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Let $A$ be a linear second order differential operator with constant coefficients defined on real-valued functions of one-variable. Suppose that we have that for an upper-semicontinuous function $u:\mathbb{R}\to\mathbb{R}$ $$Au \leq 0\mbox{ holds in the viscosity sense.}$$ Let $u^{\epsilon}(x) = \int_{-1}^1u(x-\epsilon y)\varphi(y)\mbox{dy}$, where $\varphi$ is the standard mollifier. I would like to say that $$Au^{\epsilon}\leq 0\mbox{ in the classical sense}.$$

I am convinced that this must be true in this special and simple case. I have tried a few approaches, but as is often the case with viscosity theory, the devil is in the details and I cannot seem to write something convincing. Is anyone familiar with a general theorem along the lines of ``mollification preserves sub-solutions of linear const. coeff. pde''?

Attempt at a solution: We suppose the contrary, that at some point $x_0,$ we have $Au^{\epsilon}(x_0) > 0.$ By continuity, we can extend this further to strict positivity on some open neighborhood $N$ of $x_0.$ By upper semi-continuity, I know that $u-u^{\epsilon}$ will achieve a maximum on closure $\overline{N}$. If it is a local maximum, that is great, I'm done. However, if it is not, I am looking to construct another test function, say $\psi$ such that $A\psi \geq 0$ and $u - u^{\epsilon} - \psi$ achieves a local maximum inside $N$. Clearly, $\psi$ needs to depend somehow on the convergence of $u^{\epsilon}$ to $u$. But how to construct?

Edit: problem statement has been amended to say constant coefficients. Thanks Willie Wong for pointing that out.

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Let $\chi$ be an odd, smooth function such that $\chi(x) > 0$ if $x < 0$. Let $D$ be the standard derivative operator. Let $A = \chi D^2$.

Let $u(x) = -(x+1)^2 + 1$ if $x < 0$ and $0$ otherwise. Away from $x = 0$, $u(x)$ is a bona fide subsolution, so $u(x)$ is also viscosity subsolution there. At $0$, we easily see that while $u$ is continuous, it is not differentiable. And in fact we see that by convexity that any $C^2$ function $v$ on a neighborhood of zero such that $v(0) = 0$ cannot satisfy $v(x) \geq u(x)$. Hence the viscosity condition is satisfied vacuously.

But for $0 < x < \epsilon$

$$ Au^{\epsilon}(x) = -\chi(x) \int_{x/\epsilon}^1 D^2(x + 1 - \epsilon y)^2 \phi(y) dy = -\chi(x) \int_{x/\epsilon}^1 2 \phi(y) dy > 0 $$

and so it is not a classical subsolution. So the statement as given is not true.

In general, only constant coefficient differential operators can commute with mollification. As long as you allow variable coefficients, you can potentially get into trouble: variable coefficients break translation invariance, and so the mollifying process which involves adding translated versions of the function is no longer (necessarily) taking an infinite linear combination of subsolutions, and hence the mollified function does not have to be a subsolution in general.

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I will edit the question. I only have constant coefficients, in fact. Thanks for the interesting counterexample, I'll be sure to keep that in mind in the future. – fouryear Jan 16 '13 at 14:08
I am pretty sure that if the coefficients are sufficiently smooth and satisfy some sort of ellipticity condition we can essentially freeze coefficients and obtain the same result as the constant coefficient case. But it appears that the constant coefficient case is a bit more difficult then I originally imagined. I will update if I find a trick. – Willie Wong Jan 17 '13 at 9:10
I would also be interested in the solution to something slightly simpler and I could try to take it from there. One difficulty seems to be the fact that mollification does not approximate well semi-continuous functions. After a bit of searching, I've come across the notion of "strong" semi-continuity, which seems to be enough for the difficulty to go away. But I have no such strong semi-continuity in my case. – fouryear Jan 17 '13 at 10:03
I was also trying to apply your idea of translations of sub-solutions, thinking that I'd perhaps be able to approximate the $u^{\epsilon}$ by averaging with a bunch of dirac measures and then take a limit to get the Lebesgue integral and go ahead with the contradiction, but it seems that the discontinuity set of $u$ could be too big for that idea to work. – fouryear Jan 17 '13 at 10:05

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