I am just starting to look into the theory of viscosity solutions. I can't see how the following simple argument fails. On the other hand the results seems too nice to be true.

Let us consider an elliptic PDE. Here elliptic means that $F$ is decreasing in the $D^2u$ argument: $$F(x, u(x), Du(x), D^2u(x)) = 0, x \in \mathcal{O},$$ with boundary condition $$u(x) = g(x), x \in \partial \mathcal{O}$$

Suppose that we have a viscosity solution $u$, which a priori is not supposed to be continuous (so the upper semicontinuous hull, $u^*,$ is a subsolution and the lower one, $u_*,$ a supersolution).

Suppose also that we have a nice function $F$, such that a comparison result holds true: let $v$ be an upper semicontinuous subsolution for the PDE and $w$ a lower semicontinuous supersolution. If $v \le w$ on $\partial \mathcal{O},$ then the same holds on the entire domain $\mathcal{O}.$

So if we take our solution $u,$ we get that $u^* \le u_*$ on $\partial \mathcal{O},$ iff $g$ is continuous. And in this case the consequence would be that $u^* \le u_*$ on the whole space. Thus $u$ would be continuous.

This seems a surprising result, and it's not explicitly stated in the notes I am reading. Somehow I have the feeling that all inequalities are in the wrong direction. Where am I wrong?


What notes are you reading? Usually viscosity solutions are continuous by definition (i.e., subsolutions are defined to be upper-semicontinuous, and supersolutions are defined to be lower-semicontinuous). See the User's Guide, for instance:


The definitions you quote are useful when the solution is expected to be discontinuous. As you noted, even with these definitions, if comparison holds, and the boundary data is continuous, then $u$ must be continuous.

  • $\begingroup$ Thank you for the clarification. I was reading Nizar Touzi's notes/book on stochastic optimal control. I believe that there everything is stated correctly, but to the newcomer such as myself it is important to know that the simple properties that follow are true: just to get a feeling of what we are doing. $\endgroup$ – Kore-N Dec 8 '16 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.