Routine question about derivatives of automorphic forms being L^2

Question

I consider Automorphic forms on $G = SL_2 (\mathbb{R})$, which are $\Gamma$-invariant, $K$-finite, $Z(g)$ finite, and of moderate growth. If I have such an automorphic form, which happens to be in $L^2(\Gamma \backslash G)$, is it true that if we will act on it by some element of the Lie algebra, the result will also be in $L^2$?

Thank you, Sasha

Answer 1

If you allow yourself to use the fact that $L^2$, $\mathfrak z$-finite $K$-finite automorphic forms are finite linear combinations of $L^2$ eigenfunctions, here cuspforms and constants, then on-the-right derivatives map such finite linear combinations to finite sums of the same type, which are again in $L^2$.

Thus, one would want to prove that no $L^2$ continuous spectrum can be $\mathfrak z$-finite and $K$-finite, which is a corollary of Plancherel.

Answer 2

There is a claim (Borel, Automorphic forms on $SL_2(\mathbb{R})$, 2.14) that if $f$ is $K$-finite on the right and $Z(g)$-finite, then it can be represented as convolution of itself with a smooth function $\phi$ with compact support. Then it is quite clear that if it is in $L^2$, if we take derivative, and "throw" it into the $\phi$ place, we will get a convolution of $L^2$ with a smooth function with compact support, hence again something in $L^2$.