Mixing
Is a Hahn-Banach extension always continuous?
$begingroup$
We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:
Theorem (Hahn-Banach): Let $X$ be a real vector space and $q : X to mathbb{R}$ be sublinear. Let $U subseteq X$ be a linear subspace
and $l : U to mathbb{R}$ be a linear functional that satisfies $l(u)
leq q(u)$ for all $u in U$.
Then there exists a linear extension
$L : X to mathbb{R}$ of $l$ that satisfies $L(x) leq q(x)$ for all $x in X$.
Is it true that this Hahn-Banach extension $L$ is also continuous, hence a $mathbb{R}$-linear functional?
The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.
Some of my thoughts/work so far:
I see that, if $q$ is continuous, then $L$ is also continuous:
Let $(x_n) subseteq X$ with $x_n to 0$. Then
$$
L(x_n) leq q(x_n) to 0
$$
$$
-L(x_n) = L(-x_n) leq q(-x_n) to 0
$$
So $|L(x_n)| leq max{q(x_n), q(-x_n)} to 0$. Hence, $L$ is continuous at $0$, and therefore continuous.
But a sublinear function isn't necessarily continuous... so this doesn't help, right?
From what I showed above, we see that
$$
-q(-x) leq L(x) leq q(x) qquad forall x in X.
$$
Not sure if this inequality might help.
real-analysis functional-analysis hahn-banach-theorem
asked Jan 10 at 21:29
zxmknzxmkn
340213
$endgroup$
add a comment |
$begingroup$
We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:
Theorem (Hahn-Banach): Let $X$ be a real vector space and $q : X to mathbb{R}$ be sublinear. Let $U subseteq X$ be a linear subspace
and $l : U to mathbb{R}$ be a linear functional that satisfies $l(u)
leq q(u)$ for all $u in U$.
Then there exists a linear extension
$L : X to mathbb{R}$ of $l$ that satisfies $L(x) leq q(x)$ for all $x in X$.
Is it true that this Hahn-Banach extension $L$ is also continuous, hence a $mathbb{R}$-linear functional?
The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.
Some of my thoughts/work so far:
I see that, if $q$ is continuous, then $L$ is also continuous:
Let $(x_n) subseteq X$ with $x_n to 0$. Then
$$
L(x_n) leq q(x_n) to 0
$$
$$
-L(x_n) = L(-x_n) leq q(-x_n) to 0
$$
So $|L(x_n)| leq max{q(x_n), q(-x_n)} to 0$. Hence, $L$ is continuous at $0$, and therefore continuous.
But a sublinear function isn't necessarily continuous... so this doesn't help, right?
From what I showed above, we see that
$$
-q(-x) leq L(x) leq q(x) qquad forall x in X.
$$
Not sure if this inequality might help.
real-analysis functional-analysis hahn-banach-theorem
asked Jan 10 at 21:29
zxmknzxmkn
340213
$endgroup$
2
$begingroup$
Since there is no norm given, what do you mean by continuous?
$endgroup$
– SmileyCraft
Jan 10 at 21:33
2
$begingroup$
The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 21:34
$begingroup$
The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context.
$endgroup$
– Eric Wofsey
Jan 10 at 21:35
$begingroup$
Oh, I think I was getting confused because we defined linear functionals in this course as elements of $mathcal{L}( X, mathbb{K})$, the bounded linear operators from normed space $X$ to $mathbb{K} in {mathbb{R}, mathbb{C}}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal.
$endgroup$
– zxmkn
Jan 10 at 21:50
add a comment |
$begingroup$
We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:
Theorem (Hahn-Banach): Let $X$ be a real vector space and $q : X to mathbb{R}$ be sublinear. Let $U subseteq X$ be a linear subspace
and $l : U to mathbb{R}$ be a linear functional that satisfies $l(u)
leq q(u)$ for all $u in U$.
Then there exists a linear extension
$L : X to mathbb{R}$ of $l$ that satisfies $L(x) leq q(x)$ for all $x in X$.
Is it true that this Hahn-Banach extension $L$ is also continuous, hence a $mathbb{R}$-linear functional?
The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.
Some of my thoughts/work so far:
I see that, if $q$ is continuous, then $L$ is also continuous:
Let $(x_n) subseteq X$ with $x_n to 0$. Then
$$
L(x_n) leq q(x_n) to 0
$$
$$
-L(x_n) = L(-x_n) leq q(-x_n) to 0
$$
So $|L(x_n)| leq max{q(x_n), q(-x_n)} to 0$. Hence, $L$ is continuous at $0$, and therefore continuous.
But a sublinear function isn't necessarily continuous... so this doesn't help, right?
From what I showed above, we see that
$$
-q(-x) leq L(x) leq q(x) qquad forall x in X.
$$
Not sure if this inequality might help.
real-analysis functional-analysis hahn-banach-theorem
asked Jan 10 at 21:29
zxmknzxmkn
340213
$endgroup$
We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:
Theorem (Hahn-Banach): Let $X$ be a real vector space and $q : X to mathbb{R}$ be sublinear. Let $U subseteq X$ be a linear subspace
and $l : U to mathbb{R}$ be a linear functional that satisfies $l(u)
leq q(u)$ for all $u in U$.
Then there exists a linear extension
$L : X to mathbb{R}$ of $l$ that satisfies $L(x) leq q(x)$ for all $x in X$.
Is it true that this Hahn-Banach extension $L$ is also continuous, hence a $mathbb{R}$-linear functional?
The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.
Some of my thoughts/work so far:
I see that, if $q$ is continuous, then $L$ is also continuous:
Let $(x_n) subseteq X$ with $x_n to 0$. Then
$$
L(x_n) leq q(x_n) to 0
$$
$$
-L(x_n) = L(-x_n) leq q(-x_n) to 0
$$
So $|L(x_n)| leq max{q(x_n), q(-x_n)} to 0$. Hence, $L$ is continuous at $0$, and therefore continuous.
But a sublinear function isn't necessarily continuous... so this doesn't help, right?
From what I showed above, we see that
$$
-q(-x) leq L(x) leq q(x) qquad forall x in X.
$$
Not sure if this inequality might help.
real-analysis functional-analysis hahn-banach-theorem
real-analysis functional-analysis hahn-banach-theorem
asked Jan 10 at 21:29
zxmknzxmkn
340213
asked Jan 10 at 21:29
zxmknzxmkn
340213
asked Jan 10 at 21:29
zxmknzxmkn
340213
asked Jan 10 at 21:29
zxmknzxmkn
340213
asked Jan 10 at 21:29
zxmknzxmkn
340213
340213
2
$begingroup$
Since there is no norm given, what do you mean by continuous?
$endgroup$
– SmileyCraft
Jan 10 at 21:33
2
$begingroup$
The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 21:34
$begingroup$
The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context.
$endgroup$
– Eric Wofsey
Jan 10 at 21:35
$begingroup$
Oh, I think I was getting confused because we defined linear functionals in this course as elements of $mathcal{L}( X, mathbb{K})$, the bounded linear operators from normed space $X$ to $mathbb{K} in {mathbb{R}, mathbb{C}}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal.
$endgroup$
– zxmkn
Jan 10 at 21:50
add a comment |
2
$begingroup$
Since there is no norm given, what do you mean by continuous?
$endgroup$
– SmileyCraft
Jan 10 at 21:33
2
$begingroup$
The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 21:34
$begingroup$
The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context.
$endgroup$
– Eric Wofsey
Jan 10 at 21:35
$begingroup$
Oh, I think I was getting confused because we defined linear functionals in this course as elements of $mathcal{L}( X, mathbb{K})$, the bounded linear operators from normed space $X$ to $mathbb{K} in {mathbb{R}, mathbb{C}}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal.
$endgroup$
– zxmkn
Jan 10 at 21:50
2
2
$begingroup$
Since there is no norm given, what do you mean by continuous?
$endgroup$
– SmileyCraft
Jan 10 at 21:33
$begingroup$
Since there is no norm given, what do you mean by continuous?
$endgroup$
– SmileyCraft
Jan 10 at 21:33
2
2
$begingroup$
The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 21:34
$begingroup$
The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 21:34
$begingroup$
The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context.
$endgroup$
– Eric Wofsey
Jan 10 at 21:35
$begingroup$
The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context.
$endgroup$
– Eric Wofsey
Jan 10 at 21:35
$begingroup$
Oh, I think I was getting confused because we defined linear functionals in this course as elements of $mathcal{L}( X, mathbb{K})$, the bounded linear operators from normed space $X$ to $mathbb{K} in {mathbb{R}, mathbb{C}}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal.
$endgroup$
– zxmkn
Jan 10 at 21:50
$begingroup$
Oh, I think I was getting confused because we defined linear functionals in this course as elements of $mathcal{L}( X, mathbb{K})$, the bounded linear operators from normed space $X$ to $mathbb{K} in {mathbb{R}, mathbb{C}}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal.
$endgroup$
– zxmkn
Jan 10 at 21:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The formulation of Hahn-Banach that you have does not require any topology on $X$. One obvious application is the case where $X$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $q$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Here is a counterexample: Let $X$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $q$. Evidently, $q$ is sublinear. Now, take $U = {0}$ and $l(0) = 0$. However, the only linear function $L : X to mathbb R$ satisfying $q$ is $L = q$ which is discontinuous.
answered Jan 11 at 8:14
gerwgerw
19.5k11334
$endgroup$
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
The formulation of Hahn-Banach that you have does not require any topology on $X$. One obvious application is the case where $X$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $q$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
The formulation of Hahn-Banach that you have does not require any topology on $X$. One obvious application is the case where $X$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $q$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
The formulation of Hahn-Banach that you have does not require any topology on $X$. One obvious application is the case where $X$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $q$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
The formulation of Hahn-Banach that you have does not require any topology on $X$. One obvious application is the case where $X$ does have a norm and you use the theorem to extend functionals while preserving the norm.
But the form with the seminorm $q$ is the one that allows one to obtain the geometric form of Hahn-Banach (i.e., the separation theorems) which is often the most useful.
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 22:59
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
$begingroup$
Here is a counterexample: Let $X$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $q$. Evidently, $q$ is sublinear. Now, take $U = {0}$ and $l(0) = 0$. However, the only linear function $L : X to mathbb R$ satisfying $q$ is $L = q$ which is discontinuous.
answered Jan 11 at 8:14
gerwgerw
19.5k11334
$endgroup$
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
add a comment |
$begingroup$
Here is a counterexample: Let $X$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $q$. Evidently, $q$ is sublinear. Now, take $U = {0}$ and $l(0) = 0$. However, the only linear function $L : X to mathbb R$ satisfying $q$ is $L = q$ which is discontinuous.
answered Jan 11 at 8:14
gerwgerw
19.5k11334
$endgroup$
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
add a comment |
$begingroup$
Here is a counterexample: Let $X$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $q$. Evidently, $q$ is sublinear. Now, take $U = {0}$ and $l(0) = 0$. However, the only linear function $L : X to mathbb R$ satisfying $q$ is $L = q$ which is discontinuous.
answered Jan 11 at 8:14
gerwgerw
19.5k11334
$endgroup$
Here is a counterexample: Let $X$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $q$. Evidently, $q$ is sublinear. Now, take $U = {0}$ and $l(0) = 0$. However, the only linear function $L : X to mathbb R$ satisfying $q$ is $L = q$ which is discontinuous.
answered Jan 11 at 8:14
gerwgerw
19.5k11334
answered Jan 11 at 8:14
gerwgerw
19.5k11334
answered Jan 11 at 8:14
gerwgerw
19.5k11334
answered Jan 11 at 8:14
gerwgerw
19.5k11334
19.5k11334
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
add a comment |
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Thanks, but I don't quite see how you conclude $L = q$. Could you elaborate a bit on how that follows?
$endgroup$
– zxmkn
Jan 11 at 9:24
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
$begingroup$
Oh, never mind. Now I see why $L=q$. [Suppose not, since if $L(x) < q(x)$, then $L(-x) > q(-x)$, which violates the requirement that $q$ dominate $L$.]
$endgroup$
– zxmkn
Jan 11 at 9:33
add a comment |
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2
$begingroup$
Since there is no norm given, what do you mean by continuous?
$endgroup$
– SmileyCraft
Jan 10 at 21:33
2
$begingroup$
The version of H-B you quote does not assume any topology on the space $X$. Of course, a standard example is when $X$ is normed and $q$ is the norm function. Then the extension is continuous for the norm topology.
$endgroup$
– Lord Shark the Unknown
Jan 10 at 21:34
$begingroup$
The term "functional" does not necessarily imply continuity, though sometimes this is implicit in context.
$endgroup$
– Eric Wofsey
Jan 10 at 21:35
$begingroup$
Oh, I think I was getting confused because we defined linear functionals in this course as elements of $mathcal{L}( X, mathbb{K})$, the bounded linear operators from normed space $X$ to $mathbb{K} in {mathbb{R}, mathbb{C}}$. Looking around online, though (mathworld.wolfram.com/LinearFunctional.html), I see that's a more restrictive definition that normal.
$endgroup$
– zxmkn
Jan 10 at 21:50