The Hermite differential equation is given as such $$ y'' - 2xy'+2\lambda y=0 $$

writing this in strum-liouville form you get $$-(\exp(-x^2)y')'= 2\exp(-x^2)\lambda y $$

However, in order for it to have real eigenvalues, according to strum-liouville theory, the differential operator, $$ Ly = \lambda y$$ must be self adjoint which implies that the solution must have compact support ( i.e. $ y\rightarrow 0 $ as $x \rightarrow \pm\infty $). However, we know that the solutions to the Hermite equation are the Hermite polynomials, which obviously are not compactly supported.

So I'm confused. I was wondering if anyone could tell where I'm going wrong in my logic. Also I've seen solutions to the Hermite differential equation such as $ \exp(-x^2/2)H_n $ and $H_n$, where $H_n$ is the Hermite polynomial of order $n$. Basically, what's up with that?


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