= Solution
Differentiate the expression from the source before applying any finite-dimensional result. The exponential factors cancel:
$$
\boxed{Tf=f''-2xf'-2f.}
$$
This is a <polynomial> whenever $f$ is a <polynomial>, and its <polynomial degree> is at most that of $f$. Differentiation and multiplication are linear, so $T$ is a <linear transformation> $V\to V$, and every $V_n$ is an <invariant subspace>.
For <self-adjointness>, put $w(x)=e^{-x^2}$. Then $w'=-2xw$ and $wTf=(wf')'-2wf$. <Integration by parts> therefore gives
$$
\begin{aligned}
\langle Tf,g\rangle
&=\int_{-\infty}^{\infty}\bigl((w\overline{f'})'-2w\overline f\bigr)g\,dx\\
&=-\int_{-\infty}^{\infty}w\overline{f'}g'\,dx
-2\int_{-\infty}^{\infty}w\overline f g\,dx\\
&=\langle f,Tg\rangle.
\end{aligned}
$$
The discarded endpoint terms are $w\overline{f'}g$ and $w\overline f g'$. They vanish at both infinities because exponential decay dominates every <polynomial>. The same decay justifies the intermediate integrals. Consequently
$$
\boxed{T(V_n)\subseteq V_n,\qquad
\langle Tf,g\rangle=\langle f,Tg\rangle\quad(f,g\in V).}
$$
Here the question's <self-adjointness> is in the sense of an <adjoint on an inner product space>: $T$ itself represents the adjoint on the polynomial space. Each finite-dimensional restriction $T|_{V_n}$ is a <self-adjoint operator>. No choice of operator domain in a completed <Hilbert space> is needed for this argument. The resulting <Gaussian polynomial operator> will produce <Hermite polynomials> in part (iv).
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