Mixing
Prove that there exists a linear functional $l$ definied on $H$ that is not bounded .
$begingroup$
Let $H$ be an infinite-dimensional Hilbert space . Then there exists a linear functional $l$ defined on $H$ that is not bounded .
My attempt :
Let $B={e_{alpha}}_{alpha in I}$ denote the basis of $H$ , then we can construct a countable subset of $B$ . WLOG , we say $A={e_1,...,e_n,...}$ is an orthonormal subset of $B$ . Then construct functional $l$ as follows :
$$l(e_k)=k ,,,,,,,,,text{whenever $k in N^*$ }$$
$$l(f)=0 ,,,,,,,,,text{whenever $f in B-A$ }$$
Then if $l$ is well-defined on $H$ then $||l(e_k)||=k$ is not a bounded functional .
My qusetion :
I want to show if $I' subset I$ is countable and $$f=sum_{alpha in I'} a_{alpha}e_{alpha}$$ then $$l(f)=sum_{alpha in I'} p(alpha)a_{alpha} ,,,,,,,,,,,,,text{here $p(alpha)=n $ whenever $e_{alpha}=e_n$ and $p(alpha)=0$ otherwise}$$
However , this might leads some contradiction . Consider $f_1=sum_{n=1}^{infty} frac1n e_n$ , it is obvious $f_1 in H$ but $l(f_1)=sum_{n=1}^{infty} 1$ is not well-defined .
real-analysis functional-analysis hilbert-spaces
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
$endgroup$
add a comment |
$begingroup$
Let $H$ be an infinite-dimensional Hilbert space . Then there exists a linear functional $l$ defined on $H$ that is not bounded .
My attempt :
Let $B={e_{alpha}}_{alpha in I}$ denote the basis of $H$ , then we can construct a countable subset of $B$ . WLOG , we say $A={e_1,...,e_n,...}$ is an orthonormal subset of $B$ . Then construct functional $l$ as follows :
$$l(e_k)=k ,,,,,,,,,text{whenever $k in N^*$ }$$
$$l(f)=0 ,,,,,,,,,text{whenever $f in B-A$ }$$
Then if $l$ is well-defined on $H$ then $||l(e_k)||=k$ is not a bounded functional .
My qusetion :
I want to show if $I' subset I$ is countable and $$f=sum_{alpha in I'} a_{alpha}e_{alpha}$$ then $$l(f)=sum_{alpha in I'} p(alpha)a_{alpha} ,,,,,,,,,,,,,text{here $p(alpha)=n $ whenever $e_{alpha}=e_n$ and $p(alpha)=0$ otherwise}$$
However , this might leads some contradiction . Consider $f_1=sum_{n=1}^{infty} frac1n e_n$ , it is obvious $f_1 in H$ but $l(f_1)=sum_{n=1}^{infty} 1$ is not well-defined .
real-analysis functional-analysis hilbert-spaces
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
$endgroup$
add a comment |
$begingroup$
Let $H$ be an infinite-dimensional Hilbert space . Then there exists a linear functional $l$ defined on $H$ that is not bounded .
My attempt :
Let $B={e_{alpha}}_{alpha in I}$ denote the basis of $H$ , then we can construct a countable subset of $B$ . WLOG , we say $A={e_1,...,e_n,...}$ is an orthonormal subset of $B$ . Then construct functional $l$ as follows :
$$l(e_k)=k ,,,,,,,,,text{whenever $k in N^*$ }$$
$$l(f)=0 ,,,,,,,,,text{whenever $f in B-A$ }$$
Then if $l$ is well-defined on $H$ then $||l(e_k)||=k$ is not a bounded functional .
My qusetion :
I want to show if $I' subset I$ is countable and $$f=sum_{alpha in I'} a_{alpha}e_{alpha}$$ then $$l(f)=sum_{alpha in I'} p(alpha)a_{alpha} ,,,,,,,,,,,,,text{here $p(alpha)=n $ whenever $e_{alpha}=e_n$ and $p(alpha)=0$ otherwise}$$
However , this might leads some contradiction . Consider $f_1=sum_{n=1}^{infty} frac1n e_n$ , it is obvious $f_1 in H$ but $l(f_1)=sum_{n=1}^{infty} 1$ is not well-defined .
real-analysis functional-analysis hilbert-spaces
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
$endgroup$
Let $H$ be an infinite-dimensional Hilbert space . Then there exists a linear functional $l$ defined on $H$ that is not bounded .
My attempt :
Let $B={e_{alpha}}_{alpha in I}$ denote the basis of $H$ , then we can construct a countable subset of $B$ . WLOG , we say $A={e_1,...,e_n,...}$ is an orthonormal subset of $B$ . Then construct functional $l$ as follows :
$$l(e_k)=k ,,,,,,,,,text{whenever $k in N^*$ }$$
$$l(f)=0 ,,,,,,,,,text{whenever $f in B-A$ }$$
Then if $l$ is well-defined on $H$ then $||l(e_k)||=k$ is not a bounded functional .
My qusetion :
I want to show if $I' subset I$ is countable and $$f=sum_{alpha in I'} a_{alpha}e_{alpha}$$ then $$l(f)=sum_{alpha in I'} p(alpha)a_{alpha} ,,,,,,,,,,,,,text{here $p(alpha)=n $ whenever $e_{alpha}=e_n$ and $p(alpha)=0$ otherwise}$$
However , this might leads some contradiction . Consider $f_1=sum_{n=1}^{infty} frac1n e_n$ , it is obvious $f_1 in H$ but $l(f_1)=sum_{n=1}^{infty} 1$ is not well-defined .
real-analysis functional-analysis hilbert-spaces
real-analysis functional-analysis hilbert-spaces
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
edited Jan 29 at 0:50
J.Guo
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
asked Jan 29 at 0:19
J.GuoJ.Guo
4389
4389
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This actually does not work in quite the way you are trying to make it work. The reason is that there aren't any "constructible" discontinuous linear functionals on complete normed vector spaces.
There is a reasonable way to try to proceed, which is basically what you tried to do but cleaned up a bit. Take a Schauder basis $B$ of $H$, and take a countable subset $C$ with enumeration $e_k$. Define $l$ on the span of $C$ by defining $l(e_k)=k$ and extending by linearity. Define $l$ to be $0$ on the span of $B setminus C$. Then extend by linearity again, hopefully extending $l$ to all of $H$.
Except you didn't actually extend $l$ to all of $H$: you only extended $l$ to the direct sum of the span of $C$ and the span of $B setminus C$. Because $B$ is just a Schauder basis, this direct sum is in general not all of $H$; in fact it does not even necessarily contain the closed span of $C$. To get an extension to all of $H$, you would need to be sure that for all $f in H$, if the projection of $f$ onto the span of $C$ is $sum_{k=1}^infty a_k e_k$, then $sum_{k=1}^infty k a_k$ (which is what you would expect $l(f)$ to be) converges. In fact it must converge absolutely as well. This property just doesn't hold (your example shows why), so $l$ is not defined on all of $H$.
You can circumvent this issue by taking $B$ to be a Hamel basis in the first place. Then the argument I gave works, because the direct sum of the span of $C$ and of the span of $B setminus C$ is just $H$. But a Hamel basis of an infinite dimensional normed vector space requires some kind of choice principle to "construct" it.
answered Jan 29 at 0:39
IanIan
68.9k25392
$endgroup$
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
add a comment |
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1 Answer
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$begingroup$
This actually does not work in quite the way you are trying to make it work. The reason is that there aren't any "constructible" discontinuous linear functionals on complete normed vector spaces.
There is a reasonable way to try to proceed, which is basically what you tried to do but cleaned up a bit. Take a Schauder basis $B$ of $H$, and take a countable subset $C$ with enumeration $e_k$. Define $l$ on the span of $C$ by defining $l(e_k)=k$ and extending by linearity. Define $l$ to be $0$ on the span of $B setminus C$. Then extend by linearity again, hopefully extending $l$ to all of $H$.
Except you didn't actually extend $l$ to all of $H$: you only extended $l$ to the direct sum of the span of $C$ and the span of $B setminus C$. Because $B$ is just a Schauder basis, this direct sum is in general not all of $H$; in fact it does not even necessarily contain the closed span of $C$. To get an extension to all of $H$, you would need to be sure that for all $f in H$, if the projection of $f$ onto the span of $C$ is $sum_{k=1}^infty a_k e_k$, then $sum_{k=1}^infty k a_k$ (which is what you would expect $l(f)$ to be) converges. In fact it must converge absolutely as well. This property just doesn't hold (your example shows why), so $l$ is not defined on all of $H$.
You can circumvent this issue by taking $B$ to be a Hamel basis in the first place. Then the argument I gave works, because the direct sum of the span of $C$ and of the span of $B setminus C$ is just $H$. But a Hamel basis of an infinite dimensional normed vector space requires some kind of choice principle to "construct" it.
answered Jan 29 at 0:39
IanIan
68.9k25392
$endgroup$
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
add a comment |
$begingroup$
This actually does not work in quite the way you are trying to make it work. The reason is that there aren't any "constructible" discontinuous linear functionals on complete normed vector spaces.
There is a reasonable way to try to proceed, which is basically what you tried to do but cleaned up a bit. Take a Schauder basis $B$ of $H$, and take a countable subset $C$ with enumeration $e_k$. Define $l$ on the span of $C$ by defining $l(e_k)=k$ and extending by linearity. Define $l$ to be $0$ on the span of $B setminus C$. Then extend by linearity again, hopefully extending $l$ to all of $H$.
Except you didn't actually extend $l$ to all of $H$: you only extended $l$ to the direct sum of the span of $C$ and the span of $B setminus C$. Because $B$ is just a Schauder basis, this direct sum is in general not all of $H$; in fact it does not even necessarily contain the closed span of $C$. To get an extension to all of $H$, you would need to be sure that for all $f in H$, if the projection of $f$ onto the span of $C$ is $sum_{k=1}^infty a_k e_k$, then $sum_{k=1}^infty k a_k$ (which is what you would expect $l(f)$ to be) converges. In fact it must converge absolutely as well. This property just doesn't hold (your example shows why), so $l$ is not defined on all of $H$.
You can circumvent this issue by taking $B$ to be a Hamel basis in the first place. Then the argument I gave works, because the direct sum of the span of $C$ and of the span of $B setminus C$ is just $H$. But a Hamel basis of an infinite dimensional normed vector space requires some kind of choice principle to "construct" it.
answered Jan 29 at 0:39
IanIan
68.9k25392
$endgroup$
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
add a comment |
$begingroup$
This actually does not work in quite the way you are trying to make it work. The reason is that there aren't any "constructible" discontinuous linear functionals on complete normed vector spaces.
There is a reasonable way to try to proceed, which is basically what you tried to do but cleaned up a bit. Take a Schauder basis $B$ of $H$, and take a countable subset $C$ with enumeration $e_k$. Define $l$ on the span of $C$ by defining $l(e_k)=k$ and extending by linearity. Define $l$ to be $0$ on the span of $B setminus C$. Then extend by linearity again, hopefully extending $l$ to all of $H$.
Except you didn't actually extend $l$ to all of $H$: you only extended $l$ to the direct sum of the span of $C$ and the span of $B setminus C$. Because $B$ is just a Schauder basis, this direct sum is in general not all of $H$; in fact it does not even necessarily contain the closed span of $C$. To get an extension to all of $H$, you would need to be sure that for all $f in H$, if the projection of $f$ onto the span of $C$ is $sum_{k=1}^infty a_k e_k$, then $sum_{k=1}^infty k a_k$ (which is what you would expect $l(f)$ to be) converges. In fact it must converge absolutely as well. This property just doesn't hold (your example shows why), so $l$ is not defined on all of $H$.
You can circumvent this issue by taking $B$ to be a Hamel basis in the first place. Then the argument I gave works, because the direct sum of the span of $C$ and of the span of $B setminus C$ is just $H$. But a Hamel basis of an infinite dimensional normed vector space requires some kind of choice principle to "construct" it.
answered Jan 29 at 0:39
IanIan
68.9k25392
$endgroup$
This actually does not work in quite the way you are trying to make it work. The reason is that there aren't any "constructible" discontinuous linear functionals on complete normed vector spaces.
There is a reasonable way to try to proceed, which is basically what you tried to do but cleaned up a bit. Take a Schauder basis $B$ of $H$, and take a countable subset $C$ with enumeration $e_k$. Define $l$ on the span of $C$ by defining $l(e_k)=k$ and extending by linearity. Define $l$ to be $0$ on the span of $B setminus C$. Then extend by linearity again, hopefully extending $l$ to all of $H$.
Except you didn't actually extend $l$ to all of $H$: you only extended $l$ to the direct sum of the span of $C$ and the span of $B setminus C$. Because $B$ is just a Schauder basis, this direct sum is in general not all of $H$; in fact it does not even necessarily contain the closed span of $C$. To get an extension to all of $H$, you would need to be sure that for all $f in H$, if the projection of $f$ onto the span of $C$ is $sum_{k=1}^infty a_k e_k$, then $sum_{k=1}^infty k a_k$ (which is what you would expect $l(f)$ to be) converges. In fact it must converge absolutely as well. This property just doesn't hold (your example shows why), so $l$ is not defined on all of $H$.
You can circumvent this issue by taking $B$ to be a Hamel basis in the first place. Then the argument I gave works, because the direct sum of the span of $C$ and of the span of $B setminus C$ is just $H$. But a Hamel basis of an infinite dimensional normed vector space requires some kind of choice principle to "construct" it.
answered Jan 29 at 0:39
IanIan
68.9k25392
edited Jan 29 at 0:52
answered Jan 29 at 0:39
IanIan
68.9k25392
answered Jan 29 at 0:39
IanIan
68.9k25392
answered Jan 29 at 0:39
IanIan
68.9k25392
68.9k25392
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
add a comment |
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
I didn't see that how to extend span of $C$ to the closure of span $C$ , since $l(f_1)$ defined doesn't make sense .
$endgroup$
– J.Guo
Jan 29 at 0:57
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@J.Guo That's exactly the problem, you can't extend $l$ to the closed span of $C$ because it requires you to make sense of non-convergent sums.
$endgroup$
– Ian
Jan 29 at 0:59
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@lan Thanks for the answer . I think there still has something I did not understand . Is there an example such that $H$ is hilbert space and $l$ is an unbounded functional ?
$endgroup$
– J.Guo
Jan 29 at 1:10
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
$begingroup$
@J.Guo There is no explicit example, the use of some kind of choice principle is required. Explicit examples can exist only on incomplete spaces.
$endgroup$
– Ian
Jan 29 at 1:23
add a comment |
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