I don't know much about the spectral theory of $\text{SL}_2(\mathbb{Z}) \backslash \mathbb{H}$ nor do I know much about Bessel functions, hence the following question.
Suppose $f$ is Maass form of weight zero for $\text{SL}_2(\mathbb{Z})$ with eigenvalue written $\lambda = 1/4 - \nu^2$ for $\nu \in \mathbb{C}$, which is unique up to sign, and must either be real or purely imaginary, because $\Delta$ is a symmetric, positive operator. Does there exist a $y > 0$ such that $$ K_{\nu}(y) = \frac{1}{2} \int_{0}^{\infty}{e^{-y/2( t + t^{-1})}\, t^{\nu}\, \frac{dt}{t}} \neq 0 ? $$ This is of course true if $\nu$ is real, because then the integrand is strictly positive (in fact, for all $y >0$). What if $\nu$ is purely imaginary? In that case, one has $\nu = i \kappa$ where $\kappa > 3/2 \pi^2 - 1/4 \geq 3.8$ (as is proved in Motohashi's book "Spectral theory for the Riemann zeta function", Lemma 1.4)
I am interested in the non-vanishing of these functions, because I had the following thought: Suppose $f$ is a Maass cusp form. Then we know that $$ f(z) = \sum_{m \neq 0}{a_f(m) \sqrt{y}K_{\nu}(2\pi |m|y) e(mx)}$$ for some Fourier coefficients $a_f(m)$. Suppose that $\{c(m)\}_{m \in \mathbb{Z} - \{0\}}$ is a collection of complex numbers such that also $$f(z) = \sum_{m \neq 0}{c(m) \sqrt{y}K_{\nu}(2\pi |m|y) e(mx)}\,.$$ Then, since the Fourier transform $C^{\infty}(\mathbb{R}/\mathbb{Z}) \rightarrow c_0(\mathbb{Z})$ is injective, we know that $$a_f(m) \sqrt{y}K_{\nu}(2\pi |m|y) = c(m) \sqrt{y}K_{\nu}(2\pi |m|y)$$ for all $m \neq 0$ and all $y > 0$. From which I'd like to deduce $a_f(m) = c(m)$.