Proof. The statement follows from the sectoriality 1 , the fact that \[|f(x)-f(y)|^2=|\operatorname{Re}f(x)-\operatorname{Re}f(y)|^2+|\operatorname{Im}f(x)-\operatorname{Im}f(y)|^2.\] and the definition of \(\mathcal{Q}\). ◻

Proof. Let \(f\in \ell^2(X,m)\). Let \((K_n)_{n\in \Bbb N}\) be a finite exhaustion of \((X, b, m)\), i.e. \(K_n\subset X\) is finite and connected, \(K_{n}\subset K_{n+1}\) for any \(n\in \Bbb N\) and \(\cup_{n\in \Bbb N} K_n=X\). Let \(f_n:=f\mid_{K_n}\), \(n\in \Bbb N\). Then \[\|f-f_n\|^2=\sum_{x\in X\setminus K_n}|f(x)|^2m(x)\to 0,\] as \(n\to\infty\). ◻

Proof. Due to the locally finitness of graph all the sums are finite. The equalities follow by direct computations, see e.g [12]. ◻

Proof. Let \(f\in\mathcal{D}_0\). Then there exist \((\phi_n)_{n\in \Bbb N}\subset C_c(X)\) with \(\phi_n\to f\) pointwise and \(\operatorname{Re}\mathcal{Q}(f-\phi_n)\to 0\) as \(n\to \infty\). Hence, for any \(\varepsilon>0\) there exist \(N\in \Bbb N\) such that \(\operatorname{Re}\mathcal{Q}(f-\phi_N)<\varepsilon\), i.e \[\dfrac12\sum_{x,y\in X\setminus B_1(\mathop{\mathrm{supp}}\phi_N)}|f(x)-f(y)|^2\operatorname{Re}b(x,y)\le \operatorname{Re}\mathcal{Q}(f-\phi_N)<\varepsilon,\] where \[B_1(\mathop{\mathrm{supp}}\phi_N)=\{x\in X\;\mid\; there exists y\in \mathop{\mathrm{supp}}\phi_Nsuch that x\sim y\}.\] Hence, \(\operatorname{Re}\mathcal{Q}(f)<\infty\) and \(f\in\mathcal{D}\). ◻

Proof.

• Let \(H=\mathcal{D}_0\). Let \((\phi_n)_n\in C_c(X)\) approximates \(f\) in \(\mathcal{D}_0\). Firstly note, that \(\phi_n\to f\) pointwise implies \(\operatorname{Re}\phi_n\to \operatorname{Re}f\) and \(\operatorname{Im}\phi_n\to \operatorname{Im}f\). Secondly, by Lemma 22 we have \[\operatorname{Re}\mathcal{Q}(f-\phi_n)=\operatorname{Re}\mathcal{Q}(\operatorname{Re}f-\operatorname{Re}\phi_n)+\operatorname{Re}\mathcal{Q}(\operatorname{Im}f-\operatorname{Im}\phi_n),\] and, since all summands are positive and \(\mathcal{Q}\) is sectorial, we get \(\operatorname{Re}f, \operatorname{Im}f \in~D_0\). Moreover, they can be approximated in \(\mathcal{D}_0\) by real-valued functions. The other direction follows from the fact that \(\mathcal{D}_0\) is a vector space.

• Let \(H=\ell^2(X,m)\). Then we obtain \[\|f\|^2=\sum_{x\in X}|f(x)|^2m(x)=\sum_{x\in X}(|\operatorname{Re}f(x)|^2+|\operatorname{Im}f(x)|^2)m(x)\] from where the statement follows.

• Let \(H=\overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\). By Lemma 22 and definition of \(\ell^2\)-norm we get \[\begin{align} &\operatorname{Re}\mathcal{Q}(f-\phi_n)+\|f-\phi_n\|^2=\operatorname{Re}\mathcal{Q}(\operatorname{Re}f-\operatorname{Re}\phi_n)+\operatorname{Re}\mathcal{Q}(\operatorname{Im}f-\operatorname{Im}\phi_n)\\ &+\|\operatorname{Re}f-\operatorname{Re}\phi_n\|^2+\|\operatorname{Im}f-\operatorname{Im}\phi_n\|^2, \end{align}\] from where the statement follows due to the definition of \(\|\cdot\|_\mathcal{Q}\) and positivity of all summands.

 ◻

Proof of Theorem 3. Let \(f\in \overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\). Then there exists a sequence \((\phi_n)_{n\in \Bbb N}\subset C_c(X)\) such that \[\|f-\phi_n\|_\mathcal{Q}=(\operatorname{Re}\mathcal{Q}(f-\phi_n)+\|f-\phi_n\|^2)^{\frac{1}{2}}\to 0as n\to\infty.\] Therefore, \(\phi_n\to f\) in \(\ell^2(X,m)\) and, hence, \(f\in \ell^2(X,m)\) and \(\phi_n\to f\) pointwise. Moreover, \(\operatorname{Re}\mathcal{Q}(f-\phi_n)\to 0\) and \(f\in \mathcal{D}_0\). From all the above follows \(\overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\subset \mathcal{D}_0\cap \ell^2(X,m)\).

Let \(f\in D_0\cap \ell^2(X,m)\). Then, by Lemma 5 we obtain that \(\operatorname{Re}f,\operatorname{Im}f \in D_0\cap \ell^2(X,m)\) and can be approximated in \(\mathcal{D}_0\) by real-valued functions. Hence, since \(\operatorname{Re}\mathcal{Q}_b(\cdot)=\mathcal{Q}_{\operatorname{Re}b}(\cdot)\) one can apply the reasoning for positive forms on real-weighted graph \((X,\operatorname{Re}b,m)\), see [4], to show that \(\operatorname{Re}f,\operatorname{Im}f \in \overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\) which implies \(f \in \overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\) by Lemma 5. Therefore, \(\mathcal{D}_0\cap \ell^2(X,m)\subset\overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\) and the proof is finished. ◻

Proof. The sectoriality follows from Lemma 1. The density is clear, since \(C_c(X)\subset D(Q^{(D)})\) is dense in \(\ell^2(X,m)\). Finally, the form is closed due to Theorem 3. ◻

Proof. By the definition of an operator, associated with a form, for any \(f\in D(L^{(D)}),\) and \(\phi\in C(X)\subset D(Q^{(D)})\) we have \[(L^{(D)}f\;\mid\; \phi)=Q^{(D)}(f,\phi)=\mathcal{Q}(f,\phi)=\sum_{x\in X}\mathcal{L}f(x)\overline{\phi(x)}m(x),\] where the last equality is due to the Green’s formula. Now taking \(\phi:=1_x/m\) the statement follows. ◻

Proof. Due to the Green’s formula and 5 , if a solution \(v_\alpha\) exists, we have \[r\cdot (\mathcal{L}v_\alpha+\alpha v_\alpha)(x)m(x)= ((\mathcal{L}+\alpha)v_\alpha\;\mid\;v_\alpha)=\mathcal{Q}(v_\alpha,v_\alpha)+\alpha \|v_\alpha\|^2,\] which proves 6 . Since \(\mathcal{Q}(v_\alpha,v_\alpha)=0\) if and only if \(v_\alpha\equiv 0\), the latest is the only solution of 5 for \(r=0\). Hence, the corresponding matrix is invertible and 5 has a unique solution \(v_\alpha\) for any \(r\in \Bbb R\).

To prove the estimate, let us note that by Green’s formula \[(\mathcal{L}v_\alpha+\alpha v_\alpha)(x)m(x)=((\mathcal{L}+\alpha)v_\alpha\;\mid\;1_x)=\mathcal{Q}(v_\alpha,1_x)+\alpha r\cdot m(x),\] Combining the latest with 6 we obtain: \[\label{eq::QvrQvr1} |\mathcal{Q}(v_\alpha)|+\alpha \|v_\alpha\|^2\le |r|\cdot|\mathcal{Q}(v_\alpha,1_x)|+\alpha|r|^2\cdot m(x).\tag{8}\]

Now, using the inequality \[2|z_1 z_2 | \le\varepsilon |z_1|^2 +\dfrac{1}{\varepsilon} |z_2 |^2for any z_1 , z_2 \in \Bbb C, \varepsilon>0,\] we get \[\begin{gather} \label{eq::Qvr1x} |\mathcal{Q}(v_\alpha,1_x)|\le \dfrac12\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)||1_x(y)-1_x(z)||b(y,z)|\\\le \dfrac{\varepsilon}{4}\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)|^2|b(y,z)|+\dfrac{1}{4\varepsilon}\sum_{y,z\in X}|1_x(y)-1_x(z)|^2|b(y,z)|. \end{gather}\tag{9}\] and, similarly, \[\label{eq::alphaeps} \alpha |r|\cdot m(x)\le \dfrac{\varepsilon}{2}\alpha|r|^2 \cdot m(x)+ \dfrac{\alpha}{2\varepsilon}\cdot m(x)\le \dfrac{\varepsilon}{2}\alpha \|v_\alpha\|^2+ \dfrac{\alpha}{2\varepsilon}\cdot m(x)\tag{10}\] From the other hand, note that for any \(y,z\in X\) we have due to the sectoriality 1 \[\operatorname{Re}b(y,z)\ge (1+c^2)^{-\frac{1}{2}}|b(y,z)|,\] and, hence, we obtain \[\begin{gather} |\mathcal{Q}(v_\alpha)| \ge \operatorname{Re}\mathcal{Q}(v_\alpha)=\dfrac12\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)|^2 \operatorname{Re}b(y,z)\\\ge \dfrac{1}{2(1+c^2)^{\frac{1}{2}}}\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)|^2 |b(y,z)|. \end{gather}\] Combining this with 8 , 9 , 10 and using the fact that \(1/(1+c^2)^\frac{1}{2}\le 1\) we get \[\begin{align} \label{eq::240143c241} &\dfrac{1}{2(1+c^2)^{\frac{1}{2}}}\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)|^2 |b(y,z)|+\dfrac{1}{(1+c^2)^{\frac{1}{2}}}\alpha \|v_\alpha\|^2\nonumber\\ &\le|\mathcal{Q}(v_\alpha)| +\alpha \|v_\alpha\|^2\le |r|\cdot|\mathcal{Q}(v_\alpha,1_x)|+\alpha|r|^2\cdot m(x) \\& \le\dfrac{\varepsilon |r|}{4}\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)|^2|b(y,z)|+ \dfrac{\varepsilon |r|}{2}\alpha \|v_\alpha\|^2+\dfrac{|r|}{2\varepsilon}\widetilde{C} \nonumber, \end{align}\tag{11}\] where \[\widetilde{C}:=\dfrac12\sum_{y,z\in X}|1_x(y)-1_x(z)|^2|b(y,z)|+\alpha\cdot m(x).\] Now plugging in 11 \(\varepsilon=(1+c^2)^{-\frac{1}{2}}|r|^{-1}\) we obtain from the first and the third line after simplification the following: \[\dfrac{1}{4}\sum_{y,z\in X}|v_\alpha(y)-v_\alpha(z)|^2 |b(y,z)|+\dfrac{1}{2}\alpha \|v_\alpha\|^2\\\le\dfrac{(1+c^2)|r|^2}{2}\widetilde{C},\] from where the estimate 7 follows, since \[\widetilde{C}=\frac{1}{2}\sum_{y,z\in X}|1_x(y)-1_x(z)|^2|b(y,z)|+\alpha\cdot m(x)=\sum_{\substack {y\in X:}\\{y\sim x}}|b(x,y)|+\alpha\cdot m(x).\] ◻

Proof. Let \(v_\alpha\) be as in Lemma 7 with \(r=1\). It follows immediately from 6 that \((\mathcal{L}v_\alpha+ \alpha v_\alpha)(x)\ne 0\), since \(v_\alpha\not\equiv 0\). Hence, the function \(\widehat v_\alpha\in \ell^2(K,m_K)\) defined by \[\widehat v_\alpha(y)=\dfrac{v_\alpha(y)}{(\mathcal{L}v_\alpha+\alpha v_\alpha)(x)},\] for all \(y\in K\) is the solution of 12 . To show the uniqueness firstly note, that \(\widehat v_\alpha(x)\ne 0\) for any solution \(\widehat v_\alpha\) of 12 . Indeed, if \(\widehat v_\alpha(x)=0\), then, by the definition of the Dirichlet Laplacian through the form \(\widehat Q_K\), we have \[0=((\widehat L_K+\alpha)\widehat v_\alpha\;\mid\;\widehat v_\alpha)=\widehat Q_K(\widehat v_\alpha)+\alpha \|\widehat v_\alpha\|^2,\] i.e. \(\widehat v_\alpha\equiv 0\) which is a contradiction. Then uniqueness follows from the fact, that for any \(\widehat v_\alpha\), solution of 12 , the function \[\widetilde{v}_\alpha(y)=\dfrac{\widehat v_\alpha(y)}{\widehat v_\alpha(x)},\] is the solution of 5 for \(r=1\), which is unique by Lemma 7.

Finally note that since the operator \((L_K+\alpha)\) is linear, it follows from the existence of the unique solution of 12 that it is invertible. ◻

Proof. Since \(C_c(X)\) is dense in \(\ell^2(X,m)\), the Dirichlet Laplacian \(L^{(D)}\) is an operator, associated to \(Q^{(D)}\), and, hence, a linear relation. Further, it is clear, that the operator \((L_n+\alpha)^{-1}\), \(\alpha>0\), is the needed linear relation by its definition 14 , since \(\widehat L_n\) is the operator, associated with \(\widehat Q_n\). Note that \(\widehat Q_n(u)=Q^{(D)}(u)\) for all \(u\in D(\widehat Q_{K_n})\). Now the statement is an immediate corollary of Theorem 5, since \(C_c(X)\) is a core of \(Q^{(D)}\) by Theorem 3, and, hence the sequence \(u_n\) can be taken such that \(u_n=u\) eventually. ◻

Proof. By [11] we have \[\dfrac{\operatorname{Re}a^{(s)}(x,y)}{|a^{(s)}(x,y)|}\ge \dfrac{\operatorname{Re}s}{|s|},\] which can be rewritten as \[{(\operatorname{Re}^2 a^{(s)}(x,y))}{(\operatorname{Re}^2s+\operatorname{Im}^2 s)}\ge {(\operatorname{Re}^2 s)}{(\operatorname{Re}^2a^{(s)}(x,y)+\operatorname{Im}^2 a^{(s)}(x,y))},\] which is equivalent to \[{(\operatorname{Re}^2 a^{(s)}(x,y))}{(\operatorname{Im}^2 s)}\ge {(\operatorname{Re}^2 s)}{(\operatorname{Im}^2 a^{(s)}(x,y))},\] from where immediately follows that the weight \(a^{(s)}(x,y)\) satisfies sectoriality 1 with \[c = \dfrac{|\operatorname{Im}s|}{\operatorname{Re}s}.\] ◻

Proof. We need to prove that the equation \[\dfrac{s_0}{L_{xy}s_0^2+R_{xy}s_0+D_{xy}}=b(x,y).\] has a solution in terms of \(L_{xy},R_{xy}, D_{xy}\), where all the variables are positive, for any \(b(x,y)\ne 0\), satisfying sectoriality 1 . Then the corresponding electrical network will be defined by \[a^{(s)}(x,y):= \dfrac{s}{L_{xy}s^2+R_{xy}s+D_{xy}},\] for all \(x,y\in X\) with \(b(x,y)\ne 0\) and \(a^{(s)}(x,y):=0\) otherwise.

Note, that if \(\operatorname{Im}b(x,y)=0\), then \(L_{xy}=D_{xy}=0\) and \(R_{xy}=1/b(x,y)\) is the needed solution. So assume \(\operatorname{Im}b(x,y)\ne 0\). Let us denote \(b:=b(x,y)\), omit subscripts \({xy}\) for the proof and solve: \[\dfrac{s_0}{Ls_0^2+Rs_0+D}=b.\] Assume, the denominator is not zero (which will hold if the solution will be non-negative and positive for at least one of the variables \(L,R,D\)). Without loss of generality we can assume that \(\operatorname{Re}s_0=1\) (otherwise scale \(L,R,D\)). Hence, let us write \(s_0=1+\omega i\), where \(|\omega|\ge c\). Then we should solve \[b(L(1+i\omega)^2+R(1+i\omega)+D)={1+i\omega},\] which we rewrite as \[\label{eq::beq1} L(1+i\omega)^2+R(1+i\omega)+D=\dfrac{{(1+i\omega)}\overline{b}}{|b|^2}.\tag{15}\] Let \(b_1=\operatorname{Re}b\) and \(b_2=\operatorname{Im}b\). By assumption \(0<|b_2|\le c\cdot b_1\). Then the equation 15 , being split into real and imaginary part, is equivalent to: \[\begin{cases} (1-\omega^2)L+R+D=\dfrac{b_1+\omega b_2}{|b|^2},\\ 2\omega L+\omega R=\dfrac{b_1\omega - b_2}{|b|^2}. \end{cases}\] Since there are \(2\) equations and \(3\) variables, let us assume that \(D=D_0\) is fixed (we will choose it later) and find a solution for \((L_0, R_0)\), using the Cramer’s rule. Let \[A=\begin{pmatrix} 1-\omega^2&1\\ 2\omega&\omega \end{pmatrix}, A_L= \begin{pmatrix} \dfrac{b_1+\omega b_2}{|b|^2}-D_0&1\\ \dfrac{b_1\omega - b_2}{|b|^2}&\omega \end{pmatrix},\] and \[A_R= \begin{pmatrix} 1-\omega^2&\dfrac{b_1+\omega b_2}{|b|^2}-D_0\\ 2\omega&\dfrac{b_1\omega - b_2}{|b|^2} \end{pmatrix}.\] Hence we get \(\det A=-\omega^3-\omega\) and \(\mathop{\mathrm{sgn}}(\det A)=-\mathop{\mathrm{sgn}}\omega\). Further, \[\label{eq::detAL} \det A_L=\dfrac{(1+\omega^2)b_2}{|b|^2}-\omega D_0,\tag{16}\] \[\det A_R=\dfrac{-(1+\omega^2)(b_1\omega+b_2)}{|b|^2}+2\omega D_0.\] Since \(L=\dfrac{\det A_L}{\det A}\) and \(R=\dfrac{\det A_R}{\det A}\) should be non-negative, we should require \(\mathop{\mathrm{sgn}}(\det A_L), \mathop{\mathrm{sgn}}(\det A_R)\in \{0, -\mathop{\mathrm{sgn}}\omega\}\), i.e. we should choose \(D_0\) such that \[\label{eq::D0} \dfrac{(1+\omega^2)b_2}{\omega |b|^2}\le D_0\le \dfrac{(1+\omega^2)(b_1\omega + b_2)}{2 \omega |b|^2}.\tag{17}\] Firstly note, that \(|b_1\omega|=|b_1||\omega|\ge b_1c\ge|b_2|\), while \(\mathop{\mathrm{sgn}}b_1\omega=\mathop{\mathrm{sgn}}\omega\) (since \(b_1\ge c|b_2|>0\)), and, hence, \(\mathop{\mathrm{sgn}}(b_1\omega + b_2)=\mathop{\mathrm{sgn}}\omega\) or \(\mathop{\mathrm{sgn}}(b_1\omega+b_2)=0\), i.e. the right hand side of the inequality 17 is non-negative. If \(\mathop{\mathrm{sgn}}b_2=-\mathop{\mathrm{sgn}}\omega\), then the left hand side is negative and there exists a needed \(D_0\ge 0\) (note that if \(D_0=0\), then \(L_0\ne 0\) since \(\det A_L\ne 0\) due to 16 and the assumption \(b_2\ne 0\)). So assume \(\mathop{\mathrm{sgn}}b_2=\mathop{\mathrm{sgn}}\omega\). In this case since \(b_1>0\), we obtain \[\dfrac{b_1\omega+b_2}{\omega}=\dfrac{|b_1\omega+b_2|}{|\omega|}=\dfrac{|b_1||\omega|+|b_2|}{|\omega|}\ge \dfrac{|b_1|\cdot c+|b_2|}{|\omega|}\ge \dfrac{2|b_2|}{|\omega|}= \dfrac{2b_2}{\omega},\] from where immediately follows that right-hand side of 17 is greater than or equal to the left-hand side and, hence, there exists \(D_0>0\), satisfying 17 , since \(b_2\ne 0\). ◻

Proof. We will prove the equivalences for a fixed \(x\in X\). Then the equivalence of “some” and “all” will follow from (iii). Indeed, (iii) for “some” \(x\in X\) defines a recurrence on the real-weighted graph \((X,\operatorname{Re}b,m)\), and for real-weighted graphs the equivalence of “some” and “all” is known, see, e.g. [4].

(i)\(\Rightarrow\)(ii) is clear.

(ii)\(\Rightarrow\)(i) follows from the sectoriality, Lemma 1.

(ii)\(\Leftrightarrow\)(iii) since \[\begin{align} \operatorname{Re}\mathcal{Q}(\phi)&=\operatorname{Re}\left(\frac{1}{2} \sum_{x,y\in X}|\phi(x)-\phi(y)|^2 b(x,y)\right)\\ &=\frac{1}{2} \sum_{x,y\in X}|\phi(x)-\phi(y)|^2 \operatorname{Re}b(x,y)=\mathcal{Q}_{\operatorname{Re}b}(\phi), \end{align}\] for any \(\phi\in C_c(X)\). ◻

Proof. “\(\Rightarrow\)” Let the graph \((X,b,m)\) be recurrent. Then the graph \((X,\operatorname{Re}b,m)\) is recurrent by Corollary 2. Hence, by [4], there exists a sequence \((\phi_n)_{n\in\Bbb N}\subset C_c(X)\) with \(\phi_n\to f\) pointwise and \[\mathcal{Q}_{\operatorname{Re}b}(1-\phi_n)=\dfrac12\sum_{x,y\in X}|\phi_n(x)-\phi_n(y)|^2\operatorname{Re}b(x,y)\to 0 as n\to \infty.\] Hence, \(\operatorname{Re}\mathcal{Q}(1-\phi_n)=\mathcal{Q}_{\operatorname{Re}b}(1-\phi_n)\to 0\) and \(1\in \mathcal{D}_0\).

“\(\Leftarrow\)” Let \(1\in \mathcal{D}_0\). Then there exists \((\phi_n)_{n\in\Bbb N}\subset C_c(X)\) such that \(Q(1-\phi_n)\to 0\) as \(n\to \infty\). Hence, \[\mathcal{Q}_{\operatorname{Re}b}(1-|\phi_n|)\le Q_{\operatorname{Re}b}(1-\phi_n)=\operatorname{Re}Q(1-\phi_n)\to 0 as n\to \infty,\] and the graph \((X,\operatorname{Re}b,m)\) is recurrent by [4], which by Corollary 2 implies that \((X,b,m)\) is recurrent. ◻

Proof. In [11] it is proven, that the limit of effective capacities exists in the case of electrical networks and is a holomorphic function on \(s\in \Bbb H_r\) (see Definition 3), for the ball exhaustion \((B_n)_{n\in \Bbb N\cup\{0\}}\), \(B_n(x)=\{y\in X\mid\mathop{\mathrm{dist}}(x,y)< n\}\), where by \(\mathop{\mathrm{dist}}(x,y)\) one means the lenght of the shortest path between vertices \(x,y\in X\). Using the same outline of proof, one can show that the limit exists for any exhaustion \((K_n)_{n\in \Bbb N\cup\{0\}}\). Moreover, since the capacity in this case is a holomorphic function on the parameter \(s\in \Bbb H_r\) and the limits coincide for all \(s\in \Bbb R^+\) due to the classical theory of real-weighted graph, we immediately get, that the effective capacity of electrical network does not depend on the exhaustion, i.e. the capacity is well-defined for the electrical networks.

Finally, due to Theorem 8, the capacity is well-defined for any complex-weighted graph, since the uniform limit (see again [11]) on any compact subset of \(\Bbb H_r\) implies pointwise limit. ◻

Proof. Due to Definition 5 of recurrence it is enough to prove that \(\operatorname{cap}(x)=~0\) is equivalent to \[\label{eq::reccapx} \inf \;\{\;|\mathcal{Q}(\phi)|\;|\;\phi(x)=1, \phi\in C_c(X)\;\}=0,\tag{19}\] for a fixed \(x\in X\). Then the statement for all \(x\in X\) will follow from Theorem 9.

Let \((K_n)_{n\in \Bbb N\cup \{0\}}\) be a finite exhaustion of \((X,b,m)\), \(x\in K_0\). Let \((u_n)_{n\in \Bbb N}\) be the solutions of the Dirichlet problems \[\begin{cases} u_n(x)=1,\\ \mathcal{L}_b u_n(y)=0,\; y\ne x, y\in K_n,\\ u_n(y)=0,\; y\in X\setminus K_n. \end{cases}\]

• Let \(\operatorname{cap}(x)= 0\). Then \(\mathcal{Q}(u_n)=\operatorname{cap}_{K_n}(x)\to 0, n\to \infty\), i.e. 19 holds.

• Let now 19 holds. Let us consider the Dirichlet problems on \((X,\operatorname{Re}b,m)\), i.e. \[\begin{cases} \widetilde{u}_n(x)=1,\\ \mathcal{L}_{\operatorname{Re}b} \widetilde{u}_n(a)=0,\; y\ne x, y\in K_n,\\ \widetilde{u}_n(y)=0,\; y\in X\setminus K_n. \end{cases}\] Note, that \(\mathcal{Q}_{\operatorname{Re}b}(\widetilde{u}_n)\to 0\), as \(n\to \infty\), due to the classical theory of real-weighted graphs (see [4]), since \((X, \operatorname{Re}b, m)\) is recurrent by Corollary 2. Hence \[\operatorname{Re}\mathcal{Q}_b(\widetilde{u}_n) = \mathcal{Q}_{\operatorname{Re}b}(\widetilde{u}_n)\to 0 as n\to\infty.\] Further, since all functions are finitely supported, we get by Green’s formula \[\begin{align} &\mathcal{Q}_b(u_n, \widetilde{u}_n)=(\mathcal{L}_b u_n\;|\; \widetilde{u}_n )=\mathcal{L}_b u_n(x)m(x)\\ &=\mathcal{L}_b u_n(x)u_n(x)m(x)=(\mathcal{L}_b u_n\;|\; u_n )=\mathcal{Q}_b(u_n). \end{align}\] Using 3 , we get \[\operatorname{Re}\mathcal{Q}_b(u_n)\le | \mathcal{Q}_b(u_n)|= | \mathcal{Q}_b(u_n, \widetilde{u}_n)| \le(1+c)(\operatorname{Re}\mathcal{Q}_b(u_n))^\frac{1}{2} (\operatorname{Re}\mathcal{Q}_b(\widetilde{u}_n))^\frac{1}{2},\] where \(c\) is the sectoriality constant 1 for \((X,b,m)\). Therefore, \[\operatorname{Re}\mathcal{Q}_b(u_n)\le(1+c)^2 \operatorname{Re}\mathcal{Q}(\widetilde{u}_n)=(1+c)^2 \mathcal{Q}_{\operatorname{Re}b}(\widetilde{u}_n)\to 0 as n\to\infty.\] and \(| \mathcal{Q}_b(u_n)|\le (1+c)\operatorname{Re}\mathcal{Q}_b(u_n) \to 0\) as \(n\to \infty\), where the last inequality is due to the sectoriality of \(\mathcal{Q}\).

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Proof. By resolvent equality and Theorem 6 we have: \[\label{eq::GxyeqL} G(x,y)=\lim_{\substack{\alpha\to 0,\\\alpha>0}}(L^{(D)}+\alpha)^{-1}1_x(y)=\lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}({L_n}+\alpha)^{-1}1_x(y),\tag{20}\] where \(L_n:=L_{K_n}, n\in \Bbb N\cup\{0\}\) are the Dirichlet Laplacians with respect to \(K_n\) (see 14 ), for some finite exhaustion \((K_n)_{ n\in \Bbb N\cup\{0\}}\) of \((X,b,m)\). Without loss of generality we can assume \(x\in K_0\). It is clear by 14 , that \[(L_n+\alpha)^{-1}1_x(y)= \begin{cases}(\widehat L_n+\alpha)^{-1}1_x(y),\quad y\in K_n\\ 0, otherwise, \end{cases}\] where \(\widehat L_n\) is the Laplacian on \(\ell^2(K_n,m_{K_n})\). Hence, the quantity \(({L_n}+\alpha)^{-1}\) is well-defined for any \(x, y\in X\). Moreover, for a fixed \(y\in X\) there exist \(N\in \Bbb N\) such that \(y\in K_N\) and, hence, \[\label{eq::LnN} (L_n+\alpha)^{-1}1_x(y)=(\widehat L_n+\alpha)^{-1}1_x(y)for all n>N.\tag{21}\] Therefore we need to prove the existence of the limit \[\lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}({L_n}+\alpha)^{-1}1_x(y)=\lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}({\widehat L_n}+\alpha)^{-1}1_x(y).\] By the proof of Theorem 4 we have \[\label{eq::LhateqL} (\widehat {L}_n+\alpha)^{-1}1_x(y) =\dfrac{v_{\alpha,n}(y)}{(\mathcal{L}v_{\alpha,n}+\alpha v_{\alpha,n})(x)}for all \alpha>0, n\in \Bbb N,\tag{22}\] where \(v_{\alpha,n},n\in \Bbb N\), are the solutions of the following Dirichlet problems: \[\label{eq::dirpralpha1} \begin{cases} v_{\alpha,n}(x)=1,\\ (\mathcal{L}+\alpha) v_{\alpha,n}=0 on K_n\setminus\{x\},\\ v_{\alpha,n}= 0on X\setminus K_n. \end{cases}\tag{23}\]

By Theorem 8 there exist an electrical network \((X,a^{(s)}, m)\) such that for \(s_0=1+ic\) (i.e \(\operatorname{Im}s_0/\operatorname{Re}s_0=c\)), where \(c\) is the sectoriality constant of the graph \((X,b,m)\), the following holds \[a^{(s_0)}(x,y)=b(x,y),\] for all \(x,y\in X\). Let us extend the Dirichlet problem 23 to the electrical network \((X,a^{(s)},m)\), i.e. we have \[\begin{cases} v_{\alpha,n}^{(s)}(x)=1,\\ (\mathcal{L}^{(s)} +\alpha) v_{\alpha,n}^{(s)}=0 on K_n\setminus\{x\},\\ v_{\alpha,n}^{(s)}= 0on X\setminus K_n. \end{cases}\] By Lemma 7 we obtain \[\begin{align} \label{eq::Lv1cfreq} \Big|\big(\mathcal{L}v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x)\Big|&\le \Big|\mathcal{Q}\big(v_{\alpha,n}^{(s)}\big)\Big|+\alpha \| v_{\alpha,n}\|^2\notag\\ &\le \left(1+\dfrac{|\operatorname{Im}s|^2}{(\operatorname{Re}{s})^2}\right)\left(\sum_{\substack{y\in X:\\y\sim x}}\big|a^{(s)}(x,y)\big|+\alpha\cdot m(x)\right), \end{align}\tag{24}\] where we have used Theorem 7. By simple calculations, see [11], we have: \[\label{eq::sumas} \sum_{\substack{y\in X:\\y\sim x}} a^{(s)}(x,y) \le \frac{1 + |s|^2}{\operatorname{Re}s} \sum_{\substack{y\in X:\\y\sim x}} \left( R _{xy} + L _{xy} +D_{xy} \right)^{-1},{ for any }x,y\in X.\tag{25}\] Combining this with 24 we get \[\label{eq::unifombound} \Big|\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x)\Big|\le \dfrac{|s|^2}{(\operatorname{Re}s)^2} \left(\frac{1 + |s|^2}{\operatorname{Re}s}C_1+C_2\right),\tag{26}\] where \(C_1,C_2\) do not depend on \(s\) and \(n\). Hence, the function \(\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x)\) is uniformly bounded for any \(n\in \Bbb N\) in any domain \[\{\operatorname{Re}s\ge\varepsilon, |s|\le C\}, \quad \varepsilon, C\in \Bbb R^+.\] Hence, by Montel’s theorem (see, e.g. [19]), the sequence \[\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x), \quad n\in \Bbb N,\] has a normally converging subsequence. Since the limit of the sequence is known to exist for any \(s\in \Bbb R_+\) (in this case we get a real-weighted graph), and the holomorphic function is uniquely determined by its values on the real line, we get that there exist the limit \[\label{eq::firstLim} \lim_{n\to \infty}\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x),\tag{27}\] for any \(\alpha\ge 0\), and this limit is a holomorphic function on \(\Bbb H_r\). Further, since 26 implies uniform boundness of \(\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x)\) for any \(\alpha<1\), we obtain by the same line of arguments, that there exist a holomorphic limit \[\label{eq::secondLim} \lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x).\tag{28}\] Since the limit 27 for \(\alpha=0\) and the limit 28 coincide on the real positive half-line ([4]) we have: \[\label{eq::eqLim}\lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}\big(\mathcal{L}^{(s)} v_{\alpha,n}^{(s)}+\alpha v_{\alpha,n}^{(s)}\big)(x)=\lim_{n\to\infty}\big(\mathcal{L}^{(s)} v_{0,n}^{(s)}\big)(x).\tag{29}\] Moreover, since by Green’s formula \[\operatorname{Re}\big(\mathcal{L}^{(s)} v_{0,n}^{(s)}\big)(x)=\dfrac{\operatorname{Re}\mathcal{Q}(v^{(s)}_{0,n})}{m(x)}>0,\] for all \(n>N\), for any \(s\in\Bbb H_r\), by Hurwitz theorem (see, e.g. [19]) the limit in 29 either has no zeros on the right half-plane or is identically zero. Note that due to Theorem 12, the latest case is equivalent to the recurrence of the graph \((X,b,m)\), where \(b=a^{(s)}\) for (some) all \(s\in \Bbb H_r\).

Now we apply a similar argument to show the convergence of \(v_{\alpha,n}^{(s)}(y)\), where \(y\in X\) is fixed, i.e. it is enough to prove the uniform boundness of \(v_{\alpha,n}^{(s)}(y)\) for compact subsets of \(\Bbb H_r\). Let \(x=x_0\sim x_1\sim \dots \sim x_k=y\) be a path between \(x\) and \(y\). By Cauchy–Schwarz inequality, 24 and 25 we get \[\begin{align} \label{eq::estv} \big|&v_{\alpha,n}^{(s)}(y)-1\big|^2=\big|v_{\alpha,n}^{(s)}(y)-v_{\alpha,n}^{(s)}(x)\big|^2\\ &\le \left( \sum_{i=1}^{k} \big|v_{\alpha,n}^{(s)}(x_{i})-v_{\alpha,n}^{(s)}(x_{i-1})\big| \,\sqrt{\operatorname{Re}a^{(s)}(x_{i-1},x_i)} \cdot \frac{1}{\sqrt{\operatorname{Re}a^{(s)}(x_{i-1},x_i)}} \right)^2\notag \\ &\le \left(\sum_{i=1}^{k} \big|v_{\alpha,n}^{(s)}(x_{i})-v_{\alpha,n}^{(s)}(x_{i-1})\big|^2 \operatorname{Re}a^{(s)}(x_{i-1},x_i)\right) \left (\sum_{i=1}^{k}\frac{1}{\operatorname{Re}a^{(s)}(x_{i-1},x_i)}\right)\notag\\ &\le \Big|\mathcal{Q}\big(v^{(s)}_{\alpha,n}\big)\Big|\,\left (\sum_{i=1}^{k}\frac{1}{\operatorname{Re}a^{(s)}(x_{i-1},x_i)}\right)\notag\\ &\le \left(1+\dfrac{|\operatorname{Im}s|^2}{(\operatorname{Re}{s})^2}\right)\left(\sum_{\substack{y\in X:\\y\sim x}}\big|a^{(s)}(x,y)\big|+\alpha\cdot m(x)\right)\left (\sum_{i=1}^{k}\frac{1}{\operatorname{Re}a^{(s)}(x_{i-1},x_i)}\right)\notag\\ &\le \dfrac{|s|^2}{(\operatorname{Re}s)^2} \left(\frac{1 + |s|^2}{\operatorname{Re}s}C_1+C_2\right)\left (\sum_{i=1}^{k}\frac{1}{\operatorname{Re}a^{(s)}(x_{i-1},x_i)}\right)\notag, \end{align}\tag{30}\] where \(C_1, C_2\) does not depend on \(s\). Let us estimate the last term, starting from an estimate of \(\operatorname{Re}a^{(s)}(z,w), z,w\in X, s\in \Bbb H_r\). By Definition 3 of an electrical network we have \[\operatorname{Re}a^{(s)}(z,w)=\dfrac{s}{L_{zw}s^2+R_{zw}s+D_{zw}}.\] We estimate, using \(|s|^2\ge |s|\operatorname{Re}s\) for any \(s\in \Bbb C\) in the second line: \[\begin{align} \operatorname{Re}a^{(s)}&(z,w)=\operatorname{Re}\dfrac{s(L_{zw}\overline{s}^2+R_{zw}\overline{s}+D_{zw})}{|L_{zw}s^2+R_{zw}s+D_{zw}|^2}\ge \operatorname{Re}\dfrac{s(L_{zw}\overline{s}^2+R_{zw}\overline{s}+D_{zw})}{(L_{zw}|s|^2+R_{zw}|s|+D_{zw})^2}\\ &=\operatorname{Re}\dfrac{L_{zw}|s|^2 \overline{s}+R_{zw} |s|^2+D_{zw}s}{(L_{zw}|s|^2+R_{zw}|s|+D_{zw})^2}\ge (\operatorname{Re}s) \dfrac{L_{zw}|s|^2 +R_{zw} |s|+D_{zw}}{(L_{zw}|s|^2+R_{zw}|s|+D_{zw})^2}\\ &=(\operatorname{Re}s) \dfrac{1}{L_{zw}|s|^2+R_{zw}|s|+D_{zw}}\ge \dfrac{\operatorname{Re}s}{\max \{1,|s|,|s|^2\}} \dfrac{1}{L_{zw}+R_{zw}+D_{zw}}\\ &=\dfrac{\operatorname{Re}s}{\max\{ 1,|s|^2\}} \dfrac{1}{L_{zw}+R_{zw}+D_{zw}}\ge \dfrac{\operatorname{Re}s}{ 1+|s|^2} \dfrac{1}{L_{zw}+R_{zw}+D_{zw}}. \end{align}\] since either \(|s|<1\) or \(|s|\le|s|^2\). Combining this with 30 we obtain: \[\big|v_{\alpha,n}^{(s)}(y)-1\big|^2\le \dfrac{|s|^2(1+|s|^2)}{(\operatorname{Re}s)^3} \left(\frac{1 + |s|^2}{\operatorname{Re}s}C_1+C_2\right)C_3,\] where \(C_3= \sum_{i=1}^{k}(L_{x_{i-1}x_i}+R_{x_{i-1}x_i}+D_{x_{i-1}x_i})\), \(C_1,C_2,C_3\) do not depend on \(s\), \(n\) and any \(\alpha<1\). Hence, \(v_{\alpha,n}^{(s)}(y)\) is uniformly bounded for any \(n\in \Bbb N\) in any domain \[\{\operatorname{Re}s\ge\varepsilon, |s|\le C\}, \quad\varepsilon, C\in \Bbb R^+.\]

Therefore, again by Montel’s theorem and existence of the limit on the real half-line, we conclude the existence and the equality of the limits \[\label{eq::eqLimv} \lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}v_{\alpha,n}^{(s)}(y)=\lim_{n\to\infty} v_{0,n}^{(s)}(y).\tag{31}\] Moreover, note that in the case \(\Big|\mathcal{Q}\big(v^{(s)}_{0,n}\big)\Big|\to 0\) (i.e in the case of recurrence), we have \(\lim_{n\to\infty} v_{0,n}^{(s)}(y)=1\) by the fourth line of 30 .

Finally, 22 , 29 and 31 imply \[\lim_{\substack{\alpha\to 0,\\\alpha>0}}\lim_{n\to\infty}({\widehat L_n}+\alpha)^{-1}1_x(y)= \lim_{n\to\infty}({\widehat L_n})^{-1}1_x(y)=\lim_{n\to\infty}\dfrac{v_{0,n}^{(s_0)}(y)}{\big(\mathcal{L}^{(s_0)} v_{0,n}^{(s_0)}\big)(x)},\] and the last limit is finite in the case of transient graph and is \(1/0=\infty\) in the case of recurrent graph. Due to 20 and 21 the theorem is proven. ◻

Proof. The sectoriality follows from Lemma 1. The density is clear, since \(C_c(X)\subset \mathcal{D}\) is dense in \(\ell^2(X,m)\). To prove the closedness, we use the fact that the lower semi-continuity implies closedness (see [4]). Since \(\operatorname{Re}\mathcal{Q}(f)=\mathcal{Q}_{\operatorname{Re}b}(f)\), using lower-semicontinuity of the form \(\mathcal{Q}_{\operatorname{Re}b}\) on the real-weighted graph \((X,\operatorname{Re},b,m)\) (see [4] for the details) and Lemma 2(2), we obtain that the form \(\mathcal{Q}=\mathcal{Q}_b\) is lower semi-continuos, i.e. \[\label{eq:fQlow} \operatorname{Re}\mathcal{Q}(f)\le \lim \inf_{n\to \infty} \operatorname{Re}\mathcal{Q}(f_n)\tag{32}\] for any \(f, f_n\in \mathcal{D}, n\in \Bbb N\) such that \(f_n(x)\to f(x), n\to \infty\) pointwise for all \(x\in X\). Since \(Q^{(N)}\) is a restriction of \(\mathcal{Q}\) and convergence in \(\ell^2(X,m)\) implies pointwise convergence, we get from 32 that \[\operatorname{Re}Q^{(N)}(f)\le \lim \inf_{n\to \infty} \operatorname{Re}Q^{(N)}(f_n)\] for any \(f\in \ell^2(X,m)\) and \((f_n)_{n\in \Bbb N}\) converging to \(f\) in \(\ell^2(X,m)\). Therefore, \(Q^{(N)}\) with the domain \(D(Q^{(N)})=\mathcal{D}\cap \ell^2(X,m)\) is also lower semi-continuous, and, hence, closed. ◻

Proof. By the definition of the associated operator, \(L^{(D)}\) maps to \(\ell^2(X,m)\), Hence, since its action coincides with the action of the formal Lapalcian, the inclusion \[\label{eq::incl} D(L^{(D)})\subset\{f\in D(Q^{(D)})\;\mid\; \mathcal{L}f\in \ell^2(X,m)\},\tag{33}\] always holds. This immediately implies the inclusion \[D(L^{(D)})\subset\{f\in D(Q^{(N)})\;\mid\; \mathcal{L}f\in \ell^2(X,m)\},\] since \(D(Q^{(D)})\subset D(Q^{(N)})\) by the definitions of the domains.

• Let \(D\big(Q^{(D)}\big)=D\big(Q^{(N)}\big)\). By the above we need to prove \[\{f\in D(Q^{(N)})\;\mid\; \mathcal{L}f\in \ell^2(X,m)\}\subset D(L^{(D)}).\] Let \(f\in D(Q^{(N)})=D(Q^{(D)})\). Then by Greens’ formula we have \[Q^{(D)}(f, \phi)=(\mathcal{L}f\;\mid\;\phi),\] for all \(\phi\in C_c(X)\). As \(D(Q^{(D)})=\overline{C_c(X)}^{\|\cdot\|_\mathcal{Q}}\) by Theorem 3, we conclude \[Q^{(D)}(f, g)=(\mathcal{L}f\;\mid\;g),\] for all \(g\in D(Q^{(D)})\). Hence, \(f\in D(L^{(D)})\).

• Let \(D(L^{(D)})=\{f\in D(Q^{(N)})\;\mid\; \mathcal{L}f\in \ell^2(X,m)\}.\) By the definition of the associated operator, \(L^{(N)}\) maps to \(\ell^2(X,m)\). Hence, since its action coincides with the action of the formal Lapalcian we conclude \[D(L^{(N)})\subset \{f\in D(Q^{(N)})\;\mid\; \mathcal{L}f\in \ell^2(X,m)\}=D(L^{(D)}).\] As both operators are restrictions of \(\mathcal{L}\) and \(m\)-sectorial, we conclude \(D(L^{(N)})= D(L^{(D)}).\) Hence, the associated forms also coincide.

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Proof. Recurrence\(\Rightarrow\)(i). Let \((X,b,m)\) be recurrent. By Corollary 2 the graph \((X,\operatorname{Re}b,m)\) is also recurrent. Hence, any real-valued function from \(\mathcal{D}_0\) is in \(\mathcal{D}\) (see [4]). Now (i) follows from Lemma 5 and Lemma 2.

(i)\(\Rightarrow\)(ii) since \(D\big(Q_\mu^{(D)}\big)=\mathcal{D}_0\cap\ell^2(X,\mu)\) and \(D\big(Q_\mu^{(D)}\big)=\mathcal{D}\cap\ell^2(X,\mu)\) by the definitions of the corresponding forms.

(ii)\(\Rightarrow\)Recurrence. Taking a finite measure \(\mu\) (i.e. \(\mu(X)<\infty\)), we get \(1\in \ell^2(X,\mu)\). Hence, \(1\in D(Q_\mu^{(N)})=D(Q_\mu^{(D)})\), from where follows \(1\in \mathcal{D}_0\) and recurrence by Theorem 10.

Finally, (ii)\(\Leftrightarrow\)(iii) due to Lemma 9. ◻