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
    13 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
    13 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
    13 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
    13 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
    13 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
    13 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






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked 14 hours ago









PSG

3749




3749








  • 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
    13 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
    13 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
    13 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
    13 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
    13 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
    13 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
13 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
13 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
13 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
13 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
13 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
13 hours ago












3 Answers
3






active

oldest

votes


















2














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





























    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; you'll get a contradiction.






      share|cite|improve this answer























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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

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        active

        oldest

        votes









        2














        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


























          2














          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
























            2












            2








            2






            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 12 hours ago









            mechanodroid

            26.8k62347




            26.8k62347























                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 13 hours ago









                    supinf

                    6,0791027




                    6,0791027























                        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; you'll get a contradiction.






                        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; you'll get a contradiction.






                          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; you'll get a contradiction.






                            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; you'll get a contradiction.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 11 hours ago

























                            answered 13 hours ago









                            mathcounterexamples.net

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