## Appendix B Proof of Lemma 1  
Let \(\hbar\subset\{1,\cdots,I\}\) and \(|\hbar|\leq 2\). Let \(\alpha_{\hbar}\in\mathbb{R}^{|\hbar|}=\text{vec}[\alpha_{i}\mid\forall i\in \hbar]\) and \(\mathbf{G}(\alpha_{\hbar})\in\mathbb{C}^{\text{N}_{\Omega}M\times|\hbar|}\) be the submatrix formed by indexing the columns of \(\mathbf{G}(\mathbf{\alpha})\) with \(\hbar\), i.e., \(\mathbf{G}(\alpha_{\hbar})=\left(\text{mat}\!\left[(\phi(\alpha_{i}))^{\top} \mid\forall i\in\hbar\right]\right)^{\top}\). \(\lambda_{max}\) and \(\lambda_{min}\) as \[\lambda_{max}=\max_{ \begin{subarray}{c}\hbar\subset\{1,\cdots,I\},\\ \mathrm{s.t.}|\hbar|=1,2\end{subarray} }\Big{\{}\mathrm{eig}\Big{(}(\mathbf{G}( \boldsymbol{\alpha}_{\hbar}))^{\mathrm{H}}\mathbf{G}(\boldsymbol{\alpha}_{ \hbar})\Big{)}\Big{\}} \tag{11}\]  
\[\lambda_{min}=\min_{ \begin{subarray}{c}\hbar\subset\{1,\cdots,I\},\\ \mathrm{s.t.}|\hbar|=1,2\end{subarray} }\Big{\{}\mathrm{eig}\Big{(}(\mathbf{G}( \boldsymbol{\alpha}_{\hbar}))^{\mathrm{H}}\mathbf{G}(\boldsymbol{\alpha}_{ \hbar})\Big{)}\Big{\}} \tag{12}\]  
where \(\mathrm{eig}(\bullet)\) yields all eigenvalues of a matrix. Let \(\boldsymbol{\mu}_{\hbar}\in\mathbb{C}^{|\hbar|}\) be the subvector of \(\boldsymbol{\mu}\Big{(}\) in \(\mathcal{D}_{(I,2)}\Big{)}\) indexed by \(\hbar\). Using these formulas, the inequality in Lemma 1 can be simplified to \(\Big{(}1-\delta_{(2)}\Big{)}\|\ \boldsymbol{\mu}_{\hbar}\ \|_{2}^{2}\leq\|\ C_{(2)}^{-1} \mathbf{G}(\boldsymbol{\alpha}_{\hbar})\boldsymbol{\mu}_{\hbar}\ \|_{2}^{2}\leq\Big{(}1+\delta_{(2)} \Big{)}\|\ \boldsymbol{\mu}_{\hbar}\ \|_{2}^{2}\), where \(\forall\hbar\subset\{1,\cdots,I\}\) and \(|\hbar|\leq 2\). This implies that, given an appropriate regularization factor \(C_{(2)}\), the eigenvalues of \(C_{(2)}^{-1}(\mathbf{G}(\boldsymbol{\alpha}_{\hbar}))^{\mathrm{H}}\mathbf{G}( \boldsymbol{\alpha}_{\hbar})\) need to be within the range \(\Big{[}1-\delta_{(2)},1+\delta_{(2)}\Big{]}\), where \(\delta_{(2)}\) is a constant and \(\delta_{(2)}\in(0,1),\forall\hbar\subset\{1,\cdots,I\}\), \(|\hbar|\leq 2\)[36]. Therefore, we need to prove that (i) all eigenvalues of \((\mathbf{G}(\boldsymbol{\alpha}_{\hbar}))^{\mathrm{H}}\mathbf{G}( \boldsymbol{\alpha}_{\hbar})\) are upper bounded and (ii) \(\mathbf{G}(\boldsymbol{\alpha}_{\hbar})\) is full column rank for \(\hbar\subset\{1,\cdots,I\}\), \(|\hbar|\leq 2\). * Condition (i): According to Equation (8), we know that \(\boldsymbol{\psi}_{m}(\alpha_{i};\omega)\) is finite for all \(\forall\alpha_{i}\in[s_{U},s_{D}]\). Hence, all elements and eigenvalues of \((\mathbf{G}(\boldsymbol{\alpha}_{\hbar}))^{\mathrm{H}}\mathbf{G}( \boldsymbol{\alpha}_{\hbar})\) are bounded. * Condition (ii): Our goal is to prove that \(\mathbf{G}(\boldsymbol{\alpha}_{\hbar})\) satisfies condition (ii) when \(|\hbar|\) is \(1\) and \(2\), respectively. If \(|\hbar|=1\) (i.e., \(\|\ \boldsymbol{\mu}\ \|_{0}=1\)), \(\mathbf{G}(\boldsymbol{\alpha}_{\hbar})\) reduces to a vector \(\varphi(\alpha_{i})\). In this case, \(\mathbf{G}(\boldsymbol{\alpha}_{\hbar})\) is full column rank, if \(\|\ \mathbf{G}(\boldsymbol{\alpha}_{\hbar})\ \|_{2}\
eq 0\), which is obviously satisfied. If \(|\hbar|=2\) (i.e., \(\|\ \boldsymbol{\mu}\ \|_{0}=2\)), let us assume that \(\boldsymbol{\alpha}_{\hbar}=\Big{[}\alpha_{i},\alpha_{i}^{{}^{\prime}}\Big{]}\) when \(\alpha_{i}\
eq\alpha_{i}^{{}^{\prime}}\in[s_{U},s_{D}]\). Then, \(\mathbf{G}(\boldsymbol{\alpha}_{\hbar})\) can be specified as \(\mathbf{G}(\boldsymbol{\alpha}_{\hbar})=\Big{[}\boldsymbol{\varphi}(\alpha_{ i}),\boldsymbol{\varphi}\Big{(}_{i}^{{}^{\prime}}\Big{)}\Big{]}\). Let \(\Omega=\{\omega_{1},\cdots,\omega_{N_{\Omega}}\}\) and let \(\boldsymbol{\beta}=\big{[}s_{1},\cdots,s_{M},\omega_{1},\cdots,\omega_{N_{ \Omega}}\big{]}^{\top}\). Consider \(\wp(\beta;\kappa)=\varphi(\alpha_{i})+\kappa\varphi\Big{(}\alpha_{i}^{{}^{ \prime}}\Big{)}\) (where \(\kappa\) is a constant), which is a composite function involving hyperbolic functions.