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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

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?


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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