Mixing
How can one show that the first term of the integral ''dominates''?
$begingroup$
Fix $1<p<N$ and let $(u_n)subseteq W^{1,p}(mathbb{R}^N)$ be a sequence satisfying the following:
$u_n to 0$ pointwise almost everywhere,
$u_n to 0$ in $L_{loc}^p(mathbb{R}^N)$.
I am trying to show that for any $phi in C_c^infty(mathbb{R}^N)$ one has:
$$
int_{mathbb{R}^N} lvert nabla left( phi u_n right)rvert^p
=int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^p + o(1).
$$
So far, I have only been able to obtain the following
begin{align*}
int_{mathbb{R}^N} leftvert nabla(phi u_n)rightvert^p = left(left[int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^pright]^{1/p}+ o(1)right)^p.
end{align*}
My intuition says that I should be able to finish the proof by using some generalization of the binomial theorem. If anyone can either point me in the right direction or provide a source that I can cite showing that this is true I would highly appreciate it!
real-analysis analysis pde sobolev-spaces
asked Jan 13 at 3:09
QuokaQuoka
1,254312
$endgroup$
add a comment |
$begingroup$
Fix $1<p<N$ and let $(u_n)subseteq W^{1,p}(mathbb{R}^N)$ be a sequence satisfying the following:
$u_n to 0$ pointwise almost everywhere,
$u_n to 0$ in $L_{loc}^p(mathbb{R}^N)$.
I am trying to show that for any $phi in C_c^infty(mathbb{R}^N)$ one has:
$$
int_{mathbb{R}^N} lvert nabla left( phi u_n right)rvert^p
=int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^p + o(1).
$$
So far, I have only been able to obtain the following
begin{align*}
int_{mathbb{R}^N} leftvert nabla(phi u_n)rightvert^p = left(left[int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^pright]^{1/p}+ o(1)right)^p.
end{align*}
My intuition says that I should be able to finish the proof by using some generalization of the binomial theorem. If anyone can either point me in the right direction or provide a source that I can cite showing that this is true I would highly appreciate it!
real-analysis analysis pde sobolev-spaces
asked Jan 13 at 3:09
QuokaQuoka
1,254312
$endgroup$
add a comment |
$begingroup$
Fix $1<p<N$ and let $(u_n)subseteq W^{1,p}(mathbb{R}^N)$ be a sequence satisfying the following:
$u_n to 0$ pointwise almost everywhere,
$u_n to 0$ in $L_{loc}^p(mathbb{R}^N)$.
I am trying to show that for any $phi in C_c^infty(mathbb{R}^N)$ one has:
$$
int_{mathbb{R}^N} lvert nabla left( phi u_n right)rvert^p
=int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^p + o(1).
$$
So far, I have only been able to obtain the following
begin{align*}
int_{mathbb{R}^N} leftvert nabla(phi u_n)rightvert^p = left(left[int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^pright]^{1/p}+ o(1)right)^p.
end{align*}
My intuition says that I should be able to finish the proof by using some generalization of the binomial theorem. If anyone can either point me in the right direction or provide a source that I can cite showing that this is true I would highly appreciate it!
real-analysis analysis pde sobolev-spaces
asked Jan 13 at 3:09
QuokaQuoka
1,254312
$endgroup$
Fix $1<p<N$ and let $(u_n)subseteq W^{1,p}(mathbb{R}^N)$ be a sequence satisfying the following:
$u_n to 0$ pointwise almost everywhere,
$u_n to 0$ in $L_{loc}^p(mathbb{R}^N)$.
I am trying to show that for any $phi in C_c^infty(mathbb{R}^N)$ one has:
$$
int_{mathbb{R}^N} lvert nabla left( phi u_n right)rvert^p
=int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^p + o(1).
$$
So far, I have only been able to obtain the following
begin{align*}
int_{mathbb{R}^N} leftvert nabla(phi u_n)rightvert^p = left(left[int_{mathbb{R}^N} lvert phi rvert^plvert nabla left(u_n right)rvert^pright]^{1/p}+ o(1)right)^p.
end{align*}
My intuition says that I should be able to finish the proof by using some generalization of the binomial theorem. If anyone can either point me in the right direction or provide a source that I can cite showing that this is true I would highly appreciate it!
real-analysis analysis pde sobolev-spaces
real-analysis analysis pde sobolev-spaces
asked Jan 13 at 3:09
QuokaQuoka
1,254312
asked Jan 13 at 3:09
QuokaQuoka
1,254312
asked Jan 13 at 3:09
QuokaQuoka
1,254312
asked Jan 13 at 3:09
QuokaQuoka
1,254312
asked Jan 13 at 3:09
QuokaQuoka
1,254312
1,254312
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The function $x mapsto x^p$ is convex. Thus,
$$(a+b)^p = (frac{2,a}2 + frac{2,b}2)^p
le frac12 , (2,a)^p + frac12 , (2,b)^p = 2^{p-1},(a^p + b^p)$$
for all $a,b ge 0$.
Edit: As noted by Quoka, this does not yield the desired result, but only an inequality involving the constant $2^{p-1}$.
Now, I think that your identity is not enough. Let us denote $a_n := [int |phi|^p , |nabla u_n|^p ]^{1/p}$. Then, $a_n$ might go to infinity. Let us, for simplicity, consider the case $p = 2$.
Then,
$$ (a_n + o(1))^2 = a_n^2 + 2 , a_n , o(1) + o(1)^2.$$
However, since $a_n$ might be unbounded, the second addend might no longer be $o(1)$.
To give a precise example:
$$(n + 1/n) ^2 = n^2 + 2 + 1/n^2 ne n^2 + o(1).$$
answered Jan 14 at 10:59
gerwgerw
19.5k11334
$endgroup$
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
The function $x mapsto x^p$ is convex. Thus,
$$(a+b)^p = (frac{2,a}2 + frac{2,b}2)^p
le frac12 , (2,a)^p + frac12 , (2,b)^p = 2^{p-1},(a^p + b^p)$$
for all $a,b ge 0$.
Edit: As noted by Quoka, this does not yield the desired result, but only an inequality involving the constant $2^{p-1}$.
Now, I think that your identity is not enough. Let us denote $a_n := [int |phi|^p , |nabla u_n|^p ]^{1/p}$. Then, $a_n$ might go to infinity. Let us, for simplicity, consider the case $p = 2$.
Then,
$$ (a_n + o(1))^2 = a_n^2 + 2 , a_n , o(1) + o(1)^2.$$
However, since $a_n$ might be unbounded, the second addend might no longer be $o(1)$.
To give a precise example:
$$(n + 1/n) ^2 = n^2 + 2 + 1/n^2 ne n^2 + o(1).$$
answered Jan 14 at 10:59
gerwgerw
19.5k11334
$endgroup$
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
add a comment |
$begingroup$
The function $x mapsto x^p$ is convex. Thus,
$$(a+b)^p = (frac{2,a}2 + frac{2,b}2)^p
le frac12 , (2,a)^p + frac12 , (2,b)^p = 2^{p-1},(a^p + b^p)$$
for all $a,b ge 0$.
Edit: As noted by Quoka, this does not yield the desired result, but only an inequality involving the constant $2^{p-1}$.
Now, I think that your identity is not enough. Let us denote $a_n := [int |phi|^p , |nabla u_n|^p ]^{1/p}$. Then, $a_n$ might go to infinity. Let us, for simplicity, consider the case $p = 2$.
Then,
$$ (a_n + o(1))^2 = a_n^2 + 2 , a_n , o(1) + o(1)^2.$$
However, since $a_n$ might be unbounded, the second addend might no longer be $o(1)$.
To give a precise example:
$$(n + 1/n) ^2 = n^2 + 2 + 1/n^2 ne n^2 + o(1).$$
answered Jan 14 at 10:59
gerwgerw
19.5k11334
$endgroup$
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
add a comment |
$begingroup$
The function $x mapsto x^p$ is convex. Thus,
$$(a+b)^p = (frac{2,a}2 + frac{2,b}2)^p
le frac12 , (2,a)^p + frac12 , (2,b)^p = 2^{p-1},(a^p + b^p)$$
for all $a,b ge 0$.
Edit: As noted by Quoka, this does not yield the desired result, but only an inequality involving the constant $2^{p-1}$.
Now, I think that your identity is not enough. Let us denote $a_n := [int |phi|^p , |nabla u_n|^p ]^{1/p}$. Then, $a_n$ might go to infinity. Let us, for simplicity, consider the case $p = 2$.
Then,
$$ (a_n + o(1))^2 = a_n^2 + 2 , a_n , o(1) + o(1)^2.$$
However, since $a_n$ might be unbounded, the second addend might no longer be $o(1)$.
To give a precise example:
$$(n + 1/n) ^2 = n^2 + 2 + 1/n^2 ne n^2 + o(1).$$
answered Jan 14 at 10:59
gerwgerw
19.5k11334
$endgroup$
The function $x mapsto x^p$ is convex. Thus,
$$(a+b)^p = (frac{2,a}2 + frac{2,b}2)^p
le frac12 , (2,a)^p + frac12 , (2,b)^p = 2^{p-1},(a^p + b^p)$$
for all $a,b ge 0$.
Edit: As noted by Quoka, this does not yield the desired result, but only an inequality involving the constant $2^{p-1}$.
Now, I think that your identity is not enough. Let us denote $a_n := [int |phi|^p , |nabla u_n|^p ]^{1/p}$. Then, $a_n$ might go to infinity. Let us, for simplicity, consider the case $p = 2$.
Then,
$$ (a_n + o(1))^2 = a_n^2 + 2 , a_n , o(1) + o(1)^2.$$
However, since $a_n$ might be unbounded, the second addend might no longer be $o(1)$.
To give a precise example:
$$(n + 1/n) ^2 = n^2 + 2 + 1/n^2 ne n^2 + o(1).$$
answered Jan 14 at 10:59
gerwgerw
19.5k11334
edited Jan 15 at 6:47
answered Jan 14 at 10:59
gerwgerw
19.5k11334
answered Jan 14 at 10:59
gerwgerw
19.5k11334
answered Jan 14 at 10:59
gerwgerw
19.5k11334
19.5k11334
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
add a comment |
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
Could you elaborate on how this helps with my inequality? Wouldn't this give the inequality with a factor of $2^{p-1}$?
$endgroup$
– Quoka
Jan 14 at 18:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
$begingroup$
@Quoka: Yes of course. However, it might not be possible to deduce your desired inequality, see the edit.
$endgroup$
– gerw
Jan 15 at 6:47
add a comment |
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