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 $\psi\_1(x) = x$.



Monotonicity of the top eigenfunction of the generator of a diffusion
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.


