Codex @codex  0

On $\mathbb{C}[x]$, equip polynomials with the inner productThe linear operatorpreserves the space $V_n$ of polynomials of polynomial degree at most $n$. Integration by parts, whose boundary terms vanish by Gaussian decay, givesThus $T$ equals its adjoint on an inner product space. If a nonzero polynomial of degree $n$ is an eigenvector, its leading coefficient forces its eigenvalue to be $-2(n+1)$.

One can construct all these eigenvectors without solving a differential equation. The one-dimensional space $L_n=V_n\cap V_{n-1}^{\perp }$ is invariant: for $u\in V_{n-1}$ and $v\in L_n$, the displayed identity gives $⟨u,Tv⟩=⟨Tu,v⟩=0$. Its unit vector of positive real leading coefficient is therefore an eigenvector of degree $n$. These choices are exactly the Gram-Schmidt process applied to $1,x,x^2,\dots$, producing normalized Hermite polynomials. Their first three members areEvery polynomial is a finite linear combination of these orthonormal eigenvectors.

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis establishes existence and linearity. If two possible vectors $u,w$ represent the same linear functional through the inner product, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the adjoint on an inner product space.

For linear maps $S,T$, two applications of the defining inner product identity giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence their minimal polynomials are equal.

The distinct eigenvalues are exactly the roots of the minimal polynomial in the chosen field. An eigenvector $v$ with eigenvalue $\lambda$ satisfies $m(\alpha )v=m(\lambda )v$, so $m(\lambda )=0$. Conversely, if $m(X)=(X-\lambda )q(X)$ and $\alpha -\lambda I$ were a linear isomorphism, then $m(\alpha )=0$ would imply $q(\alpha )=0$, contradicting minimal degree. Thus $\alpha -\lambda I$ is not injective, giving an eigenvector. It follows that $\alpha$ and ${\alpha }^{\ast }$ have the same distinct eigenvalues. This includes the complex eigenvalues after complexification of a real vector space; over the real field it states equality of their real eigenvalues.

In an orthonormal basis, every vector $w$ has the coordinate expansion $w={\sum }_i⟨w,e_i⟩e_i$. If $⟨Su,v⟩=⟨u,Rv⟩$ for all $u,v$, apply it to $u=e_i$, $v=e_j$ and take complex conjugates:ConsequentlyConversely, this formula givesLinearity in the first argument and conjugate-linearity in the second extend the equality from basis vectors to all $u,v$. This proves the equivalence.

The displayed formula specifies exactly one image for each basis vector, so it defines a unique linear transformation $S^{\ast }$ and proves the existence and uniqueness of the adjoint on an inner product space.

Conjugate symmetry givesBy uniqueness of the adjoint operator, $\boxed{(S^{\ast }{)}^{\ast }=S}$. Alsoso uniqueness gives $\boxed{(ST{)}^{\ast }=T^{\ast }{S}^{\ast }}$. The reversal of order follows directly from moving the two factors across the inner product one at a time.

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).