 Solution

If $f\in V_n\setminus V_{n-1}$, write $f(x)=a_n{x}^n+\cdots$ with $a_n\ne 0$. In the expression $Tf=f^{\prime \prime }-2x{f}^{\prime }-2f$, the first term has polynomial degree at most $n-2$, while the other two contributeThus, if $Tf=\lambda f$, comparison of the leading coefficients givesFor $n=0$ take $V_{-1}=\{0\}$; the same result follows from $T1=-2$. This identifies the eigenvalue conditional on the existence of the eigenvector; existence is established next.