Equivalent conditions for a topological vector space to admit an inner-product
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Let $V$ be a topological vector space.
Inspired by this list
- Every inner-product induces a (unique?) norm
- Every norm induces a (unique?) metric
- Every metric induces a (unique?) uniform structure
- Every uniform structure induces a (unique?) topology,
I'm trying to complete the list
- $V$ admits an inner-product iff ???
- $V$ admits a norm iff it is Hausdorff and has a convex bounded neighborhood of zero (Kolmogorov's theorem)
- $V$ admits a metric iff it is Hausdorff and has a countable base of neighborhoods of zero
- $V$ admits a (unique) uniform structure, for free
and the list
- $V$ is a Euclidean space iff ???
- $V$ is a Hilbert space iff it has a complete inner-product and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a Banach space iff it has a complete norm and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a ??? iff it has a complete metric
Namely, what are equivalent conditions for $V$ to admit an inner-product? How do Euclidean spaces $textbfR^n$ fit into this scheme? Does $V$ have a special name when it has a complete metric, or are these spaces not important enough? In the first list, is the induced structure always unique?
general-topology functional-analysis topological-vector-spaces
add a comment |Â
up vote
4
down vote
favorite
Let $V$ be a topological vector space.
Inspired by this list
- Every inner-product induces a (unique?) norm
- Every norm induces a (unique?) metric
- Every metric induces a (unique?) uniform structure
- Every uniform structure induces a (unique?) topology,
I'm trying to complete the list
- $V$ admits an inner-product iff ???
- $V$ admits a norm iff it is Hausdorff and has a convex bounded neighborhood of zero (Kolmogorov's theorem)
- $V$ admits a metric iff it is Hausdorff and has a countable base of neighborhoods of zero
- $V$ admits a (unique) uniform structure, for free
and the list
- $V$ is a Euclidean space iff ???
- $V$ is a Hilbert space iff it has a complete inner-product and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a Banach space iff it has a complete norm and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a ??? iff it has a complete metric
Namely, what are equivalent conditions for $V$ to admit an inner-product? How do Euclidean spaces $textbfR^n$ fit into this scheme? Does $V$ have a special name when it has a complete metric, or are these spaces not important enough? In the first list, is the induced structure always unique?
general-topology functional-analysis topological-vector-spaces
This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/â¦
â Arnaud D.
Jul 4 '17 at 9:23
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $V$ be a topological vector space.
Inspired by this list
- Every inner-product induces a (unique?) norm
- Every norm induces a (unique?) metric
- Every metric induces a (unique?) uniform structure
- Every uniform structure induces a (unique?) topology,
I'm trying to complete the list
- $V$ admits an inner-product iff ???
- $V$ admits a norm iff it is Hausdorff and has a convex bounded neighborhood of zero (Kolmogorov's theorem)
- $V$ admits a metric iff it is Hausdorff and has a countable base of neighborhoods of zero
- $V$ admits a (unique) uniform structure, for free
and the list
- $V$ is a Euclidean space iff ???
- $V$ is a Hilbert space iff it has a complete inner-product and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a Banach space iff it has a complete norm and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a ??? iff it has a complete metric
Namely, what are equivalent conditions for $V$ to admit an inner-product? How do Euclidean spaces $textbfR^n$ fit into this scheme? Does $V$ have a special name when it has a complete metric, or are these spaces not important enough? In the first list, is the induced structure always unique?
general-topology functional-analysis topological-vector-spaces
Let $V$ be a topological vector space.
Inspired by this list
- Every inner-product induces a (unique?) norm
- Every norm induces a (unique?) metric
- Every metric induces a (unique?) uniform structure
- Every uniform structure induces a (unique?) topology,
I'm trying to complete the list
- $V$ admits an inner-product iff ???
- $V$ admits a norm iff it is Hausdorff and has a convex bounded neighborhood of zero (Kolmogorov's theorem)
- $V$ admits a metric iff it is Hausdorff and has a countable base of neighborhoods of zero
- $V$ admits a (unique) uniform structure, for free
and the list
- $V$ is a Euclidean space iff ???
- $V$ is a Hilbert space iff it has a complete inner-product and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a Banach space iff it has a complete norm and the scalars are $Bbb R$ or $Bbb C$
- $V$ is a ??? iff it has a complete metric
Namely, what are equivalent conditions for $V$ to admit an inner-product? How do Euclidean spaces $textbfR^n$ fit into this scheme? Does $V$ have a special name when it has a complete metric, or are these spaces not important enough? In the first list, is the induced structure always unique?
general-topology functional-analysis topological-vector-spaces
edited Aug 10 at 23:45
asked Jul 4 '17 at 9:15
étale-cohomology
9151617
9151617
This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/â¦
â Arnaud D.
Jul 4 '17 at 9:23
add a comment |Â
This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/â¦
â Arnaud D.
Jul 4 '17 at 9:23
This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/â¦
â Arnaud D.
Jul 4 '17 at 9:23
This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/â¦
â Arnaud D.
Jul 4 '17 at 9:23
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I'll assume TVS includes that $X$ is $T_1$ (so Tychonoff)
Some random relevant facts:
A normed space $(X,||.||)$ has its norm induced by an inner product iff the paralellogram law holds:
$$forall x,y in X: 2||x||^2 + 2||y||^2 = ||x-y||^2 +||x+y||^2$$
e.g. see this question and its answers. The inner product is uniquely determined by the norm.
A TVS is normable iff it is locally convex and it has a base at $0$ of bounded sets (in the sense that $A$ is bounded iff for every neighbourhood $V$ of $0$ there is some scalar $c$ with $A subseteq cV$). See any good textbook. The norm can always be scaled etc. so it's not quite unique.
$V$ is a Euclidean space iff it has finite dimension (if the scalar field is $mathbbR$ or $mathbbC$, of course).
A TVS that has a complete and compatible metric is called a Fréchet space. They can be topologically characterised by $V$ with a countable local base at $0$ which is also uniformly complete in the induced uniformity. See these notes for more info.
The uniformity on $V$ is never unique (but it exists for all $T_1$ TVS's). Spaces with unique uniformities are "almost-compact" and TVS's never are.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I'll assume TVS includes that $X$ is $T_1$ (so Tychonoff)
Some random relevant facts:
A normed space $(X,||.||)$ has its norm induced by an inner product iff the paralellogram law holds:
$$forall x,y in X: 2||x||^2 + 2||y||^2 = ||x-y||^2 +||x+y||^2$$
e.g. see this question and its answers. The inner product is uniquely determined by the norm.
A TVS is normable iff it is locally convex and it has a base at $0$ of bounded sets (in the sense that $A$ is bounded iff for every neighbourhood $V$ of $0$ there is some scalar $c$ with $A subseteq cV$). See any good textbook. The norm can always be scaled etc. so it's not quite unique.
$V$ is a Euclidean space iff it has finite dimension (if the scalar field is $mathbbR$ or $mathbbC$, of course).
A TVS that has a complete and compatible metric is called a Fréchet space. They can be topologically characterised by $V$ with a countable local base at $0$ which is also uniformly complete in the induced uniformity. See these notes for more info.
The uniformity on $V$ is never unique (but it exists for all $T_1$ TVS's). Spaces with unique uniformities are "almost-compact" and TVS's never are.
add a comment |Â
up vote
2
down vote
accepted
I'll assume TVS includes that $X$ is $T_1$ (so Tychonoff)
Some random relevant facts:
A normed space $(X,||.||)$ has its norm induced by an inner product iff the paralellogram law holds:
$$forall x,y in X: 2||x||^2 + 2||y||^2 = ||x-y||^2 +||x+y||^2$$
e.g. see this question and its answers. The inner product is uniquely determined by the norm.
A TVS is normable iff it is locally convex and it has a base at $0$ of bounded sets (in the sense that $A$ is bounded iff for every neighbourhood $V$ of $0$ there is some scalar $c$ with $A subseteq cV$). See any good textbook. The norm can always be scaled etc. so it's not quite unique.
$V$ is a Euclidean space iff it has finite dimension (if the scalar field is $mathbbR$ or $mathbbC$, of course).
A TVS that has a complete and compatible metric is called a Fréchet space. They can be topologically characterised by $V$ with a countable local base at $0$ which is also uniformly complete in the induced uniformity. See these notes for more info.
The uniformity on $V$ is never unique (but it exists for all $T_1$ TVS's). Spaces with unique uniformities are "almost-compact" and TVS's never are.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I'll assume TVS includes that $X$ is $T_1$ (so Tychonoff)
Some random relevant facts:
A normed space $(X,||.||)$ has its norm induced by an inner product iff the paralellogram law holds:
$$forall x,y in X: 2||x||^2 + 2||y||^2 = ||x-y||^2 +||x+y||^2$$
e.g. see this question and its answers. The inner product is uniquely determined by the norm.
A TVS is normable iff it is locally convex and it has a base at $0$ of bounded sets (in the sense that $A$ is bounded iff for every neighbourhood $V$ of $0$ there is some scalar $c$ with $A subseteq cV$). See any good textbook. The norm can always be scaled etc. so it's not quite unique.
$V$ is a Euclidean space iff it has finite dimension (if the scalar field is $mathbbR$ or $mathbbC$, of course).
A TVS that has a complete and compatible metric is called a Fréchet space. They can be topologically characterised by $V$ with a countable local base at $0$ which is also uniformly complete in the induced uniformity. See these notes for more info.
The uniformity on $V$ is never unique (but it exists for all $T_1$ TVS's). Spaces with unique uniformities are "almost-compact" and TVS's never are.
I'll assume TVS includes that $X$ is $T_1$ (so Tychonoff)
Some random relevant facts:
A normed space $(X,||.||)$ has its norm induced by an inner product iff the paralellogram law holds:
$$forall x,y in X: 2||x||^2 + 2||y||^2 = ||x-y||^2 +||x+y||^2$$
e.g. see this question and its answers. The inner product is uniquely determined by the norm.
A TVS is normable iff it is locally convex and it has a base at $0$ of bounded sets (in the sense that $A$ is bounded iff for every neighbourhood $V$ of $0$ there is some scalar $c$ with $A subseteq cV$). See any good textbook. The norm can always be scaled etc. so it's not quite unique.
$V$ is a Euclidean space iff it has finite dimension (if the scalar field is $mathbbR$ or $mathbbC$, of course).
A TVS that has a complete and compatible metric is called a Fréchet space. They can be topologically characterised by $V$ with a countable local base at $0$ which is also uniformly complete in the induced uniformity. See these notes for more info.
The uniformity on $V$ is never unique (but it exists for all $T_1$ TVS's). Spaces with unique uniformities are "almost-compact" and TVS's never are.
answered Jul 4 '17 at 11:09
Henno Brandsma
91.7k342100
91.7k342100
add a comment |Â
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This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/â¦
â Arnaud D.
Jul 4 '17 at 9:23