Laplace-eigenfunctions on the torus with high vanishing order

We use the sum-of-squares theorem from number theory to construct eigenfunctions of the Laplacian on the d-dimensional torus, d ≥ 2, which vanish to any prescribed order at some point. These functions are then applied to provide a negative answer (in dimension d ≥ 2) to a question in the context of quantitative unique continuation for spectral projectors of Schrödinger operators, asked by Egidi and Veselić in [EV16]. For a compact, connected manifold X with C Riemannian metric, the Laplace (or Laplace-Beltrami) operator ∆ on X is a non-positive, self-adjoint operator. One says that f : X → C vanishes to order N > 0 at x0 ∈ X if lim sup δ→0 supy∈Bδ(x0)|f(x)| δN <∞. In [DF88], Donnelly and Feffermann proved that if f is a nonzero eigenfunction of −∆ on X with eigenvalue λ it will vanish to order at most c √ λ at any point x ∈ X where c is a constant that depends only on X. A complementary question is: Given any vanishing order N , can we find eigenfunctions vanishing to order at least N at some point x0? In dimension d = 1, for instance if X is an interval with its end points identified, this is not possible as soon as N > 1. This follows immediately from the fact that all nonzero eigenfunctions are of the form α sin(ωx+ y0) + β cos(ωx + y0) and vanish to order at most 1 at any point. However, in dimension d ≥ 2 it is indeed possible to find eigenfunctions of high vanishing order as our first result, Theorem 1, shows. Let us denote by T = [−π, π) the d-dimensional torus, i.e. opposite sides of [−π, π) are identified, and by ∆ the Laplace operator on T. Theorem 1. Let d ≥ 2. For every N > 0 there is a nonzero function f ∈ L2(Td) with −∆f = λf for some λ > 0 such that f vanishes to order at least N at 0. By translation invariance on T, we may replace 0 by any point x0 ∈ T. Proof. For λ ≥ 0 let Iλ := {k ∈ Z : |k|2 = λ} and for k ∈ Z let ψk(x) := exp(ik · x) ∈ L2(Td). A function f ∈ L2(Td) is an eigenfunction to the eigenvalue λ ≥ 0 if and only if f is of the form f = ∑