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Consider in 1D the operator given by $$ \mathcal{L} = \frac{d^2}{dx^2} - V'(x)\frac{d}{dx}, $$ where $V(x)$ is a convex, sufficiently quickly growing potential, so that $\mathcal{L}$ has a complete set of eigenfunctions $\psi_n(x)$ (orthogonal with respect to a stationary distribution $\rho(x)dx$).

Let $\psi_1$ be the first nonconstant eigenfunction. Is it true that $\psi_1$ is monotone? This is the case in the simple case of e.g. $V(x)=x^2$, where the eigenfunctions are the Hermite polynomials and $\psi_1(x) = x$.

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  • $\begingroup$ Why do you think it should be true? $\endgroup$
    – Giorgio Metafune
    Commented Feb 17, 2022 at 15:45
  • $\begingroup$ Not much more than wishful thinking, I'd be happy to see a counterexample. $\endgroup$
    – Michal Kotowski
    Commented Feb 17, 2022 at 19:11
  • $\begingroup$ It seems that it can be true. Call $A$ your operator and $B=DA$, where $D$ is the derivative. If $Au=\lambda u$ with $\lambda \neq 0$, then $u$ is not a constant and $Bu'=\lambda u'$ with $u' \neq 0$. Conversely, from this last we get $\lambda u-Au= c$ and then $(\lambda-A)(u-c/\lambda)=0$. Also $u-c/\lambda$ cannot be zero otherwise $u'=0$. Of course we should check that $u,u'$ belong to the right spaces simultaneusly, etc...This means that $\sigma (A)=\sigma (B) \cup \{0\}$ and then the first nonzero eigenvalue of $A$ is the first of $B$ which has a positive eigenfunction, namely $u'$. $\endgroup$
    – Giorgio Metafune
    Commented Feb 17, 2022 at 23:24
  • $\begingroup$ I'm sorry, I have trouble following your derivation - why are you claiming that $Bu' = \lambda u'$? $\endgroup$
    – Michal Kotowski
    Commented Feb 18, 2022 at 10:02
  • $\begingroup$ The notation is unclear, sorry. I have this in mind: if for example $Au=u''-x^3 u'=\lambda u$, then $u'''-x^3 u''-3x^2 u'=\lambda u'$ so that $B=D^2-x^3D-3x^2$. $\endgroup$
    – Giorgio Metafune
    Commented Feb 18, 2022 at 10:32

1 Answer 1

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This is rather a way to get a (positive) answer than a complete one, some missing details will be clear in a moment.

Let me use $A$ for $\mathcal L$ and $B=A-V''$. The symmetrizing measure (for both) is $e^{-V}\, dx$. Both operators are negative and self-adjoint in $L^2(e^{-V}\, dx)$ and the first eigenvalue of $A$ is zero with eigenfunction $1$.

Assume that $\lambda<0$ is an eigenvalue for $A$ with eigenfunction $u$. Then $u$ is not constant and $u'$ is an eigenfunction of $B$ corresponding to the same $\lambda$, by differentiating. Of course, domain problems arise in this point and require some assumptions on $V$.

Conversely, if $0 \neq u'$ is an eigenfunction of $B$, then $\lambda u-Au=c$ and $(\lambda-A)(u-\frac{c}{\lambda})=0$ and $u-\frac{c}{\lambda} \neq 0$ since $u' \neq 0$. Domain and integrability problems here arise again.

Summing up, and assuming compactness of both resolvents, $\sigma (B)= \sigma (A) \setminus \{0\}$ ($0$ is not an eigenvalue of $B$, since $V'' >0$). But then the second eigenvalue of $A$ is the first of $B$ which, by general arguments, is simple and has a positive eigenfunction.

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