Consider a connected (we define connected components by defining the set of vertices where every vertex has one neighbour) sublattice $V$ of the square lattice $V \subset\mathbb{Z}^2.$

On this we define the discrete Laplacian as $T:\ell^2(V) \rightarrow \ell^2(V)$ by

$$ (Tf)(x)=\sum_{y \text{ neighbour of } x}(f(y)-f(x)).$$

Now, my question is: Is there an infinite(!) connected sublattice $V \subset\mathbb{Z}^2$ on which this objects has eigenfunctions?

Why do I ask: It is easy to see that on $\mathbb{Z}^2$ by using the Fourier transform for example, the spectrum is purely absolutely continuous.

I would like to understand whether this is because $\mathbb{Z}^2$ is infinite (and so every eigenfunction would dissolve to infinity) or whether this is because $\mathbb{Z}^2$ is translational invariant.

Why the assumptions: If the sublattice is finite, we obtain eigenfunctions because this operator is just a matrix.

If the sublattice was not connected, it could have a finite connected component.

So to exclude this, I stated the assumptions.