When is a topological vector space “inner product-able”?












5












$begingroup$


This is a follow-up to my question here. A topological vector space is normable, i.e. its topology is induced by some norm on the vector space, if and only if it is Hausdorff and the $0$ vector has a bounded convex neighborhood. My question is, under what circumstances is a topological vector space “inner product-able”, i.e. when is its topology induced by some inner product on the vector space?



Note that this is not the same as asking under what circumstances a norm is induced by some inner product. The answer to that question is when the norm obeys the parallelogram law. But we could have a situation where the topology is induced by multiple norms, one which is not induced by any inner product and another which is induced by some inner product. In any case, another way to phrase my question is, under what circumstances is a given norm equivalent to some norm obeying the parallelogram law?










share|cite|improve this question









$endgroup$












  • $begingroup$
    What does 'bounded' mean in your first paragraph?
    $endgroup$
    – JonathanZ
    Dec 18 '18 at 16:03






  • 2




    $begingroup$
    @JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $lambda U = {lambda x: xin U}$ where $lambda$ is some real number greater than 1.
    $endgroup$
    – Keshav Srinivasan
    Dec 18 '18 at 18:19












  • $begingroup$
    Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:59
















5












$begingroup$


This is a follow-up to my question here. A topological vector space is normable, i.e. its topology is induced by some norm on the vector space, if and only if it is Hausdorff and the $0$ vector has a bounded convex neighborhood. My question is, under what circumstances is a topological vector space “inner product-able”, i.e. when is its topology induced by some inner product on the vector space?



Note that this is not the same as asking under what circumstances a norm is induced by some inner product. The answer to that question is when the norm obeys the parallelogram law. But we could have a situation where the topology is induced by multiple norms, one which is not induced by any inner product and another which is induced by some inner product. In any case, another way to phrase my question is, under what circumstances is a given norm equivalent to some norm obeying the parallelogram law?










share|cite|improve this question









$endgroup$












  • $begingroup$
    What does 'bounded' mean in your first paragraph?
    $endgroup$
    – JonathanZ
    Dec 18 '18 at 16:03






  • 2




    $begingroup$
    @JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $lambda U = {lambda x: xin U}$ where $lambda$ is some real number greater than 1.
    $endgroup$
    – Keshav Srinivasan
    Dec 18 '18 at 18:19












  • $begingroup$
    Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:59














5












5








5


3



$begingroup$


This is a follow-up to my question here. A topological vector space is normable, i.e. its topology is induced by some norm on the vector space, if and only if it is Hausdorff and the $0$ vector has a bounded convex neighborhood. My question is, under what circumstances is a topological vector space “inner product-able”, i.e. when is its topology induced by some inner product on the vector space?



Note that this is not the same as asking under what circumstances a norm is induced by some inner product. The answer to that question is when the norm obeys the parallelogram law. But we could have a situation where the topology is induced by multiple norms, one which is not induced by any inner product and another which is induced by some inner product. In any case, another way to phrase my question is, under what circumstances is a given norm equivalent to some norm obeying the parallelogram law?










share|cite|improve this question









$endgroup$




This is a follow-up to my question here. A topological vector space is normable, i.e. its topology is induced by some norm on the vector space, if and only if it is Hausdorff and the $0$ vector has a bounded convex neighborhood. My question is, under what circumstances is a topological vector space “inner product-able”, i.e. when is its topology induced by some inner product on the vector space?



Note that this is not the same as asking under what circumstances a norm is induced by some inner product. The answer to that question is when the norm obeys the parallelogram law. But we could have a situation where the topology is induced by multiple norms, one which is not induced by any inner product and another which is induced by some inner product. In any case, another way to phrase my question is, under what circumstances is a given norm equivalent to some norm obeying the parallelogram law?







general-topology examples-counterexamples normed-spaces inner-product-space topological-vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 15:56









Keshav SrinivasanKeshav Srinivasan

2,33021445




2,33021445












  • $begingroup$
    What does 'bounded' mean in your first paragraph?
    $endgroup$
    – JonathanZ
    Dec 18 '18 at 16:03






  • 2




    $begingroup$
    @JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $lambda U = {lambda x: xin U}$ where $lambda$ is some real number greater than 1.
    $endgroup$
    – Keshav Srinivasan
    Dec 18 '18 at 18:19












  • $begingroup$
    Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:59


















  • $begingroup$
    What does 'bounded' mean in your first paragraph?
    $endgroup$
    – JonathanZ
    Dec 18 '18 at 16:03






  • 2




    $begingroup$
    @JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $lambda U = {lambda x: xin U}$ where $lambda$ is some real number greater than 1.
    $endgroup$
    – Keshav Srinivasan
    Dec 18 '18 at 18:19












  • $begingroup$
    Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:59
















$begingroup$
What does 'bounded' mean in your first paragraph?
$endgroup$
– JonathanZ
Dec 18 '18 at 16:03




$begingroup$
What does 'bounded' mean in your first paragraph?
$endgroup$
– JonathanZ
Dec 18 '18 at 16:03




2




2




$begingroup$
@JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $lambda U = {lambda x: xin U}$ where $lambda$ is some real number greater than 1.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 18:19






$begingroup$
@JonathanZ A subset $B$ of a topological vector space $X$ is said to be bounded if every neighborhood $U$ of the $0$ vector can be stretched to include $B$ as a subset. Stretching the set $U$ means taking the set $lambda U = {lambda x: xin U}$ where $lambda$ is some real number greater than 1.
$endgroup$
– Keshav Srinivasan
Dec 18 '18 at 18:19














$begingroup$
Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179
$endgroup$
– SmileyCraft
Dec 20 '18 at 19:59




$begingroup$
Apparently, any Banach space in which every closed subspace is complemented is isomorphic to a Hilbert space ieeexplore.ieee.org/document/325179
$endgroup$
– SmileyCraft
Dec 20 '18 at 19:59










1 Answer
1






active

oldest

votes


















4





+150







$begingroup$

In his dissertation (Topological characterization of an inner product space), Howard Lambert proves the following:



Theorem 12 Let $X$ be a topological linear space such that there exists a Hamel basis $H$ with the property that the set
begin{equation}
C
=
left{x : x=sum_{i=1}^n alpha_i x_i, x_iin H ,sum_{i=1}^n |alpha_i|^2<1,n text{ arbitrary} right}
end{equation}


is open and bounded, then $X$ is innerproductable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 20:08












  • $begingroup$
    It is not only for finite dimensional spaces.
    $endgroup$
    – Dunham
    Dec 20 '18 at 21:05











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045323%2fwhen-is-a-topological-vector-space-inner-product-able%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4





+150







$begingroup$

In his dissertation (Topological characterization of an inner product space), Howard Lambert proves the following:



Theorem 12 Let $X$ be a topological linear space such that there exists a Hamel basis $H$ with the property that the set
begin{equation}
C
=
left{x : x=sum_{i=1}^n alpha_i x_i, x_iin H ,sum_{i=1}^n |alpha_i|^2<1,n text{ arbitrary} right}
end{equation}


is open and bounded, then $X$ is innerproductable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 20:08












  • $begingroup$
    It is not only for finite dimensional spaces.
    $endgroup$
    – Dunham
    Dec 20 '18 at 21:05
















4





+150







$begingroup$

In his dissertation (Topological characterization of an inner product space), Howard Lambert proves the following:



Theorem 12 Let $X$ be a topological linear space such that there exists a Hamel basis $H$ with the property that the set
begin{equation}
C
=
left{x : x=sum_{i=1}^n alpha_i x_i, x_iin H ,sum_{i=1}^n |alpha_i|^2<1,n text{ arbitrary} right}
end{equation}


is open and bounded, then $X$ is innerproductable.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 20:08












  • $begingroup$
    It is not only for finite dimensional spaces.
    $endgroup$
    – Dunham
    Dec 20 '18 at 21:05














4





+150







4





+150



4




+150



$begingroup$

In his dissertation (Topological characterization of an inner product space), Howard Lambert proves the following:



Theorem 12 Let $X$ be a topological linear space such that there exists a Hamel basis $H$ with the property that the set
begin{equation}
C
=
left{x : x=sum_{i=1}^n alpha_i x_i, x_iin H ,sum_{i=1}^n |alpha_i|^2<1,n text{ arbitrary} right}
end{equation}


is open and bounded, then $X$ is innerproductable.






share|cite|improve this answer









$endgroup$



In his dissertation (Topological characterization of an inner product space), Howard Lambert proves the following:



Theorem 12 Let $X$ be a topological linear space such that there exists a Hamel basis $H$ with the property that the set
begin{equation}
C
=
left{x : x=sum_{i=1}^n alpha_i x_i, x_iin H ,sum_{i=1}^n |alpha_i|^2<1,n text{ arbitrary} right}
end{equation}


is open and bounded, then $X$ is innerproductable.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 19:50









DunhamDunham

2,084614




2,084614












  • $begingroup$
    Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 20:08












  • $begingroup$
    It is not only for finite dimensional spaces.
    $endgroup$
    – Dunham
    Dec 20 '18 at 21:05


















  • $begingroup$
    Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 20:08












  • $begingroup$
    It is not only for finite dimensional spaces.
    $endgroup$
    – Dunham
    Dec 20 '18 at 21:05
















$begingroup$
Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
$endgroup$
– SmileyCraft
Dec 20 '18 at 20:08






$begingroup$
Doesn't this basically follow directly directly from the definitions? And does this theorem only apply to finite dimensional vector spaces?
$endgroup$
– SmileyCraft
Dec 20 '18 at 20:08














$begingroup$
It is not only for finite dimensional spaces.
$endgroup$
– Dunham
Dec 20 '18 at 21:05




$begingroup$
It is not only for finite dimensional spaces.
$endgroup$
– Dunham
Dec 20 '18 at 21:05


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045323%2fwhen-is-a-topological-vector-space-inner-product-able%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix