Exercise 4.33 in Brezis functional analysis.
up vote
2
down vote
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Fix a function $phi in C_c(mathbb{R}), phinotequiv0,$ and consider the family of functions
$mathcal{F} = {phi_n:ninmathbb N},$ where $phi_n(x) = phi(x+n), xinmathbb{R}.$
The problem is to prove that $mathcal{F}$ does not have compact closure in $L^p(mathbb{R}) (1le p <infty),$ but there is a theorem that the closure $mathcal{F}|_{Omega}$ in $L^p(Omega)$ is compact for any finite-measure subset $Omegainmathbb{R}.$ Thus this problem is intended to show that the theorem is not applicable to infinite-measure sets, in particular, $mathbb{R}.$
I think I need to induce contradiction assuming $mathcal{F}$ has compact closure in $L^p(mathbb{R}),$ but I cannot come up with the next step.
Any hint or idea would be appreciated. Thanks in advance.
functional-analysis lp-spaces
edited Nov 25 at 5:31
Aweygan
13.1k21441
asked Nov 25 at 5:22
Euduardo
576
add a comment |
up vote
2
down vote
favorite
Fix a function $phi in C_c(mathbb{R}), phinotequiv0,$ and consider the family of functions
$mathcal{F} = {phi_n:ninmathbb N},$ where $phi_n(x) = phi(x+n), xinmathbb{R}.$
The problem is to prove that $mathcal{F}$ does not have compact closure in $L^p(mathbb{R}) (1le p <infty),$ but there is a theorem that the closure $mathcal{F}|_{Omega}$ in $L^p(Omega)$ is compact for any finite-measure subset $Omegainmathbb{R}.$ Thus this problem is intended to show that the theorem is not applicable to infinite-measure sets, in particular, $mathbb{R}.$
I think I need to induce contradiction assuming $mathcal{F}$ has compact closure in $L^p(mathbb{R}),$ but I cannot come up with the next step.
Any hint or idea would be appreciated. Thanks in advance.
functional-analysis lp-spaces
edited Nov 25 at 5:31
Aweygan
13.1k21441
asked Nov 25 at 5:22
Euduardo
576
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Fix a function $phi in C_c(mathbb{R}), phinotequiv0,$ and consider the family of functions
$mathcal{F} = {phi_n:ninmathbb N},$ where $phi_n(x) = phi(x+n), xinmathbb{R}.$
The problem is to prove that $mathcal{F}$ does not have compact closure in $L^p(mathbb{R}) (1le p <infty),$ but there is a theorem that the closure $mathcal{F}|_{Omega}$ in $L^p(Omega)$ is compact for any finite-measure subset $Omegainmathbb{R}.$ Thus this problem is intended to show that the theorem is not applicable to infinite-measure sets, in particular, $mathbb{R}.$
I think I need to induce contradiction assuming $mathcal{F}$ has compact closure in $L^p(mathbb{R}),$ but I cannot come up with the next step.
Any hint or idea would be appreciated. Thanks in advance.
functional-analysis lp-spaces
edited Nov 25 at 5:31
Aweygan
13.1k21441
asked Nov 25 at 5:22
Euduardo
576
Fix a function $phi in C_c(mathbb{R}), phinotequiv0,$ and consider the family of functions
$mathcal{F} = {phi_n:ninmathbb N},$ where $phi_n(x) = phi(x+n), xinmathbb{R}.$
The problem is to prove that $mathcal{F}$ does not have compact closure in $L^p(mathbb{R}) (1le p <infty),$ but there is a theorem that the closure $mathcal{F}|_{Omega}$ in $L^p(Omega)$ is compact for any finite-measure subset $Omegainmathbb{R}.$ Thus this problem is intended to show that the theorem is not applicable to infinite-measure sets, in particular, $mathbb{R}.$
I think I need to induce contradiction assuming $mathcal{F}$ has compact closure in $L^p(mathbb{R}),$ but I cannot come up with the next step.
Any hint or idea would be appreciated. Thanks in advance.
functional-analysis lp-spaces
functional-analysis lp-spaces
edited Nov 25 at 5:31
Aweygan
13.1k21441
asked Nov 25 at 5:22
Euduardo
576
edited Nov 25 at 5:31
Aweygan
13.1k21441
asked Nov 25 at 5:22
Euduardo
576
edited Nov 25 at 5:31
Aweygan
13.1k21441
edited Nov 25 at 5:31
Aweygan
13.1k21441
edited Nov 25 at 5:31
Aweygan
13.1k21441
13.1k21441
asked Nov 25 at 5:22
Euduardo
576
asked Nov 25 at 5:22
Euduardo
576
asked Nov 25 at 5:22
Euduardo
576
576
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
Hint: We have $operatorname{supp}phisubset [-N,N]$ for some $Ninmathbb N$. Can the sequence ${phi_{nN}}_{ninmathbb N}$ have a convergent subsequence?
answered Nov 25 at 5:31
Aweygan
13.1k21441
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
1
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: We have $operatorname{supp}phisubset [-N,N]$ for some $Ninmathbb N$. Can the sequence ${phi_{nN}}_{ninmathbb N}$ have a convergent subsequence?
answered Nov 25 at 5:31
Aweygan
13.1k21441
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
1
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
add a comment |
up vote
2
down vote
accepted
Hint: We have $operatorname{supp}phisubset [-N,N]$ for some $Ninmathbb N$. Can the sequence ${phi_{nN}}_{ninmathbb N}$ have a convergent subsequence?
answered Nov 25 at 5:31
Aweygan
13.1k21441
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
1
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: We have $operatorname{supp}phisubset [-N,N]$ for some $Ninmathbb N$. Can the sequence ${phi_{nN}}_{ninmathbb N}$ have a convergent subsequence?
answered Nov 25 at 5:31
Aweygan
13.1k21441
Hint: We have $operatorname{supp}phisubset [-N,N]$ for some $Ninmathbb N$. Can the sequence ${phi_{nN}}_{ninmathbb N}$ have a convergent subsequence?
answered Nov 25 at 5:31
Aweygan
13.1k21441
answered Nov 25 at 5:31
Aweygan
13.1k21441
answered Nov 25 at 5:31
Aweygan
13.1k21441
answered Nov 25 at 5:31
Aweygan
13.1k21441
13.1k21441
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
1
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
add a comment |
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
1
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
Your answer makes me feel I am stupid... Thank you Aweygan!
– Euduardo
Nov 25 at 5:33
1
1
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
You're welcome, but don't feel stupid!
– Aweygan
Nov 25 at 5:36
add a comment |
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