Prove $L^1(mathbb{R})cap L^2(mathbb{R})$ has empty interior both in $L^1(mathbb{R})$ and in $L^2(mathbb{R})$
$begingroup$
I suppose it has interior, so we can find an open ball of radius epsilon in $L^1(mathbb{R})cap L^2(mathbb{R})$ and so that ball is contained in both the sets, so we can write:
$$B_varepsilon={u(x)in L^1(mathbb{R}):||u(x)||_{L^1}<varepsilon}={u(x)in L^2(mathbb{R}):||u(x)||_{L^2}<varepsilon}$$
But now I can't find the contradiction. I tried to compute the norms but I can't end with a contradiction. I can't find it at all. I don't know if this is the right way.
functional-analysis lebesgue-measure
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
asked Dec 2 '18 at 11:59
james wattjames watt
34610
$endgroup$
add a comment |
$begingroup$
I suppose it has interior, so we can find an open ball of radius epsilon in $L^1(mathbb{R})cap L^2(mathbb{R})$ and so that ball is contained in both the sets, so we can write:
$$B_varepsilon={u(x)in L^1(mathbb{R}):||u(x)||_{L^1}<varepsilon}={u(x)in L^2(mathbb{R}):||u(x)||_{L^2}<varepsilon}$$
But now I can't find the contradiction. I tried to compute the norms but I can't end with a contradiction. I can't find it at all. I don't know if this is the right way.
functional-analysis lebesgue-measure
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
asked Dec 2 '18 at 11:59
james wattjames watt
34610
$endgroup$
$begingroup$
Maybe this can help math.stackexchange.com/questions/2560595/…. It is not exactly your question, but it is close and can be adapted I think.
$endgroup$
– zwim
Dec 2 '18 at 12:10
1
$begingroup$
Interior where? Please edit the question so that it makes sense.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 12:23
$begingroup$
@zwim Thanks man.
$endgroup$
– james watt
Dec 2 '18 at 12:26
add a comment |
$begingroup$
I suppose it has interior, so we can find an open ball of radius epsilon in $L^1(mathbb{R})cap L^2(mathbb{R})$ and so that ball is contained in both the sets, so we can write:
$$B_varepsilon={u(x)in L^1(mathbb{R}):||u(x)||_{L^1}<varepsilon}={u(x)in L^2(mathbb{R}):||u(x)||_{L^2}<varepsilon}$$
But now I can't find the contradiction. I tried to compute the norms but I can't end with a contradiction. I can't find it at all. I don't know if this is the right way.
functional-analysis lebesgue-measure
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
asked Dec 2 '18 at 11:59
james wattjames watt
34610
$endgroup$
I suppose it has interior, so we can find an open ball of radius epsilon in $L^1(mathbb{R})cap L^2(mathbb{R})$ and so that ball is contained in both the sets, so we can write:
$$B_varepsilon={u(x)in L^1(mathbb{R}):||u(x)||_{L^1}<varepsilon}={u(x)in L^2(mathbb{R}):||u(x)||_{L^2}<varepsilon}$$
But now I can't find the contradiction. I tried to compute the norms but I can't end with a contradiction. I can't find it at all. I don't know if this is the right way.
functional-analysis lebesgue-measure
functional-analysis lebesgue-measure
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
asked Dec 2 '18 at 11:59
james wattjames watt
34610
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
asked Dec 2 '18 at 11:59
james wattjames watt
34610
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
edited Dec 2 '18 at 13:36
Andrés E. Caicedo
65k8158246
65k8158246
asked Dec 2 '18 at 11:59
james wattjames watt
34610
asked Dec 2 '18 at 11:59
james wattjames watt
34610
asked Dec 2 '18 at 11:59
james wattjames watt
34610
34610
$begingroup$
Maybe this can help math.stackexchange.com/questions/2560595/…. It is not exactly your question, but it is close and can be adapted I think.
$endgroup$
– zwim
Dec 2 '18 at 12:10
1
$begingroup$
Interior where? Please edit the question so that it makes sense.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 12:23
$begingroup$
@zwim Thanks man.
$endgroup$
– james watt
Dec 2 '18 at 12:26
add a comment |
$begingroup$
Maybe this can help math.stackexchange.com/questions/2560595/…. It is not exactly your question, but it is close and can be adapted I think.
$endgroup$
– zwim
Dec 2 '18 at 12:10
1
$begingroup$
Interior where? Please edit the question so that it makes sense.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 12:23
$begingroup$
@zwim Thanks man.
$endgroup$
– james watt
Dec 2 '18 at 12:26
$begingroup$
Maybe this can help math.stackexchange.com/questions/2560595/…. It is not exactly your question, but it is close and can be adapted I think.
$endgroup$
– zwim
Dec 2 '18 at 12:10
$begingroup$
Maybe this can help math.stackexchange.com/questions/2560595/…. It is not exactly your question, but it is close and can be adapted I think.
$endgroup$
– zwim
Dec 2 '18 at 12:10
1
1
$begingroup$
Interior where? Please edit the question so that it makes sense.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 12:23
$begingroup$
Interior where? Please edit the question so that it makes sense.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 12:23
$begingroup$
@zwim Thanks man.
$endgroup$
– james watt
Dec 2 '18 at 12:26
$begingroup$
@zwim Thanks man.
$endgroup$
– james watt
Dec 2 '18 at 12:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $f$ is an interior point in $L^{2}$. Consider $f+frac 1 {nx} I_{x>1}$. Show that this sequence converges to $f$ in $L^{2}$ and none of these functions are in $L^{1}$. This proves that interior in $L^{2}$ is empty. For interior in $L^{1}$ consider $f+frac 1 {nsqrt x} I_{0<x<1}$.
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
2
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
2
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
2
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
|
show 7 more comments
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1 Answer
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$begingroup$
Suppose $f$ is an interior point in $L^{2}$. Consider $f+frac 1 {nx} I_{x>1}$. Show that this sequence converges to $f$ in $L^{2}$ and none of these functions are in $L^{1}$. This proves that interior in $L^{2}$ is empty. For interior in $L^{1}$ consider $f+frac 1 {nsqrt x} I_{0<x<1}$.
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
2
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
2
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
2
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
|
show 7 more comments
$begingroup$
Suppose $f$ is an interior point in $L^{2}$. Consider $f+frac 1 {nx} I_{x>1}$. Show that this sequence converges to $f$ in $L^{2}$ and none of these functions are in $L^{1}$. This proves that interior in $L^{2}$ is empty. For interior in $L^{1}$ consider $f+frac 1 {nsqrt x} I_{0<x<1}$.
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
2
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
2
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
2
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
|
show 7 more comments
$begingroup$
Suppose $f$ is an interior point in $L^{2}$. Consider $f+frac 1 {nx} I_{x>1}$. Show that this sequence converges to $f$ in $L^{2}$ and none of these functions are in $L^{1}$. This proves that interior in $L^{2}$ is empty. For interior in $L^{1}$ consider $f+frac 1 {nsqrt x} I_{0<x<1}$.
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
Suppose $f$ is an interior point in $L^{2}$. Consider $f+frac 1 {nx} I_{x>1}$. Show that this sequence converges to $f$ in $L^{2}$ and none of these functions are in $L^{1}$. This proves that interior in $L^{2}$ is empty. For interior in $L^{1}$ consider $f+frac 1 {nsqrt x} I_{0<x<1}$.
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Dec 2 '18 at 12:28
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
52.9k32055
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
2
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
2
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
2
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
|
show 7 more comments
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
2
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
2
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
2
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
Nice, I'm trying to understand all the things you put here. If I'll have other doubts I'll write here. Thanks.
$endgroup$
– james watt
Dec 2 '18 at 12:57
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
$begingroup$
@jameswatt Most welcome I will try to clear all your doubts.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 23:11
2
2
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
$begingroup$
@jameswatt Since $f$ is already in $L^{1}$ we cannot have the sum in $L^{1}$ unless the second term is in $L^{1}$. But $frac 1 {nx} I_{x>1}$ is not in $L^{1}$ for any $n$. BTW $int_1^{infty} frac 1 x dx=infty$, not zero.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 23:21
2
2
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
$begingroup$
@soundwave $lim infty$ is not $0$. Besides, there was no limit; we had to just compute the integral for fixed $n$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 5:15
2
2
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
$begingroup$
@jameswatt You have used the inequality $|a+b| >|b|$ which is false. Your proof is almost same as mine but you have made a mistake in the proof.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 23:18
|
show 7 more comments
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$begingroup$
Maybe this can help math.stackexchange.com/questions/2560595/…. It is not exactly your question, but it is close and can be adapted I think.
$endgroup$
– zwim
Dec 2 '18 at 12:10
1
$begingroup$
Interior where? Please edit the question so that it makes sense.
$endgroup$
– Kavi Rama Murthy
Dec 2 '18 at 12:23
$begingroup$
@zwim Thanks man.
$endgroup$
– james watt
Dec 2 '18 at 12:26