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In the proof of the Sobolev extension theorem for $W^{1,p}$ spaces where $p\in [1,\infty )$, in Evans' PDE book, he reduces proving the theorem down to proving it for smooth functions on a open set with flat boundary. Then he uses higher-order reflection to solve this case. I understand this proof quite well, but I have one question: is it necessary to approximate the given function with smooth functions? Can't you just reflect the original function $f\in W^{1,p}$?

If this is not possible, it seems that I have an incorrect intuition of weak derivatives.

Evans left $p=\infty $ as an exercise, and I know how to do this in another way, but I thought that this, reflecting the original function, if possible, should be the intended way.

Pedro
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Gheehyun Nahm
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    Can't you just reflect the original function $f \in W^{1,p}$? A posteriori, the answer is "yes, you can". If you take some $f \in W^{1,p}$, approximate it with smooth $f_k$, extend them by reflection to $\bar{f}_k$, then take the limit $\bar{f}$, it turns out that $\bar{f}$ is the original function $f$ extended by reflection. – Michał Miśkiewicz Dec 27 '17 at 20:09
  • @MichałMiśkiewicz I was asking about "just" reflecting, not using higher-order reflections. – Gheehyun Nahm Dec 28 '17 at 01:39
  • Ah, this wasn't clear to me; I thought the problem is with approximation, not with higher order reflections. Anyway, my comment above applies to this case just as well. – Michał Miśkiewicz Dec 28 '17 at 09:29

1 Answers1

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You can extend the original function by reflection. Problem is, how will you prove that the extended function is in $W^{1,p}$? Computing the derivatives pointwise a.e. is not enough, one must verify the definition of weak derivatives. It can be done but is a chore.

The advantage of temporarily switching to smooth functions is that smoothness is verified pointwise. To show something is $C^1$ we only need to check the derivatives at each point individually. Simply put, classical derivatives are easier to handle than weak derivatives.

But the case $p=\infty$ is different. Here smooth approximation is not available, and we have to extend the original function. Fortunately, in this case there is a pointwise characterization that helps: $W^{1,\infty}$ is precisely the space of Lipschitz functions (see relation between $W^{1,\infty}$ and $C^{0,1}$). The Lipschitz condition of the extended function is easy to verify by checking two cases (two points in one closed halfspace, two points in different halfspaces). But actually, in this there is no need for higher order reflection: one can simply let $$\tilde u(x_1,\dots, x_{n-1} ,x_n) = u(x_1,\dots, x_{n-1} , |x_n|)$$ which is Lipschitz, being the composition of Lipschitz functions. The "straightening the boundary" step works by observing that both $\Phi$ and $\Psi$ are Lipschitz continuous, as they are smooth (see $C^1$ function on compact set is Lipschitz for example).

  • I was wondering whether you can extend the original function, for any $p\in [1,\infty ]$, by reflection (the simple version, not the higher order one). (Is this what you meant in your first paragraph, not the higher order one?) Furthermore, I thought that verifying the definition of weak derivatives is not that complicated, especially in this case. I agree that switching to smooth functions is simpler, though. – Gheehyun Nahm Dec 27 '17 at 18:22
  • You are right, higher order reflections are not needed. Everything still works if you choose to use the ordinary reflection - the only problem is that extending a $C^1$ function by reflection produces a function that doesn't lie in $C^1$. – Michał Miśkiewicz Dec 27 '17 at 20:06
  • @OrekiHoutarou As Michał Miśkiewicz said above, ordinary reflection works for all $W^{1,p}$, it's just that the proof that Evans presents (via $C^1$ approximation) requires higher order. –  Dec 27 '17 at 21:58
  • Thank you for clarifying! I was worried that I had a serious problem with my understanding on weak derivatives, which I should not have. – Gheehyun Nahm Dec 28 '17 at 01:32
  • @CrazyIvan Sorry about asking almost the same thing again (which feels quite stupid), but it bothers me that to me, it seems that the same proof (simply reflecting the original function) works for all $W^{n,p}$. Is this correct? I have doubt in myself since Evans seems to prefer the higher order method to prove for general $n$, which needs more work.. – Gheehyun Nahm Dec 28 '17 at 14:25
  • @Oreki No, simple reflection does not work for spaces with $n > 1$. For simplicity take one dimensional case and $u(x) = x$ for $x >0$. The simple reflection, which is $|x|$, is not in $W^{2,p}$ for any $p$. Its first derivative is not absolutely continuous, hence the second order weak derivative does not exist. –  Dec 28 '17 at 15:09
  • @CrazyIvan Oh, my question was stupid indeed; thanks! – Gheehyun Nahm Dec 28 '17 at 15:49