Is an arbitrary Brownian path a viscosity solution of every differential equation? 


My intuition is that a path of Brownian motion is so ill-behaved that it not only does not have derivatives anywhere but it also does not have (local) subdifferentials and superdifferentials. Equivalently, there exists no positive measure of points at which one can append either a smooth underestimator or a smooth overestimator the Brownian path. (An overrestimator would be a smooth function that lies above the Brownian path but touches it at one point. It is a test function in the definition of viscosity solution.)


The conclusion that an arbitrary Brownian path solves every differential equation seems nonsensical to me. Am I missing something in the definition of a viscosity solution?



Is an arbitrary Brownian-motion path a viscosity solution of every differential equation?
Yes, I think you're missing something in the definition.
Quoting from Wikipedia's definition,



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> An equation $ H(x,u,Du,D^2 u) = 0 $ in a domain $ \Omega $ is defined to be ''degenerate elliptic'' if [...]
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In particular,



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> Any first order equation is degenerate elliptic.
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So let's consider the equation $u(x)+1=0$. Next,



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> An upper semicontinuous function $u$ in $\Omega$ is defined to be a *subsolution* of a degenerate elliptic equation in the ''viscosity sense'' if for any point $x\_0 \in \Omega$ and any $C^2$ function $\phi$ such that $\phi(x\_0) = u(x\_0)$ and $\phi \geq u$ in a neighborhood of $x\_0$, we have $ H(x\_0,\phi(x\_0),D\phi(x\_0),D^2 \phi(x\_0)) \leq 0 $.
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Let $H(x,u,Du,D^2u)=u(x)+1$; then this would say $\phi(x\_0)\le -1$.



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> A continuous function $u$ is a viscosity solution of the PDE if it is both a viscosity supersolution and a viscosity subsolution.
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Note that Brownian motion has a countable dense set of local maxima $x\_n$, $n\in\mathbb N$. At such points we can take $\phi$ to be a constant function. If Brownian motion $u(x)$ is a viscosity solution to $u(x)+1=0$ then in particular it is a subsolution, which implies that $u(x)\le -1$ for all $x$. But Brownian motion is not, in any reasonable sense, bounded above by -1.


