Characterization of boundedness of a linear operator under compactness
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The following is an exercise from An Introduction to Banach Space Theory by Robert E. Megginson.
Suppose that $T$ is a linear operator from a normed space $X$ into a normed space $Y$. Prove that the following are equivalent.
a) The operator $T$ is continuous.
b) The set $T(K)$ is a compact subset of $Y$ whenever $K$ is a compact subset of $X$.
c) The set $T(K)$ is a weakly compact subset of $Y$ whenever $K$ is a weakly compact subset of $X$.
The idea that I have is that, to prove that b) implies a), let $x_{n}rightarrow 0$, and consider the sequentially compact set $K={x_{n}:n=1,2,...}cup{0}$, then $K$ is also compact, and hence $T(K)={T(x_{n}):n=1,2,...}cup{0}$ is sequentially compact, it seems that we cannot say immediately that $T(x_{n})rightarrow 0$, this is where I got stuck.
To prove that a) and c) are equivalent, I have only the idea that norm-to-norm continuous is equivalent to weak-to-weak continuous, but what can I do next?
The space $X$ has no approximate property, so one cannot use some sort of approximated by compact operators to argue this exercise.
Edit:
Part of the answer has been addressed here. That is, by Mazur compactness theorem, weakly compact is equivalent to weakly sequentially compact, so the statement c) can be rephrased as the following one.
The set $T(K)$ is sequentially compact subset of $Y$ whenever $K$ is a weakly sequentially compact.
To show that c) implies a). Assume that some sequence ${x_{n}}$ is such that $x_{n}rightarrow 0$ but $|T(x_{n})|rightarrowinfty$, then the set $K={x_{n}:n=1,2,...}cup{0}$ is weakly sequentially compact which implies by assumption that $T(K)={T(x_{n}): n=1,2,...}cup{0}$ is also weakly sequentially compact. In particular, the sequence ${T(x_{n})}$ has a weakly convergent sequence and hence the subsequence is bounded, this contradicts that $|T(x_{n})|rightarrowinfty$.
In the same fashion, one proves that b) implies a).
To show that a) implies c). As I have said, $T$ is weak-tp-weak continuous. Assume that $K$ is a weakly compact subset of $X$ but $T(K)$ is not, then by sequential characterization again, one finds some sequence ${T(x_{n})}$ such that $T(x_{n})$ has no weakly convergent subsequence, where ${x_{n}}subseteq K$. But ${x_{n}}$ has a weakly convergent subsequence, say, ${x_{n_{k}}}$, then by weak-to-weak continuity of $T$, ${T(x_{n_{k}})}$ is weakly convergent, this is a contradiction.
In the same fashion, one proves that a) implies b).
functional-analysis operator-theory
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
$endgroup$
add a comment |
$begingroup$
The following is an exercise from An Introduction to Banach Space Theory by Robert E. Megginson.
Suppose that $T$ is a linear operator from a normed space $X$ into a normed space $Y$. Prove that the following are equivalent.
a) The operator $T$ is continuous.
b) The set $T(K)$ is a compact subset of $Y$ whenever $K$ is a compact subset of $X$.
c) The set $T(K)$ is a weakly compact subset of $Y$ whenever $K$ is a weakly compact subset of $X$.
The idea that I have is that, to prove that b) implies a), let $x_{n}rightarrow 0$, and consider the sequentially compact set $K={x_{n}:n=1,2,...}cup{0}$, then $K$ is also compact, and hence $T(K)={T(x_{n}):n=1,2,...}cup{0}$ is sequentially compact, it seems that we cannot say immediately that $T(x_{n})rightarrow 0$, this is where I got stuck.
To prove that a) and c) are equivalent, I have only the idea that norm-to-norm continuous is equivalent to weak-to-weak continuous, but what can I do next?
The space $X$ has no approximate property, so one cannot use some sort of approximated by compact operators to argue this exercise.
Edit:
Part of the answer has been addressed here. That is, by Mazur compactness theorem, weakly compact is equivalent to weakly sequentially compact, so the statement c) can be rephrased as the following one.
The set $T(K)$ is sequentially compact subset of $Y$ whenever $K$ is a weakly sequentially compact.
To show that c) implies a). Assume that some sequence ${x_{n}}$ is such that $x_{n}rightarrow 0$ but $|T(x_{n})|rightarrowinfty$, then the set $K={x_{n}:n=1,2,...}cup{0}$ is weakly sequentially compact which implies by assumption that $T(K)={T(x_{n}): n=1,2,...}cup{0}$ is also weakly sequentially compact. In particular, the sequence ${T(x_{n})}$ has a weakly convergent sequence and hence the subsequence is bounded, this contradicts that $|T(x_{n})|rightarrowinfty$.
In the same fashion, one proves that b) implies a).
To show that a) implies c). As I have said, $T$ is weak-tp-weak continuous. Assume that $K$ is a weakly compact subset of $X$ but $T(K)$ is not, then by sequential characterization again, one finds some sequence ${T(x_{n})}$ such that $T(x_{n})$ has no weakly convergent subsequence, where ${x_{n}}subseteq K$. But ${x_{n}}$ has a weakly convergent subsequence, say, ${x_{n_{k}}}$, then by weak-to-weak continuity of $T$, ${T(x_{n_{k}})}$ is weakly convergent, this is a contradiction.
In the same fashion, one proves that a) implies b).
functional-analysis operator-theory
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
$endgroup$
$begingroup$
What's the problem with $T(x_n) to 0$? If you assume $T$ to be continous and $x_n to 0$ then you get $T(x_n) to 0$.
$endgroup$
– eddie
Dec 28 '18 at 21:18
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Sorry, I wrote the wrong a) and b).
$endgroup$
– user284331
Dec 28 '18 at 21:19
$begingroup$
Anyway, it seems that I have figured out the solution.
$endgroup$
– user284331
Dec 28 '18 at 21:23
$begingroup$
Yes it's not necessary to show that $Tx_n to 0$, it sufficies to show that $Tx_n$ is bounded.
$endgroup$
– eddie
Dec 28 '18 at 21:32
add a comment |
$begingroup$
The following is an exercise from An Introduction to Banach Space Theory by Robert E. Megginson.
Suppose that $T$ is a linear operator from a normed space $X$ into a normed space $Y$. Prove that the following are equivalent.
a) The operator $T$ is continuous.
b) The set $T(K)$ is a compact subset of $Y$ whenever $K$ is a compact subset of $X$.
c) The set $T(K)$ is a weakly compact subset of $Y$ whenever $K$ is a weakly compact subset of $X$.
The idea that I have is that, to prove that b) implies a), let $x_{n}rightarrow 0$, and consider the sequentially compact set $K={x_{n}:n=1,2,...}cup{0}$, then $K$ is also compact, and hence $T(K)={T(x_{n}):n=1,2,...}cup{0}$ is sequentially compact, it seems that we cannot say immediately that $T(x_{n})rightarrow 0$, this is where I got stuck.
To prove that a) and c) are equivalent, I have only the idea that norm-to-norm continuous is equivalent to weak-to-weak continuous, but what can I do next?
The space $X$ has no approximate property, so one cannot use some sort of approximated by compact operators to argue this exercise.
Edit:
Part of the answer has been addressed here. That is, by Mazur compactness theorem, weakly compact is equivalent to weakly sequentially compact, so the statement c) can be rephrased as the following one.
The set $T(K)$ is sequentially compact subset of $Y$ whenever $K$ is a weakly sequentially compact.
To show that c) implies a). Assume that some sequence ${x_{n}}$ is such that $x_{n}rightarrow 0$ but $|T(x_{n})|rightarrowinfty$, then the set $K={x_{n}:n=1,2,...}cup{0}$ is weakly sequentially compact which implies by assumption that $T(K)={T(x_{n}): n=1,2,...}cup{0}$ is also weakly sequentially compact. In particular, the sequence ${T(x_{n})}$ has a weakly convergent sequence and hence the subsequence is bounded, this contradicts that $|T(x_{n})|rightarrowinfty$.
In the same fashion, one proves that b) implies a).
To show that a) implies c). As I have said, $T$ is weak-tp-weak continuous. Assume that $K$ is a weakly compact subset of $X$ but $T(K)$ is not, then by sequential characterization again, one finds some sequence ${T(x_{n})}$ such that $T(x_{n})$ has no weakly convergent subsequence, where ${x_{n}}subseteq K$. But ${x_{n}}$ has a weakly convergent subsequence, say, ${x_{n_{k}}}$, then by weak-to-weak continuity of $T$, ${T(x_{n_{k}})}$ is weakly convergent, this is a contradiction.
In the same fashion, one proves that a) implies b).
functional-analysis operator-theory
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
$endgroup$
The following is an exercise from An Introduction to Banach Space Theory by Robert E. Megginson.
Suppose that $T$ is a linear operator from a normed space $X$ into a normed space $Y$. Prove that the following are equivalent.
a) The operator $T$ is continuous.
b) The set $T(K)$ is a compact subset of $Y$ whenever $K$ is a compact subset of $X$.
c) The set $T(K)$ is a weakly compact subset of $Y$ whenever $K$ is a weakly compact subset of $X$.
The idea that I have is that, to prove that b) implies a), let $x_{n}rightarrow 0$, and consider the sequentially compact set $K={x_{n}:n=1,2,...}cup{0}$, then $K$ is also compact, and hence $T(K)={T(x_{n}):n=1,2,...}cup{0}$ is sequentially compact, it seems that we cannot say immediately that $T(x_{n})rightarrow 0$, this is where I got stuck.
To prove that a) and c) are equivalent, I have only the idea that norm-to-norm continuous is equivalent to weak-to-weak continuous, but what can I do next?
The space $X$ has no approximate property, so one cannot use some sort of approximated by compact operators to argue this exercise.
Edit:
Part of the answer has been addressed here. That is, by Mazur compactness theorem, weakly compact is equivalent to weakly sequentially compact, so the statement c) can be rephrased as the following one.
The set $T(K)$ is sequentially compact subset of $Y$ whenever $K$ is a weakly sequentially compact.
To show that c) implies a). Assume that some sequence ${x_{n}}$ is such that $x_{n}rightarrow 0$ but $|T(x_{n})|rightarrowinfty$, then the set $K={x_{n}:n=1,2,...}cup{0}$ is weakly sequentially compact which implies by assumption that $T(K)={T(x_{n}): n=1,2,...}cup{0}$ is also weakly sequentially compact. In particular, the sequence ${T(x_{n})}$ has a weakly convergent sequence and hence the subsequence is bounded, this contradicts that $|T(x_{n})|rightarrowinfty$.
In the same fashion, one proves that b) implies a).
To show that a) implies c). As I have said, $T$ is weak-tp-weak continuous. Assume that $K$ is a weakly compact subset of $X$ but $T(K)$ is not, then by sequential characterization again, one finds some sequence ${T(x_{n})}$ such that $T(x_{n})$ has no weakly convergent subsequence, where ${x_{n}}subseteq K$. But ${x_{n}}$ has a weakly convergent subsequence, say, ${x_{n_{k}}}$, then by weak-to-weak continuity of $T$, ${T(x_{n_{k}})}$ is weakly convergent, this is a contradiction.
In the same fashion, one proves that a) implies b).
functional-analysis operator-theory
functional-analysis operator-theory
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
edited Dec 28 '18 at 21:27
user284331
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
asked Dec 28 '18 at 20:39
user284331user284331
35.4k31646
35.4k31646
$begingroup$
What's the problem with $T(x_n) to 0$? If you assume $T$ to be continous and $x_n to 0$ then you get $T(x_n) to 0$.
$endgroup$
– eddie
Dec 28 '18 at 21:18
$begingroup$
Sorry, I wrote the wrong a) and b).
$endgroup$
– user284331
Dec 28 '18 at 21:19
$begingroup$
Anyway, it seems that I have figured out the solution.
$endgroup$
– user284331
Dec 28 '18 at 21:23
$begingroup$
Yes it's not necessary to show that $Tx_n to 0$, it sufficies to show that $Tx_n$ is bounded.
$endgroup$
– eddie
Dec 28 '18 at 21:32
add a comment |
$begingroup$
What's the problem with $T(x_n) to 0$? If you assume $T$ to be continous and $x_n to 0$ then you get $T(x_n) to 0$.
$endgroup$
– eddie
Dec 28 '18 at 21:18
$begingroup$
Sorry, I wrote the wrong a) and b).
$endgroup$
– user284331
Dec 28 '18 at 21:19
$begingroup$
Anyway, it seems that I have figured out the solution.
$endgroup$
– user284331
Dec 28 '18 at 21:23
$begingroup$
Yes it's not necessary to show that $Tx_n to 0$, it sufficies to show that $Tx_n$ is bounded.
$endgroup$
– eddie
Dec 28 '18 at 21:32
$begingroup$
What's the problem with $T(x_n) to 0$? If you assume $T$ to be continous and $x_n to 0$ then you get $T(x_n) to 0$.
$endgroup$
– eddie
Dec 28 '18 at 21:18
$begingroup$
What's the problem with $T(x_n) to 0$? If you assume $T$ to be continous and $x_n to 0$ then you get $T(x_n) to 0$.
$endgroup$
– eddie
Dec 28 '18 at 21:18
$begingroup$
Sorry, I wrote the wrong a) and b).
$endgroup$
– user284331
Dec 28 '18 at 21:19
$begingroup$
Sorry, I wrote the wrong a) and b).
$endgroup$
– user284331
Dec 28 '18 at 21:19
$begingroup$
Anyway, it seems that I have figured out the solution.
$endgroup$
– user284331
Dec 28 '18 at 21:23
$begingroup$
Anyway, it seems that I have figured out the solution.
$endgroup$
– user284331
Dec 28 '18 at 21:23
$begingroup$
Yes it's not necessary to show that $Tx_n to 0$, it sufficies to show that $Tx_n$ is bounded.
$endgroup$
– eddie
Dec 28 '18 at 21:32
$begingroup$
Yes it's not necessary to show that $Tx_n to 0$, it sufficies to show that $Tx_n$ is bounded.
$endgroup$
– eddie
Dec 28 '18 at 21:32
add a comment |
1 Answer
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Proof of b) implies a): if $T$ is not continuous there exists ${y_n}$ such that $|Ty_n| >n^{2}|y_n|$. Let $x_n =frac {y_n} {n |y_n|}$. Then $x_n to 0$ and $|T(x_n)|>n$. Hence ${0,x_1,x_2,cdots}$ is a compact set whose image is not bounded.
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
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add a comment |
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$begingroup$
Proof of b) implies a): if $T$ is not continuous there exists ${y_n}$ such that $|Ty_n| >n^{2}|y_n|$. Let $x_n =frac {y_n} {n |y_n|}$. Then $x_n to 0$ and $|T(x_n)|>n$. Hence ${0,x_1,x_2,cdots}$ is a compact set whose image is not bounded.
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Proof of b) implies a): if $T$ is not continuous there exists ${y_n}$ such that $|Ty_n| >n^{2}|y_n|$. Let $x_n =frac {y_n} {n |y_n|}$. Then $x_n to 0$ and $|T(x_n)|>n$. Hence ${0,x_1,x_2,cdots}$ is a compact set whose image is not bounded.
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Proof of b) implies a): if $T$ is not continuous there exists ${y_n}$ such that $|Ty_n| >n^{2}|y_n|$. Let $x_n =frac {y_n} {n |y_n|}$. Then $x_n to 0$ and $|T(x_n)|>n$. Hence ${0,x_1,x_2,cdots}$ is a compact set whose image is not bounded.
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
Proof of b) implies a): if $T$ is not continuous there exists ${y_n}$ such that $|Ty_n| >n^{2}|y_n|$. Let $x_n =frac {y_n} {n |y_n|}$. Then $x_n to 0$ and $|T(x_n)|>n$. Hence ${0,x_1,x_2,cdots}$ is a compact set whose image is not bounded.
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Dec 28 '18 at 23:27
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
74.8k53270
add a comment |
add a comment |
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$begingroup$
What's the problem with $T(x_n) to 0$? If you assume $T$ to be continous and $x_n to 0$ then you get $T(x_n) to 0$.
$endgroup$
– eddie
Dec 28 '18 at 21:18
$begingroup$
Sorry, I wrote the wrong a) and b).
$endgroup$
– user284331
Dec 28 '18 at 21:19
$begingroup$
Anyway, it seems that I have figured out the solution.
$endgroup$
– user284331
Dec 28 '18 at 21:23
$begingroup$
Yes it's not necessary to show that $Tx_n to 0$, it sufficies to show that $Tx_n$ is bounded.
$endgroup$
– eddie
Dec 28 '18 at 21:32