Continuity of Linear Operator Between Hilbert SpacesNorm of adjoint operator in Hilbert spaceSesquilinear...
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Continuity of Linear Operator Between Hilbert Spaces
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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 5 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 5 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
5 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.
4 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
4 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
4 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 5 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 5 hours ago
JackJack
887
asked 5 hours ago
JackJack
887
edited 4 hours ago
Jack
asked 5 hours ago
JackJack
887
asked 5 hours ago
JackJack
887
asked 5 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
5 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.
4 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
4 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
4 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
5 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.
4 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
4 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
4 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
5 hours ago
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
$endgroup$
– Andres Mejia
5 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.
4 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.
4 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
4 hours ago
$begingroup$
Thank you. Do you have a hint?
$endgroup$
– Jack
4 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
4 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
$endgroup$
– Jack
4 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 5 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.
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
$endgroup$
– Jack
11 mins ago
$begingroup$
Do you have a link to the corollary of Uniform Boundedness which you refer to?
$endgroup$
– Jack
10 mins ago
|
show 1 more comment
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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 5 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.
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
$endgroup$
– Jack
11 mins ago
$begingroup$
Do you have a link to the corollary of Uniform Boundedness which you refer to?
$endgroup$
– Jack
10 mins ago
|
show 1 more 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 5 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.
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
$endgroup$
– Jack
11 mins ago
$begingroup$
Do you have a link to the corollary of Uniform Boundedness which you refer to?
$endgroup$
– Jack
10 mins ago
|
show 1 more 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 5 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 5 hours ago
jmerryjmerry
14.3k1629
edited 2 hours ago
answered 5 hours ago
jmerryjmerry
14.3k1629
answered 5 hours ago
jmerryjmerry
14.3k1629
answered 5 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.
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
$endgroup$
– Jack
11 mins ago
$begingroup$
Do you have a link to the corollary of Uniform Boundedness which you refer to?
$endgroup$
– Jack
10 mins ago
|
show 1 more 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.
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
$endgroup$
– Jack
11 mins ago
$begingroup$
Do you have a link to the corollary of Uniform Boundedness which you refer to?
$endgroup$
– Jack
10 mins 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.
4 hours ago
$begingroup$
@Jack no, that won't rescue the proof.
$endgroup$
– Umberto P.
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
New proof attempt. Please check if you can.
$endgroup$
– Jack
4 hours ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
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– Jack
11 mins ago
$begingroup$
I am not familiar with that. I am only aware that given a family of linear operators $(T_{alpha})$ are such that $T_{alpha}:Erightarrow F$, where $E,F$ are Banach spaces. If $sup_{alphain A}|T_{alpha}x|<infty$ then $sup_{alphain A}|T_{alpha}|<infty$.
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– Jack
11 mins ago
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Do you have a link to the corollary of Uniform Boundedness which you refer to?
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– Jack
10 mins ago
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Do you have a link to the corollary of Uniform Boundedness which you refer to?
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– Jack
10 mins ago
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2
$begingroup$
The last inequality basically implies that the norm of T is bounded or that it is continuous
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– Andres Mejia
5 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!
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– Umberto P.
4 hours ago
$begingroup$
Thank you. Do you have a hint?
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– Jack
4 hours ago
$begingroup$
Third attempt made. Although not sure if this holds either.
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– Jack
4 hours ago