Let $d \geq 2$ and $K$ some positive integer. Consider distinct points $\theta_1, \ldots, \theta_K\in \mathbb{T}^d$ and (not necessarily distinct) $z_1, \ldots, z_K \in \mathbb{C}$, does there exist an eigenfunction of the Laplacian $f: \mathbb{T}^d \to\mathbb{C}$ such that $f(\theta_i) = z_i$ for all $i$?

When $d = 1$ the answer is in general in the negative. Every eigenfunction has the form $f(\theta) = A_+ e^{2\pi i n \theta} + A_- e^{2\pi i (-n) \theta} $, so for fixed $\theta_1, \ldots, \theta_K$, the set $\{ (f(\theta_1), \ldots, f(\theta_K) \}$ is an at-most 2 (complex) dimensional subspace of $\mathbb{C}^K$ so generic $(z_1, \ldots, z_K)$ cannot be achieved, when $K \geq 3$.

A different phenomenon happens when one has higher dimensional eigenspaces. For example, take $d = 2$. Then the functions with eigenvalue 65 is given in general as $$ f(\theta_x, \theta_y) = \sum A_{n_x, n_y} e^{2\pi i (n_x \theta_x + n_y\theta_y)} $$ where the sum ranges over $$(n_x, n_y)\in \{ (-8,-1), (-8,1), (-1,-8), \ldots, (8,1), (-7,-4), \ldots (7,4)\}$$ so for generic angles $\theta_1, \ldots, \theta_K$, the corresponding set $\{ (f(\theta_1), \ldots, f(\theta_K))\}$ is expected to be a $r_2(65) = 16$ dimensional subspace of $\mathbb{C}^K$. So one may expect, at least generically, the extension problem is solvable up to $K = 16$, using just the eigenvalue 65 subspace.

Using that the sums of square function $r_k$ is unbounded for $k \geq 2$, a similar argument leads to the expectation that the question posed at the beginning of this post has a positive answer for generic $(\theta_1, \ldots, \theta_K)$ (in some suitable sense of the word).

Question 1: is this indeed the case?

Question 2: how big is this generic set?

Remark 1: this question is inspired by this other queston

Remark 2: I'm not really sure what the correct tags should be, feel free to edit.

  • $\begingroup$ If I may interrupt, you have moved quite tangentially from my question, which was actually a analysis question rather than number theory. The possible misunderstanding stems from the word "weak" in "weak eigenfunction". being ignored. But if you are independently interested in this question, its great to see my question inadverdently sparked this number theory question. My intention is not really in this question. $\endgroup$
    – Rajesh D
    Aug 2 '18 at 4:20
  • 2
    $\begingroup$ @RajeshDachiraju: I did write "inspired by..." $\endgroup$ Aug 2 '18 at 4:23
  • $\begingroup$ I wouldn't say exactly related, but a useful read, although elementary, this one: math.stackexchange.com/a/2287154/2987 $\endgroup$
    – Rajesh D
    Aug 2 '18 at 4:30
  • $\begingroup$ Finally I agree that all weak solutions are $C^{\infty}$, so this current question seems more appropriate than the parent one. mathoverflow.net/q/307356/14414 $\endgroup$
    – Rajesh D
    Aug 2 '18 at 16:23
  • $\begingroup$ But, this question is not for me, as I don't think I can handle number theory. $\endgroup$
    – Rajesh D
    Aug 2 '18 at 16:26

(EDITED) You certainly can't have it for all $\theta$. Consider $n=2$. An eigenfunction $f(\theta_1,\theta_2) = \exp(i k_1 \theta_1 + i k_2 \theta_2)$ for eigenvalue $\lambda = -k_1^2 - k_2^2$ will have $f(0,0) + f(\pi,\pi) = f(0,\pi) + f(\pi,0)$ unless $k_1$ and $k_2$ are both odd (i.e. $\lambda \equiv 2 \mod 4$), in which case $f(0,0) + f(\pi,\pi) = -f(0,\pi) - f(\pi,0)$.


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