are linear functionals on C[0, 1] bounded and thus continuous
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
asked 14 hours ago
Nikola
736719
add a comment |
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
asked 14 hours ago
Nikola
736719
add a comment |
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
asked 14 hours ago
Nikola
736719
I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!
functional-analysis hahn-banach-theorem
functional-analysis hahn-banach-theorem
asked 14 hours ago
Nikola
736719
asked 14 hours ago
Nikola
736719
edited 8 hours ago
asked 14 hours ago
Nikola
736719
asked 14 hours ago
Nikola
736719
asked 14 hours ago
Nikola
736719
736719
add a comment |
add a comment |
1 Answer
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No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered 14 hours ago
supinf
6,0791027
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered 14 hours ago
supinf
6,0791027
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
add a comment |
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered 14 hours ago
supinf
6,0791027
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
add a comment |
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered 14 hours ago
supinf
6,0791027
No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.
answered 14 hours ago
supinf
6,0791027
answered 14 hours ago
supinf
6,0791027
answered 14 hours ago
supinf
6,0791027
answered 14 hours ago
supinf
6,0791027
6,0791027
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
add a comment |
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation.
– Nikola
14 hours ago
1
1
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
@Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral.
– David C. Ullrich
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
Then boundedness of the functional is just presumed in the book I"m using.
– Nikola
11 hours ago
add a comment |
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