Proof. By orthogonality,

$$ \begin{align*} C^*(a,b)&\cong C^*(a)\oplus C^*(b)\\ &\cong C_0(\sigma(a)\setminus \{0\})\oplus C_0(\sigma(b)\setminus \{0\}) \quad\\ &\cong C_0( (\sigma(a) \setminus \{0\}) \sqcup (\sigma(b) \setminus \{0\} )). \end{align*} $$

Since $\sigma (a)$ and $\sigma (b)$ are compact, it follows from (2-1) that the disjoint union $(\sigma (a) \setminus \{0\}) \sqcup (\sigma (b) \setminus \{0\} )$ is homeomorphic to $U := (\sigma (a) \cup \sigma (b)) \setminus \{0\} \subseteq \mathbb {C}$ . Under the identification $\Phi \ \colon C^*(a,b) \xrightarrow {\cong } C_0(U)$ , we find

$$ \begin{align*} \Phi(a)(z)&=\begin{cases} z\quad\text{if }z\in\sigma(a),\\ 0\quad\text{if }z\in \sigma(b), \end{cases}\\ \Phi(b)(z)&=\begin{cases} 0\quad\text{if }z\in\sigma(a),\\ z\quad\text{if }z\in \sigma(b), \end{cases}\\ \Phi(a+b)(z)&=z \quad \text{for all } z \in U. \end{align*} $$

Since $\Phi (a+b)$ separates points and vanishes nowhere in U, it generates $C_0(U)$ as a $C^*$ -algebra (by the locally compact version of the Stone–Weierstrass theorem). Hence, $a,b \in C^*(a+b)$ as desired.

Proof. Let F be the subgraph of E consisting only of the edges $E^1_{\mathrm {sink}} \cup E^1_{\mathrm {bdry}}$ along with their ranges and sources, together with $E^0_{\mathrm {sink}}$ . Since F is entrance complete (in the sense that for each edge $e \in F^1$ , we have $r_E^{-1}(r_E(e)) \subseteq F^1$ ), the family

$$ \begin{align*} \{p_v : v \in E^0_{\mathrm{sink}} \cup E^0_{\mathrm{bdry}}\} \cup \{s_e : e \in E^1_{\mathrm{sink}} \cup E^1_{\mathrm{bdry}}\} \end{align*} $$

is a Cuntz–Krieger F-family in $C^*(E)$ , so there is a $*$ -homomorphism ${\varphi \ \colon \!C^*(F) \to B}$ . Clearly, $\varphi $ is surjective; since F is acyclic, the Cuntz–Krieger uniqueness theorem [Reference Fowler, Laca and Raeburn6, Theorem 2] shows that $\varphi $ is injective and hence an isomorphism. (As noted earlier, this paper uses the opposite convention for the direction of edge partial isometries to that used in [Reference Fowler, Laca and Raeburn6, Reference Rangaswamy and Tomforde19].)

Note that F is acyclic and has countably many paths, since the maximum path length in F is $2$ . By [Reference Rangaswamy and Tomforde19, Theorem 6.5], $C^*(F)$ has an elementary composition series (of countable length). It then follows from [Reference Huruya10] that every self-adjoint element of ${C^*(F) \cong B}$ has countable spectrum, as required.

Proof. By orthogonality of range projections of edges and Lemma 4.2, we have that ${a^*d = a^*(b+c) = d^*(b+c) = 0}$ . (In (4-2), the sum defining a contains terms of the form $\alpha _{y,k}s_{y,k}s^*_{y,k-1}$ for $k \ge 2$ , so $a^*d$ indeed vanishes.) Hence,

$$ \begin{align*} g^*g = a^*a + (b+c)^*(b+c) + d^*d. \end{align*} $$

The positive elements $a^*a + (b+c)^*(b+c)$ and $d^*d$ are orthogonal, and, away from 0, they have disjoint spectra by condition (iii) above. Therefore, $d^*d$ belongs to $C^*(g)$ by Lemma 2.5. It follows that $A(E) \subseteq C^*(g)$ by condition (i) above.

We now show that all partial isometries associated with interior edges belong to $C^*(g)$ . Let $e \in E^1_{\mathrm {int}}$ . Then, $s_e s_e ^* \in A(E) \subseteq C^*(g)$ , so $s_e s_e^* g \in C^*(g)$ . Observe that

$$ \begin{align*} s_e s_e^* g = s_e s_e^* d = \epsilon_e s_e \xi^{\frac{1}{2}} \end{align*} $$

by orthogonality of range projections. Next, by condition (ii) above, there exists ${h \in A(E) \subseteq C^*(g)}$ such that $\xi ^{\frac 12}h =s_e^*s_e$ . So,

$$ \begin{align*} s_e s_e^* g h = \epsilon_e s_e \xi^{\frac{1}{2}}h = \epsilon_e s_e s_e^* s_e = \epsilon_e s_e \end{align*} $$

is an element of $C^*(g)$ . Since $\epsilon _e>0$ , we get that $s_e \in C^*(g)$ , as desired.

Proof. Recall from Notation 4.1 that Y denotes the set of vertices $s(E^1_{\mathrm {bdry}}) \subseteq E^0_{\mathrm {int}}$ and that the set of boundary edges is equal to $\{e_{y,n}\}_{y \in Y, n \ge 1}$ . Our first claim is that for any $y \in Y$ , the partial isometry $s_{y,1}$ belongs to $C^*(g, a_1)$ . By the construction of $\xi $ and d in (4-4), there exists an element $h \in A(E)$ such that $\xi ^{\frac 12}h= s_{y,1}^* s_{y,1}$ . By Lemma 4.2,

$$ \begin{align*} a_1h = \alpha_{y,1} s_{y,1}s_{y,1}^* s_{y,1} = \alpha_{y,1} s_{y,1}. \end{align*} $$

Observe that $a_1h$ belongs to $C^*(g, a_1)$ by Proposition 5.1 and, therefore, $s_{y,1}$ also belongs to $C^*(g, a_1)$ since $\alpha _{y,1}>0$ . This proves the first claim.

Next, since $\xi \in A(E) \subseteq C^*(g)$ (Proposition 5.1), we deduce that

$$ \begin{align*} a = \sum_{\substack{y \in Y \\ n \ge 1}} \alpha_{y,n+1} s_{y,n+1} s_{y,n}^* = a_1 - \bigg( \sum_{y \in Y} \alpha_{y,1} s_{y,1} \bigg)\, \xi^{\frac{1}{2}} \end{align*} $$

belongs to $C^*(g, a_1)$ . Fix $y \in Y$ and note that ${a s_{y,n} = \alpha _{y,n+1} s_{y,n+1}}$ for any $n \ge 1$ by Lemma 4.2. It follows from the first claim that $s_{y,2} \in C^*(g, a_1)$ since $\alpha _{y,2}> 0$ . Continuing inductively, we deduce that $s_{y,n} \in C^*(g, a_1)$ for any $n \ge 1$ . Since $y \in Y$ is arbitrary, we have shown that all partial isometries associated with boundary edges belong to $C^*(g, a_1)$ , which finishes the proof.

Proof. For any $v \in E^0_{\mathrm {bdry}}$ ,

$$ \begin{align*} \|p_v a_1\| &\le \sum_{y\in Y} \alpha_{y,1} \|p_v s_{y,1}\| + \sum_{\substack{y \in Y \\ n \ge 1}} \alpha_{y,n+1} \|p_vs_{y,n+1} s^*_{y,n}\|\\ &= \sum_{\substack{y \in Y, n \ge 1\\ r(e_{y,n}) = v}} \alpha_{y,n}. \end{align*} $$

Therefore, by (4-1),

$$ \begin{align*} \sum_{\ell \ge m} \sum_{v \in V_\ell} \|p_v a_1\| \le \delta_m \end{align*} $$

for any m, as desired. Taking $m = 1$ gives that $\sum _{v \in E^0_{\mathrm {bdry}}} \|p_v a_1\| \le \delta _1 < \infty $ . Hence, the sum $\sum _{v \in E^0_{\mathrm {bdry}}} p_v a_1$ converges absolutely in norm. Since all edges in $\{e_{y,n}\}_{y \in Y, n \ge 1}$ have range in $E^0_{\mathrm {bdry}}$ , it is clear from (5-2) that this sum is equal to $a_1$ .

Proof. For $k \ge 1$ , we have the following by Lemma 4.2:

(5-5) $$ \begin{align} \frac{1}{\gamma_{M,N}^k}g_{M,N}^k &= \ \frac{1}{\gamma_{M,N}^k}a_M^k + \frac{1}{\gamma_{M,N}^k} \sum_{m \ge M} \delta_m^k q_m + \sum_{(m,n) \ge (M,N)} \frac{\gamma_{m,n}^k-\delta_m^k}{\gamma_{M,N}^k} t_{m,n}t_{m,n}^* \nonumber\\&\quad + \sum_{(m,n) \ge (M,N)} \frac{\beta_{m,n} t_{m,n} }{\gamma_{M,N}^k} \bigg( \sum_{j=1}^{k} \gamma_{m,n}^{k-j} \ a_M^{j-1} \bigg). \end{align} $$

By assumption, we have $\delta _M < \gamma _{M,N}$ and $\|a_M\| \le \delta _M$ by Lemma 5.3, so the first term converges to $0$ as $k \to \infty $ . The second term converges to 0 by the monotone convergence theorem, since $\delta _m \le \delta _M < \gamma _{M,N}$ for all $m \ge M$ . For the third and fourth terms, the sums over $(m,n) \ge (M, N+1)$ converge to 0 as $k \to \infty $ . Indeed,

$$ \begin{align*} \bigg\|\sum_{(m,n) \ge (M,N+1)} \frac{\gamma_{m,n}^k-\delta_m^k}{\gamma_{M,N}^k} t_{m,n}t_{m,n}^*\bigg\| \le \max_{(m,n) \ge (M,N+1)} \bigg( \frac{\gamma_{m,n}^k-\delta_m^k}{\gamma_{M,N}^k}\bigg) \le \frac{\gamma_{M,N+1}^k}{\gamma_{M,N}^k} \end{align*} $$

since the projections $t_{m,n}t_{m,n}^*$ are mutually orthogonal and the double sequence $(\gamma _{m,n})$ is decreasing with respect to the lexicographic ordering. Note that $\sum _{j=1}^\infty \gamma _{M,N+1}^{-j} \|a_M\|^{j-1}$ converges since $\delta _M < \gamma _{M,N+1}$ . Then, for $(m,n) \ge (M,N+1)$ ,

$$ \begin{align*} \bigg\|\frac{\beta_{m,n} t_{m,n}}{\gamma_{M,N}^k} \sum_{j=1}^{k} \gamma_{m,n}^{k-j} \ a_M^{j-1}\bigg\| &\le \frac{\beta_{m,n} \gamma_{M,N+1}^k}{\gamma_{M,N}^k} \sum_{j=1}^\infty \gamma_{M,N+1}^{-j} \|a_M\|^{j-1}\\ &\le C_{M,N} \beta_{m,n} \bigg(\frac{\gamma_{M,N+1}}{\gamma_{M,N}}\bigg)^k, \end{align*} $$

where $C_{M,N}$ is a constant independent of $m,n$ and k. It is then clear that the sum of these terms over $(m,n) \ge (M,N+1)$ converges to 0 as $k \to \infty $ . Overall, when taking the limit $k \to \infty $ in (5-5), all that remains is the limit of the summands $(m,n) = (M,N)$ from the third and fourth terms. Therefore,

(5-6) $$ \begin{align} \frac{1}{\gamma_{M,N}^k}g_{M,N}^k \to t_{M,N}t_{M,N}^* + \beta_{M,N} t_{M,N} \sum_{j=1}^\infty \gamma_{M,N}^{-j} a_M^{j-1} =: y \end{align} $$

as $k \to \infty $ , so $y \in C^*(g_{M,N})$ . Write

$$ \begin{align*} z := \beta_{M,N} \sum_{j=1}^\infty \gamma_{M,N}^{-j} a_M^{j-1}, \end{align*} $$

so $y = t_{M,N}t_{M,N}^* + t_{M,N} z$ . Then, $ yy^* = t_{M,N}t_{M,N}^* + t_{M,N} z z^* t_{M,N}^*. $ Since $\|a_M\| \le \delta _M$ and $\beta _{M,N} = \tfrac 12 (\gamma _{M,N} - \delta _M)$ ,

$$ \begin{align*} \|z\| \leq \frac{\beta_{M,N}}{\gamma_{M,N} - \|a_M\|} \leq \frac{1}{2}. \end{align*} $$

Hence, $yy^*$ is a small perturbation of the projection $t_{M,N}t_{M,N}^*$ , so that $ \lim _{k\to \infty } (yy^*)^{1/k} = t_{M,N}t_{M,N}^*$ . Explicitly, $yy^*$ is the sum of a projection $t_{M,N}t_{M,N}^*$ and a positive element ${x := t_{M,N} z z^* t_{M,N}^*}$ that is contained in $t_{M,N}t_{M,N}^*$ and has norm strictly less than 1. By working in the commutative $C^*$ -algebra $C^*(x, t_{M,N}t_{M,N}^*)$ , it is clear that $(x + t_{M,N}t_{M,N}^*)^{1/k} \to t_{M,N}t_{M,N}^*$ as $k \to \infty $ . Thus, $t_{M,N}t_{M,N}^* \in C^*(g_{M,N})$ . Finally, note that

$$ \begin{align*} t_{M,N}t_{M,N}^* g_{M,N} = \beta_{M,N} t_{M,N} + \gamma_{M,N} t_{M,N}t_{M,N}^* \end{align*} $$

so $t_{M,N} \in C^*(g_{M,N})$ , as claimed.

Next, consider the limit of $({1}/{\delta _M^k})(g_{M+1,1} + \delta _M q_M)^k$ as $k \to \infty $ . The elements $g_{M+1,1}$ and $q_M$ are orthogonal by Lemma 4.2; hence, $(g_{M+1,1} + \delta _M q_M)^k = g_{M+1,1}^k + \delta _M^k q_M$ . By (5-6), the limit $\lim _{k \to \infty }({1}/({\gamma _{M+1,1}^k}))g_{M+1,1}^k$ exists; hence, $\lim _{k \to \infty }({1}/{\delta _{M}^k})g_{M+1,1}^k = 0$ since $\gamma _{M+1,1} < \delta _M$ . It follows that $\lim _{k \to \infty } ({1}/{\delta _M^k})(g_{M+1,1} + \delta _M q_M)^k = q_M$ . So, indeed $q_m \in C^*(g_{M+1,1} + \delta _M q_M)$ , as claimed.

Proof. Fix $m \ge 1$ . Let us show, by induction on n, that $t_{m,n} \in C^*(g_{m,1})$ for all $n \ge 1$ . By Lemma 5.4, the partial isometry $t_{m,1}$ belongs to $C^*(g_{m,1})$ . Assume that ${t_{m,1}, \ldots , t_{m,N} \in C^*(g_{m,1})}$ for some $N \ge 1$ . Then,

$$ \begin{align*} g_{m,N+1} = g_{m,1} - \sum_{n=1}^N \beta_{m,n} t_{m,n} - \sum_{n=1}^N (\gamma_{m,n} - \delta_{m}) t_{m,n} t_{m,n}^* \end{align*} $$

also belongs to $C^*(g_{m,1})$ . By Lemma 5.4, the partial isometry $t_{m,N+1}$ belongs to $C^*(g_{m,N+1})$ and, hence, to $C^*(g_{m,1})$ . Therefore, by induction on n, we have that ${t_{m,n} \in C^*(g_{m,1})}$ for all $n \ge 1$ , as claimed.

Next, let us show, by induction on m, that $q_m$ and $g_{m,1}$ belong to $C^*(g_{1,1})$ for any $m \ge 1$ . Assume that $g_{m,1} \in C^*(g_{1,1})$ for some $m \ge 1$ . By the claim above, the partial isometry $t_f$ belongs to $C^*(g_{1,1})$ for any edge f terminating at the sink $w_m$ . By considering $t_f^* t_f$ for such edges, it follows that $p_v$ belongs to $C^*(g_{1,1})$ for any vertex $v \in V_m$ (since for any vertex $v \in V_m$ , there exists an edge whose source is equal to v and whose range is equal to $w_m$ ). By Lemma 4.2, we have $p_v g_{1,1} = p_v a_1$ for any $v \in E^0_{\mathrm {bdry}}$ and, therefore, $\sum _{v \in V_m} p_v a_1 \in C^*(g_{1,1})$ , where we have used Lemma 5.3 to show that this sum converges. By the first claim, it then follows that

$$ \begin{align*} g_{m+1,1} + \delta_m q_m = g_{m,1} - \sum_{n\ge 1} \beta_{m,n} t_{m,n} - \sum_{n \ge 1} (\gamma_{m,n} - \delta_{m}) t_{m,n} t_{m,n}^* -\sum_{v \in V_m} p_v a_1 \end{align*} $$

also belongs to $C^*(g_{1,1})$ . By Lemma 5.4, the projection $q_m$ belongs to the $C^*$ -subalgebra $C^*(g_{m+1,1} + \delta _m q_m)$ and, hence, to $C^*(g_{1,1})$ . Subtracting $\delta _m q_m$ from the expression above shows that $g_{m+1,1}$ belongs to $C^*(g_{1,1})$ . Therefore, by induction on m, the elements $q_m$ and $g_{m,1}$ belong to $C^*(g_{1,1})$ for any $m \ge 1$ . By the first claim, it follows that ${t_{m,n} \in C^*(g_{1,1})}$ for all $m,n \ge 1$ .

We have shown that for any sink edge f, the partial isometry $t_f$ lies in $C^*(g_{1,1})$ . Therefore, by considering $t_f^* t_f$ for such edges, it follows that $p_v$ belongs to $C^*(g_{1,1})$ for any boundary vertex v. By Lemma 4.2, we have $p_v g_{1,1} = p_v a_1 \in C^*(g_{1,1})$ for any $v \in E^0_{\mathrm {bdry}}$ and, therefore, $a_1$ belongs to $C^*(g_{1,1})$ since $a_1 = \sum _{v \in E^0_{\mathrm {bdry}}} p_v a_1$ (Lemma 5.3). This completes the proof of the proposition.

Proof. By Proposition 5.1, all partial isometries associated with interior edges belong to $C^*(g)$ . By Proposition 5.5, all projections and partial isometries associated with sinks and sink edges belong to $C^*(g)$ . Also, the element $a_1$ defined in (5-2) belongs to $C^*(g)$ by Proposition 5.5 and, therefore, all partial isometries associated with boundary edges belong to $C^*(g)$ by Lemma 5.2. To conclude, recall that every graph $C^*$ -algebra is generated by its edge partial isometries and sink projections.