Uncountable sum of vectors in a Hilbert Space












4














I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?






functional-analysis hilbert-spaces







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  • 2




    Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
    – Kavi Rama Murthy
    2 hours ago










  • Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
    – PSG
    2 hours ago






  • 1




    If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
    – Kavi Rama Murthy
    2 hours ago


















4














I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?






functional-analysis hilbert-spaces







share|cite|improve this question


















  • 2




    Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
    – Kavi Rama Murthy
    2 hours ago










  • Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
    – PSG
    2 hours ago






  • 1




    If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
    – Kavi Rama Murthy
    2 hours ago
















4












4








4







I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?






functional-analysis hilbert-spaces







share|cite|improve this question













I am currently reading Hilbert Spaces and confused about a thing. Say, $C={e_alpha : alphainmathcal{A}}$ be a complete orthonormal set of a Hilbert Space $H$, possibly uncountable. Is $sum_{alphainmathcal{A}}e_alpha$ well defined ? I think it should be, is there some kind of convergence needed for these sums?






functional-analysis hilbert-spaces




functional-analysis hilbert-spaces






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asked 2 hours ago









PSG

3709




3709








  • 2




    Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
    – Kavi Rama Murthy
    2 hours ago










  • Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
    – PSG
    2 hours ago






  • 1




    If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
    – Kavi Rama Murthy
    2 hours ago
















  • 2




    Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
    – Kavi Rama Murthy
    2 hours ago










  • Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
    – PSG
    2 hours ago






  • 1




    If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
    – Kavi Rama Murthy
    2 hours ago










2




2




Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
2 hours ago




Even when $mathcal A$ is countably infinite the sum $sum e_n$ does not exist either in the norm or weakly.
– Kavi Rama Murthy
2 hours ago












Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
2 hours ago




Oh, I see if it was a well defined vector, then the norm would be in $mathbf{R}$, which is not the case here. Is that ok? @KaviRamaMurthy
– PSG
2 hours ago




1




1




If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
2 hours ago






If $sum e_n=x$, say in weak topology, then we have $sum langle y,e_nrangle =langle y,xrangle$ for all $y$. You get a contradiction by taking $y=sum frac 1 j e_j$.
– Kavi Rama Murthy
2 hours ago












3 Answers
3






active

oldest

votes


















5














In an infinite dimensional Hilbert space the term
$$
sum_{ain mathcal A} e_a
$$
is not well defined.
Even for countable $mathcal A$ this sum does not converge.



For defining uncountable sums it is usually required that at most countable many summands are nonzero
and that the countable sum over the nonzero entries converges absolutely.






share|cite|improve this answer





























    2














    The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?



    The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:



    $$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$



    Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.



    This has interesting consequences.




    1. Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.

    2. For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.






    share|cite|improve this answer





























      1














      The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
      $$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$



      However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.



      Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.



      For any $F_0 subseteq F subseteq A$ finite we have
      $$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
      left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$



      so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        5














        In an infinite dimensional Hilbert space the term
        $$
        sum_{ain mathcal A} e_a
        $$
        is not well defined.
        Even for countable $mathcal A$ this sum does not converge.



        For defining uncountable sums it is usually required that at most countable many summands are nonzero
        and that the countable sum over the nonzero entries converges absolutely.






        share|cite|improve this answer


























          5














          In an infinite dimensional Hilbert space the term
          $$
          sum_{ain mathcal A} e_a
          $$
          is not well defined.
          Even for countable $mathcal A$ this sum does not converge.



          For defining uncountable sums it is usually required that at most countable many summands are nonzero
          and that the countable sum over the nonzero entries converges absolutely.






          share|cite|improve this answer
























            5












            5








            5






            In an infinite dimensional Hilbert space the term
            $$
            sum_{ain mathcal A} e_a
            $$
            is not well defined.
            Even for countable $mathcal A$ this sum does not converge.



            For defining uncountable sums it is usually required that at most countable many summands are nonzero
            and that the countable sum over the nonzero entries converges absolutely.






            share|cite|improve this answer












            In an infinite dimensional Hilbert space the term
            $$
            sum_{ain mathcal A} e_a
            $$
            is not well defined.
            Even for countable $mathcal A$ this sum does not converge.



            For defining uncountable sums it is usually required that at most countable many summands are nonzero
            and that the countable sum over the nonzero entries converges absolutely.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            supinf

            6,0541027




            6,0541027























                2














                The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?



                The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:



                $$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$



                Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.



                This has interesting consequences.




                1. Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.

                2. For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.






                share|cite|improve this answer


























                  2














                  The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?



                  The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:



                  $$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$



                  Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.



                  This has interesting consequences.




                  1. Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.

                  2. For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?



                    The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:



                    $$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$



                    Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.



                    This has interesting consequences.




                    1. Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.

                    2. For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.






                    share|cite|improve this answer












                    The background of your question is how do you define $displaystyle sum_{i in I} c_i$ in a Banach space for any set $I$?



                    The usual way is to say that the sum $displaystyle sum_{i in I} c_alpha$ of a set $mathcal C = {c_i ; i in I}$ of vectors exists when following Cauchy criteria is met:



                    $$(forall epsilon > 0) , (exists J_0 in mathcal F(I)) , (forall K in mathcal F(I setminus J_0)) , leftVert displaystyle sum_{k in K} c_k rightVert< epsilon $$



                    Where $mathcal F(A)$ is defined as the sets of finite subsets of $A$.



                    This has interesting consequences.




                    1. Such a sum $displaystyle sum_{i in I} c_alpha$ does make sense only if the set of non zero elements of $mathcal C$ is at most countable.

                    2. For a family of uncountable vectors of norm equal to $1$, which is your initial question, the sum cannot exist: take $epsilon = 1/2$ in the definition above.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    mathcounterexamples.net

                    24.8k21753




                    24.8k21753























                        1














                        The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
                        $$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$



                        However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.



                        Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.



                        For any $F_0 subseteq F subseteq A$ finite we have
                        $$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
                        left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$



                        so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.






                        share|cite|improve this answer


























                          1














                          The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
                          $$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$



                          However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.



                          Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.



                          For any $F_0 subseteq F subseteq A$ finite we have
                          $$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
                          left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$



                          so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
                            $$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$



                            However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.



                            Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.



                            For any $F_0 subseteq F subseteq A$ finite we have
                            $$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
                            left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$



                            so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.






                            share|cite|improve this answer












                            The sum $sum_{alphain A} v_alpha$ does make sense. It is defined to converge to $L in H$ if
                            $$forall varepsilon > 0 ,exists F_0 subseteq A text{ finite such that }forall F subseteq A text{ finite}, F supseteq F_0 text{ we have} left|sum_{alphain F}v_alpha - Lright| < varepsilon$$



                            However, the sum $sum_{alphain A} e_alpha$ of an orthonormal set only converges if $A$ is finite.



                            Indeed, assume that $sum_{alphain A} e_alpha = L$. For $varepsilon = frac12$ there exists $F_0 subseteq A$ finite such that for all $F subseteq A$ finite, $F supseteq F_0$ implies $left|sum_{alphain F} e_alpha -Lright| < frac12$.



                            For any $F_0 subseteq F subseteq A$ finite we have
                            $$sqrt{|F setminus F_0|}= left|sum_{alphain Fsetminus F_0}e_alpharight| = left|sum_{alphain F}e_alpha - sum_{alphain F_0}e_alpharight| le
                            left|sum_{alphain F}e_alpha - Lright| +left|L- sum_{alphain F_0}e_alpharight| < 1$$



                            so $F = F_0$. Therefore it necessarily holds $F_0 = A$ so $A$ is finite.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 34 mins ago









                            mechanodroid

                            26.8k62347




                            26.8k62347






























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                                濃尾地震

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                                濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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