= Solution
For square-integrable wavefunctions with the stated decay, multiplication by the real coordinate $x$ is <Hermitian operator>[Hermitian] because
$$
\langle\psi,x\chi\rangle=\langle x\psi,\chi\rangle.
$$
For $p_x=-i\hbar\partial_x$, <integration by parts> gives
$$
\langle\psi,p_x\chi\rangle
=\langle p_x\psi,\chi\rangle
-i\hbar\int_{-\infty}^{\infty}[\overline\psi\chi]_{x=-\infty}^{x=\infty}\,dy,
$$
and the boundary term vanishes. The same arguments apply to $y$ and $p_y$.
If $F$ and $G$ are Hermitian on a common suitable domain, then $(FG)^\dagger=GF$, so
$$
\left[\frac12(FG+GF)\right]^\dagger
=\frac12(FG+GF).
$$
Because $x$ commutes with $p_y$ and $y$ with $p_x$,
$$
L=xp_y-yp_x
$$
is Hermitian. Using $p_xx=xp_x-i\hbar$ and its $y$ analogue,
$$
D=\frac12(xp_x+p_xx+yp_y+p_yy)
=-i\hbar(x\partial_x+y\partial_y+1),
$$
so $D$ is Hermitian as well.
Finally set $R=x\partial_y-y\partial_x$ and $S=x\partial_x+y\partial_y$. Direct evaluation on a smooth test function gives $[R,S]=0$: rotations commute with radial dilations. The scalar term in $D$ also commutes with everything, and therefore
$$
\boxed{[L,D]=0}.
$$
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