Proof. Since h is vanishing at infinity with respect to $\tau $ , for each $n\in \mathbb {N}$ , there exists a finite set $F_n\subset G$ such that $\| h(g)\| _{2,\tau }<1/n$ for $g\notin F_n$ . Since $\phi $ is $L_2$ -compact, there exists a sequence of finite-rank operators $\{\phi _n\}_{n\in \mathbb {N}}$ on A such that $\| \widetilde {\phi }-\widetilde {\phi }_n\| _{2,\tau }<1/n$ . Define $T_n$ on $A\rtimes _{\alpha , r}G$ by

$$ \begin{align*} T_n\bigg(\sum_{g\in G} a_g\delta_g\bigg) = \sum_{g\in F_n}\phi_n(a_g)h(g)\delta_{g}, \end{align*} $$

where $\sum _{g\in G} a_g\delta _g \in A\rtimes _{\text {alg}}G$ . The induced map $\widetilde {T}_n$ on $L^2(A\rtimes _{\alpha , r}G, \widetilde {\tau })$ is finite-rank as $\phi _n$ is so. For $x=\sum _{g\in G}a_g\delta _g \in A\rtimes _{\text {alg}}G$ ,

$$ \begin{align*} \Phi(x)-T_n(x) = \sum_{g\in F_n} (\phi-\phi_n)(a_g)h(g)\delta_g +\sum_{g\notin F_n}\phi(a_g)h(g)\delta_g. \end{align*} $$

Denote these two sums by $M_1$ and $M_2$ , respectively. We estimate each term separately and use the inequality $\| \widetilde {\Phi }(x)-\widetilde {T}_n(x)\| ^2_{2,\tilde {\tau }} \leq 2(\| M_1\| ^2_{2, \tilde {\tau }} + \| M_2\| ^2_{2, \tilde {\tau }})$ . For $M_1$ , we compute

$$ \begin{align*} \| M_1\|^2_{2,\tilde{\tau}} &=\tau\bigg(\sum_{g\in F_n} h^*(g)(\phi-\phi_n)^*(a_g)(\phi-\phi_n)(a_g)h(g)\bigg) \\ &= \sum_{g\in F_n} \tau(h^*(g)(\phi-\phi_n)^*(a_g)(\phi-\phi_n)(a_g)h(g)) \\ &\leq \sum_{g\in F_n}\tau((\phi-\phi_n)^*(a_g)(\phi-\phi_n)(a_g)) \tau( h^*(g)h(g))\\ &= \sum_{g\in F_n} \| (\widetilde{\phi}-\widetilde{\phi}_n)(a_g)\|^2_{2,\tau} \| h(g)\|^2_{2,\tau}. \end{align*} $$

Since $\| \widetilde {\phi }-\widetilde {\phi }_n\| _{2,\tau }<1/n$ and h is bounded, the last term above is dominated by

$$ \begin{align*} \sum_{g\in F_n} \frac{1}{n^2}\| a_g\|^2_{2,\tau} K \leq \frac{K}{n^2}\sum_{g\in F_n}\tau(a^*_ga_g) = \frac{K}{n^2}\| x\|^2_{2,\tilde{\tau}}, \end{align*} $$

where $K=\sup _{g\in G}\| h(g)\| ^2_{2,\tau }$ . For $M_2$ , using $\| h(g)\| _{2, \tau } \leq {1}/{n}$ for $g\notin F$ and the boundedness of $\widetilde {\phi }$ , a similar computation gives

$$ \begin{align*} \| M_2\|^2_{2,\tilde{\tau}} \leq \sum_{g\notin F_n}\frac{1}{n^2} \| \widetilde{\phi}\|^2_{2,\tau} \| a_g\|^2_{2,\tau} \leq \frac{1}{n^2} \| \widetilde{\phi}\|^2_{2,\tau} \| x\|^2_{2,\tilde{\tau}}. \end{align*} $$

Therefore, $\| \widetilde {\Phi }(x)-\widetilde {T}_n(x)\| _{2,\tilde {\tau }} \leq ({1}/{n})(2(K+\| \widetilde {\phi }\| ^2_{2,\tau }))^{{1}/{2}}\| x\| _{2,\tilde {\tau }}$ . Thus, $\widetilde {\Phi }$ is compact.

Proof. Since $(A, \tau )$ has the Haagerup property, there exists a $\tau $ -decreasing sequence of unital completely positive maps $\{\phi _n\}_{n\in \mathbb {N}}$ on A satisfying conditions (1) and (2) of Definition 2.2. The Haagerup property of $\alpha $ yields a sequence of bounded positive definite functions $\{h_n\colon G\to Z(A)\}_{n\in \mathbb {N}} \subseteq C_{0,\tau }(G,A)$ such that $h_n(g)\in D_g$ for $g\in G$ and $h_n\to 1$ pointwise with respect to $\tau $ . For $n\in \mathbb {N}$ , define a completely positive map $\Phi _n$ on $A\rtimes _{\alpha , r}G$ corresponding to $\phi _n$ and $h_n$ as in (3.2). The sequence $\{\Phi _n\}_{n\in \mathbb {N}}$ is $\widetilde {\tau }$ -decreasing since $\{\phi _n\}_{n\in \mathbb {N}}$ is $\tau $ -decreasing. By Lemma 3.3, $\Phi _n$ is $L_2$ -compact for all $n\in \mathbb {N}$ . Let $f=\sum _{g\in F}a_g\delta _g \in A\rtimes _{\text {alg}}G$ for some finite set $F\subset G$ . Then,

$$ \begin{align*} \| \Phi_n(f)-f\|^2_{2,\tilde{\tau}} &=\bigg\|\sum_{g\in F}\phi_n(a_g)h_n(g)\delta_g-a_g\delta_{g}\bigg\|^2_{2,\tilde{\tau}}\\ &= \tau\bigg(\sum_{g\in F} (\phi_n(a_g)h_n(g)-a_g)^*(\phi_n(a_g)h_n(g)-a_g)\bigg)\\ &=\sum_{g\in F}\| \phi_n(a_g)h_n(g)-a_g\|^2_{2, \tau}\\ &\leq 2 \sum_{g\in F} (\| \phi_n(a_g)h_n(g)-\phi_n(a_g)\|^2_{2, \tau} + \| \phi_n(a_g)-a_g\|^2_{2, \tau})\\ &\leq 2 \sum_{g\in F} (\| \phi_n(a_g)\|^2_{2,\tau}\| h_n(g)-1\|^2_{2, \tau} + \| \phi_n(a_g)-a_g\|^2_{2, \tau})\\ &\leq 2 \sum_{g\in F} (K^2\| h_n(g)-1\|^2_{2, \tau} + \| \phi_n(a_g)-a_g\|^2_{2, \tau}) \to 0\quad \text{as } n\to \infty, \end{align*} $$

where $K=\sup _{g\in F}\| \phi _n(a_g)\| _{2,\tau }\leq \sup _{g\in F} \| a_g\| _{2, \tau }$ as each $\phi _n$ is unital. Therefore, $\Phi _n$ converges pointwise to the identity map on $L^2(A\rtimes _{\alpha , r}G, \widetilde {\tau })$ . This completes the proof.

Proof. Since $(A\rtimes _{\alpha , r}G, \widetilde {\tau })$ has the Haagerup property, there exists a sequence $\{\Phi _n\}_{n\in \mathbb {N}}$ of unital completely positive maps on $A\rtimes _{\alpha , r}G$ satisfying the conditions of Definition 2.2. Recall the conditional expectation $E\colon A\rtimes _{\alpha , r}G\to A$ from (2.1). For $n\in \mathbb {N}$ , define a unital completely positive map on A by

$$ \begin{align*} \phi_n(a)=E\circ \Phi_n(a) \quad \text{for } a\in A. \end{align*} $$

Then, $\tau \circ \phi _n \leq \tau $ as $\widetilde {\tau }\circ \Phi _n \leq \widetilde {\tau }$ . Let $\epsilon>0$ . Since $\Phi _n$ is $L_2$ -compact, there exists a finite-rank operator $R_n$ on $A\rtimes _{\alpha , r}G$ such that

$$ \begin{align*} \| \Phi_n(f)-R_n(f)\|_{2,\tilde{\tau}} \leq \epsilon \| f\|_{2, \tilde{\tau}} \end{align*} $$

for $f\in A\rtimes _{\alpha , r}G$ . Now, $E\circ R_n$ is also a finite-rank operator on A and

$$ \begin{align*} \| \phi_n(a)-E\circ R_n(a)\|_{2,\tau} & = \| E\circ\Phi_n(a)-E\circ R_n(a)\|_{2,\tau} \\ & \leq \| \Phi_n(a)- R_n(a)\|_{2,\tilde{\tau}}\leq \epsilon \| a\|_{2, \tilde{\tau}} = \epsilon\| a\|_{2,\tau} \quad \text{for } a\in A. \end{align*} $$

For $a\in A$ ,

$$ \begin{align*} \| \phi_n(a)-a\|^2_{2,\tau} &= \tau( (\phi_n(a)-a)^*(\phi_n(a)-a)) \\ &= \tau( (E(\phi_n(a)-a))^*(E(\phi_n(a)-a)))\\ &\leq \tau\circ E( (\phi_n(a)-a)^*(\phi_n(a)-a) ) \\ &= \| \Phi_n(a)-a\|^2_{2,\tilde{\tau}} \to 0 \quad \text{as } n\to \infty. \end{align*} $$

Therefore, $(A, \tau )$ has the Haagerup property.

Proof. (1) $\implies $ (2). Lemma 3.5 ensures that $(A, \tau )$ has the Haagerup property.

Since $(A\rtimes _{\alpha , r}G, \widetilde {\tau })$ has the Haagerup property, there exists a sequence $\{\Phi _n\}_{n\in \mathbb {N}}$ of unital completely positive maps on $A\rtimes _{\alpha , r}G$ satisfying the conditions of Definition 2.2. To prove that G has the Haagerup property, we define $\eta _n\colon G\to \mathbb {C}$ by

$$ \begin{align*} \eta_n(g) = \widetilde{\tau}(\Phi_n(1_g\delta_g)(1_{g}\delta_{g})^*), \end{align*} $$

where $1_g$ is the unit of $D_g$ , $g\in G$ and $n\in \mathbb {N}$ . Now, $\eta _n(e) =\widetilde {\tau }(\Phi _n(1_e\delta _e)1_e\delta _e) =1 $ . For $g_1,g_2,\ldots ,g_k\in G$ and $c_1,c_2,\ldots , c_k\in \mathbb {C}$ ,

$$ \begin{align*} \sum_{i, j=1}^{k} c_{i}\overline{c_j}\eta_n(g_j^{-1}g_i)&= \sum_{i, j=1}^{k} c_{i}\overline{c_j}\widetilde{\tau}(\Phi_n(1_{g_j^{-1}g_i}\delta_{g_{j}^{-1}g_i}) (1_{g_j^{-1}g_i}\delta_{g_j^{-1}g_i})^*)\\ &= \sum_{i, j=1}^{k} c_{i}\overline{c_j}\widetilde{\tau}(\Phi_n (1_{g_j^{-1}g_i}\delta_{g_{j}^{-1}g_i}) (1_{g_i^{-1}g_j}\delta_{g_{i}^{-1}g_j})) \\ &=\sum_{i, j=1}^{k} \widetilde{\tau}(c_i\overline{c_j} \Phi_n((1_{g_j^{-1}}\delta_{g_{j}^{-1}})(1_{g_i}\delta_{g_i})) (1_{g_i^{-1}}\delta_{g^{-1}_i}1_{g_j}\delta_{g_j}))\\ &= \sum_{i, j=1}^{k} \widetilde{\tau}(\overline{c_j}(1_{g_j}\delta_{g_j}) \Phi_n((1_{g_j^{-1}}\delta_{g_{j}^{-1}})(1_{g_i}\delta_{g_i})) (c_{i}1_{g_i^{-1}}\delta_{g^{-1}_i}))\\ &= \widetilde{\tau}\,\bigg(\sum_{i, j=1}^{k} (\overline{c_j}(1_{g_j}\delta_{g_j}) \Phi_n((1_{g_j^{-1}}\delta_{g_{j}^{-1}})(1_{g_i}\delta_{g_i})) (c_{i}1_{g_i^{-1}}\delta_{g^{-1}_i}))\bigg) \geq 0. \end{align*} $$

Therefore, $\eta _n$ is positive definite on G. For $g\in G$ ,

$$ \begin{align*} |\eta_n(g)-1| &= |\widetilde{\tau}(\Phi_n(1_g\delta_g)1_{g^{-1}}\delta_{g^{-1}}) -1| \\ &= |\widetilde{\tau}(\Phi_n(1_g\delta_g)1_{g^{-1}}\delta_{g^{-1}}) -\widetilde{\tau}(1_g\delta_g\cdot 1_{g^{-1}}\delta_{g^{-1}})| \\ &= |\widetilde{\tau}((\Phi_n(1_g\delta_g) - 1_g\delta_g) 1_{g^{-1}}\delta_{g^{-1}})| \\ &\leq |\widetilde{\tau}(\Phi_n(1_g\delta_g) - 1_g\delta_g)|\\ & \leq \| \Phi_n(1_g\delta_g) - 1_g\delta_g\|_{2,\tilde{\tau}} \to 0 \quad \text{as } n\to 0. \end{align*} $$

Therefore, $\eta _n\to 1$ pointwise. Moreover, $\eta _n$ is vanishing at infinity as $\Phi _n$ is $L_2$ -compact. Hence, G has the Haagerup property.

(2) $\implies $ (3). Since G has the Haagerup property, there exists a sequence of positive linear functions $\{\eta _n\}_{n\in \mathbb {N}}$ such that $\eta _n(e)=1$ , $\eta _n$ is vanishing at infinity for all $n $ and $\eta _n\to 1$ pointwise. Let $1_g$ be the unit element of $D_{g}$ for $g\in G$ . For $n\in \mathbb {N}$ , we define $h_n(g)=\eta _n(g)1_g$ . Then, the sequence $\{h_n\}_{n\in \mathbb {N}}$ fulfils the requirement of Definition 3.1 for the partial action $\alpha $ . Therefore, $\alpha $ has the Haagerup property.

(3) $\implies $ (1). This follows from Proposition 3.4.

Proof. Since $\tau _{n+1}\circ \phi _n=\tau _n$ , each map $U_n$ is a well-defined isometry:

$$ \begin{align*} \|U_n(a+N_{\tau_n})\|_{2,\tau_{n+1}}^2 = \tau_{n+1}(\phi_n(a)^*\phi_n(a)) = \tau_n(a^*a) = \|a+N_{\tau_n}\|_{2,\tau_n}^2. \end{align*} $$

Hence, we may identify each $L^2(A^{(n)},\tau _n)$ as a subspace of $L^2(A^{(n+1)},\tau _{n+1})$ . Set

$$ \begin{align*} \mathcal{H}_0=\bigcup_{n\geq 1} L^2(A^{(n)},\tau_n) \end{align*} $$

and let $\mathcal H$ be the Hilbert space completion of $\mathcal {H}_0$ . Then, $\mathcal H=\varinjlim L^2(A^{(n)},\tau _n)$ . Let $\lambda _n\colon A^{(n)}\to A$ denote the canonical embedding of the inductive limit. Define a linear map $\pi _0 \colon \bigcup _{n\geq 1}\lambda _n(A^{(n)})\to \mathcal {H}_0$ by

$$ \begin{align*} \pi_0(\lambda_n(a))=a+N_{\tau_n}\in L^2(A^{(n)},\tau_n)\subseteq \mathcal{H}_0. \end{align*} $$

If $\lambda _n(a)=\lambda _m(b)$ in A, then choose some $k\geq n,m$ with $\phi _{n,k}(a)=\phi _{m,k}(b)$ , where $\phi _{n,k}\colon A^{(n)} \to A^{(k)}$ is given by $\phi _{n,k}=\phi _k\circ \phi _{k-1}\circ \cdots \circ \phi _{n+1}\circ \phi _n$ . Similarly, $\phi _{m,k}$ . Analogously, we can define $U_{n,k}$ and $U_{m,k}$ . Since

$$ \begin{align*} U_{n,k}(a+N_{\tau_n}) =\phi_{n,k}(a)+N_{\tau_k} =\phi_{m,k}(b)+N_{\tau_k} =U_{m,k}(b+N_{\tau_m}), \end{align*} $$

the map $\pi _0$ is well defined. Moreover,

$$ \begin{align*} \| \pi_0(\lambda_n(a))\|_{2,\tau_n}^2 =\tau_n(a^*a) =\tau(\lambda_n(a)^*\lambda_n(a)), \end{align*} $$

so $ \pi _0 $ preserves the inner product induced by $\tau $ . Thus, $\pi _0$ extends uniquely to an isometry $\pi \colon L^2(A,\tau )\to \mathcal H$ . Since $\bigcup _{n\geq 1}\lambda _n(A^{(n)})$ is dense in A and $\mathcal H_0$ is dense in $\mathcal H$ , the range of $\pi $ is dense in $\mathcal H$ . Hence, $\pi $ is unitary.

Proof. Let $\mathcal H=L^2(A,\tau )$ and $\mathcal {H}_n=L^2(A^{(n)},\tau _n)$ , viewed as a closed subspace of $\mathcal H$ . Let $P_n$ denote the orthogonal projection of $\mathcal H$ onto $\mathcal H_n$ . Then, the operator $T^{\prime }$ can be written as $T^{\prime } = T P_n$ . Thus, $T^{\prime }$ acts on $\mathcal H_n$ as T and zero on $\mathcal H_n^{\perp }$ . Since T is compact on the Hilbert space $\mathcal H_n$ , there exists a sequence of finite-rank operators $T_m$ on $\mathcal H_n$ such that

$$ \begin{align*} \| T-T_m\| \to 0 \quad \text{as} \ m\to \infty. \end{align*} $$

Define $T^{\prime }_m = T_m P_n$ on $\mathcal H$ . Then, each $T^{\prime }_m$ has finite-rank in $\mathcal H$ , because its range is contained in the finite dimensional subspace $\text {range}(T_m)\subseteq H_n$ . Moreover,

$$ \begin{align*} \| T^{\prime}-T^{\prime}_m\| = \| (T-T_m)P_n\| \leq \| T-T_m\| \to 0 \quad \text{as } \ m\to \infty. \end{align*} $$

Hence, $T^{\prime }$ is compact.

Proof. Since each $(A^{(n)}, \tau _n)$ has the Haagerup property, there exists a sequence $\{\varphi _{n,k}\}_{k\in \mathbb {N}}$ of unital completely positive maps on $A^{(n)}$ such that $\tau _n\circ \varphi _{n,k}\leq \tau _n$ for all $k\in \mathbb {N}$ . Moreover, $\varphi _{n,k}$ is $L_2$ -compact for all $k\in \mathbb {N}$ and $\| \varphi _{n,k}(a)-a\| _{2,\tau _n}\to 0$ as $k\to \infty $ for all $a\in A^{(n)}$ . We extend $\varphi _{n,k}$ to a unital completely positive map $\psi _{n,k}$ on A satisfying $\psi _{n,k}(\lambda _n(a)) = \lambda _n(\varphi _{n,k}(a))$ for all $a\in A^{(n)}$ . Then, $\psi _{n,k}$ is $\tau $ -decreasing, that is, $\tau \circ \psi _{n,k}\leq \tau $ . Set $\psi _n =\psi _{n,n}$ for $n\in \mathbb {N}$ . Lemma 4.2 ensures that $\psi _n$ is $L_2$ -compact for all $n\in \mathbb {N}$ . Let $x=\lambda _n(a)\in \bigcup _{m}\lambda _m(A^{(m)})$ , where $a\in A^{(n)}$ . Then,

$$ \begin{align*} \| \psi_n(x)-x\|_{2,\tau} = \| \psi_{n,n}(\lambda_n(a))-\lambda_n(a)\|_{2, \tau} & = \| \lambda_n(\varphi_{n,n}(a))-\lambda_n(a)\|_{2, \tau} \\ & = \| \varphi_{n,n}(a)-a\|_{2, \tau_n} \to 0 \quad \text{as } n\to \infty. \end{align*} $$

After passing to the limit, we conclude that $\| \psi _n(x)-x\| _{2,\tau } \to 0$ as $n\to \infty $ for all $x\in A=\varinjlim A^{(n)}$ . Therefore, $(A, \tau )$ has the Haagerup property.

Proof. Since G is amenable, by [Reference Hossain10, Theorem 3.6],

$$ \begin{align*} A\rtimes_{\alpha, r} G\cong \varinjlim A^{(n)}\rtimes_{\alpha^{(n)}, r}G. \end{align*} $$

As $\{\tau _n\}_{n\in \mathbb {N}}$ is a G-invariant sequence of tracial states, A has a G-invariant tracial state $\tau $ and $\widetilde {\tau } = \varinjlim \widetilde {\tau }_n$ , where $\widetilde {\tau }= \tau \circ E$ (see [Reference Hossain10, Proposition 4.10]). Since each $\alpha _n$ has the Haagerup property and $(A^{(n)}, \tau _n)$ has the Haagerup property, it follows from Proposition 3.4 that $( A^{(n)}\rtimes _{\alpha ^{(n)},r}G, \widetilde {\tau }_n)$ has the Haagerup property. Therefore, Proposition 4.3 ensures that $(A\rtimes _{\alpha , r}G, \widetilde {\tau })$ has the Haagerup property.