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Jan
17
comment How to show that $u(t,x)\in C([0,T],H^{1}(\mathbb{R}^{n}))$?
Do you have a PDE that $u$ satisfies? Evans' PDE book has results of this kind.
Jan
17
comment Smoothening viscosity solution to classical solution
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.
Jan
17
comment Smoothening viscosity solution to classical solution
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.
Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
comment Smoothening viscosity solution to classical solution
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.
Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
comment Can I use my powers for good?
I agree with the overall insights mentioned here. Connections are supremely important!
Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
revised Smoothening viscosity solution to classical solution
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Jan
16
asked Smoothening viscosity solution to classical solution
Jan
16
comment Question on proof of deformation lemma
It stops as soon as it hits the boundary of $A$ at energy $c-\hat{\epsilon}$. What is meant by "may or may not enter $A$" is that we do not know that $I$ even decreases all the way to $c - \hat{\epsilon}$. In the former case, we don't care. In the latter case, $I$ will not decrease beyond $c - \hat\epsilon$ since as soon as that happens $\eta\in\partial A\subset A$ and the evolution stops. Even though the boundary of $A$ is technically in $A$, the next inequality works. The trajectory $\eta(t,u)$ does not enter $A$ in the sense that its energy, is always greater than $c-\hat\epsilon$
Jan
15
comment Question on proof of deformation lemma
Yes, $ t\mapsto \eta(t,u)$ is constant as soon as it enters $A$, which it may or may not do. The reason it is constant is because of the fact that $\frac{d\eta}{dt}$ is 0 on $A$, which follows just from the definition (look closely at how $W$ and $g$ are defined). This is not a contradiction, it is direct.
Jan
14
awarded  Scholar
Jan
14
accepted How to solve this (difficult) simple equation in one variable?
Jan
14
awarded  Editor
Jan
14
revised Question on proof of deformation lemma
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Jan
14
answered Question on proof of deformation lemma