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!



X-harmonic and mean value property
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. 


