A sequence of norm 1 converging weakly under weakly compact operator
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Let $X,Y$ be Banach spaces and $T : X rightarrow Y$ a bounded operator that is weakly compact.
How to prove that there exists $epsilon > 0$ and a sequence $(x_n)_n subset X$ such that $|x_n| = 1, |T(x_n)| geq epsilon$ forall $n in mathbb N$, and the sequence $(T(x_n))_n$ is congerging weakly to $0$ ?
All what I can say is that I can extract a subsequence from $(Tx_n)_n$ which converges to a $y in Y$.
Thanks for any help
functional-analysis
asked Aug 20 at 13:01
Kébir J
255
add a comment |Â
up vote
1
down vote
favorite
Let $X,Y$ be Banach spaces and $T : X rightarrow Y$ a bounded operator that is weakly compact.
How to prove that there exists $epsilon > 0$ and a sequence $(x_n)_n subset X$ such that $|x_n| = 1, |T(x_n)| geq epsilon$ forall $n in mathbb N$, and the sequence $(T(x_n))_n$ is congerging weakly to $0$ ?
All what I can say is that I can extract a subsequence from $(Tx_n)_n$ which converges to a $y in Y$.
Thanks for any help
functional-analysis
asked Aug 20 at 13:01
Kébir J
255
I'm not sure you always can. In $Y=ell_1$, a sequence is weakly null if and only if it is norm null.
– David Mitra
Aug 20 at 13:09
Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true.
– Kébir J
Aug 20 at 13:50
There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind?
– David Mitra
Aug 20 at 13:59
I suppose I can extract a subsequence for which $|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent.
– Kébir J
Aug 20 at 14:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X,Y$ be Banach spaces and $T : X rightarrow Y$ a bounded operator that is weakly compact.
How to prove that there exists $epsilon > 0$ and a sequence $(x_n)_n subset X$ such that $|x_n| = 1, |T(x_n)| geq epsilon$ forall $n in mathbb N$, and the sequence $(T(x_n))_n$ is congerging weakly to $0$ ?
All what I can say is that I can extract a subsequence from $(Tx_n)_n$ which converges to a $y in Y$.
Thanks for any help
functional-analysis
asked Aug 20 at 13:01
Kébir J
255
Let $X,Y$ be Banach spaces and $T : X rightarrow Y$ a bounded operator that is weakly compact.
How to prove that there exists $epsilon > 0$ and a sequence $(x_n)_n subset X$ such that $|x_n| = 1, |T(x_n)| geq epsilon$ forall $n in mathbb N$, and the sequence $(T(x_n))_n$ is congerging weakly to $0$ ?
All what I can say is that I can extract a subsequence from $(Tx_n)_n$ which converges to a $y in Y$.
Thanks for any help
functional-analysis
asked Aug 20 at 13:01
Kébir J
255
asked Aug 20 at 13:01
Kébir J
255
asked Aug 20 at 13:01
Kébir J
255
asked Aug 20 at 13:01
Kébir J
255
255
I'm not sure you always can. In $Y=ell_1$, a sequence is weakly null if and only if it is norm null.
– David Mitra
Aug 20 at 13:09
Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true.
– Kébir J
Aug 20 at 13:50
There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind?
– David Mitra
Aug 20 at 13:59
I suppose I can extract a subsequence for which $|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent.
– Kébir J
Aug 20 at 14:23
add a comment |Â
I'm not sure you always can. In $Y=ell_1$, a sequence is weakly null if and only if it is norm null.
– David Mitra
Aug 20 at 13:09
Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true.
– Kébir J
Aug 20 at 13:50
There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind?
– David Mitra
Aug 20 at 13:59
I suppose I can extract a subsequence for which $|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent.
– Kébir J
Aug 20 at 14:23
I'm not sure you always can. In $Y=ell_1$, a sequence is weakly null if and only if it is norm null.
– David Mitra
Aug 20 at 13:09
I'm not sure you always can. In $Y=ell_1$, a sequence is weakly null if and only if it is norm null.
– David Mitra
Aug 20 at 13:09
Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true.
– Kébir J
Aug 20 at 13:50
Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true.
– Kébir J
Aug 20 at 13:50
There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind?
– David Mitra
Aug 20 at 13:59
There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind?
– David Mitra
Aug 20 at 13:59
I suppose I can extract a subsequence for which $|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent.
– Kébir J
Aug 20 at 14:23
I suppose I can extract a subsequence for which $|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent.
– Kébir J
Aug 20 at 14:23
add a comment |Â
1 Answer
1
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1
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As in the paper you referenced in the comments, we'll assume $T$ is not compact.
Then, we may (and do) choose a sequence $(x_n)_n$ in $S_X$ such that
no subsequence of $(Tx_n)_n$ is norm convergent. Now choose a subsequence $(x_n_k)_k$ of $(x_n)_n$ so that $(T x_n_k)_k$ is weakly convergent.
Now, $(T x_n_k)_k$ is not norm-Cauchy. So we find $epsilon>0$ and subsequences $(p_k)$ and $(q_k)$ of $(n_k)$ with
$$Vert T( x_p_k- x_q_k)Vert>epsilon,quad k=1,2,ldots.tag1$$
Set $y_k= x_p_k- x_q_k$.
Then $(Ty_k)$ is weakly null (since $(T(x_n_k))_k$ is weakly convergent).
Since $T$ is bounded and since (1) holds, it follows that $(Vert y_kVert)_k$ is bounded away from $0$. From this, it follows that $bigl(T( y_k/Vert y_kVert)bigr)_k$ is weakly null. Appealing to (1) again, $biglVert T(y_k/Vert y_k Vert)bigrVert>epsilon/2 $ for each $k$.
So, $(y_k/Vert y_kVert)$ is the sought after sequence.
answered Aug 20 at 15:52
David Mitra
61.9k694157
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As in the paper you referenced in the comments, we'll assume $T$ is not compact.
Then, we may (and do) choose a sequence $(x_n)_n$ in $S_X$ such that
no subsequence of $(Tx_n)_n$ is norm convergent. Now choose a subsequence $(x_n_k)_k$ of $(x_n)_n$ so that $(T x_n_k)_k$ is weakly convergent.
Now, $(T x_n_k)_k$ is not norm-Cauchy. So we find $epsilon>0$ and subsequences $(p_k)$ and $(q_k)$ of $(n_k)$ with
$$Vert T( x_p_k- x_q_k)Vert>epsilon,quad k=1,2,ldots.tag1$$
Set $y_k= x_p_k- x_q_k$.
Then $(Ty_k)$ is weakly null (since $(T(x_n_k))_k$ is weakly convergent).
Since $T$ is bounded and since (1) holds, it follows that $(Vert y_kVert)_k$ is bounded away from $0$. From this, it follows that $bigl(T( y_k/Vert y_kVert)bigr)_k$ is weakly null. Appealing to (1) again, $biglVert T(y_k/Vert y_k Vert)bigrVert>epsilon/2 $ for each $k$.
So, $(y_k/Vert y_kVert)$ is the sought after sequence.
answered Aug 20 at 15:52
David Mitra
61.9k694157
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
add a comment |Â
up vote
1
down vote
accepted
As in the paper you referenced in the comments, we'll assume $T$ is not compact.
Then, we may (and do) choose a sequence $(x_n)_n$ in $S_X$ such that
no subsequence of $(Tx_n)_n$ is norm convergent. Now choose a subsequence $(x_n_k)_k$ of $(x_n)_n$ so that $(T x_n_k)_k$ is weakly convergent.
Now, $(T x_n_k)_k$ is not norm-Cauchy. So we find $epsilon>0$ and subsequences $(p_k)$ and $(q_k)$ of $(n_k)$ with
$$Vert T( x_p_k- x_q_k)Vert>epsilon,quad k=1,2,ldots.tag1$$
Set $y_k= x_p_k- x_q_k$.
Then $(Ty_k)$ is weakly null (since $(T(x_n_k))_k$ is weakly convergent).
Since $T$ is bounded and since (1) holds, it follows that $(Vert y_kVert)_k$ is bounded away from $0$. From this, it follows that $bigl(T( y_k/Vert y_kVert)bigr)_k$ is weakly null. Appealing to (1) again, $biglVert T(y_k/Vert y_k Vert)bigrVert>epsilon/2 $ for each $k$.
So, $(y_k/Vert y_kVert)$ is the sought after sequence.
answered Aug 20 at 15:52
David Mitra
61.9k694157
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As in the paper you referenced in the comments, we'll assume $T$ is not compact.
Then, we may (and do) choose a sequence $(x_n)_n$ in $S_X$ such that
no subsequence of $(Tx_n)_n$ is norm convergent. Now choose a subsequence $(x_n_k)_k$ of $(x_n)_n$ so that $(T x_n_k)_k$ is weakly convergent.
Now, $(T x_n_k)_k$ is not norm-Cauchy. So we find $epsilon>0$ and subsequences $(p_k)$ and $(q_k)$ of $(n_k)$ with
$$Vert T( x_p_k- x_q_k)Vert>epsilon,quad k=1,2,ldots.tag1$$
Set $y_k= x_p_k- x_q_k$.
Then $(Ty_k)$ is weakly null (since $(T(x_n_k))_k$ is weakly convergent).
Since $T$ is bounded and since (1) holds, it follows that $(Vert y_kVert)_k$ is bounded away from $0$. From this, it follows that $bigl(T( y_k/Vert y_kVert)bigr)_k$ is weakly null. Appealing to (1) again, $biglVert T(y_k/Vert y_k Vert)bigrVert>epsilon/2 $ for each $k$.
So, $(y_k/Vert y_kVert)$ is the sought after sequence.
answered Aug 20 at 15:52
David Mitra
61.9k694157
As in the paper you referenced in the comments, we'll assume $T$ is not compact.
Then, we may (and do) choose a sequence $(x_n)_n$ in $S_X$ such that
no subsequence of $(Tx_n)_n$ is norm convergent. Now choose a subsequence $(x_n_k)_k$ of $(x_n)_n$ so that $(T x_n_k)_k$ is weakly convergent.
Now, $(T x_n_k)_k$ is not norm-Cauchy. So we find $epsilon>0$ and subsequences $(p_k)$ and $(q_k)$ of $(n_k)$ with
$$Vert T( x_p_k- x_q_k)Vert>epsilon,quad k=1,2,ldots.tag1$$
Set $y_k= x_p_k- x_q_k$.
Then $(Ty_k)$ is weakly null (since $(T(x_n_k))_k$ is weakly convergent).
Since $T$ is bounded and since (1) holds, it follows that $(Vert y_kVert)_k$ is bounded away from $0$. From this, it follows that $bigl(T( y_k/Vert y_kVert)bigr)_k$ is weakly null. Appealing to (1) again, $biglVert T(y_k/Vert y_k Vert)bigrVert>epsilon/2 $ for each $k$.
So, $(y_k/Vert y_kVert)$ is the sought after sequence.
answered Aug 20 at 15:52
David Mitra
61.9k694157
answered Aug 20 at 15:52
David Mitra
61.9k694157
answered Aug 20 at 15:52
David Mitra
61.9k694157
answered Aug 20 at 15:52
David Mitra
61.9k694157
61.9k694157
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
add a comment |Â
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
Thank you. I was very very far from this.
– Kébir J
Aug 20 at 16:48
add a comment |Â
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I'm not sure you always can. In $Y=ell_1$, a sequence is weakly null if and only if it is norm null.
– David Mitra
Aug 20 at 13:09
Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true.
– Kébir J
Aug 20 at 13:50
There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind?
– David Mitra
Aug 20 at 13:59
I suppose I can extract a subsequence for which $|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent.
– Kébir J
Aug 20 at 14:23