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1d
comment Find a cut-off function in a ball.
In this case $f$ is Lipschitz. So it has a derivative which is defined almost everywhere. That's the meaning of $f'$.
1d
comment Find a cut-off function in a ball.
For your information, $(\phi_\delta'\star f)=(\phi_\delta\star f')$. To answer your question, note that $f$ is a continuous cut-off function, so it remains only to regularize it, which I did by using convolution. The length $\epsilon$ is used only to guarantee some space for the regularization process.
1d
comment Find a cut-off function in a ball.
Nice, you are almost there. For the derivative, note that if $\eta(x)=(\phi_\delta\star f)(x)$ then, $\eta'(x)=(\phi_\delta\star f')(x)$. Can you conclude now?
2d
comment Find a cut-off function in a ball.
Yes, I will be here. Take your time and try to understand it first, then you come back here.
2d
comment Find a cut-off function in a ball.
Well, try it first. If you not try, you will not understand it. I suggest you to draw a picture of $f$ and understand why I choose $\epsilon$ and $\delta$ as I did.
2d
revised Find a cut-off function in a ball.
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2d
answered Find a cut-off function in a ball.
2d
revised Find a cut-off function in a ball.
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2d
revised Find a cut-off function in a ball.
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2d
comment Find a cut-off function in a ball.
Have you tried something?
May
20
comment Laplacian operator on $L^2(\Omega)$?
When you apply the Laplacian operator, you have to use two derivatives. Thus, if by $\Delta$ in $L^2(\Omega)$, you mean in the distributional sense then, $\Delta: L^2(\Omega)\to H^{-2}(\Omega)$.
May
19
comment Is Cantor set closed?
You should watch the video again. He does not say such a thing.
May
18
comment Does weak convergence in $W^{1,p}$ imply strong convergence in $L^q$?
There is only contnuous embedding in this case.
May
18
comment Does weak convergence in $W^{1,p}$ imply strong convergence in $L^q$?
Do you know Rellich-Kondrachov compactness theorem?
May
18
answered Related question to : If $u \in H^1(U)$, then $Du=0$ a.e. on the set $\{u=0\}$
May
16
comment Showing Sobolev space $W^{1,2}$ is a Hilbert space
Yes, that's right.
May
16
comment Showing Sobolev space $W^{1,2}$ is a Hilbert space
Because $f_n$ is a Cauchy sequence in $W^{1,2}$.
May
15
answered Showing Sobolev space $W^{1,2}$ is a Hilbert space
May
15
comment Showing Sobolev space $W^{1,2}$ is a Hilbert space
If the convergence is only in $L^2$ then, it is no true. If you want to prove that $W^{1,2}$ is closed, you have to assume that the convergence is in $W^{1,2}$ (by definiton).
May
15
comment Showing Sobolev space $W^{1,2}$ is a Hilbert space
I suppose that $W=W^{1,2}$ and that you mean $f_n\to f$ in $W^{1,2}$?