In my current work, I use the eigenfunctions of the Laplacian: $\varphi_k, \lambda_k \in H^1_0, \mathbb{R}^+$ satisfying
\begin{align*}
-\Delta \varphi_k = \lambda_k \varphi_k.
\end{align*}
It is well known that $\{\varphi_k\}_{k=1}^\infty$ provides an orthonormal basis for $L^2$, and is also orthogonal in $H^1$. Hence, any function $f \in L^2$ can be expressed as $f = \sum_{k=1}^\infty f_k \varphi_k$ where $f_k = (f, \varphi_k)$.
Unfortunately, there are not many properties which can be derived from this eigenfunction expansion. Besides the fact that the squared $L^2$ norm of $f$ is simply $\sum_{k=1}^\infty f_k^2$, and the $H^1$ norm is $\sum_{k=1}^\infty f_k^2 \lambda_k$, the connection between regularity and expansion is tenuous at best.
I was excited about the paper Eigenfunction Expansions of Analytic Functions by Seeley. In it, the author claimed to have derived a theorem giving necessary and sufficient conditions on analyticity and the eigenfunction expansion: a function $f$ is analytic iff $\sum_{k=1}^\infty s^{\sqrt\lambda_k} f_k^2 < \infty$ or $\{s^{\sqrt{\lambda_k}} |f_k| \}$ is bounded for some $s > 1$. Unfortunately, I don’t think it is an iff.
In particular, on a square with $f = 1$. We know the coefficients are
\begin{align*}
f_{ij} = \frac{2 \left((-1)^i-1\right) \left((-1)^j-1\right)}{\pi ^2 i j} \approx \frac{1}{ij}
\end{align*}
and so the theorem is stating that
\begin{align*}
s^{\sqrt{\lambda_{mm}}} f_{mm} &= \frac{2 \left((-1)^m-1\right)^2 s^{\sqrt{2} \pi m}}{\pi ^2 m^2} \\
&\approx \frac{s^{\sqrt{2} \pi m}}{m^2} \to \infty
\end{align*}
as $m\to \infty$ which is clearly unbounded.
Rather, more conditions needs to be imposed on the theorem. It is not hard to show that any function satisfying $\sum_{k=1}^\infty s^{\sqrt\lambda_k} f_k^2 < \infty$ is in any $\mathbb{H}^s := \{ u \in L^2 | \sum_{k=1}^\infty u_k^2 \lambda_k^s < \infty \}$ space for $s \ge 0$. Coincidentally, we know $\mathbb{H}^s = H^s_0$ for $1 > s > 1/2$ by an interpolation argument, meaning that at the minimum, our functions need to vanish at the boundary.