Timeline for Is an arbitrary Brownian-motion path a viscosity solution of every differential equation?

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when toggle format	what		by	license	comment
May 1, 2015 at 19:51	comment	added	Bjørn Kjos-Hanssen		@RomansPancs you can do both, which gives me even more imaginary internet points :D
May 1, 2015 at 19:34	comment	added	Romans Pancs		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).
May 1, 2015 at 19:32	vote	accept	Romans Pancs
May 1, 2015 at 17:50	comment	added	Romans Pancs		Thank you very much, Nate. This is exactly what I have been puzzling over.
Apr 26, 2015 at 17:50	comment	added	Nate Eldredge		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.
Apr 26, 2015 at 17:34	history	answered	Bjørn Kjos-Hanssen	CC BY-SA 3.0