Let $\mathbb{H}$ be the upper half plane. For $g=\left(\begin{array}{cc}a&b\\c&d\end{array}\right) \in SL_2(\mathbb{R})$ and $z \in \mathbb{H}$, set $J(g,z)=cz+d$. Let $SO(2)$ be the special orthogonal group. For a positive integer $k$, define $H_k(\mathbb{H})$ to be $$ H_k(\mathbb{H})=\{f:\mathbb{H} \stackrel{holomorphic}{\longrightarrow} \mathbb{C}|f(xz)=J(x,z)^kf(z),{}^\forall x \in SO(2)\}. $$
I guess that $H_k(\mathbb{H})$ is zero. But I have no proof of that. Could you tell me the proof that $H_k(\mathbb{H})$ is zero (or not)? Is this well-known? If so, tell me related publications for such functions.
For all $c,d\in \Bbb{R}^2-(0,0)$ and $\Im(z)>0$ $$f(\frac{dz-c}{cz+d})= (\frac{cz+d}{\sqrt{c^2+d^2}})^kf(z)$$
By analytic continuation it stays true for all $\Im(z) > 0,c,d\in \Bbb{C}^2, cz+d\ne 0, c^2+d^2\ne 0,\Im(\frac{dz-c}{cz+d}) >0$.
The point is that if $z = ir, r > 0, f(z)\ne 0$ then the RHS has a singularity at $c=1,d=i$ while the LHS doesn't.
So it must be that $f$ vanishes on the imaginary axis, whence it vanishes everywhere.