# Tensor Completion Based on Triple Tubal Nuclear Norm
## 2 Notations and Preliminaries
### Tensor Singular Value Decomposition  
Firstly, five block-based operators, i.e., bvec, bvfold, bdiag, bdfoldand bcirc [1], are introduced. Given a tensor \(\mathbf{\mathcal{T}}\in\mathbb{R}^{n_{1}\times n_{2}\times n_{3}}\), the block vectorizing and its opposite operation are defined as follows:  
\[\mathrm{bvec}(\mathbf{\mathcal{T}}):= \begin{bmatrix}\mathbf{\mathcal{T}}^{(1)}\\ \mathbf{\mathcal{T}}^{(2)}\\ \vdots\\ \mathbf{\mathcal{T}}^{(n_{3})}\end{bmatrix} ,\quad\mathrm{bvfold}(\mathrm{bvec}( \mathbf{\mathcal{T}}))=\mathbf{\mathcal{T}},\]  
the block diag matrix and its opposite operation:  
\[\mathrm{bdiag}(\mathbf{\mathcal{T}}):= \begin{bmatrix}\mathbf{\mathcal{T}}^{(1)}\\ &\ddots\\ &&\mathbf{\mathcal{T}}^{(n_{3})}\end{bmatrix} ,\quad\mathrm{bdfold}(\mathrm{bdiag}( \mathbf{\mathcal{T}}))=\mathbf{\mathcal{T}},\]  
and the block circulant matrix as follows:  
\[\mathrm{bcirc}(\mathbf{\mathcal{T}}):= \begin{bmatrix}\mathbf{\mathcal{T}}^{(1)}&\mathbf{ \mathcal{T}}^{(n_{3})}&\cdots&\mathbf{\mathcal{T}}^{(2)}\\ \mathbf{\mathcal{T}}^{(2)}&\mathbf{\mathcal{T}}^{(1)}&\cdots&\mathbf{\mathcal{T}}^{(3)}\\ \vdots&\ddots&\ddots&\vdots\\ \mathbf{\mathcal{T}}^{(n_{3})}&\mathbf{\mathcal{T}}^{(n_{3}-1)}&\cdots&\mathbf{\mathcal{T}} ^{(1)}\end{bmatrix} .\]  
Based on the five operators defined above, we are able to give the definition of the tensor t-product. **Definition 1** (\(t\)-product [1]). : _Let \(\mathbf{\mathcal{T}}_{1}\in\mathbb{R}^{n_{1}\times n_{2}\times n_{3}}\) and \(\mathbf{\mathcal{T}}_{2}\in\mathbb{R}^{n_{2}\times n_{4}\times n_{3}}\). The t-product of \(\mathbf{\mathcal{T}}_{1}\) and \(\mathbf{\mathcal{T}}_{2}\) is a tensor \(\mathbf{\mathcal{T}}\) of size \(n_{1}\times n_{4}\times n_{3}\):_  
\[\mathbf{\mathcal{T}}=\mathbf{\mathcal{T}}_{1}\ast\mathbf{\mathcal{T}}_{2}=:\mathrm{bvfold }\{\mathrm{bcirc}(\mathbf{\mathcal{T}}_{1})\mathrm{bvec}(\mathbf{\mathcal{T}}_{2})\}. \tag{1}\]  
Viewing a 3D tensor \(\mathbf{\mathcal{T}}\in\mathbb{R}^{n_{1}\times n_{2}\times n_{3}}\) as an \(n_{1}\times n_{2}\) matrix of tubes, the tensor t-product is analogous to the matrix multiplication by replacing scalar multiplication with the vector circular convolution between the tubes, as follows:  
\[\mathbf{\mathcal{T}}(i,j,:)=\sum_{k=1}^{n_{2}}\mathbf{\mathcal{T}}_{1}(i,k,:)\bullet \mathbf{\mathcal{T}}_{2}(k,j,:), \tag{2}\]  
where \(\bullet\) denotes the circular convolution [1] between two tube vector \(\mathbf{x},\mathbf{y}\in\mathbb{R}^{n_{3}}\) defined as:  
\[(\mathbf{x}\bullet\mathbf{y})_{j}=\sum_{k=1}^{n_{3}}\mathbf{x}_{k}\mathbf{y}_{1+(j-k)\mathrm{ mod}n_{3}}.\]  
Due to the relationship between the circular convolution and the DFT, the t-product in the original domain is equivalent to matrix multiplication of the frontal slices in the Fourier domain [1], i.e.,  
\[\mathbf{\widetilde{\mathcal{T}}}^{(k)}=\mathbf{\widetilde{\mathcal{T}}}_{1}^{(k)}\mathbf{ \widetilde{\mathcal{T}}}_{2}^{(k)},\quad k=1,\cdots,n_{3}. \tag{3}\]  
The tensor transpose, identity tensor, f-diagonal tensor and orthogonal tensor are further defined. **Definition 2** (Tensor transpose [1]). : _Let \(\mathbf{\mathcal{T}}\) be a tensor of size \(n_{1}\times n_{2}\times n_{3}\); then \(\mathbf{\mathcal{T}}^{\top}\) is the \(n_{2}\times n_{1}\times n_{3}\) tensor obtained by transposing each of the frontal slices and then reversing the order of transposed frontal Slices 2 through \(n_{3}\)._  
**Definition 3** (Identity tensor [1]). : _The identity tensor \(\mathbf{\mathcal{I}}\in\mathbb{R}^{n_{1}\times n_{1}\times n_{3}}\) is a tensor whose first frontal slice is the \(n_{1}\times n_{1}\) identity matrix, and all other frontal slices are zero.A_  
**Definition 4** (F-diagonal tensor [1]). : _A tensor is called f-diagonal if each frontal slice of the tensor is a diagonal matrix._  
**Definition 5** (Orthogonal tensor [1]).