Recently I have read many papers which focusing on the image enhancement. The description,



> 
> $D$ are the Toeplitz matrices from the discrete gradient with forward difference
> 
> 
> 


occurs many times.


An example is shown below:


To minimize an object function,


$$ F(T)=\sum\_x\left((T(x)-L(x))^2 + \lambda\sum\_{d\in\lbrace h,v\rbrace}A\_d(x)\left(\partial\_dT(x)\right)^2\right), $$


where $L(x)$ and $A\_d(x)$ is known. So to minimize this function, we differentiating it and setting the derivative to 0, the problem is simplified to:


$$t=\left(I + \sum\_{d\in\lbrace h, v\rbrace}D\_d^TDiag(a\_d)D\_d\right)^{-1}l,$$


where $m$ is the vectorized version of $M$ and the operator $Diag(m)$ is to construct a diagonal matrix using vector $m$, and **$D$ are the Toeplitz matrices from the discrete gradient with forward difference** .


So my question is, what on earth is $D$ ? How to calculate this $D$ ? And what does the transpose (or adjoint) of this $D^T$ mean ?



What is the "Toeplitz matrices from the discrete gradient operators with forward difference"
A Toeplitz matrix is a band matrix in which each descending diagonal from left to right has constant elements. The derivative of a function can be represented in a finite-difference calculation as the product of a Toeplitz matrix and equally spaced values of $f$. One distinguishes forward differences and backwards differences, the forward difference is given by
$$D[f](x)=f(x+\delta x)-f(x),$$
corresponding to a matrix with elements (in the $4\times 4$ case)
$$D=\begin{pmatrix}
-1&1&0&0\\
0&-1&1&0\\
0&0&-1&1\\
1&0&0&-1
\end{pmatrix}
$$
The backward difference is given by $-D^\top$, minus the transpose of the forward difference,
$$-D^\top[f](x)=f(x)-f(x-\delta x).$$


