Proof: $(i)$ Each component $i$ is Y-realistic and therefore meromorphic on $\mathbb{C}$, which causes $\mathbf{Y}_{b}$ to be meromorphic on $\mathbb{C}$. Meromorphic functions on a domain form a field of fractions, which makes $\mathbf{Y}=\mathbf{A}\mathbf{Y}_{b}\mathbf{A}^{t}$ meromorphic on $\mathbb{C}$.

According to Definition 1, the fact that component $i$ is Y-realistic implies that there exists $\alpha_{i}$ such that $Y_{i}\!\left(s\right)+Y_{i}^{*}\!\left(s\right) \succeq \alpha_{i}\mathbf{Id}$ whenever $\Re(s)\ge0$ and $|s|$ is large enough. Let $\alpha_{{\rm min}}>0$ be the infimum of the $\alpha_{i}$. Then $\alpha_{{\rm min}}>0$ and we have that the branch admittance matrix $\mathbf{Y}_{b}\!\left(s\right)+\mathbf{Y}_{b}^{*}\!\left(s\right)\succeq\alpha_{\min}\mathbf{Id}$ for all $s$ with $\Re(s)\geq0$ and $|s|\geq K$, for some $K$.

Since the incident matrix $\mathbf{A}$ has full rank, by the assumed connectivity of the graph of the circuit, the symmetric matrix $\mathbf{A}\,\mathbf{A}^{t}$ is nonsingular so there exist $\lambda>0$ such that $\mathbf{A}\,\mathbf{A}^{t}\!\succeq\!\lambda\,{\bf Id}$. Letting $\alpha=\alpha_{{\rm min}}\,\lambda$, we deduce from (2.5) and the realness of $\mathbf{A}$ that assertion $(i)$ holds.

$(ii)$ Assume that for some $s_0$ satisfying $\Re(s_0)\geq0$ and $|s_0|\geq K$, the matrix $\mathbf{Y}(s_0)$ fails to be invertible. Then, there is a non zero complex vector $v$ such that ${\bf Y}(s_0) v=0$, thus also $v^{*}{\bf Y}(s_0) v=0$. Taking the real part, we get $v^{*}({\bf Y}(s_0)+{\bf Y}^{*}(s_0))v=0$ which contradicts the first item. The determinant of $Y$ is therefore not the zero function, and assertion $(ii)$ holds.

$(iii)$ Combining assertion $\left(i\right)$ with the fact that $\mathbf{Y}+\mathbf{Y}^{*}=\mathbf{Y}(\mathbf{Z}+\mathbf{Z}^{*})\mathbf{Y}^{*}$, we have for $\Re(s)\geq0$ and $|s|\geq K$ that

Letting $u\!\left(s\right)$ be a maximizing vector of $\mathbf{Z}^{*}\!\left(s\right)$ with unit norm, i.e. such that $\|Z^*(s) u(s)\|=\|Z(s)\|$, we get

hence $\|\mathbf{Z}\!\left(s\right)\|\leq 2/\alpha$.

∎

We have proven that the branch admittance ${\bf Y}$ matrix of the circuit is Y-realistic. Working with the branch impedance matrix ${\bf Z}$ instead, it can be similarly proven that ${\bf Z}$ is Z-realistic. In fact, a linearized circuit made of active and passive realistic components is realistic in the sense of Definition 1. The dual of Proposition 1 can be proven, changing $Z$ in $Y$ and Y-realistic into Z-realistic.

Proof: The meromorphic property of $G$ is a consequence of $(ii)$ and its boundedness a consequence of $(iii)$.

∎

Recall that $\mathcal{H}^{\infty}$ is the space of bounded analytic functions in the right half-plane $\Pi^+$. Recall also that a rational function is said to be strictly proper if it vanishes at infinity, i.e. if either the degree of the numerator is strictly less than the degree of the denominator or else it is the zero function.

Proof: Let $\overline{D(0,K)}$ denote the closed disk centered at $0$ of radius $K$. By the previous lemma, the unstable poles of $G\left(s\right)$ (if any) lie in the compact set $\overline{\Pi}^+\cap\overline{D(0,K)}$. Since $G\left(s\right)$ is a meromorphic function on $\mathbb{C}$, it can only have finitely many poles on a compact set. Hence $G\left(s\right)$ has at most finitely many unstable poles, as announced.

Number these unstable poles as $s_{1}$…$s_{N}$, with respective multiplicities $\nu_{1}$,…$\nu_{N}$. Let $\mathfrak{r}_{j}=p_{\nu_{j}-1}(s)/(s-s_{j})^{\nu_{j}}$ be the principal part of $G\left(s\right)$ at $s_{j}$, where $p_{\nu_{j}-1}(s)$ is a polynomial of degree $\nu_{j}-1$. It is a strictly proper rational function. Set $\mathfrak{r}=\sum_{j=1}^{N}\mathfrak{r}_{j}$; if $N=0$ the sum is empty and we have that $\mathfrak{r}=0$. By construction, $G-\mathfrak{r}$ is a meromorphic function with no poles in the closed right half-plane. Moreover it is bounded there for large $|s|$, because so is $G\left(s\right)$ by Lemma Lemma 1 and so is the strictly proper rational function $\mathfrak{r}$. Hence $G-\mathfrak{r}\in \mathcal{H}^{\infty}$, as was to be shown.

∎