Weak convergence and pointwise convergence of norm
1
Assume that $f_nrightharpoonup f$ weakly in some $L^p$ space, and $|f_n|rightarrow|f|$ pointwise. Does this imply that $f_nrightarrow f$ pointwise? (or a subsequence?)
analysis lp-spaces weak-convergence
edited Nov 27 at 9:35
Masacroso
12.9k41746
asked Nov 27 at 8:45
P.R
164
add a comment |
1
Assume that $f_nrightharpoonup f$ weakly in some $L^p$ space, and $|f_n|rightarrow|f|$ pointwise. Does this imply that $f_nrightarrow f$ pointwise? (or a subsequence?)
analysis lp-spaces weak-convergence
edited Nov 27 at 9:35
Masacroso
12.9k41746
asked Nov 27 at 8:45
P.R
164
I'm not really familiar with weak convergence, so correct me if I'm wrong. If $f_n(x)=1-1/n$ and we fix $x_0$ to define $g_n(x)=f_n(x)$ and $g_n(x_0)=-f_n(x_0)$ then does $g_n$ converg weakly to the constant $1$? If so then that would be a counterexample.
– freakish
Nov 27 at 8:59
could you please hint the proof?
– P.R
Nov 27 at 10:05
add a comment |
1
1
1
Assume that $f_nrightharpoonup f$ weakly in some $L^p$ space, and $|f_n|rightarrow|f|$ pointwise. Does this imply that $f_nrightarrow f$ pointwise? (or a subsequence?)
analysis lp-spaces weak-convergence
edited Nov 27 at 9:35
Masacroso
12.9k41746
asked Nov 27 at 8:45
P.R
164
Assume that $f_nrightharpoonup f$ weakly in some $L^p$ space, and $|f_n|rightarrow|f|$ pointwise. Does this imply that $f_nrightarrow f$ pointwise? (or a subsequence?)
analysis lp-spaces weak-convergence
analysis lp-spaces weak-convergence
edited Nov 27 at 9:35
Masacroso
12.9k41746
asked Nov 27 at 8:45
P.R
164
edited Nov 27 at 9:35
Masacroso
12.9k41746
asked Nov 27 at 8:45
P.R
164
edited Nov 27 at 9:35
Masacroso
12.9k41746
edited Nov 27 at 9:35
Masacroso
12.9k41746
edited Nov 27 at 9:35
Masacroso
12.9k41746
12.9k41746
asked Nov 27 at 8:45
P.R
164
asked Nov 27 at 8:45
P.R
164
asked Nov 27 at 8:45
P.R
164
164
I'm not really familiar with weak convergence, so correct me if I'm wrong. If $f_n(x)=1-1/n$ and we fix $x_0$ to define $g_n(x)=f_n(x)$ and $g_n(x_0)=-f_n(x_0)$ then does $g_n$ converg weakly to the constant $1$? If so then that would be a counterexample.
– freakish
Nov 27 at 8:59
could you please hint the proof?
– P.R
Nov 27 at 10:05
add a comment |
I'm not really familiar with weak convergence, so correct me if I'm wrong. If $f_n(x)=1-1/n$ and we fix $x_0$ to define $g_n(x)=f_n(x)$ and $g_n(x_0)=-f_n(x_0)$ then does $g_n$ converg weakly to the constant $1$? If so then that would be a counterexample.
– freakish
Nov 27 at 8:59
could you please hint the proof?
– P.R
Nov 27 at 10:05
I'm not really familiar with weak convergence, so correct me if I'm wrong. If $f_n(x)=1-1/n$ and we fix $x_0$ to define $g_n(x)=f_n(x)$ and $g_n(x_0)=-f_n(x_0)$ then does $g_n$ converg weakly to the constant $1$? If so then that would be a counterexample.
– freakish
Nov 27 at 8:59
I'm not really familiar with weak convergence, so correct me if I'm wrong. If $f_n(x)=1-1/n$ and we fix $x_0$ to define $g_n(x)=f_n(x)$ and $g_n(x_0)=-f_n(x_0)$ then does $g_n$ converg weakly to the constant $1$? If so then that would be a counterexample.
– freakish
Nov 27 at 8:59
could you please hint the proof?
– P.R
Nov 27 at 10:05
could you please hint the proof?
– P.R
Nov 27 at 10:05
add a comment |
1 Answer
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Consider $f_n(x)=-mathbf{1}_{mathbb{Q} cap [0,1]}$ and $f(x)=mathbf{1}_{mathbb{Q} cap [0,1]}$.
Then $|f_n(x)|=|f(x)|$ and $f_n rightharpoonup f$ since
$$forall phi in L^q, int (f_n-f)phi=0,$$
as $f_n-f$ is non-zero over a set of $0$ measure.
answered Nov 27 at 10:27
nicomezi
4,0741820
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
1
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
add a comment |
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Consider $f_n(x)=-mathbf{1}_{mathbb{Q} cap [0,1]}$ and $f(x)=mathbf{1}_{mathbb{Q} cap [0,1]}$.
Then $|f_n(x)|=|f(x)|$ and $f_n rightharpoonup f$ since
$$forall phi in L^q, int (f_n-f)phi=0,$$
as $f_n-f$ is non-zero over a set of $0$ measure.
answered Nov 27 at 10:27
nicomezi
4,0741820
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
1
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
add a comment |
1
Consider $f_n(x)=-mathbf{1}_{mathbb{Q} cap [0,1]}$ and $f(x)=mathbf{1}_{mathbb{Q} cap [0,1]}$.
Then $|f_n(x)|=|f(x)|$ and $f_n rightharpoonup f$ since
$$forall phi in L^q, int (f_n-f)phi=0,$$
as $f_n-f$ is non-zero over a set of $0$ measure.
answered Nov 27 at 10:27
nicomezi
4,0741820
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
1
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
add a comment |
1
1
1
Consider $f_n(x)=-mathbf{1}_{mathbb{Q} cap [0,1]}$ and $f(x)=mathbf{1}_{mathbb{Q} cap [0,1]}$.
Then $|f_n(x)|=|f(x)|$ and $f_n rightharpoonup f$ since
$$forall phi in L^q, int (f_n-f)phi=0,$$
as $f_n-f$ is non-zero over a set of $0$ measure.
answered Nov 27 at 10:27
nicomezi
4,0741820
Consider $f_n(x)=-mathbf{1}_{mathbb{Q} cap [0,1]}$ and $f(x)=mathbf{1}_{mathbb{Q} cap [0,1]}$.
Then $|f_n(x)|=|f(x)|$ and $f_n rightharpoonup f$ since
$$forall phi in L^q, int (f_n-f)phi=0,$$
as $f_n-f$ is non-zero over a set of $0$ measure.
answered Nov 27 at 10:27
nicomezi
4,0741820
answered Nov 27 at 10:27
nicomezi
4,0741820
answered Nov 27 at 10:27
nicomezi
4,0741820
answered Nov 27 at 10:27
nicomezi
4,0741820
4,0741820
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
1
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
add a comment |
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
1
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
Everything is basically the 0 function in your example I think.
– P.R
Nov 27 at 22:01
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
As an element of $L^p$ space yes. But pointwise convergence makes little sense on those spaces, that is what this answer is telling you. You need to make further assumptions if you want to avoid such cases.
– nicomezi
Nov 28 at 9:04
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
I think a more interesting example would be $f_n$ not converging to $f$ almost everywhere
– Sorin Tirc
Nov 28 at 9:13
1
1
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
It would be interesting indeed, since there is no such example. Indeed, it would mean that $f_n to -f$ on a set of non zero measure (where $f$ is non identically $0$). Hence you cannot have weak convergence of $f_n$ to $f$. Just consider $phi$ as the indicator function of this set. (or a part of it if is has infinite measure) @SorinTirc
– nicomezi
Nov 28 at 11:58
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
Yup you are right @nicomezi hats off for your solution in this case :D
– Sorin Tirc
Nov 28 at 14:30
add a comment |
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I'm not really familiar with weak convergence, so correct me if I'm wrong. If $f_n(x)=1-1/n$ and we fix $x_0$ to define $g_n(x)=f_n(x)$ and $g_n(x_0)=-f_n(x_0)$ then does $g_n$ converg weakly to the constant $1$? If so then that would be a counterexample.
– freakish
Nov 27 at 8:59
could you please hint the proof?
– P.R
Nov 27 at 10:05