Continuity of Linear Operator Between Hilbert SpacesNorm of adjoint operator in Hilbert spaceSesquilinear...
Define, (actually define) the "stability" and "energy" of a compound
Why do Australian milk farmers need to protest supermarkets' milk price?
Do I need to be arrogant to get ahead?
Creature kill and resurrect effects on the stack interaction?
Can a druid choose the size of its wild shape beast?
In a future war, an old lady is trying to raise a boy but one of the weapons has made everyone deaf
How to use deus ex machina safely?
If the DM rolls initiative once for a group of monsters, how do end-of-turn effects work?
Identifying the interval from A♭ to D♯
Can I use USB data pins as power source
How to make healing in an exploration game interesting
Why would a flight no longer considered airworthy be redirected like this?
Most cost effective thermostat setting: consistent temperature vs. lowest temperature possible
My adviser wants to be the first author
Why doesn't using two cd commands in bash script execute the second command?
It's a yearly task, alright
Is it possible to upcast ritual spells?
Gantt Chart like rectangles with log scale
Min function accepting varying number of arguments in C++17
Co-worker team leader wants to inject his friend's awful software into our development. What should I say to our common boss?
Recruiter wants very extensive technical details about all of my previous work
How to deal with a cynical class?
If curse and magic is two sides of the same coin, why the former is forbidden?
What approach do we need to follow for projects without a test environment?
Continuity of Linear Operator Between Hilbert Spaces
Norm of adjoint operator in Hilbert spaceSesquilinear forms on Hilbert spacesGradient of inner product in Hilbert spaceDissipativity for Hilbert spacesA self-adjoint operator on a Hilbert spaceComplementary slackness in Hilbert spacesProof that every bounded linear operator between hilbert spaces has an adjoint.Proof explanation related to the operator matricesShowing that $exists x in H : |A(x)| = |A|_mathcal{L}$ if $H$ is Hilbert and $A in mathcal{L}_c(X,Y)$.Why is this operator symmetric? A question concerning a paper from Brezis and Crandall
$begingroup$
Note: Please do not give a solution; I am curious to understand why my solution is incorrect, and would prefer guidance to help me complete the question myself. Thank you.
Let $mathcal{H}$ be a Hilbert space, and suppose that $Tintext{Hom}(mathcal{H},mathcal{H})$. Suppose that there exists an operator $tilde{T}:mathcal{H}rightarrowmathcal{H}$ such that,
begin{align}
langle Tx,yrangle =langle x,tilde{T}yrangle,
end{align}
$forall x,yinmathcal{H}$. Show that $T$ is continuous.
My current solution is as follows:
Assume for all $delta>0$ there exists $n>Ninmathbb{N}$ such that,
begin{align}
|x_{n}-x|<delta.
end{align}
Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle &= |Tx_{n}-Tx|^{2}\
&leq|Tx_{n}-Tx|=|T(x_{n}-x)|\
&leq|T||x_{n}-x|rightarrow 0text{ as }nrightarrowinfty.
end{align}
What am I doing wrong? I notice I do not use the existence of $tilde{T}$.
Second Attempt:
Assume $langle x_{n},xrangle rightarrow langle x,xrangle$ as $nrightarrowinfty$. Then, given $langle Tx,yrangle = langle x,tilde{T}yrangle$,
begin{align}
langle Tx_{n},yrangle &= langle x_{n},tilde{T}yranglerightarrow_{nrightarrowinfty}langle x,tilde{T}yrangle=langle Tx,yrangle.
end{align}
Therefore, $Tx_{n}rightarrow Tx$ as $nrightarrowinfty$.
Third Attempt:
Assume $|x_{n}-x|rightarrow 0$ as $nrightarrowinfty$. Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle=langle x_{n}-x,x_{n}-xrangle=|x_{n}-x|^{2}.
end{align}
By assumption $|x_{n}-x|^{2}rightarrow 0$ as $nrightarrowinfty$. Hence,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle = |Tx_{n}-Tx|^{2}rightarrow 0text{ as }nrightarrowinfty.
end{align}
Therefore, $T$ is continuous.
functional-analysis continuity hilbert-spaces
asked 4 hours ago
JackJack
887
$endgroup$
add a comment |
$begingroup$
Note: Please do not give a solution; I am curious to understand why my solution is incorrect, and would prefer guidance to help me complete the question myself. Thank you.
Let $mathcal{H}$ be a Hilbert space, and suppose that $Tintext{Hom}(mathcal{H},mathcal{H})$. Suppose that there exists an operator $tilde{T}:mathcal{H}rightarrowmathcal{H}$ such that,
begin{align}
langle Tx,yrangle =langle x,tilde{T}yrangle,
end{align}
$forall x,yinmathcal{H}$. Show that $T$ is continuous.
My current solution is as follows:
Assume for all $delta>0$ there exists $n>Ninmathbb{N}$ such that,
begin{align}
|x_{n}-x|<delta.
end{align}
Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle &= |Tx_{n}-Tx|^{2}\
&leq|Tx_{n}-Tx|=|T(x_{n}-x)|\
&leq|T||x_{n}-x|rightarrow 0text{ as }nrightarrowinfty.
end{align}
What am I doing wrong? I notice I do not use the existence of $tilde{T}$.
Second Attempt:
Assume $langle x_{n},xrangle rightarrow langle x,xrangle$ as $nrightarrowinfty$. Then, given $langle Tx,yrangle = langle x,tilde{T}yrangle$,
begin{align}
langle Tx_{n},yrangle &= langle x_{n},tilde{T}yranglerightarrow_{nrightarrowinfty}langle x,tilde{T}yrangle=langle Tx,yrangle.
end{align}
Therefore, $Tx_{n}rightarrow Tx$ as $nrightarrowinfty$.
Third Attempt:
Assume $|x_{n}-x|rightarrow 0$ as $nrightarrowinfty$. Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle=langle x_{n}-x,x_{n}-xrangle=|x_{n}-x|^{2}.
end{align}
By assumption $|x_{n}-x|^{2}rightarrow 0$ as $nrightarrowinfty$. Hence,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle = |Tx_{n}-Tx|^{2}rightarrow 0text{ as }nrightarrowinfty.
end{align}
Therefore, $T$ is continuous.
functional-analysis continuity hilbert-spaces
asked 4 hours ago
JackJack
887
$endgroup$
2
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
4 hours ago
1
$begingroup$
Comment on the second attempt: you showed that $Tx_n to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying!
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
3 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
3 hours ago
add a comment |
$begingroup$
Note: Please do not give a solution; I am curious to understand why my solution is incorrect, and would prefer guidance to help me complete the question myself. Thank you.
Let $mathcal{H}$ be a Hilbert space, and suppose that $Tintext{Hom}(mathcal{H},mathcal{H})$. Suppose that there exists an operator $tilde{T}:mathcal{H}rightarrowmathcal{H}$ such that,
begin{align}
langle Tx,yrangle =langle x,tilde{T}yrangle,
end{align}
$forall x,yinmathcal{H}$. Show that $T$ is continuous.
My current solution is as follows:
Assume for all $delta>0$ there exists $n>Ninmathbb{N}$ such that,
begin{align}
|x_{n}-x|<delta.
end{align}
Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle &= |Tx_{n}-Tx|^{2}\
&leq|Tx_{n}-Tx|=|T(x_{n}-x)|\
&leq|T||x_{n}-x|rightarrow 0text{ as }nrightarrowinfty.
end{align}
What am I doing wrong? I notice I do not use the existence of $tilde{T}$.
Second Attempt:
Assume $langle x_{n},xrangle rightarrow langle x,xrangle$ as $nrightarrowinfty$. Then, given $langle Tx,yrangle = langle x,tilde{T}yrangle$,
begin{align}
langle Tx_{n},yrangle &= langle x_{n},tilde{T}yranglerightarrow_{nrightarrowinfty}langle x,tilde{T}yrangle=langle Tx,yrangle.
end{align}
Therefore, $Tx_{n}rightarrow Tx$ as $nrightarrowinfty$.
Third Attempt:
Assume $|x_{n}-x|rightarrow 0$ as $nrightarrowinfty$. Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle=langle x_{n}-x,x_{n}-xrangle=|x_{n}-x|^{2}.
end{align}
By assumption $|x_{n}-x|^{2}rightarrow 0$ as $nrightarrowinfty$. Hence,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle = |Tx_{n}-Tx|^{2}rightarrow 0text{ as }nrightarrowinfty.
end{align}
Therefore, $T$ is continuous.
functional-analysis continuity hilbert-spaces
asked 4 hours ago
JackJack
887
$endgroup$
Note: Please do not give a solution; I am curious to understand why my solution is incorrect, and would prefer guidance to help me complete the question myself. Thank you.
Let $mathcal{H}$ be a Hilbert space, and suppose that $Tintext{Hom}(mathcal{H},mathcal{H})$. Suppose that there exists an operator $tilde{T}:mathcal{H}rightarrowmathcal{H}$ such that,
begin{align}
langle Tx,yrangle =langle x,tilde{T}yrangle,
end{align}
$forall x,yinmathcal{H}$. Show that $T$ is continuous.
My current solution is as follows:
Assume for all $delta>0$ there exists $n>Ninmathbb{N}$ such that,
begin{align}
|x_{n}-x|<delta.
end{align}
Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle &= |Tx_{n}-Tx|^{2}\
&leq|Tx_{n}-Tx|=|T(x_{n}-x)|\
&leq|T||x_{n}-x|rightarrow 0text{ as }nrightarrowinfty.
end{align}
What am I doing wrong? I notice I do not use the existence of $tilde{T}$.
Second Attempt:
Assume $langle x_{n},xrangle rightarrow langle x,xrangle$ as $nrightarrowinfty$. Then, given $langle Tx,yrangle = langle x,tilde{T}yrangle$,
begin{align}
langle Tx_{n},yrangle &= langle x_{n},tilde{T}yranglerightarrow_{nrightarrowinfty}langle x,tilde{T}yrangle=langle Tx,yrangle.
end{align}
Therefore, $Tx_{n}rightarrow Tx$ as $nrightarrowinfty$.
Third Attempt:
Assume $|x_{n}-x|rightarrow 0$ as $nrightarrowinfty$. Then,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle=langle x_{n}-x,x_{n}-xrangle=|x_{n}-x|^{2}.
end{align}
By assumption $|x_{n}-x|^{2}rightarrow 0$ as $nrightarrowinfty$. Hence,
begin{align}
langle Tx_{n}-Tx,Tx_{n}-Txrangle = |Tx_{n}-Tx|^{2}rightarrow 0text{ as }nrightarrowinfty.
end{align}
Therefore, $T$ is continuous.
functional-analysis continuity hilbert-spaces
functional-analysis continuity hilbert-spaces
asked 4 hours ago
JackJack
887
asked 4 hours ago
JackJack
887
edited 3 hours ago
Jack
asked 4 hours ago
JackJack
887
asked 4 hours ago
JackJack
887
asked 4 hours ago
JackJack
887
887
2
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
4 hours ago
1
$begingroup$
Comment on the second attempt: you showed that $Tx_n to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying!
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
3 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
3 hours ago
add a comment |
2
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
4 hours ago
1
$begingroup$
Comment on the second attempt: you showed that $Tx_n to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying!
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
3 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
3 hours ago
2
2
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
4 hours ago
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
4 hours ago
1
1
$begingroup$
Comment on the second attempt: you showed that $Tx_n to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying!
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
Comment on the second attempt: you showed that $Tx_n to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying!
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
3 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
3 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
3 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The problem is that we can't assume that $T$ has a finite norm. Before we add that condition about having an adjoint map $tilde{T}$, we're simply assuming that $T$ is a linear map.
In fact, a linear map between normed vector spaces is continuous if and only if it has a finite operator norm. You assumed the statement we were trying to prove.
Second attempt: The assumption here should have been that $x_nto x$, as in the others. Then, yes, $langle Tx_n,yrangle to langle Tx,yrangle$ for each $y$. This is real progress. But, as stated in the comments, it's weak convergence rather than convergence in norm. Not quite there.
Third attempt: No, $langle Tu,Turangle$ is not equal to $langle u,urangle$ - it's equal to $langle u,tilde{T}Turangle$, and you don't know what $tilde{T}T$ does. This is not helpful.
All, right, lets go back to the attempt that made some progress. Are you familiar with the uniform boundedness principle? One consequence of that theorem is that any sequence of points in a Hilbert space that converges weakly is bounded. Can we use this to ensure that $T$ is a bounded operator?
answered 4 hours ago
jmerryjmerry
14.3k1629
$endgroup$
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
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',
autoActivateHeartbeat: false,
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%2f3149977%2fcontinuity-of-linear-operator-between-hilbert-spaces%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
$begingroup$
The problem is that we can't assume that $T$ has a finite norm. Before we add that condition about having an adjoint map $tilde{T}$, we're simply assuming that $T$ is a linear map.
In fact, a linear map between normed vector spaces is continuous if and only if it has a finite operator norm. You assumed the statement we were trying to prove.
Second attempt: The assumption here should have been that $x_nto x$, as in the others. Then, yes, $langle Tx_n,yrangle to langle Tx,yrangle$ for each $y$. This is real progress. But, as stated in the comments, it's weak convergence rather than convergence in norm. Not quite there.
Third attempt: No, $langle Tu,Turangle$ is not equal to $langle u,urangle$ - it's equal to $langle u,tilde{T}Turangle$, and you don't know what $tilde{T}T$ does. This is not helpful.
All, right, lets go back to the attempt that made some progress. Are you familiar with the uniform boundedness principle? One consequence of that theorem is that any sequence of points in a Hilbert space that converges weakly is bounded. Can we use this to ensure that $T$ is a bounded operator?
answered 4 hours ago
jmerryjmerry
14.3k1629
$endgroup$
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
add a comment |
$begingroup$
The problem is that we can't assume that $T$ has a finite norm. Before we add that condition about having an adjoint map $tilde{T}$, we're simply assuming that $T$ is a linear map.
In fact, a linear map between normed vector spaces is continuous if and only if it has a finite operator norm. You assumed the statement we were trying to prove.
Second attempt: The assumption here should have been that $x_nto x$, as in the others. Then, yes, $langle Tx_n,yrangle to langle Tx,yrangle$ for each $y$. This is real progress. But, as stated in the comments, it's weak convergence rather than convergence in norm. Not quite there.
Third attempt: No, $langle Tu,Turangle$ is not equal to $langle u,urangle$ - it's equal to $langle u,tilde{T}Turangle$, and you don't know what $tilde{T}T$ does. This is not helpful.
All, right, lets go back to the attempt that made some progress. Are you familiar with the uniform boundedness principle? One consequence of that theorem is that any sequence of points in a Hilbert space that converges weakly is bounded. Can we use this to ensure that $T$ is a bounded operator?
answered 4 hours ago
jmerryjmerry
14.3k1629
$endgroup$
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
add a comment |
$begingroup$
The problem is that we can't assume that $T$ has a finite norm. Before we add that condition about having an adjoint map $tilde{T}$, we're simply assuming that $T$ is a linear map.
In fact, a linear map between normed vector spaces is continuous if and only if it has a finite operator norm. You assumed the statement we were trying to prove.
Second attempt: The assumption here should have been that $x_nto x$, as in the others. Then, yes, $langle Tx_n,yrangle to langle Tx,yrangle$ for each $y$. This is real progress. But, as stated in the comments, it's weak convergence rather than convergence in norm. Not quite there.
Third attempt: No, $langle Tu,Turangle$ is not equal to $langle u,urangle$ - it's equal to $langle u,tilde{T}Turangle$, and you don't know what $tilde{T}T$ does. This is not helpful.
All, right, lets go back to the attempt that made some progress. Are you familiar with the uniform boundedness principle? One consequence of that theorem is that any sequence of points in a Hilbert space that converges weakly is bounded. Can we use this to ensure that $T$ is a bounded operator?
answered 4 hours ago
jmerryjmerry
14.3k1629
$endgroup$
The problem is that we can't assume that $T$ has a finite norm. Before we add that condition about having an adjoint map $tilde{T}$, we're simply assuming that $T$ is a linear map.
In fact, a linear map between normed vector spaces is continuous if and only if it has a finite operator norm. You assumed the statement we were trying to prove.
Second attempt: The assumption here should have been that $x_nto x$, as in the others. Then, yes, $langle Tx_n,yrangle to langle Tx,yrangle$ for each $y$. This is real progress. But, as stated in the comments, it's weak convergence rather than convergence in norm. Not quite there.
Third attempt: No, $langle Tu,Turangle$ is not equal to $langle u,urangle$ - it's equal to $langle u,tilde{T}Turangle$, and you don't know what $tilde{T}T$ does. This is not helpful.
All, right, lets go back to the attempt that made some progress. Are you familiar with the uniform boundedness principle? One consequence of that theorem is that any sequence of points in a Hilbert space that converges weakly is bounded. Can we use this to ensure that $T$ is a bounded operator?
answered 4 hours ago
jmerryjmerry
14.3k1629
edited 1 hour ago
answered 4 hours ago
jmerryjmerry
14.3k1629
answered 4 hours ago
jmerryjmerry
14.3k1629
answered 4 hours ago
jmerryjmerry
14.3k1629
14.3k1629
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
add a comment |
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
So is the idea for me to use $tilde{T}$ to cancel out the operator norm in my final inequality?
$endgroup$
– Jack
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
3 hours ago
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.
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%2f3149977%2fcontinuity-of-linear-operator-between-hilbert-spaces%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
2
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
4 hours ago
1
$begingroup$
Comment on the second attempt: you showed that $Tx_n to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying!
$endgroup$
– Umberto P.
3 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
3 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
3 hours ago