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The paper by Liberzon and Brockett, "Spectral analysis of Fokker–Planck and related operators arising from linear stochastic differential equations." SIAM Journal on Control and Optimization 38.5 (2000): 1453-1467" studies the spectrum of the Fokker–Planck operator with linear drift, in $\mathbf{R}$. Proposition 2 states that the eigenvalues of the operator $L_{\sigma}:=\sigma\partial_{xx}+x\partial_x+1$ for $\sigma>0$ are all complex numbers with real parts less than $1/2$. The domain is chosen to be a dense subset of $L^2(\mathbf{R})$.

The proof that all negative integers are eigenvalues is straightforward by looking at the action of the operator on (modified) Hermite functions $u_k(x)$, which are the usual Hermite functions with some appropriate scalings. The relationship is:

$L_{\sigma}u_k=-ku_k+\sqrt{k(k-1}u_{k-2}$, which shows that the matrix is upper triangular in the ${u_k}$ basis, with $-k$ on the diagonals.

However the paper doesn't provide detail for the proving the:

Main claim : All complex numbers with real part less than 0.5 are eigenvalues.

Proof sketch from paper: The paper says that using the above relation of action on $u_k$, one can see (but I don't see how) that a scalar $\lambda$ will be an eigenvalue if one of the following series converges:

  1. $\sum_{n=1}^{\infty} \dfrac{\lambda^2(\lambda+2)^2\dotsm(\lambda+2n-2)^2}{(2n)!}$

or

  1. $\sum_{n=1}^{\infty} \dfrac{(\lambda+1)^2(\lambda+3)^2\dotsm(\lambda+2n-1)^2}{(2n+1)!}$

which is then shown (as proved by Giorgio Metafune in comments also) to give the condition $\operatorname{Real}(\lambda)<0.5$.

Can anyone prove the above claim or give any hints ?

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    $\begingroup$ Writing $\sum_n a_n$, I checked (for the first series) that $a_{n+1}/a_n=(1+(\lambda-3/2)/n+O(n^{-2}))$. Taking logarithms and summing you get $a_n \equiv Cn^{\lambda-3/2}$ which gives the result. $\endgroup$
    – Giorgio Metafune
    Commented May 31, 2023 at 10:12
  • $\begingroup$ @GiorgioMetafune Thanks, I see how we can derive the condition on $\lambda$ from the series formula. Do you any thoughts on the main claim ? $\endgroup$
    – mathamphetamine
    Commented May 31, 2023 at 14:19
  • $\begingroup$ It shoud follow by writing $f=\sum_k c_k u_k$ and computing $Lf$ by the recursion formulas. Then $Lf=\lambda f$ gives $c_{k+2}\sqrt{(k+2)(k+1)}=(\lambda+k)c_k$ and, after choosing $c_0$ or $c_1$ one imposes that the $(c_k)$ are square summable. $\endgroup$
    – Giorgio Metafune
    Commented May 31, 2023 at 17:16

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