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.

didwrite "inspired by..." $\endgroup$