If $T_nto T$ strongly and $x_nto x$ weakly, must $T_nx_nto Tx$ weakly?
$begingroup$
Let $X$ and $Y$ be Banach spaces, and let ${T_n}subset L(X,Y)$, where $L(X,Y)$ denotes the space of bounded linear operators from X to Y.If $T_nto T$ strongly and $x_nto x$ weakly, must $T_nx_nto Tx$ weakly?
For this question, we need to show for $forall gin Y^*$, $g(T_nx_n)to g(Tx)$. And I try to break it into two parts:
$|g(T_nx_n)- g(Tx)|leq|g(T_nx_n)-g(T_nx)|+|g(T_nx)-g(Tx)|$. The second piece will goes to zero as $gin Y^*$ and $T_nto T$ strongly. But I have difficulty to control the first piece, or any counter example?
functional-analysis convergence operator-theory banach-spaces weak-convergence
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
asked Jan 12 at 20:35
Irene YuIrene Yu
132
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be Banach spaces, and let ${T_n}subset L(X,Y)$, where $L(X,Y)$ denotes the space of bounded linear operators from X to Y.If $T_nto T$ strongly and $x_nto x$ weakly, must $T_nx_nto Tx$ weakly?
For this question, we need to show for $forall gin Y^*$, $g(T_nx_n)to g(Tx)$. And I try to break it into two parts:
$|g(T_nx_n)- g(Tx)|leq|g(T_nx_n)-g(T_nx)|+|g(T_nx)-g(Tx)|$. The second piece will goes to zero as $gin Y^*$ and $T_nto T$ strongly. But I have difficulty to control the first piece, or any counter example?
functional-analysis convergence operator-theory banach-spaces weak-convergence
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
asked Jan 12 at 20:35
Irene YuIrene Yu
132
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be Banach spaces, and let ${T_n}subset L(X,Y)$, where $L(X,Y)$ denotes the space of bounded linear operators from X to Y.If $T_nto T$ strongly and $x_nto x$ weakly, must $T_nx_nto Tx$ weakly?
For this question, we need to show for $forall gin Y^*$, $g(T_nx_n)to g(Tx)$. And I try to break it into two parts:
$|g(T_nx_n)- g(Tx)|leq|g(T_nx_n)-g(T_nx)|+|g(T_nx)-g(Tx)|$. The second piece will goes to zero as $gin Y^*$ and $T_nto T$ strongly. But I have difficulty to control the first piece, or any counter example?
functional-analysis convergence operator-theory banach-spaces weak-convergence
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
asked Jan 12 at 20:35
Irene YuIrene Yu
132
$endgroup$
Let $X$ and $Y$ be Banach spaces, and let ${T_n}subset L(X,Y)$, where $L(X,Y)$ denotes the space of bounded linear operators from X to Y.If $T_nto T$ strongly and $x_nto x$ weakly, must $T_nx_nto Tx$ weakly?
For this question, we need to show for $forall gin Y^*$, $g(T_nx_n)to g(Tx)$. And I try to break it into two parts:
$|g(T_nx_n)- g(Tx)|leq|g(T_nx_n)-g(T_nx)|+|g(T_nx)-g(Tx)|$. The second piece will goes to zero as $gin Y^*$ and $T_nto T$ strongly. But I have difficulty to control the first piece, or any counter example?
functional-analysis convergence operator-theory banach-spaces weak-convergence
functional-analysis convergence operator-theory banach-spaces weak-convergence
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
asked Jan 12 at 20:35
Irene YuIrene Yu
132
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
asked Jan 12 at 20:35
Irene YuIrene Yu
132
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
edited Jan 13 at 14:38
Davide Giraudo
126k16150261
126k16150261
asked Jan 12 at 20:35
Irene YuIrene Yu
132
asked Jan 12 at 20:35
Irene YuIrene Yu
132
asked Jan 12 at 20:35
Irene YuIrene Yu
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
The statement is not true. Take $X=Y=H=l^2(mathbb{N})$ as a Hilbert space. Let $(e_n)_{n=1}^infty$ be an orthonormal basis where $e_n$ is defined by
$
e_n(j) = delta_n(j).
$ Let $$
S(x_1,x_2,x_3,ldots,x_nldots)=(x_2,x_3,x_4,ldots)
$$ be the unilateral shift on $l^2$. Now, note that $$
|S^nx|^2=sum_{j>n}|x_j|^2to 0
$$ as $ntoinfty$ for all $x=(x_1,x_2,ldots)in l^2$, and that $e_{n+1}to 0$ weakly. This says that $T_n =S^n$ and $x_n = e_{n+1}$ satisfies the given assumption, however, $$
S^n e_{n+1}=e_1
$$ does not converge to $0$ weakly. (Note: But if we require $T_nto T$ in operator norm, then the conclusion is true.)
answered Jan 13 at 14:30
SongSong
11.1k628
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The statement is not true. Take $X=Y=H=l^2(mathbb{N})$ as a Hilbert space. Let $(e_n)_{n=1}^infty$ be an orthonormal basis where $e_n$ is defined by
$
e_n(j) = delta_n(j).
$ Let $$
S(x_1,x_2,x_3,ldots,x_nldots)=(x_2,x_3,x_4,ldots)
$$ be the unilateral shift on $l^2$. Now, note that $$
|S^nx|^2=sum_{j>n}|x_j|^2to 0
$$ as $ntoinfty$ for all $x=(x_1,x_2,ldots)in l^2$, and that $e_{n+1}to 0$ weakly. This says that $T_n =S^n$ and $x_n = e_{n+1}$ satisfies the given assumption, however, $$
S^n e_{n+1}=e_1
$$ does not converge to $0$ weakly. (Note: But if we require $T_nto T$ in operator norm, then the conclusion is true.)
answered Jan 13 at 14:30
SongSong
11.1k628
$endgroup$
add a comment |
$begingroup$
The statement is not true. Take $X=Y=H=l^2(mathbb{N})$ as a Hilbert space. Let $(e_n)_{n=1}^infty$ be an orthonormal basis where $e_n$ is defined by
$
e_n(j) = delta_n(j).
$ Let $$
S(x_1,x_2,x_3,ldots,x_nldots)=(x_2,x_3,x_4,ldots)
$$ be the unilateral shift on $l^2$. Now, note that $$
|S^nx|^2=sum_{j>n}|x_j|^2to 0
$$ as $ntoinfty$ for all $x=(x_1,x_2,ldots)in l^2$, and that $e_{n+1}to 0$ weakly. This says that $T_n =S^n$ and $x_n = e_{n+1}$ satisfies the given assumption, however, $$
S^n e_{n+1}=e_1
$$ does not converge to $0$ weakly. (Note: But if we require $T_nto T$ in operator norm, then the conclusion is true.)
answered Jan 13 at 14:30
SongSong
11.1k628
$endgroup$
add a comment |
$begingroup$
The statement is not true. Take $X=Y=H=l^2(mathbb{N})$ as a Hilbert space. Let $(e_n)_{n=1}^infty$ be an orthonormal basis where $e_n$ is defined by
$
e_n(j) = delta_n(j).
$ Let $$
S(x_1,x_2,x_3,ldots,x_nldots)=(x_2,x_3,x_4,ldots)
$$ be the unilateral shift on $l^2$. Now, note that $$
|S^nx|^2=sum_{j>n}|x_j|^2to 0
$$ as $ntoinfty$ for all $x=(x_1,x_2,ldots)in l^2$, and that $e_{n+1}to 0$ weakly. This says that $T_n =S^n$ and $x_n = e_{n+1}$ satisfies the given assumption, however, $$
S^n e_{n+1}=e_1
$$ does not converge to $0$ weakly. (Note: But if we require $T_nto T$ in operator norm, then the conclusion is true.)
answered Jan 13 at 14:30
SongSong
11.1k628
$endgroup$
The statement is not true. Take $X=Y=H=l^2(mathbb{N})$ as a Hilbert space. Let $(e_n)_{n=1}^infty$ be an orthonormal basis where $e_n$ is defined by
$
e_n(j) = delta_n(j).
$ Let $$
S(x_1,x_2,x_3,ldots,x_nldots)=(x_2,x_3,x_4,ldots)
$$ be the unilateral shift on $l^2$. Now, note that $$
|S^nx|^2=sum_{j>n}|x_j|^2to 0
$$ as $ntoinfty$ for all $x=(x_1,x_2,ldots)in l^2$, and that $e_{n+1}to 0$ weakly. This says that $T_n =S^n$ and $x_n = e_{n+1}$ satisfies the given assumption, however, $$
S^n e_{n+1}=e_1
$$ does not converge to $0$ weakly. (Note: But if we require $T_nto T$ in operator norm, then the conclusion is true.)
answered Jan 13 at 14:30
SongSong
11.1k628
answered Jan 13 at 14:30
SongSong
11.1k628
answered Jan 13 at 14:30
SongSong
11.1k628
answered Jan 13 at 14:30
SongSong
11.1k628
11.1k628
add a comment |
add a comment |
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