Let $Af=f''(x)-xf'(x)$ and $\mu$ be the Gaussian measure, i.e, $\,d\mu(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. We consider the following eigenvalue problem $$Af=\lambda f,$$ where $f\in C^\infty(\mathbb{R})\cap L^p(\mu)$ for all $p>1$. Show that any solution to the above problem must be a polynomial.
Note: I know of one way (which uses Orstein-Uhlenbeck process) where this result comes as a by-product. I would like to see if it can be done directly.
The function $f$ will be a polynomial when $\lambda\in\mathbb N$. That can be seen from the fact that if we substitute a series $y=\sum_0^\infty a_nx^n$, we get the recursions $$\tag1 a_{2n+2}=\frac{(2n-\lambda)(2n-2-\lambda)\cdots(2-\lambda)\lambda}{(2n+2)!}\,a_0, $$ $$\tag2 a_{2n+1}=\frac{(2n-1-\lambda)(2n-3-\lambda)\cdots(1-\lambda)}{(2n+1)!}\,a_1, $$ that can only stop when $a_0=0$ and $\lambda\in\mathbb N$ is odd, or when $a_1=0$ and $\lambda\in\mathbb N$ is even.
If, on the other hand, $\lambda\not\in \mathbb N$, either $\lambda<0$ or $\lambda\in(2k,2k+2)$ for some $k\in \mathbb N\cap \{0\}$. Consider separately the analytic solutions $\sum_na_{2n}x^{2n}$ and $\sum_na_{2n-1}x^{2n-1}$. Then, with $r_k=(2k+2-\lambda)\cdots(2-\lambda)$, and $x>0$, $$ \left|\sum_{n\geq k}a_{2n}x^{2n}\right| =|r_k|\,\sum_{n\geq k}\frac{(2n-2-\lambda)\cdots(2k+4-\lambda)\,x^{2n}}{(2n)!} \geq|r_k|\,\sum_{n\geq k}\frac{x^{2n}}{(2n)!}. $$ So the solution behaves like an exponential, and for $p$ big enough the power will overcome the $e^{-x^2}$ from the measure, and the solution cannot be in $L^2(\mu)$. A similar argument can be applied to the other solution.
So the only eigenvalues are the positive integers, and the eigenvectors are polynomials satisfying the recursions $(1)$ or $(2)$.