$\ell^1$-bound on graph laplacian with weight Consider the $\mathbb Z^2$ lattice, we then define for $u=(u_{ij})_{i,j \in \mathbb Z}$ the discrete Laplacian
$$(\Delta u)_{i,j}=u_{i+1,j}+u_{i-1,j}+ u_{i,j+1}+u_{i,j-1}$$
and the weight which pushes the mass down at every point
$$ (T u)_{i,j}=\frac{1}{\sqrt{i^2+j^2+1}} u_{i,j}.$$
We then define $v_{ij}=\delta_{i,0} \delta_{j,0}$ (so $v_{0,0}=1$ and $v_{i,j}=0$ otherwise).
It is then easy to study bounds on
$$\sum_{ij} \vert ((T \Delta)^n v)_{i,j} \vert$$
that are exponential in $n$. Here, I look at the $n$-fold composition of the discrete Laplacian with $T$ and take the $\ell^1$ norm over the $\mathbb Z^2$ lattice. I would like to know if it is possible to obtain a polynomial bound on that sum in $n$ or if this sum scales indeed exponentially?
Please let me know if there are any questions!
 A: Direct calculation shows $((T\Delta)^2 v)_{0,0} = 2\sqrt{2}$.
This implies that
$$((T\Delta)^{2n}v)_{0,0} \ge (2\sqrt{2})^n$$
for $n = 1,2,\ldots$.
You might be interested in studying the iterates of $P_{\lambda} = \frac{1}{4}\lambda T\Delta = \lambda T\Delta'$ where $\Delta' = \frac{1}{4}\Delta$ and $\lambda > 0$ (I prefer $\Delta'$ to $\Delta$ since it does not increase the $L^1$-norm of any function).
In this notation your original question was about $P_4$.
The same calculation as above shows that $(P_{\lambda}^2 v)_{0,0} = \frac{\lambda^2}{4\sqrt{2}}$ so that there is exponential growth if $\lambda > 2\sqrt[4]{2}$.
I think I can show that there is exponential contraction of the $L^1$ norm if $\lambda < \sqrt[4]{2}$ (I have no idea what happens in between $\sqrt[4]{2}$ and $2\sqrt[4]{2}$).
Call a pair $(m,n) \in \mathbb{Z}^2$ a black square if $m = n\text{ (mod 2)}$ and a white square otherwise.
Call a function $f:\mathbb{Z}^2 \to \mathbb{R}$ black if it is zero on all white squares, and white if it is zero on all black squares.
Now $T$ does not increase the $L^1$ norm of any function.  Furthermore it preserves the subspace of white functions and the subspace of black functions.  Finallay, $T$ contracts the $L^1$ norm of all white functions by at least $\frac{1}{\sqrt{2}}$.
On the other hand $\Delta'$ sends white functions to black ones and vice versa.  And never increases the $L^1$ norm.
These observations imply that if $f$ is black then
$$|P_{\lambda}^{2n}f|_1 = \lambda^{2n}|(T\Delta')^{2n} f|_1 \le \left(\frac{\lambda^{2}}{\sqrt{2}}\right)^n|f|_1,$$
for all $n$.
This gives exponential contraction if $\lambda < \sqrt[4]{2}$.
