 Solution

Construct compatible choices in the nested spaces, rather than choosing unrelated orthonormal bases for different $V_n$. Set $V_{-1}=\{0\}$ and defineThe orthogonal complement is taken for the inner product from part (i). Since $\dim V_n=n+1$ and $\dim V_{n-1}=n$, the space $L_n$ has dimension one and $V_n=V_{n-1}\oplus L_n$.

It is an invariant subspace. Indeed, if $v\in L_n$ and $u\in V_{n-1}$, part (ii) givesbecause $Tu\in V_{n-1}$. Also $Tv\in V_n$. Hence $Tv\in L_n$. Any nonzero vector in this one-dimensional space is consequently an eigenvector. Choose a unit vector $g_n\in L_n$. It has polynomial degree exactly $n$: membership in $V_{n-1}$ would make it orthogonal to itself and thus zero. Part (iii) now supplies its eigenvalue.

For $m<n$, $g_m\in V_{n-1}$ and $g_n\perp V_{n-1}$, so the choices are pairwise orthogonal. We have provedAn explicit compatible construction is the Gram-Schmidt process on $1,x,x^2,\dots$:The leading term of $r_n$ is $x^n$, so it is never zero. This fixes the positive real leading coefficient and yields the normalized Hermite polynomials. For example, the Gaussian integral givesEvery polynomial is a finite linear combination of these eigenvectors, since $g_0,\dots ,g_n$ form an orthonormal basis of $V_n$ for each $n$.