Mixing
Metric projection from space of bounded functions to finite-dimensional linear space
$begingroup$
Apologies if the answer is obvious or should be easy to find, but so far I've had no luck.
Let $X$ be a subspace of $mathbb{R^k}$ for a finite $k$ and let $mathcal{B}(X)$ be the Banach space of bounded real-valued functions on $X$. Let $Y$ be a finite-dimensional linear subspace of $mathcal{B}(X)$. I know that there must exist a bounded linear projection from $mathcal{B}(X)$ to $Y$, and I know that the metric projection (i.e., projection that achieves to minimum distance from $Y$) exists. What I want to know is a. whether the metric projection is continuous, and b. whether it is linear. And, if not, can one find a bounded linear projection that achieves a distance arbitrarily close to that of the metric projection?
functional-analysis metric-spaces banach-spaces approximation-theory projection
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
$endgroup$
add a comment |
$begingroup$
Apologies if the answer is obvious or should be easy to find, but so far I've had no luck.
Let $X$ be a subspace of $mathbb{R^k}$ for a finite $k$ and let $mathcal{B}(X)$ be the Banach space of bounded real-valued functions on $X$. Let $Y$ be a finite-dimensional linear subspace of $mathcal{B}(X)$. I know that there must exist a bounded linear projection from $mathcal{B}(X)$ to $Y$, and I know that the metric projection (i.e., projection that achieves to minimum distance from $Y$) exists. What I want to know is a. whether the metric projection is continuous, and b. whether it is linear. And, if not, can one find a bounded linear projection that achieves a distance arbitrarily close to that of the metric projection?
functional-analysis metric-spaces banach-spaces approximation-theory projection
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
$endgroup$
add a comment |
$begingroup$
Apologies if the answer is obvious or should be easy to find, but so far I've had no luck.
Let $X$ be a subspace of $mathbb{R^k}$ for a finite $k$ and let $mathcal{B}(X)$ be the Banach space of bounded real-valued functions on $X$. Let $Y$ be a finite-dimensional linear subspace of $mathcal{B}(X)$. I know that there must exist a bounded linear projection from $mathcal{B}(X)$ to $Y$, and I know that the metric projection (i.e., projection that achieves to minimum distance from $Y$) exists. What I want to know is a. whether the metric projection is continuous, and b. whether it is linear. And, if not, can one find a bounded linear projection that achieves a distance arbitrarily close to that of the metric projection?
functional-analysis metric-spaces banach-spaces approximation-theory projection
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
$endgroup$
Apologies if the answer is obvious or should be easy to find, but so far I've had no luck.
Let $X$ be a subspace of $mathbb{R^k}$ for a finite $k$ and let $mathcal{B}(X)$ be the Banach space of bounded real-valued functions on $X$. Let $Y$ be a finite-dimensional linear subspace of $mathcal{B}(X)$. I know that there must exist a bounded linear projection from $mathcal{B}(X)$ to $Y$, and I know that the metric projection (i.e., projection that achieves to minimum distance from $Y$) exists. What I want to know is a. whether the metric projection is continuous, and b. whether it is linear. And, if not, can one find a bounded linear projection that achieves a distance arbitrarily close to that of the metric projection?
functional-analysis metric-spaces banach-spaces approximation-theory projection
functional-analysis metric-spaces banach-spaces approximation-theory projection
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
edited Feb 2 at 21:21
SecretlyAnEconomist
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
asked Feb 2 at 20:00
SecretlyAnEconomistSecretlyAnEconomist
42028
42028
add a comment |
add a comment |
1 Answer
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$begingroup$
I just want to say first up, you'd need to know from the outset that $Y$ is a Chebyshev (meaning: admits unique metric projections) subspace, as in general, $Y$ may not be Chebyshev. If $X$ is an infinite set, then $mathcal{B}(X)$ is not a reflexive space, which means (by James' theorem) that $mathcal{B}(X)$ admits an antiproximinal (meaning no metric projections exist from points outside the set) hyperplane. You'd need some extra assumptions, such as $Y$ is finite-dimensional.
If you assume $Y$ is Chebyshev, unfortunately no, there is no guarantee that the metric projection is continuous or linear. If I remember correctly, metric projections onto a closed subspace always being a linear map is a property exclusively of Hilbert Spaces. As for continuity, Lindenstrauss published an example of a Chebyshev subspace of a Banach space that is discontinuous. If I remember correctly, it was a subspace of $C[0, 1]$, so not quite $mathcal{B}(X)$, but I think it's highly dubious to expect the projection map to be continuous.
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
$endgroup$
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
1
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
1
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
I just want to say first up, you'd need to know from the outset that $Y$ is a Chebyshev (meaning: admits unique metric projections) subspace, as in general, $Y$ may not be Chebyshev. If $X$ is an infinite set, then $mathcal{B}(X)$ is not a reflexive space, which means (by James' theorem) that $mathcal{B}(X)$ admits an antiproximinal (meaning no metric projections exist from points outside the set) hyperplane. You'd need some extra assumptions, such as $Y$ is finite-dimensional.
If you assume $Y$ is Chebyshev, unfortunately no, there is no guarantee that the metric projection is continuous or linear. If I remember correctly, metric projections onto a closed subspace always being a linear map is a property exclusively of Hilbert Spaces. As for continuity, Lindenstrauss published an example of a Chebyshev subspace of a Banach space that is discontinuous. If I remember correctly, it was a subspace of $C[0, 1]$, so not quite $mathcal{B}(X)$, but I think it's highly dubious to expect the projection map to be continuous.
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
$endgroup$
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
1
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
1
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
add a comment |
$begingroup$
I just want to say first up, you'd need to know from the outset that $Y$ is a Chebyshev (meaning: admits unique metric projections) subspace, as in general, $Y$ may not be Chebyshev. If $X$ is an infinite set, then $mathcal{B}(X)$ is not a reflexive space, which means (by James' theorem) that $mathcal{B}(X)$ admits an antiproximinal (meaning no metric projections exist from points outside the set) hyperplane. You'd need some extra assumptions, such as $Y$ is finite-dimensional.
If you assume $Y$ is Chebyshev, unfortunately no, there is no guarantee that the metric projection is continuous or linear. If I remember correctly, metric projections onto a closed subspace always being a linear map is a property exclusively of Hilbert Spaces. As for continuity, Lindenstrauss published an example of a Chebyshev subspace of a Banach space that is discontinuous. If I remember correctly, it was a subspace of $C[0, 1]$, so not quite $mathcal{B}(X)$, but I think it's highly dubious to expect the projection map to be continuous.
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
$endgroup$
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
1
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
1
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
add a comment |
$begingroup$
I just want to say first up, you'd need to know from the outset that $Y$ is a Chebyshev (meaning: admits unique metric projections) subspace, as in general, $Y$ may not be Chebyshev. If $X$ is an infinite set, then $mathcal{B}(X)$ is not a reflexive space, which means (by James' theorem) that $mathcal{B}(X)$ admits an antiproximinal (meaning no metric projections exist from points outside the set) hyperplane. You'd need some extra assumptions, such as $Y$ is finite-dimensional.
If you assume $Y$ is Chebyshev, unfortunately no, there is no guarantee that the metric projection is continuous or linear. If I remember correctly, metric projections onto a closed subspace always being a linear map is a property exclusively of Hilbert Spaces. As for continuity, Lindenstrauss published an example of a Chebyshev subspace of a Banach space that is discontinuous. If I remember correctly, it was a subspace of $C[0, 1]$, so not quite $mathcal{B}(X)$, but I think it's highly dubious to expect the projection map to be continuous.
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
$endgroup$
I just want to say first up, you'd need to know from the outset that $Y$ is a Chebyshev (meaning: admits unique metric projections) subspace, as in general, $Y$ may not be Chebyshev. If $X$ is an infinite set, then $mathcal{B}(X)$ is not a reflexive space, which means (by James' theorem) that $mathcal{B}(X)$ admits an antiproximinal (meaning no metric projections exist from points outside the set) hyperplane. You'd need some extra assumptions, such as $Y$ is finite-dimensional.
If you assume $Y$ is Chebyshev, unfortunately no, there is no guarantee that the metric projection is continuous or linear. If I remember correctly, metric projections onto a closed subspace always being a linear map is a property exclusively of Hilbert Spaces. As for continuity, Lindenstrauss published an example of a Chebyshev subspace of a Banach space that is discontinuous. If I remember correctly, it was a subspace of $C[0, 1]$, so not quite $mathcal{B}(X)$, but I think it's highly dubious to expect the projection map to be continuous.
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
answered Feb 2 at 20:54
Theo BenditTheo Bendit
20.9k12355
20.9k12355
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
1
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
1
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
add a comment |
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
1
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
1
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
$begingroup$
Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev.
$endgroup$
– SecretlyAnEconomist
Feb 2 at 21:22
1
1
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
$begingroup$
Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued.
$endgroup$
– Theo Bendit
Feb 3 at 5:37
1
1
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
$begingroup$
If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general.
$endgroup$
– Theo Bendit
Feb 3 at 5:41
add a comment |
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