Mixing
proving an interpolation inequality in $L^p$ norms
$begingroup$
Let $1le p le infty$. Prove that for all $epsilon >0$ there exists a constant $C>0$ such that
$$|u'|_{L^p(mathbb{R})}le epsilon |u''|_{L^p(mathbb{R})}+C|u|_{L^p(mathbb{R})} ;; ; forall uin W^{2,p}(mathbb{R}).$$
This is a special case of an interpolation inequality, see here http://conteudo.icmc.usp.br/pessoas/andcarva/sobolew.pdf , theorem 2.
I need help to prove this. In any case the first step is to choose a continuous representative function for u. Then, I have the following 2 ideas:
1) To do it similarly as to prove theorem 2. Apply the mean value theorem to u to receive an estimation for $|u'|$. But the first problem I have here is, that the mean value theorem is local. But here the domain is $mathbb{R}$. Therefore, i don't know how to proceed.
2) To use the fundamental theorem of anlysis applied to $u'$, to receive something like $$u'(delta+x)=u'(x)+int_x^{x+delta} u''(z)dz.$$ Furthermore, convexity of $|cdot |^p$ and Jensens inequality https://de.wikipedia.org/wiki/Jensensche_Ungleichung might be important, but I am not completely sure how to math that.
The second idea may be more promising. I appreciant any help.
Edit: There are other variants of this inequality available and meanwhile I have discovered Interpolation inequality and Proving an interpolation inequality for $C^2_b$ functions which seem to be a variant of the inequality above (however, the norms are different).
analysis inequality pde sobolev-spaces calculus-of-variations
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
$endgroup$
add a comment |
$begingroup$
Let $1le p le infty$. Prove that for all $epsilon >0$ there exists a constant $C>0$ such that
$$|u'|_{L^p(mathbb{R})}le epsilon |u''|_{L^p(mathbb{R})}+C|u|_{L^p(mathbb{R})} ;; ; forall uin W^{2,p}(mathbb{R}).$$
This is a special case of an interpolation inequality, see here http://conteudo.icmc.usp.br/pessoas/andcarva/sobolew.pdf , theorem 2.
I need help to prove this. In any case the first step is to choose a continuous representative function for u. Then, I have the following 2 ideas:
1) To do it similarly as to prove theorem 2. Apply the mean value theorem to u to receive an estimation for $|u'|$. But the first problem I have here is, that the mean value theorem is local. But here the domain is $mathbb{R}$. Therefore, i don't know how to proceed.
2) To use the fundamental theorem of anlysis applied to $u'$, to receive something like $$u'(delta+x)=u'(x)+int_x^{x+delta} u''(z)dz.$$ Furthermore, convexity of $|cdot |^p$ and Jensens inequality https://de.wikipedia.org/wiki/Jensensche_Ungleichung might be important, but I am not completely sure how to math that.
The second idea may be more promising. I appreciant any help.
Edit: There are other variants of this inequality available and meanwhile I have discovered Interpolation inequality and Proving an interpolation inequality for $C^2_b$ functions which seem to be a variant of the inequality above (however, the norms are different).
analysis inequality pde sobolev-spaces calculus-of-variations
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
$endgroup$
$begingroup$
Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist?
$endgroup$
– Idun E.
Jan 16 at 2:59
$begingroup$
thank you. I fixed it
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 10:32
$begingroup$
The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand?
$endgroup$
– MaoWao
Jan 16 at 15:42
$begingroup$
@MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 16:15
add a comment |
$begingroup$
Let $1le p le infty$. Prove that for all $epsilon >0$ there exists a constant $C>0$ such that
$$|u'|_{L^p(mathbb{R})}le epsilon |u''|_{L^p(mathbb{R})}+C|u|_{L^p(mathbb{R})} ;; ; forall uin W^{2,p}(mathbb{R}).$$
This is a special case of an interpolation inequality, see here http://conteudo.icmc.usp.br/pessoas/andcarva/sobolew.pdf , theorem 2.
I need help to prove this. In any case the first step is to choose a continuous representative function for u. Then, I have the following 2 ideas:
1) To do it similarly as to prove theorem 2. Apply the mean value theorem to u to receive an estimation for $|u'|$. But the first problem I have here is, that the mean value theorem is local. But here the domain is $mathbb{R}$. Therefore, i don't know how to proceed.
2) To use the fundamental theorem of anlysis applied to $u'$, to receive something like $$u'(delta+x)=u'(x)+int_x^{x+delta} u''(z)dz.$$ Furthermore, convexity of $|cdot |^p$ and Jensens inequality https://de.wikipedia.org/wiki/Jensensche_Ungleichung might be important, but I am not completely sure how to math that.
The second idea may be more promising. I appreciant any help.
Edit: There are other variants of this inequality available and meanwhile I have discovered Interpolation inequality and Proving an interpolation inequality for $C^2_b$ functions which seem to be a variant of the inequality above (however, the norms are different).
analysis inequality pde sobolev-spaces calculus-of-variations
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
$endgroup$
Let $1le p le infty$. Prove that for all $epsilon >0$ there exists a constant $C>0$ such that
$$|u'|_{L^p(mathbb{R})}le epsilon |u''|_{L^p(mathbb{R})}+C|u|_{L^p(mathbb{R})} ;; ; forall uin W^{2,p}(mathbb{R}).$$
This is a special case of an interpolation inequality, see here http://conteudo.icmc.usp.br/pessoas/andcarva/sobolew.pdf , theorem 2.
I need help to prove this. In any case the first step is to choose a continuous representative function for u. Then, I have the following 2 ideas:
1) To do it similarly as to prove theorem 2. Apply the mean value theorem to u to receive an estimation for $|u'|$. But the first problem I have here is, that the mean value theorem is local. But here the domain is $mathbb{R}$. Therefore, i don't know how to proceed.
2) To use the fundamental theorem of anlysis applied to $u'$, to receive something like $$u'(delta+x)=u'(x)+int_x^{x+delta} u''(z)dz.$$ Furthermore, convexity of $|cdot |^p$ and Jensens inequality https://de.wikipedia.org/wiki/Jensensche_Ungleichung might be important, but I am not completely sure how to math that.
The second idea may be more promising. I appreciant any help.
Edit: There are other variants of this inequality available and meanwhile I have discovered Interpolation inequality and Proving an interpolation inequality for $C^2_b$ functions which seem to be a variant of the inequality above (however, the norms are different).
analysis inequality pde sobolev-spaces calculus-of-variations
analysis inequality pde sobolev-spaces calculus-of-variations
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
edited Jan 16 at 15:33
TheAppliedTheoreticalOne
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
asked Jan 14 at 23:08
TheAppliedTheoreticalOneTheAppliedTheoreticalOne
717
717
$begingroup$
Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist?
$endgroup$
– Idun E.
Jan 16 at 2:59
$begingroup$
thank you. I fixed it
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 10:32
$begingroup$
The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand?
$endgroup$
– MaoWao
Jan 16 at 15:42
$begingroup$
@MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 16:15
add a comment |
$begingroup$
Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist?
$endgroup$
– Idun E.
Jan 16 at 2:59
$begingroup$
thank you. I fixed it
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 10:32
$begingroup$
The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand?
$endgroup$
– MaoWao
Jan 16 at 15:42
$begingroup$
@MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 16:15
$begingroup$
Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist?
$endgroup$
– Idun E.
Jan 16 at 2:59
$begingroup$
Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist?
$endgroup$
– Idun E.
Jan 16 at 2:59
$begingroup$
thank you. I fixed it
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 10:32
$begingroup$
thank you. I fixed it
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 10:32
$begingroup$
The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand?
$endgroup$
– MaoWao
Jan 16 at 15:42
$begingroup$
The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand?
$endgroup$
– MaoWao
Jan 16 at 15:42
$begingroup$
@MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 16:15
$begingroup$
@MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 16:15
add a comment |
1 Answer
1
active
oldest
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$begingroup$
In fact, I think you only need to prove for $uin C_c^{infty}(mathbb{R})$ which is a dense subset of $W^{1,2}(mathbb{R})$ (But NOT dense in $W^{1,2}[0,100]$). Then the desired result holds by pushing limit. I hope I remembered it right. :)
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
$endgroup$
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In fact, I think you only need to prove for $uin C_c^{infty}(mathbb{R})$ which is a dense subset of $W^{1,2}(mathbb{R})$ (But NOT dense in $W^{1,2}[0,100]$). Then the desired result holds by pushing limit. I hope I remembered it right. :)
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
$endgroup$
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
add a comment |
$begingroup$
In fact, I think you only need to prove for $uin C_c^{infty}(mathbb{R})$ which is a dense subset of $W^{1,2}(mathbb{R})$ (But NOT dense in $W^{1,2}[0,100]$). Then the desired result holds by pushing limit. I hope I remembered it right. :)
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
$endgroup$
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
add a comment |
$begingroup$
In fact, I think you only need to prove for $uin C_c^{infty}(mathbb{R})$ which is a dense subset of $W^{1,2}(mathbb{R})$ (But NOT dense in $W^{1,2}[0,100]$). Then the desired result holds by pushing limit. I hope I remembered it right. :)
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
$endgroup$
In fact, I think you only need to prove for $uin C_c^{infty}(mathbb{R})$ which is a dense subset of $W^{1,2}(mathbb{R})$ (But NOT dense in $W^{1,2}[0,100]$). Then the desired result holds by pushing limit. I hope I remembered it right. :)
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
answered Jan 14 at 23:37
Zixiao_LiuZixiao_Liu
938
938
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
add a comment |
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
Could you explain how you would push the Limits? Since $C^infty_c(mathbb{R})$ is dense, we finde a converving subsequence with regards to $W^{2,k}$. But Since $mathbb{R}$ is not bounded, this does not imply convergence in $L^p$, if I remeber it correctly. So how does ist work?
$endgroup$
– Idun E.
Jan 16 at 11:51
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
$begingroup$
@IdunE. $C_c^infty(mathbb{R})$ is dense in $W^{2,p}(mathbb{R})$. Since both sides of the inequality are continuous w.r.t. to the $W^{2,p}$-norm, it suffices to prove it on a dense subset. The case $p=infty$ has to be treated separately, I think (if it's even true at all).
$endgroup$
– MaoWao
Jan 16 at 15:44
add a comment |
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$begingroup$
Should it not rather be $W^{2,p}$ otherwise the summands don’t Need to exist?
$endgroup$
– Idun E.
Jan 16 at 2:59
$begingroup$
thank you. I fixed it
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 10:32
$begingroup$
The first pdf you link gives a proof (they even prove it first for the one-dimensional case). Is there anything you don't understand?
$endgroup$
– MaoWao
Jan 16 at 15:42
$begingroup$
@MaoWao in the meantime I have been succesful proving it with idea 2! I am going to post my solution on MSE as soon as possible
$endgroup$
– TheAppliedTheoreticalOne
Jan 16 at 16:15