Characterization of weak convergence in $ell^p$
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1
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I am a tutor for functional analysis and the students are supposed to show the here to be found characterization for weak convergence in $ell^p$, $1<p<infty$.
I am fully aware of a correct solution but I found the following approach and did not see whether this is fixable.
Let $varphi in (ell^p)'simeq ell^q$ be arbitrary.
Then we can write $varphi(z)=sum_{i=1}^infty z^{(i)}y^{(i)}$ for suitable $y=(y^{(i)})in ell^q$.
Then we have (EDIT:Note that $(x_n)$ is bounded in $ell^p$):
$$
lim_{nto infty} varphi(x_n)=lim_{nto infty} sum_{i=1}^infty x_n^{(i)}y^{(i)}.
$$
Now, the student is trying get the limit inside of the integral, saying that the sum converges uniformly and the limit is bounded independently of $n$.
Nevertheless, I am not aware of a theorem which would allow use, moreover, the counterexample for the case $p=1$ shows that for general $p$ this is not valid.
Now, my question is whether we can fix this for $p>1$. The first thing that came to my mind was to try to apply dominated convergence but I did not manage to do so.
I am happy for any insights also for showing that this cannot not be fixed.
sequences-and-series functional-analysis lp-spaces
asked Nov 24 at 16:28
Jonas Lenz
505212
add a comment |
up vote
1
down vote
favorite
I am a tutor for functional analysis and the students are supposed to show the here to be found characterization for weak convergence in $ell^p$, $1<p<infty$.
I am fully aware of a correct solution but I found the following approach and did not see whether this is fixable.
Let $varphi in (ell^p)'simeq ell^q$ be arbitrary.
Then we can write $varphi(z)=sum_{i=1}^infty z^{(i)}y^{(i)}$ for suitable $y=(y^{(i)})in ell^q$.
Then we have (EDIT:Note that $(x_n)$ is bounded in $ell^p$):
$$
lim_{nto infty} varphi(x_n)=lim_{nto infty} sum_{i=1}^infty x_n^{(i)}y^{(i)}.
$$
Now, the student is trying get the limit inside of the integral, saying that the sum converges uniformly and the limit is bounded independently of $n$.
Nevertheless, I am not aware of a theorem which would allow use, moreover, the counterexample for the case $p=1$ shows that for general $p$ this is not valid.
Now, my question is whether we can fix this for $p>1$. The first thing that came to my mind was to try to apply dominated convergence but I did not manage to do so.
I am happy for any insights also for showing that this cannot not be fixed.
sequences-and-series functional-analysis lp-spaces
asked Nov 24 at 16:28
Jonas Lenz
505212
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am a tutor for functional analysis and the students are supposed to show the here to be found characterization for weak convergence in $ell^p$, $1<p<infty$.
I am fully aware of a correct solution but I found the following approach and did not see whether this is fixable.
Let $varphi in (ell^p)'simeq ell^q$ be arbitrary.
Then we can write $varphi(z)=sum_{i=1}^infty z^{(i)}y^{(i)}$ for suitable $y=(y^{(i)})in ell^q$.
Then we have (EDIT:Note that $(x_n)$ is bounded in $ell^p$):
$$
lim_{nto infty} varphi(x_n)=lim_{nto infty} sum_{i=1}^infty x_n^{(i)}y^{(i)}.
$$
Now, the student is trying get the limit inside of the integral, saying that the sum converges uniformly and the limit is bounded independently of $n$.
Nevertheless, I am not aware of a theorem which would allow use, moreover, the counterexample for the case $p=1$ shows that for general $p$ this is not valid.
Now, my question is whether we can fix this for $p>1$. The first thing that came to my mind was to try to apply dominated convergence but I did not manage to do so.
I am happy for any insights also for showing that this cannot not be fixed.
sequences-and-series functional-analysis lp-spaces
asked Nov 24 at 16:28
Jonas Lenz
505212
I am a tutor for functional analysis and the students are supposed to show the here to be found characterization for weak convergence in $ell^p$, $1<p<infty$.
I am fully aware of a correct solution but I found the following approach and did not see whether this is fixable.
Let $varphi in (ell^p)'simeq ell^q$ be arbitrary.
Then we can write $varphi(z)=sum_{i=1}^infty z^{(i)}y^{(i)}$ for suitable $y=(y^{(i)})in ell^q$.
Then we have (EDIT:Note that $(x_n)$ is bounded in $ell^p$):
$$
lim_{nto infty} varphi(x_n)=lim_{nto infty} sum_{i=1}^infty x_n^{(i)}y^{(i)}.
$$
Now, the student is trying get the limit inside of the integral, saying that the sum converges uniformly and the limit is bounded independently of $n$.
Nevertheless, I am not aware of a theorem which would allow use, moreover, the counterexample for the case $p=1$ shows that for general $p$ this is not valid.
Now, my question is whether we can fix this for $p>1$. The first thing that came to my mind was to try to apply dominated convergence but I did not manage to do so.
I am happy for any insights also for showing that this cannot not be fixed.
sequences-and-series functional-analysis lp-spaces
sequences-and-series functional-analysis lp-spaces
asked Nov 24 at 16:28
Jonas Lenz
505212
asked Nov 24 at 16:28
Jonas Lenz
505212
edited Nov 24 at 18:30
asked Nov 24 at 16:28
Jonas Lenz
505212
asked Nov 24 at 16:28
Jonas Lenz
505212
asked Nov 24 at 16:28
Jonas Lenz
505212
505212
add a comment |
add a comment |
1 Answer
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If the argument would be valid, it would allow to show that the sequence of sequences $x_n = n , e_n$, where $e_n$ is the n'th unit sequence, would converge weakly to zero in $ell^p$.
Indeed, $lim_{ntoinfty} x_n^{(i) }= 0$ for all fixed $n$. Thus, the wrong argument would imply
$$lim_{ntoinfty}varphi(x_n) = 0.$$
However, $x_n$ cannot converge weakly since it is not bounded.
answered Nov 24 at 17:25
gerw
19k11133
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
add a comment |
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If the argument would be valid, it would allow to show that the sequence of sequences $x_n = n , e_n$, where $e_n$ is the n'th unit sequence, would converge weakly to zero in $ell^p$.
Indeed, $lim_{ntoinfty} x_n^{(i) }= 0$ for all fixed $n$. Thus, the wrong argument would imply
$$lim_{ntoinfty}varphi(x_n) = 0.$$
However, $x_n$ cannot converge weakly since it is not bounded.
answered Nov 24 at 17:25
gerw
19k11133
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
add a comment |
up vote
0
down vote
If the argument would be valid, it would allow to show that the sequence of sequences $x_n = n , e_n$, where $e_n$ is the n'th unit sequence, would converge weakly to zero in $ell^p$.
Indeed, $lim_{ntoinfty} x_n^{(i) }= 0$ for all fixed $n$. Thus, the wrong argument would imply
$$lim_{ntoinfty}varphi(x_n) = 0.$$
However, $x_n$ cannot converge weakly since it is not bounded.
answered Nov 24 at 17:25
gerw
19k11133
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
add a comment |
up vote
0
down vote
up vote
0
down vote
If the argument would be valid, it would allow to show that the sequence of sequences $x_n = n , e_n$, where $e_n$ is the n'th unit sequence, would converge weakly to zero in $ell^p$.
Indeed, $lim_{ntoinfty} x_n^{(i) }= 0$ for all fixed $n$. Thus, the wrong argument would imply
$$lim_{ntoinfty}varphi(x_n) = 0.$$
However, $x_n$ cannot converge weakly since it is not bounded.
answered Nov 24 at 17:25
gerw
19k11133
If the argument would be valid, it would allow to show that the sequence of sequences $x_n = n , e_n$, where $e_n$ is the n'th unit sequence, would converge weakly to zero in $ell^p$.
Indeed, $lim_{ntoinfty} x_n^{(i) }= 0$ for all fixed $n$. Thus, the wrong argument would imply
$$lim_{ntoinfty}varphi(x_n) = 0.$$
However, $x_n$ cannot converge weakly since it is not bounded.
answered Nov 24 at 17:25
gerw
19k11133
answered Nov 24 at 17:25
gerw
19k11133
answered Nov 24 at 17:25
gerw
19k11133
answered Nov 24 at 17:25
gerw
19k11133
19k11133
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
add a comment |
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
Thanks, but we have a boundedness condition on $(x_n)$. So we already know that the sequence $(x_n)$ is bounded. Maybe I should have written this explicitly.
– Jonas Lenz
Nov 24 at 18:15
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
But then, it is not clear what exactly you are asking. Jokingly, we could do the following: (DANGER: circlurar reasoning): The assumptions on the sequence $x_n$ implies the weak convergence $x_n rightharpoonup x$. Hence, $sum_i lim_n x_n^i y^i = varphi(x) = lim_n varphi(x_n) = lim_n sum_i x_n^i y^i$. Thus, we can interchange the sum and the limit.
– gerw
Nov 24 at 18:24
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I am aware of the fact, that the assumptions imply the weak convergence. My question is about whether we can fix the above approach to try to get the limit inside the sum by using some nice argument for $1<p<infty$.
– Jonas Lenz
Nov 24 at 18:30
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I think you misunderstood my comment. What I tried to say was the following: You can fix the argument (but this fix is totally silly!) by first showing $x_n rightharpoonup x$ (by using other arguments). Then, you can use this weak convergence to interchange the sum and the limit and, thus, you arrive (again!) at the weak convergence $x_n rightharpoonup x$. In this spirit: In the above situation, one can interchange the sum and the limit (and this is equivalent to the weak convergence), but the student did fail to give the correct argument.
– gerw
Nov 24 at 18:34
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
I see that point. Then I probably formulated my question badly. I am interested in knowing whether it is possible to get the limit inside without showing weak convergence first and then saying, ah now that it converges weakly, we get the limit inside. Or in other words, is there a way that the above approach helps to show weak convergence.
– Jonas Lenz
Nov 24 at 18:39
add a comment |
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