Mixing
Why an operator $A:D(A)to F$ is called “unbounded”?
$begingroup$
Let $E,F$ Banach spaces.
We says that a linear operator $O: Eto F$ is bounded if $$|Ox|_Fleq C|x|_E,$$
for a certain $C>0$. This definition makes sense because this is equivalent at $$|O|_{L(E,V)}=sup_{xneq 0}frac{|Ox|_F}{|x|_E}<C,$$ and thus, $O$ is bounded in $(L(E,V),|cdot |_{L(E,V)})$.We call an unbounded operator a linear operator $A:D(A)to F$ where $D(A)$ is a subspace of $E$. $D(A)$ is call the domain of $A$.
Remark : 1) If $|Au|_{F}leq C|u|_E$ for all $uin D(A)$, we says that $A$ is bounded. So, an unbounded operator can be bounded.
Question : What is the motivation to call it "unbounded" ? I really don't get it, and it sounds so confusing... (this definition come from Brezis : Analyse fonctionelle, théorie et application.) Moreover, it's not said that $D(A)$ must be proper, i.e. $D(A)=E$ could happen... so all linear operator $A:Eto F$ is call unbounded... so every bounded operator $Eto F$ are unbounded. This really don't make sense for me !
functional-analysis
asked Jan 6 at 14:50
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Let $E,F$ Banach spaces.
We says that a linear operator $O: Eto F$ is bounded if $$|Ox|_Fleq C|x|_E,$$
for a certain $C>0$. This definition makes sense because this is equivalent at $$|O|_{L(E,V)}=sup_{xneq 0}frac{|Ox|_F}{|x|_E}<C,$$ and thus, $O$ is bounded in $(L(E,V),|cdot |_{L(E,V)})$.We call an unbounded operator a linear operator $A:D(A)to F$ where $D(A)$ is a subspace of $E$. $D(A)$ is call the domain of $A$.
Remark : 1) If $|Au|_{F}leq C|u|_E$ for all $uin D(A)$, we says that $A$ is bounded. So, an unbounded operator can be bounded.
Question : What is the motivation to call it "unbounded" ? I really don't get it, and it sounds so confusing... (this definition come from Brezis : Analyse fonctionelle, théorie et application.) Moreover, it's not said that $D(A)$ must be proper, i.e. $D(A)=E$ could happen... so all linear operator $A:Eto F$ is call unbounded... so every bounded operator $Eto F$ are unbounded. This really don't make sense for me !
functional-analysis
asked Jan 6 at 14:50
NewMathNewMath
4059
$endgroup$
$begingroup$
Your definition of unbounded seems to encompass every linear operator. Are you sure you're reading it correctly?
$endgroup$
– D. Brogan
Jan 6 at 14:57
$begingroup$
In wiki the definition is the same. They don't say that $D(A)$ is a proper subspace of $E$ (they btw insist on the fact that the operator may be not defined on all $E$, but don't impose the condition $D(A)neq E$).@D.Brogan
$endgroup$
– NewMath
Jan 6 at 15:00
$begingroup$
To quote the top of the same Wikipedia page: The term "unbounded operator" can be misleading, since "unbounded" should sometimes be understood as "not necessarily bounded".
$endgroup$
– Hans Lundmark
Jan 6 at 15:11
add a comment |
$begingroup$
Let $E,F$ Banach spaces.
We says that a linear operator $O: Eto F$ is bounded if $$|Ox|_Fleq C|x|_E,$$
for a certain $C>0$. This definition makes sense because this is equivalent at $$|O|_{L(E,V)}=sup_{xneq 0}frac{|Ox|_F}{|x|_E}<C,$$ and thus, $O$ is bounded in $(L(E,V),|cdot |_{L(E,V)})$.We call an unbounded operator a linear operator $A:D(A)to F$ where $D(A)$ is a subspace of $E$. $D(A)$ is call the domain of $A$.
Remark : 1) If $|Au|_{F}leq C|u|_E$ for all $uin D(A)$, we says that $A$ is bounded. So, an unbounded operator can be bounded.
Question : What is the motivation to call it "unbounded" ? I really don't get it, and it sounds so confusing... (this definition come from Brezis : Analyse fonctionelle, théorie et application.) Moreover, it's not said that $D(A)$ must be proper, i.e. $D(A)=E$ could happen... so all linear operator $A:Eto F$ is call unbounded... so every bounded operator $Eto F$ are unbounded. This really don't make sense for me !
functional-analysis
asked Jan 6 at 14:50
NewMathNewMath
4059
$endgroup$
Let $E,F$ Banach spaces.
We says that a linear operator $O: Eto F$ is bounded if $$|Ox|_Fleq C|x|_E,$$
for a certain $C>0$. This definition makes sense because this is equivalent at $$|O|_{L(E,V)}=sup_{xneq 0}frac{|Ox|_F}{|x|_E}<C,$$ and thus, $O$ is bounded in $(L(E,V),|cdot |_{L(E,V)})$.We call an unbounded operator a linear operator $A:D(A)to F$ where $D(A)$ is a subspace of $E$. $D(A)$ is call the domain of $A$.
Remark : 1) If $|Au|_{F}leq C|u|_E$ for all $uin D(A)$, we says that $A$ is bounded. So, an unbounded operator can be bounded.
Question : What is the motivation to call it "unbounded" ? I really don't get it, and it sounds so confusing... (this definition come from Brezis : Analyse fonctionelle, théorie et application.) Moreover, it's not said that $D(A)$ must be proper, i.e. $D(A)=E$ could happen... so all linear operator $A:Eto F$ is call unbounded... so every bounded operator $Eto F$ are unbounded. This really don't make sense for me !
functional-analysis
functional-analysis
asked Jan 6 at 14:50
NewMathNewMath
4059
asked Jan 6 at 14:50
NewMathNewMath
4059
edited Jan 6 at 14:54
NewMath
asked Jan 6 at 14:50
NewMathNewMath
4059
asked Jan 6 at 14:50
NewMathNewMath
4059
asked Jan 6 at 14:50
NewMathNewMath
4059
4059
$begingroup$
Your definition of unbounded seems to encompass every linear operator. Are you sure you're reading it correctly?
$endgroup$
– D. Brogan
Jan 6 at 14:57
$begingroup$
In wiki the definition is the same. They don't say that $D(A)$ is a proper subspace of $E$ (they btw insist on the fact that the operator may be not defined on all $E$, but don't impose the condition $D(A)neq E$).@D.Brogan
$endgroup$
– NewMath
Jan 6 at 15:00
$begingroup$
To quote the top of the same Wikipedia page: The term "unbounded operator" can be misleading, since "unbounded" should sometimes be understood as "not necessarily bounded".
$endgroup$
– Hans Lundmark
Jan 6 at 15:11
add a comment |
$begingroup$
Your definition of unbounded seems to encompass every linear operator. Are you sure you're reading it correctly?
$endgroup$
– D. Brogan
Jan 6 at 14:57
$begingroup$
In wiki the definition is the same. They don't say that $D(A)$ is a proper subspace of $E$ (they btw insist on the fact that the operator may be not defined on all $E$, but don't impose the condition $D(A)neq E$).@D.Brogan
$endgroup$
– NewMath
Jan 6 at 15:00
$begingroup$
To quote the top of the same Wikipedia page: The term "unbounded operator" can be misleading, since "unbounded" should sometimes be understood as "not necessarily bounded".
$endgroup$
– Hans Lundmark
Jan 6 at 15:11
$begingroup$
Your definition of unbounded seems to encompass every linear operator. Are you sure you're reading it correctly?
$endgroup$
– D. Brogan
Jan 6 at 14:57
$begingroup$
Your definition of unbounded seems to encompass every linear operator. Are you sure you're reading it correctly?
$endgroup$
– D. Brogan
Jan 6 at 14:57
$begingroup$
In wiki the definition is the same. They don't say that $D(A)$ is a proper subspace of $E$ (they btw insist on the fact that the operator may be not defined on all $E$, but don't impose the condition $D(A)neq E$).@D.Brogan
$endgroup$
– NewMath
Jan 6 at 15:00
$begingroup$
In wiki the definition is the same. They don't say that $D(A)$ is a proper subspace of $E$ (they btw insist on the fact that the operator may be not defined on all $E$, but don't impose the condition $D(A)neq E$).@D.Brogan
$endgroup$
– NewMath
Jan 6 at 15:00
$begingroup$
To quote the top of the same Wikipedia page: The term "unbounded operator" can be misleading, since "unbounded" should sometimes be understood as "not necessarily bounded".
$endgroup$
– Hans Lundmark
Jan 6 at 15:11
$begingroup$
To quote the top of the same Wikipedia page: The term "unbounded operator" can be misleading, since "unbounded" should sometimes be understood as "not necessarily bounded".
$endgroup$
– Hans Lundmark
Jan 6 at 15:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are not the first person who is confused. See Confused about Domain of Unbounded Operators (Hilbert Spaces).
Also see the dicussion in https://ncatlab.org/nlab/show/unbounded+operator with its reference to the "red herring principle" https://ncatlab.org/nlab/show/red+herring+principle.
$endgroup$
add a comment |
$begingroup$
I agree that the terminology is a little confusing. Here 'linear operators' and 'unbounded linear operators' are seen as separate concepts, where the difference is rather subtle. Let me rewrite the two definitions:
Let $E,F$ be Banach spaces.
A linear operator from $E$ to $F$ is a linear map $A:Eto F$;
An unbounded linear operator from $E$ to $F$ is a linear map $A:D(A)to F$ where $D(A)$ is a subspace of $E$.
As you can see in both cases we are referring to operators from $E$ to $F$ but the meaning is slightly different - in the first case the domain is actually $E$, whereas in the second case it isn't. So we could say that the main difference lies in the perspective - the second definition implies that you are first taking a large space $E$ and a map $A$, and then you realize that $A$ cannot be defined on the whole $E$ but only on a subspace. In practice, this often makes things simpler when treating actually unbounded operators.
For instance, differential operators are typically defined as unbounded operators on large spaces of functions, such as $L^2(Omega)$ or $C(Omega)$, but their actual domain is smaller because not all of these functions are differentiable.
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
$endgroup$
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
$begingroup$
As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:25
$begingroup$
In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:27
$begingroup$
But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
$endgroup$
– NewMath
Jan 6 at 15:48
$begingroup$
Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:54
|
show 2 more comments
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2 Answers
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2 Answers
2
active
oldest
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oldest
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$begingroup$
You are not the first person who is confused. See Confused about Domain of Unbounded Operators (Hilbert Spaces).
Also see the dicussion in https://ncatlab.org/nlab/show/unbounded+operator with its reference to the "red herring principle" https://ncatlab.org/nlab/show/red+herring+principle.
$endgroup$
add a comment |
$begingroup$
You are not the first person who is confused. See Confused about Domain of Unbounded Operators (Hilbert Spaces).
Also see the dicussion in https://ncatlab.org/nlab/show/unbounded+operator with its reference to the "red herring principle" https://ncatlab.org/nlab/show/red+herring+principle.
$endgroup$
add a comment |
$begingroup$
You are not the first person who is confused. See Confused about Domain of Unbounded Operators (Hilbert Spaces).
Also see the dicussion in https://ncatlab.org/nlab/show/unbounded+operator with its reference to the "red herring principle" https://ncatlab.org/nlab/show/red+herring+principle.
$endgroup$
You are not the first person who is confused. See Confused about Domain of Unbounded Operators (Hilbert Spaces).
Also see the dicussion in https://ncatlab.org/nlab/show/unbounded+operator with its reference to the "red herring principle" https://ncatlab.org/nlab/show/red+herring+principle.
answered Jan 6 at 15:13
community wiki
Paul Frost
add a comment |
add a comment |
$begingroup$
I agree that the terminology is a little confusing. Here 'linear operators' and 'unbounded linear operators' are seen as separate concepts, where the difference is rather subtle. Let me rewrite the two definitions:
Let $E,F$ be Banach spaces.
A linear operator from $E$ to $F$ is a linear map $A:Eto F$;
An unbounded linear operator from $E$ to $F$ is a linear map $A:D(A)to F$ where $D(A)$ is a subspace of $E$.
As you can see in both cases we are referring to operators from $E$ to $F$ but the meaning is slightly different - in the first case the domain is actually $E$, whereas in the second case it isn't. So we could say that the main difference lies in the perspective - the second definition implies that you are first taking a large space $E$ and a map $A$, and then you realize that $A$ cannot be defined on the whole $E$ but only on a subspace. In practice, this often makes things simpler when treating actually unbounded operators.
For instance, differential operators are typically defined as unbounded operators on large spaces of functions, such as $L^2(Omega)$ or $C(Omega)$, but their actual domain is smaller because not all of these functions are differentiable.
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
$endgroup$
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
$begingroup$
As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:25
$begingroup$
In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:27
$begingroup$
But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
$endgroup$
– NewMath
Jan 6 at 15:48
$begingroup$
Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:54
|
show 2 more comments
$begingroup$
I agree that the terminology is a little confusing. Here 'linear operators' and 'unbounded linear operators' are seen as separate concepts, where the difference is rather subtle. Let me rewrite the two definitions:
Let $E,F$ be Banach spaces.
A linear operator from $E$ to $F$ is a linear map $A:Eto F$;
An unbounded linear operator from $E$ to $F$ is a linear map $A:D(A)to F$ where $D(A)$ is a subspace of $E$.
As you can see in both cases we are referring to operators from $E$ to $F$ but the meaning is slightly different - in the first case the domain is actually $E$, whereas in the second case it isn't. So we could say that the main difference lies in the perspective - the second definition implies that you are first taking a large space $E$ and a map $A$, and then you realize that $A$ cannot be defined on the whole $E$ but only on a subspace. In practice, this often makes things simpler when treating actually unbounded operators.
For instance, differential operators are typically defined as unbounded operators on large spaces of functions, such as $L^2(Omega)$ or $C(Omega)$, but their actual domain is smaller because not all of these functions are differentiable.
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
$endgroup$
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
$begingroup$
As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:25
$begingroup$
In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:27
$begingroup$
But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
$endgroup$
– NewMath
Jan 6 at 15:48
$begingroup$
Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:54
|
show 2 more comments
$begingroup$
I agree that the terminology is a little confusing. Here 'linear operators' and 'unbounded linear operators' are seen as separate concepts, where the difference is rather subtle. Let me rewrite the two definitions:
Let $E,F$ be Banach spaces.
A linear operator from $E$ to $F$ is a linear map $A:Eto F$;
An unbounded linear operator from $E$ to $F$ is a linear map $A:D(A)to F$ where $D(A)$ is a subspace of $E$.
As you can see in both cases we are referring to operators from $E$ to $F$ but the meaning is slightly different - in the first case the domain is actually $E$, whereas in the second case it isn't. So we could say that the main difference lies in the perspective - the second definition implies that you are first taking a large space $E$ and a map $A$, and then you realize that $A$ cannot be defined on the whole $E$ but only on a subspace. In practice, this often makes things simpler when treating actually unbounded operators.
For instance, differential operators are typically defined as unbounded operators on large spaces of functions, such as $L^2(Omega)$ or $C(Omega)$, but their actual domain is smaller because not all of these functions are differentiable.
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
$endgroup$
I agree that the terminology is a little confusing. Here 'linear operators' and 'unbounded linear operators' are seen as separate concepts, where the difference is rather subtle. Let me rewrite the two definitions:
Let $E,F$ be Banach spaces.
A linear operator from $E$ to $F$ is a linear map $A:Eto F$;
An unbounded linear operator from $E$ to $F$ is a linear map $A:D(A)to F$ where $D(A)$ is a subspace of $E$.
As you can see in both cases we are referring to operators from $E$ to $F$ but the meaning is slightly different - in the first case the domain is actually $E$, whereas in the second case it isn't. So we could say that the main difference lies in the perspective - the second definition implies that you are first taking a large space $E$ and a map $A$, and then you realize that $A$ cannot be defined on the whole $E$ but only on a subspace. In practice, this often makes things simpler when treating actually unbounded operators.
For instance, differential operators are typically defined as unbounded operators on large spaces of functions, such as $L^2(Omega)$ or $C(Omega)$, but their actual domain is smaller because not all of these functions are differentiable.
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
answered Jan 6 at 15:19
Lorenzo QuarisaLorenzo Quarisa
3,304519
3,304519
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
$begingroup$
As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:25
$begingroup$
In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:27
$begingroup$
But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
$endgroup$
– NewMath
Jan 6 at 15:48
$begingroup$
Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:54
|
show 2 more comments
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
$begingroup$
As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:25
$begingroup$
In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:27
$begingroup$
But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
$endgroup$
– NewMath
Jan 6 at 15:48
$begingroup$
Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 15:54
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
$begingroup$
But in your second definition, you don't avoid $D(A)neq E$, do you ? As $E$ is a subspace of it self... all operator $Eto F$ is therefore unbounded...
$endgroup$
– NewMath
Jan 6 at 15:22
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As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
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– Lorenzo Quarisa
Jan 6 at 15:25
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As others have pointed out, using the definition of 'unbounded linear operator' as we have given, does not necessarily imply that the considered operator IS unbounded. However, IN PRACTICE this definition is only used for unbounded operators (which is precisely why the definition is called 'unbounded linear operator'!). Hence your remark of taking $D(A)=E$ can be seen as a 'stretch' to the original meaning intended for the definition. In practice, whether you require $D(A)neq E$ or not, nothing changes.
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– Lorenzo Quarisa
Jan 6 at 15:25
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In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
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– Lorenzo Quarisa
Jan 6 at 15:27
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In my post I tried to explain why this definition of 'unbounded linear operator' is useful to be applied to linear operators that are actually unbounded (see the example of differential operators).
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– Lorenzo Quarisa
Jan 6 at 15:27
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But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
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– NewMath
Jan 6 at 15:48
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But by "unbounded" you mean "not continuous" or defined on a proper subspace ?
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– NewMath
Jan 6 at 15:48
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Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
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– Lorenzo Quarisa
Jan 6 at 15:54
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Again, the source of the confusion is that we are giving two meanings to the word 'unbounded' in two different contexts. In one context the word 'unbounded' simply means that we are defining the operator on a smaller subspace than the original space (by the way this is just a definition, not a theorem, so it cannot be contested!), and in the other context it means that the operator is not continuous. The link between the two is simply that the act of defining the operator on a smaller subspace is consistently used in analysis for operators which are not continuous.
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– Lorenzo Quarisa
Jan 6 at 15:54
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Your definition of unbounded seems to encompass every linear operator. Are you sure you're reading it correctly?
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– D. Brogan
Jan 6 at 14:57
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In wiki the definition is the same. They don't say that $D(A)$ is a proper subspace of $E$ (they btw insist on the fact that the operator may be not defined on all $E$, but don't impose the condition $D(A)neq E$).@D.Brogan
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– NewMath
Jan 6 at 15:00
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To quote the top of the same Wikipedia page: The term "unbounded operator" can be misleading, since "unbounded" should sometimes be understood as "not necessarily bounded".
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– Hans Lundmark
Jan 6 at 15:11