## 2 Problem Formulation  
Consider that the values \(Z\left(x_{1}\right),Z\left(x_{2}\right),\cdots Z\left(x_{n}\right)\) represent readings of a spatial process \(Z\) at locations \(x_{1},x_{2},\cdots x_{n}\) at a given time instant, and they can be collected and arranged in a vector \(\mathbf{f}=\left[Z\left(x_{1}\right),Z\left(x_{2}\right),\cdots,Z\left(x_{n} \right)\right]^{T}\) to form the network data. The network data \(\mathbf{f}\in\mathbb{R}^{n}\) is assumed to be composed as a linear combination of few atoms from a dictionary \(\mathbf{\Phi}\in\mathbb{R}^{n\times d}\), _i.e._:  
\[\mathbf{f}=\mathbf{\Phi}\mathbf{x} \tag{1}\]  
where \(\mathbf{x}\in\mathbb{R}^{d}\) is expected to be sparse, \(\left\|\mathbf{x}\right\|_{0}\ll n\). The dictionary \(\mathbf{\Phi}\) is the \(n\times d\) matrix with \(\mathrm{rank}\left(\mathbf{\Phi}\right)=n\leq d\). The dictionary is said to be redundant or overcomplete whenever \(n<d\). The network data \(\mathbf{f}\) contains the available data \(\mathbf{f}_{a}\in\mathbb{R}^{m}\) and the missing data \(\mathbf{f}_{p}\in\mathbb{R}^{n-m}\). After reordering:  
\[\mathbf{f}= \begin{bmatrix}\mathbf{f}_{a}\\ \mathbf{f}_{p}\end{bmatrix} \!\!,\ \ \mathbf{f}_{a}\in\mathbb{R}^{m},\, \mathbf{f}_{p}\in\mathbb{R}^{n-m} \tag{2}\]  
According to the indices of the available data \(\mathbf{f}_{a}\) and the missing data \(\mathbf{f}_{p}\), the rows of \(\mathbf{\Phi}\) are partitioned into two parts as:  
\[\mathbf{\Phi}= \begin{pmatrix}\mathbf{A}\\ \mathbf{B}\end{pmatrix} \!\!,\ \ \ \ \mathbf{A}\in\mathbb{R}^{m\times d},\ \ \mathbf{B}\in\mathbb{R}^{(n-m)\times d} \tag{3}\]  
With this partition, the Equation (2) can be regrouped as:  
\[\mathbf{\Phi}\mathbf{x}=\mathbf{f}\Rightarrow \begin{cases}\mathbf{A}\mathbf{ x}=\mathbf{f}_{a}\\ \mathbf{B}\mathbf{x}=\mathbf{f}_{p}\end{cases}  \tag{4}\]To recover **f**, we can find the solution **x\({}^{*}\)** first by solving:  
\[\textbf{A}\textbf{x}=\textbf{f}_{a} \tag{5}\]  
and then plug it into:  
\[\textbf{B}\textbf{x}=\textbf{f}_{p} \tag{6}\]  
to get **f\({}_{p}\)**. However, Equation (5) is under-determined since \(m<d\), thus more than one solutions are possible to satisfy it. Since **x** is sparse, we can employ sparsity to regularize the solution by solving the \(\ell_{{}_{1}}\)-minimization problem:  
\[\arg\min_{\textbf{x}}\left\|\textbf{x}\right\|_{\textbf{h}}\quad\text{ s.t. }\textbf{A}\textbf{x}=\textbf{f}_{a} \tag{7}\]  
Now suppose **f** is \(k\) -sparse, _i.e._, it can be represented as a weighted combination of \(k\) columns of dictionary \(\boldsymbol{\Phi}\). Given the support of coefficient vector **x** is \(S_{\textbf{x}}=\left\{i:x_{i}\
eq 0\right\}\), the cardinality of **x\(:\left|S_{\textbf{x}}\right|=k\)**, and \(S_{\textbf{x}}\) is the set of index of nonzero entry in **x**, accordingly, the available data is:  
\[\textbf{f}_{a}=\sum_{j\in S_{\textbf{x}}}x_{j}A_{j} \tag{8}\]  
where \(A_{j}\) stands for the \(j\)th column of **A**. For simplicity, we assume all columns of \(\boldsymbol{\Phi}\) are orthogonal to each other. Due to some entries of \(f\) are missing, the columns in **A** is shorter than the columns in \(\boldsymbol{\Phi}\). and some columns of **A** are correlated. Suppose there is another nonzero vector \(\tilde{\textbf{x}}\in\mathbb{R}^{d}\) satisfying **A\(\tilde{\textbf{x}}=\textbf{f}_{a}\)** the support of \(\tilde{\textbf{x}}\) is \(S_{\tilde{\textbf{x}}}\) with \(\left|S_{\tilde{\textbf{x}}}\right|=\tilde{k}\). If the \(i\)th (\(i\in S_{\textbf{x}}\)) column in **A** is correlated with other columns, \(\textbf{f}_{a}\) can also represented by weighted combinations of \(\tilde{k}\) column of **A** :  
\[\textbf{f}_{a}=\sum_{j\in S_{\tilde{\textbf{x}}}}\tilde{x}_{j}A_{j} \tag{9}\]  
if \(\sum_{j\in S_{\tilde{\textbf{x}}}}\left|\tilde{x}_{j}\right|\leq\sum_{i\in S_{ \textbf{x}}}\left|x_{i}\right|\), the \(\ell_{{}_{1}}\)-minimization algorithm will choose solution \(\tilde{\textbf{x}}\), and thus leading to a wrong estimation.