One of the definitions of the discrete (weighted) $p$-Laplacian is the following:
$$\Delta_{p,w}u(x):=\sum_y |u(y)-u(x)|^{p-2}(u(y)-u(x))w(x,y).$$
Consider the one dimensional case. Then the free Laplacian corresponds $w(x,y)=1$ if and only if $|y-x|=1$.
In the the continuous case one sees that the bi-Laplacian, that is the operator $$\Delta^2:=\Delta \circ \Delta,$$ is $\textbf{equal}$ to the the 4-Laplacian (defined in the standard way http://en.wikipedia.org/wiki/P-Laplacian ).
It seems that it is $\textbf{not true in the the discrete case}$, i.e., the discrete 4-Laplacian is not equal to the discrete bi-Laplacian. Am I right or do I make a mistake somewhere? If indeed there is no equality between the bi-Laplacian and the 4-Laplacian in the discrete case , how should one understand it (what is the reason, how does it relate to the continuous case etc.)?