacuously.

Taking $\epsilon =1$ shows that the compact set $K$ is a Kazhdan set. If a strongly continuous representation of $K$ has almost invariant vectors, its almost invariance supplies a $(K,1)$-invariant unit vector; step 3.1 gives a nonzero invariant vector. Hence $K$ has property (T), and the explicit average estimate and invariance were proved in steps 2.1–2.2. AC is used for the normalized Haar probability and the sequence of finite simple approximants; no further Choice use occurs.