Cumulant generating function and largest eigenvalue of operator I am working on a recent paper (arXiv:1805.02887) about an application of large deviation theory to the statistical mechanics of active matter and am a bit bewildered by a result dropped in appendix F (p.9, right column), without any reference or indication.
With $q$ such that
$$
\frac{\text{d}q}{\text{d}t} = D_r \left[\frac{1}{2q} - q\right] + \sqrt{D_r} \xi,
$$
where $\xi$ is a Gaussian white noise with zero mean and unit variance, and $D_r$ is a constant, they claim that the cumulant generating function of the time-averaged value of $q$,
$$
f(k) = \lim_{t \rightarrow \infty} \frac{1}{t} \, \log \left\langle \exp\left(- k \int_0^t \text{d}\tau \, q(\tau)\right)\right\rangle,
$$
is the largest eigenvalue of the following operator:
$$
L_k[\cdot] = -\frac{\partial}{\partial q} \left[D_r \left(\frac{1}{2q} - q\right) \cdot \right] + \frac{D_r}{2} \frac{\partial^2}{\partial q^2}[\cdot] - kq.
$$
How can this be? Where does this result come from?
Thank you for your help!
 A: Reformulating a large deviation problem of an integrated dynamical quantity as an eigenproblem is a common and useful method. 
Consider
$$
\epsilon(t) = \int_0^t \text{d}\tau \, q(\tau),
$$
such that
$$
\dot{\epsilon} = q,
$$
we then have that the joint probability distribution of $q$ and $\epsilon$ satisfies the Fokker-Planck equation
$$
\frac{\partial}{\partial t} P(q, \epsilon) = \mathcal{L} P(q, \epsilon) - \frac{\partial}{\partial \epsilon} (q P(q, \epsilon)),
$$
where
$$
\mathcal{L} : \bullet \mapsto - \frac{\partial}{\partial q} \left[D_r \left(\frac{1}{2q} - q\right) \bullet\right] + \frac{D_r}{2} \frac{\partial^2}{\partial q^2} \bullet,
$$
is the Fokker-Planck operator inferred from the Langevin equation satisfied by $q$ (see Wikipedia).
We introduce the biased measure
$$
P_k(q) = \int \text{d}\epsilon \, \exp(- k \epsilon) P(q, \epsilon) = \left<\exp\left(-k \int_0^t \text{d}\tau \, q(\tau)\right)\right>,
$$
then we have from the previous Fokker-Planck equation
$$
\begin{aligned}
\frac{\partial}{\partial t} P_k(q) &= \frac{\partial}{\partial t} \int \text{d}\epsilon \, \exp(- k \epsilon) P(q, \epsilon)\\
&= \mathcal{L} P_k(q) - q \int \text{d}\epsilon \, \exp(-k\epsilon) \frac{\partial}{\partial \epsilon} P(q, \epsilon)\\
&= (\mathcal{L} - k q) P_k(q),
\end{aligned}
$$
where we have performed an integration by parts to get to the last line.
We call the operator
$$
\mathscr{W}_{k, q} = \mathcal{L} - k q : \bullet \mapsto - \frac{\partial}{\partial q} \left[D_r \left(\frac{1}{2q} - q\right) \bullet\right] + \frac{D_r}{2} \frac{\partial^2}{\partial q^2} \bullet - k q \bullet
$$
the tilted generator. What can then be shown is that the SCGF
$$
f(k) = \lim_{t \rightarrow \infty} \frac{1}{t} \log\left<\exp\left(-k \int_0^t \text{d}\tau \, q(\tau)\right)\right>
$$
is the largest eigenvalue of the following eigenproblem (see Touchette, 2017, arXiv:1705.06492 and Jack, 2019, arXiv:1910.09883 (§ II.E))
$$
\mathscr{W}_{k,q} P_k = \Lambda P_k,
$$
which is the assertion of the question.
