 Solution

Differentiate the expression from the source before applying any finite-dimensional result. The exponential factors cancel: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 $w^{\prime }=-2xw$ and $wTf=(w{f}^{\prime }{)}^{\prime }-2wf$. Integration by parts therefore givesThe discarded endpoint terms are $w\overline{f^{\prime }}g$ and $w\overline{f}{g}^{\prime }$. They vanish at both infinities because exponential decay dominates every polynomial. The same decay justifies the intermediate integrals. ConsequentlyHere 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).