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On a complex inner product space $V$, with the inner product conjugate-linear in its first argument, an everywhere-defined linear map $T:V\to V$ admits an adjoint on $V$ if there is a linear map $T^{\ast }:V\to V$ satisfyingThe adjoint, when it exists, is unique: the difference of two candidate values at $y$ is orthogonal to every vector and hence to itself. In a finite-dimensional vector space, it always exists, and its matrix in an orthonormal basis is the conjugate transpose. In a space that is not complete, existence must instead be checked.

The identity $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y\in V$ supplies such an adjoint directly, namely $T^{\ast }=T$. This algebraic statement should be distinguished from self-adjointness of a linear operator on a completed Hilbert space, which additionally compares the domains of an operator and its Hilbert-space adjoint. The Gaussian polynomial operator is a useful example where the identity holds on the entire polynomial inner product space and can be used on every finite-dimensional invariant polynomial subspace.

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).