The limit of a strongly convergent sequence of linear bounded operators from a Banach space to a normed space
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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
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0
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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
54.6k11444
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 |
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
54.6k11444
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
54.6k11444
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
54.6k11444
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
54.6k11444
answered Nov 20 at 1:02
Tsemo Aristide
54.6k11444
answered Nov 20 at 1:02
Tsemo Aristide
54.6k11444
answered Nov 20 at 1:02
Tsemo Aristide
54.6k11444
54.6k11444
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 |
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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