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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?

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    $\begingroup$ Maybe you could post the precise definition of "viscosity solution" which you are using. Looking at Wikipedia's definition there is no mention of requiring a positive-measure set of points for anything. $\endgroup$
    – Nate Eldredge
    Commented Apr 26, 2015 at 16:20

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Yes, I think you're missing something in the definition. Quoting from Wikipedia's definition,

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

In particular,

Any first order equation is degenerate elliptic.

So let's consider the equation $u(x)+1=0$. Next,

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 $.

Let $H(x,u,Du,D^2u)=u(x)+1$; then this would say $\phi(x_0)\le -1$.

A continuous function $u$ is a viscosity solution of the PDE if it is both a viscosity supersolution and a viscosity subsolution.

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.

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    $\begingroup$ Alternatively, consider the equation $u'(x)+1=0$, corresponding to $H(x,u,Du, D^2 u) = Du+1$. Let $x_0$ be a local maximum of Brownian motion (or even, say, the global maximum on $[0,1]$, which almost surely is contained in $(0,1)$). Then setting $\phi(x) = B(x_0)$, we have $\phi(x) \ge B(x)$ on a neighborhood of $x_0$, yet $H(x_0, \phi(x_0), \phi'(x_0), \phi''(x_0)) = \phi'(x_0)+1 = 1$ since $\phi$ is a constant function. So we do not have $H(x_0, \phi(x_0), \phi'(x_0), \phi''(x_0)) \le 0$ and thus $B$ is not a subsolution. $\endgroup$
    – Nate Eldredge
    Commented Apr 26, 2015 at 17:50
  • $\begingroup$ Thank you very much, Nate. This is exactly what I have been puzzling over. $\endgroup$
    – Romans Pancs
    Commented May 1, 2015 at 17:50
  • $\begingroup$ Just did it, Bjørn. It is the first question I ever asked here, so it took me a while to find out how to accept the answer (by clicking on the check, not the arrows). $\endgroup$
    – Romans Pancs
    Commented May 1, 2015 at 19:34
  • $\begingroup$ @RomansPancs you can do both, which gives me even more imaginary internet points :D $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented May 1, 2015 at 19:51

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