From the definition, \(\Phi_{y,u}(w)\) is real rational and positive semi-definite for \(|z|=1\). The closed-loop transfer function \(\mathcal{C}(z)\) is stable and minimum phase. Therefore, \(\mathcal{C}(z)\) is analytic in \(|z|\geq 1\). Since \(\mathcal{G},\mathcal{H}\) are strictly proper, we have \(\mathcal{G}(\infty)=0,\,\mathcal{H}(\infty)=0\). On the other hand, since \(\mathcal{K}\) is proper and rational, \(\mathcal{K}(\infty)\) exists. Hence
\begin{align} \lim_{z\rightarrow\infty}\mathcal{C}(z)=\begin{bmatrix}0&I\\ 0&\mathcal{K}(\infty)\end{bmatrix}.\notag \\ \end{align}Assume that both \(\mathcal{C},~{}D=\begin{bmatrix}Q&0\\ 0&R\end{bmatrix}\) and \(\hat{\mathcal{C}},~{}\hat{D}=\begin{bmatrix}\hat{Q}&0\\ 0&\hat{R}\end{bmatrix}\) give the same \(\Phi_{y,u}\) satisfying
\(D\) and \(\hat{D}\) are block diagonal and positive definite matrices;
both \(\mathcal{C}\) and \(\hat{\mathcal{C}}\) are stable and minimum phase,
then there exists a paraunitary matrix \(\mathcal{V}(z)\) such that \cite{Anderson_1982}
\begin{align} \label{eq:c1c2} \label{eq:c1c2}\hat{\mathcal{C}}(z)=\mathcal{C}(z)\mathcal{V}(z), \\ \label{eq:q1q2}\hat{D}=\mathcal{V}(z)D\mathcal{V}^{*}(z).\\ \end{align}From (\ref{eq:c1c2}), since both \(\mathcal{C}(z)\) and \(\hat{\mathcal{C}}(z)\) are stable and minimum phase, \(\mathcal{V}(z)\) is stable and minimum phase, which implies that \(\mathcal{V}(z)\) is a constant matrix independent of \(z\) \cite{Anderson_1969, Hayden_2014}. Therefore, we denote it simply as \(V\). Take \(z\rightarrow\infty\) on both sides of (\ref{eq:c1c2}) yields
\begin{equation} \begin{bmatrix}0&I\\ 0&\hat{\mathcal{K}}(\infty)\end{bmatrix}=\begin{bmatrix}0&I\\ 0&\mathcal{K}(\infty)\end{bmatrix}\lim_{z\rightarrow\infty}V,\\ \end{equation}which leads to
\begin{align} V_{21}=0,\,V_{22}=I.\\ \end{align}Since \(VV^{*}=I\), we have \(V_{12}=0\) and \(V_{11}V_{11}^{*}=I\). As a result, (\ref{eq:c1c2}) and (\ref{eq:q1q2}) imply that
\begin{aligned} & \hat{\mathcal{C}}=\mathcal{C}\begin{bmatrix}V_{11}&0\\ 0&I\end{bmatrix}\Leftrightarrow\left\{\begin{array}{r@{\;=\;}l}\hat{\mathcal{C}}_{11}\;=\;&\mathcal{C}_{11}V_{11}\\ \hat{\mathcal{C}}_{12}\;=\;&\mathcal{C}_{12}\\ \hat{\mathcal{C}}_{21}\;=\;&\mathcal{C}_{21}V_{11}\\ \hat{\mathcal{C}}_{22}\;=\;&\mathcal{C}_{22}\end{array}\right.~{}~{}\text{and} \\ & \hat{D}=\begin{bmatrix}V^{*}_{11}&0\\ 0&I\end{bmatrix}D\begin{bmatrix}V_{11}&0\\ 0&I\end{bmatrix}\Leftrightarrow\left\{\begin{array}{r@{\;=\;}l}\hat{Q}\;=\;&V^{*}_{11}QV_{11}\\ \hat{R}\;=\;&R\end{array}\right..\nonumber \\ \end{aligned}∎
Assume that \((X,\,\tilde{S})\) is the optimal solution for (\ref{eq:optimization3}). Since \(\tilde{S}\geq g_{X}(\tilde{S})\) and \(g_{X}\) is monotonically non-decreasing in \(\tilde{S}\), we know that
\begin{align} \tilde{S}\geq g_{X}(\tilde{S})\geq g_{X}^{(2)}(\tilde{S})\geq\dots\geq 0,\notag \\ \end{align}where
\begin{align} g_{X}^{(1)}(\tilde{S})\triangleq g_{X}(\tilde{S}),\,g_{X}^{(n+1)}(\tilde{S})\triangleq g_{X}\left(g_{X}^{(n)}(\tilde{S})\right).\notag \\ \end{align}Since \(g_{X}^{(n)}(\tilde{S})\) is monotonically decreasing and positive semidefinite, it will converge to a matrix \(\tilde{S}^{*}=g_{X}(\tilde{S}^{*})\leq\tilde{S}\). Therefore, \((X,\,\tilde{S}^{*})\) is also the optimal solution of (\ref{eq:optimization3}), which finishes the proof. ∎