The limit of a strongly convergent sequence of linear bounded operators from a Banach space to a normed space
up vote
1
down vote
favorite
Let $X$ be a Banach space and $Y$ be a normed space.
If the sequence ${T_n}$ of bounded linear operators from $X$ into $Y$ is strongly convergent. Then there exists a bounded linear bounded operator $T:x rightarrow Y$ st
$lim_{nrightarrow infty} T_n(x)=T(x)$ $forall x in X$
The Proof.
I don't understand how the author deduced that $T$ is bounded.
why did he write $|Tx-T_nx| leq |T-T_n| |x| <epsilon$
all that we so far know about the operator $(T-T_n)$ is that it is a linear operator, it is not bounded so we can write this inequality $|Tx-T_nx| leq |T-T_n| |x| <epsilon$ furthermore he writes $|T-T_n| |x| <epsilon$ but the assumption said $T_n rightarrow $T$ strongly not uniformly.
I'm confused about this part
Can anyone help?
functional-analysis linear-transformations proof-explanation banach-spaces
asked Nov 20 at 0:53
HybridAlien
2008
add a comment |
up vote
1
down vote
favorite
Let $X$ be a Banach space and $Y$ be a normed space.
If the sequence ${T_n}$ of bounded linear operators from $X$ into $Y$ is strongly convergent. Then there exists a bounded linear bounded operator $T:x rightarrow Y$ st
$lim_{nrightarrow infty} T_n(x)=T(x)$ $forall x in X$
The Proof.
I don't understand how the author deduced that $T$ is bounded.
why did he write $|Tx-T_nx| leq |T-T_n| |x| <epsilon$
all that we so far know about the operator $(T-T_n)$ is that it is a linear operator, it is not bounded so we can write this inequality $|Tx-T_nx| leq |T-T_n| |x| <epsilon$ furthermore he writes $|T-T_n| |x| <epsilon$ but the assumption said $T_n rightarrow $T$ strongly not uniformly.
I'm confused about this part
Can anyone help?
functional-analysis linear-transformations proof-explanation banach-spaces
asked Nov 20 at 0:53
HybridAlien
2008
I'm confused, what notion of strong convergence is the author using? Usually we say a sequence of bounded operators converges strongly if it converges to a bounded operator in the strong operator topology. By this meaning, the claim is a tatology.
– Aweygan
Nov 20 at 2:56
Yes he defines strong convergence as you said. Let $X$ and $Y$ be normed spaces, and $T_n : X → Y$ and $T : X → Y$ are bounded linear operators. We say that: $T_n$ converges strongly to $T$ if $T_nx → T x$ for all $x ∈ X$
– HybridAlien
Nov 20 at 3:13
So the statement of the theorem is absurd.
– HybridAlien
Nov 20 at 3:18
1
I believe the theorem and proof (minus the errors already point out) work if you change "${T_n}$ is strongly convergent" to "For each $xin X$, ${T_nx}$ is convergent in $Y$". Then it's a standard exercise in a first functional analysis class.
– Aweygan
Nov 20 at 3:29
1
Regardless, if these aren't lecture notes for a class you're currently taking, I'd recommend switching to some other reference material. If these are lecture notes, I'd recommend supplementing them with some other standard references.
– Aweygan
Nov 20 at 3:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a Banach space and $Y$ be a normed space.
If the sequence ${T_n}$ of bounded linear operators from $X$ into $Y$ is strongly convergent. Then there exists a bounded linear bounded operator $T:x rightarrow Y$ st
$lim_{nrightarrow infty} T_n(x)=T(x)$ $forall x in X$
The Proof.
I don't understand how the author deduced that $T$ is bounded.
why did he write $|Tx-T_nx| leq |T-T_n| |x| <epsilon$
all that we so far know about the operator $(T-T_n)$ is that it is a linear operator, it is not bounded so we can write this inequality $|Tx-T_nx| leq |T-T_n| |x| <epsilon$ furthermore he writes $|T-T_n| |x| <epsilon$ but the assumption said $T_n rightarrow $T$ strongly not uniformly.
I'm confused about this part
Can anyone help?
functional-analysis linear-transformations proof-explanation banach-spaces
asked Nov 20 at 0:53
HybridAlien
2008
Let $X$ be a Banach space and $Y$ be a normed space.
If the sequence ${T_n}$ of bounded linear operators from $X$ into $Y$ is strongly convergent. Then there exists a bounded linear bounded operator $T:x rightarrow Y$ st
$lim_{nrightarrow infty} T_n(x)=T(x)$ $forall x in X$
The Proof.
I don't understand how the author deduced that $T$ is bounded.
why did he write $|Tx-T_nx| leq |T-T_n| |x| <epsilon$
all that we so far know about the operator $(T-T_n)$ is that it is a linear operator, it is not bounded so we can write this inequality $|Tx-T_nx| leq |T-T_n| |x| <epsilon$ furthermore he writes $|T-T_n| |x| <epsilon$ but the assumption said $T_n rightarrow $T$ strongly not uniformly.
I'm confused about this part
Can anyone help?
functional-analysis linear-transformations proof-explanation banach-spaces
functional-analysis linear-transformations proof-explanation banach-spaces
asked Nov 20 at 0:53
HybridAlien
2008
asked Nov 20 at 0:53
HybridAlien
2008
asked Nov 20 at 0:53
HybridAlien
2008
asked Nov 20 at 0:53
HybridAlien
2008
asked Nov 20 at 0:53
HybridAlien
2008
2008
I'm confused, what notion of strong convergence is the author using? Usually we say a sequence of bounded operators converges strongly if it converges to a bounded operator in the strong operator topology. By this meaning, the claim is a tatology.
– Aweygan
Nov 20 at 2:56
Yes he defines strong convergence as you said. Let $X$ and $Y$ be normed spaces, and $T_n : X → Y$ and $T : X → Y$ are bounded linear operators. We say that: $T_n$ converges strongly to $T$ if $T_nx → T x$ for all $x ∈ X$
– HybridAlien
Nov 20 at 3:13
So the statement of the theorem is absurd.
– HybridAlien
Nov 20 at 3:18
1
I believe the theorem and proof (minus the errors already point out) work if you change "${T_n}$ is strongly convergent" to "For each $xin X$, ${T_nx}$ is convergent in $Y$". Then it's a standard exercise in a first functional analysis class.
– Aweygan
Nov 20 at 3:29
1
Regardless, if these aren't lecture notes for a class you're currently taking, I'd recommend switching to some other reference material. If these are lecture notes, I'd recommend supplementing them with some other standard references.
– Aweygan
Nov 20 at 3:30
add a comment |
I'm confused, what notion of strong convergence is the author using? Usually we say a sequence of bounded operators converges strongly if it converges to a bounded operator in the strong operator topology. By this meaning, the claim is a tatology.
– Aweygan
Nov 20 at 2:56
Yes he defines strong convergence as you said. Let $X$ and $Y$ be normed spaces, and $T_n : X → Y$ and $T : X → Y$ are bounded linear operators. We say that: $T_n$ converges strongly to $T$ if $T_nx → T x$ for all $x ∈ X$
– HybridAlien
Nov 20 at 3:13
So the statement of the theorem is absurd.
– HybridAlien
Nov 20 at 3:18
1
I believe the theorem and proof (minus the errors already point out) work if you change "${T_n}$ is strongly convergent" to "For each $xin X$, ${T_nx}$ is convergent in $Y$". Then it's a standard exercise in a first functional analysis class.
– Aweygan
Nov 20 at 3:29
1
Regardless, if these aren't lecture notes for a class you're currently taking, I'd recommend switching to some other reference material. If these are lecture notes, I'd recommend supplementing them with some other standard references.
– Aweygan
Nov 20 at 3:30
I'm confused, what notion of strong convergence is the author using? Usually we say a sequence of bounded operators converges strongly if it converges to a bounded operator in the strong operator topology. By this meaning, the claim is a tatology.
– Aweygan
Nov 20 at 2:56
I'm confused, what notion of strong convergence is the author using? Usually we say a sequence of bounded operators converges strongly if it converges to a bounded operator in the strong operator topology. By this meaning, the claim is a tatology.
– Aweygan
Nov 20 at 2:56
Yes he defines strong convergence as you said. Let $X$ and $Y$ be normed spaces, and $T_n : X → Y$ and $T : X → Y$ are bounded linear operators. We say that: $T_n$ converges strongly to $T$ if $T_nx → T x$ for all $x ∈ X$
– HybridAlien
Nov 20 at 3:13
Yes he defines strong convergence as you said. Let $X$ and $Y$ be normed spaces, and $T_n : X → Y$ and $T : X → Y$ are bounded linear operators. We say that: $T_n$ converges strongly to $T$ if $T_nx → T x$ for all $x ∈ X$
– HybridAlien
Nov 20 at 3:13
So the statement of the theorem is absurd.
– HybridAlien
Nov 20 at 3:18
So the statement of the theorem is absurd.
– HybridAlien
Nov 20 at 3:18
1
1
I believe the theorem and proof (minus the errors already point out) work if you change "${T_n}$ is strongly convergent" to "For each $xin X$, ${T_nx}$ is convergent in $Y$". Then it's a standard exercise in a first functional analysis class.
– Aweygan
Nov 20 at 3:29
I believe the theorem and proof (minus the errors already point out) work if you change "${T_n}$ is strongly convergent" to "For each $xin X$, ${T_nx}$ is convergent in $Y$". Then it's a standard exercise in a first functional analysis class.
– Aweygan
Nov 20 at 3:29
1
1
Regardless, if these aren't lecture notes for a class you're currently taking, I'd recommend switching to some other reference material. If these are lecture notes, I'd recommend supplementing them with some other standard references.
– Aweygan
Nov 20 at 3:30
Regardless, if these aren't lecture notes for a class you're currently taking, I'd recommend switching to some other reference material. If these are lecture notes, I'd recommend supplementing them with some other standard references.
– Aweygan
Nov 20 at 3:30
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Me too I don't understand, you cannot want to show that an operator is bounded and in the argument, use the fact that it is bounded. Another argument: $T(x)=lim_nT_n(x)$ implies that $|T(x)|=lim_n|T_n(x)|leq |T_n||x|leq k|x|$. The $k$ is the $k$ you have defined by using the uniform boundedness principle.
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005777%2fthe-limit-of-a-strongly-convergent-sequence-of-linear-bounded-operators-from-a-b%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Me too I don't understand, you cannot want to show that an operator is bounded and in the argument, use the fact that it is bounded. Another argument: $T(x)=lim_nT_n(x)$ implies that $|T(x)|=lim_n|T_n(x)|leq |T_n||x|leq k|x|$. The $k$ is the $k$ you have defined by using the uniform boundedness principle.
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
add a comment |
up vote
0
down vote
Me too I don't understand, you cannot want to show that an operator is bounded and in the argument, use the fact that it is bounded. Another argument: $T(x)=lim_nT_n(x)$ implies that $|T(x)|=lim_n|T_n(x)|leq |T_n||x|leq k|x|$. The $k$ is the $k$ you have defined by using the uniform boundedness principle.
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
add a comment |
up vote
0
down vote
up vote
0
down vote
Me too I don't understand, you cannot want to show that an operator is bounded and in the argument, use the fact that it is bounded. Another argument: $T(x)=lim_nT_n(x)$ implies that $|T(x)|=lim_n|T_n(x)|leq |T_n||x|leq k|x|$. The $k$ is the $k$ you have defined by using the uniform boundedness principle.
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
Me too I don't understand, you cannot want to show that an operator is bounded and in the argument, use the fact that it is bounded. Another argument: $T(x)=lim_nT_n(x)$ implies that $|T(x)|=lim_n|T_n(x)|leq |T_n||x|leq k|x|$. The $k$ is the $k$ you have defined by using the uniform boundedness principle.
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
answered Nov 20 at 1:02
Tsemo Aristide
55.1k11444
55.1k11444
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
add a comment |
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
Your argument is flawed; the claim that "$lim_n|T_nx|leq|T_k||x|$ for some $k$" is not true in general.
– Aweygan
Nov 20 at 3:02
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005777%2fthe-limit-of-a-strongly-convergent-sequence-of-linear-bounded-operators-from-a-b%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I'm confused, what notion of strong convergence is the author using? Usually we say a sequence of bounded operators converges strongly if it converges to a bounded operator in the strong operator topology. By this meaning, the claim is a tatology.
– Aweygan
Nov 20 at 2:56
Yes he defines strong convergence as you said. Let $X$ and $Y$ be normed spaces, and $T_n : X → Y$ and $T : X → Y$ are bounded linear operators. We say that: $T_n$ converges strongly to $T$ if $T_nx → T x$ for all $x ∈ X$
– HybridAlien
Nov 20 at 3:13
So the statement of the theorem is absurd.
– HybridAlien
Nov 20 at 3:18
1
I believe the theorem and proof (minus the errors already point out) work if you change "${T_n}$ is strongly convergent" to "For each $xin X$, ${T_nx}$ is convergent in $Y$". Then it's a standard exercise in a first functional analysis class.
– Aweygan
Nov 20 at 3:29
1
Regardless, if these aren't lecture notes for a class you're currently taking, I'd recommend switching to some other reference material. If these are lecture notes, I'd recommend supplementing them with some other standard references.
– Aweygan
Nov 20 at 3:30