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?







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  • This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/…
    – Arnaud D.
    Jul 4 '17 at 9:23














up vote
4
down vote

favorite
1












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?







share|cite|improve this question






















  • This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/…
    – Arnaud D.
    Jul 4 '17 at 9:23












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





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?







share|cite|improve this question














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?









share|cite|improve this question













share|cite|improve this question




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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/…
    – Arnaud D.
    Jul 4 '17 at 9:23
















  • This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/…
    – Arnaud D.
    Jul 4 '17 at 9:23















This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/…
– Arnaud D.
Jul 4 '17 at 9:23




This should already answer (at least) one of your question marks : math.stackexchange.com/questions/21792/…
– Arnaud D.
Jul 4 '17 at 9:23










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.






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    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.






    share|cite|improve this answer
























      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.






      share|cite|improve this answer






















        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jul 4 '17 at 11:09









        Henno Brandsma

        91.7k342100




        91.7k342100






















             

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