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I am trying to write a complete proof of the Feynman Kac formula in the multi-dimensional case. My starting point was the proof of the univariate form on wikipedia, at https://en.wikipedia.org/wiki/Feynman–Kac_formula. I've managed to write my multi-dimensional proof (see this link), but it has two gaps in it, which are also present in the wikipedia proof, and which I haven't managed to close. They are:

  1. The proof uses Ito's Lemma twice on the solution function $u$ of a function thereof. For Ito's Lemma to be applicable, the function must be twice-differentiable. But I cannot see a way to prove that $u$ is twice-differentiable.
  2. The proof actually shows that, IF there is a solution, it must be of the form shown in the Feynman Kac formula. It does not prove that that formula is necessarily a solution, ie that any solution exists.

Can anybody suggest how to plug these gaps, or post a link to a better proof?

Thank you very much.

PS I searched the site before writing this. There are two other questions on here asking about proofs of Feynman Kac. But one is asking for a multi-dimensional version, whereas I would be happy with a univariate version (I know how to convert it to muti-dimensional myself), and has not received an answer. The other is asking about a specific aspect of the proof rather than for a complete proof.

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  • $\begingroup$ No. I didn't know that was a different place. When I follow the link, they both have the StackExchange labels at the top of the page, so they seem to be part of the same site. What's the difference between Mathematics StackExchange and math_overflow? Sorry if that's a dumb question. I find this site very confusing. $\endgroup$ Commented Mar 6, 2016 at 5:03
  • $\begingroup$ Ah, OK. I found the answer about the diff between MathSE and mathoverflow at meta.math.stackexchange.com/questions/41/…. You're right, it sounds like that may be a good place to try. I'll do that, but keep an eye on this as well in case anybody here can answer it. $\endgroup$ Commented Mar 6, 2016 at 9:59
  • $\begingroup$ The suggestion to move this to MO has been noted. I am collecting more informed opinions (than what the team of moderators can muster up). Leaving this comment here so that others will see it. $\endgroup$ Commented Mar 8, 2016 at 14:16
  • $\begingroup$ The premise of the proposition is that $u$ satisfies the parabolic partial differential equation (PDE) and therefore is twice differentiable with respect to $x$ and the second derivative thereof is continuous. The existence and uniqueness of the solution of the PDE is the property of the PDE which can be found in any decent (parabolic) PDE textbook. So it is also implied. These allows the proof to go through without a problem. $\endgroup$
    – Hans
    Commented Jul 4, 2018 at 6:37
  • $\begingroup$ @Hans. Yes that's right. I realised that some time after I posted the above, so question 1 is dissolved. My question number 2 remains though. I'd appreciate any link to an online proof. $\endgroup$ Commented Jul 5, 2018 at 8:39

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