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Thank you. Is it possible to extend Riemann's method to more general domains ? For instance, on a concave domain ? In that case some segments of $x=x_0$ or $y=y_0$ will not be connected.
I mostly agree with the answer and the previous comment but, it should be emphasized that a theory has to be renormalized even if it doesn't have divergencies.
Or, by the way, if you know the more-or-less standard proof of the path integral solution of the Schrödinger equation (any modern text on Quantum Mechanics and/or Quantum Field Theory), you can transform the Schrödinger operator into the Fokker-Planck operator and relate the two solutions. It should be possible to put it in the same language that you are writing it.
The proof is essentially the same as for the path integral solution of a transition amplitude in QM. You consider a small time and find the solution for such an interval; then you prove that the solutions "convolute", that is, if you have a solution for an interval (t,t′) and (t′,t′′), then you can construct the solution for (t,t′′). In this context that convolution is the Chapman–Kolmogorov equation, but for Markovian processes it is simply an identity on the probability distributions. Section 4.4.2 of Risken's The Fokker-Planck equation sketches the proof (In a different notation).
@DmitriPanov What the previous inequality shows is that, for any constant singular values $\sigma_1$, $\sigma_2$, there are domains $E$ that cannot be obtained from an arbitrary $D$ (in particular, the unit disk). And in principle yes, I am restricted to $\mathbb{C}^\infty$ curves. Thank you !
@DmitriPanov Yes, the result I'm citing is from this* work: "For each smoothly bounded Jordan domain $D$, there exists a constant $C = C_D$ such that $\lambda_1(\partial E) \leq C \lambda_2 (E)/\iota(E)$ for every Jordan domain $E$ which the image of $D$ under a cps-homeomorphism" (In our language a cps-homeom. is a mapping with constant singular values. And $\iota(E)$ is the inradios, i.e., the radius of the greatest inscribed circle. $\lambda_1$ the length and $\lambda_2$ the area). Not every $E$ satisfies such inequality given $D$. *doi.org/10.1155/S0161171203204166 (theorem 3.1)
@DmitriPanov Regarding the proof the lemma, why is that the disk bounded by $f(\eta)$ contains in its interior the whole image $f(D')$ ? Also, could you please explain the inequality $l(\gamma_{yx})\le \frac{1}{\sigma_1}l([yx])$ ? Thank you !