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We know in elliptic equation theory(or related area) that harmonic function has mean value property. Roughly speaking, harmonic function function at point x is equal to its average on the spherical surface(or ball) centered at x. Furthermore, a locally integrable function which satisfies mean value property is smooth and actually harmonic. More can be seen at http://en.wikipedia.org/wiki/Harmonic_function

Now, for given Ito diffusion $X_t$, a function $f$ is called $X$-harmonic in open connected set $D\subset R^n$ if: i) $f$ is locally bounded and measurable on $D$; ii). $f(x)=E^x[f(X_{\tau_{U}})]$ for all $x\in D$ and all bounded open sets $U$ with $\bar{U}\subset D$. Here $\tau_{U}$ is the exit time of $x$ with respect to $U$.

This definition starts with an 'average property' which is similar to(or stronger than) the 'mean value property'.

My question is: if $f$ is $X$-harmonic, is it true that $f$ must be continuous?

My guess is that it is not necessary $C^2$ because Ito diffusion might well contain degenerated(semi-elliptic) case. But very much likely it is continuous.

Thanks for any suggestions!

Appendix: 1. Please also comment on whether my 'formulation' is correct.

  1. I wanted to argue like this: $f$ is continuous w.r.t $x$(with fixed $t$ certainly) because $X_t^x$ is continuous w.r.t its initial data $x$ because of the property of ito diffusion. And $E^x[f(X_t)]=E[f(X_t^x)]$ is continuous because of the modulus continuity of $L^1$ integral.

Does this seem to be on the right track?

Appendix2:

Sorry, seemingly it's clear in Oksendal's book. The answer is not necessarily continuous. But if the generator of Ito diffusion is uniformly elliptic, it should be so. Thanks again!

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up vote 2 down vote accepted

In Oksendal's book, the transition probablity or generating semigroup is not fully discussed. Think of X-harmonicity and the mean value property (MVP) together in lemma 9.2.4 in Oksendal: the MVP is w.r.t to the measure on the boundary,$Q^{x}(dy)$ when $dy\in\partial D$. Therefore, to show that $f$ is continous in $x$ is equivalent to showing that the boundary measure $Q^{x}(dy)$ is continuous (in some sense, say, total variation) in $x$ .

You can not assume that $f$ is continous since it usually only has to be locally bounded and meassurable. To show $f$ is (locally) $C^2$ is equivalent to showing that the family $\{Q^{x}(dy):x\in D\}$ satisfy corresponding property.

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