Question on Hilbert Space












1












$begingroup$


$S$ is a non-empty subset of a Hilbert Space $H$, show that the set of all linear combinations of vectors in $S$ is dense in $HLeftrightarrow S^{bot}={0}$



I have been able to show $Rightarrow$ direction : Call the set of all linear combinations of vectors in $S$ as $S_1$ Say, $vin S^{bot}, exists$ a sequence $v_n$ in $S_1$ such that $v_nrightarrow v$ as $nrightarrowinfty$, $v_nbot v $
$forall ninmathbf{N}$ $therefore lim_{nrightarrowinfty}|v_n-v|^2=lim_{nrightarrowinfty} |v_n|^2+|v|^2=2|v|^2=0 Rightarrow v=0$



I am struggling with the other implication, We have $S^{botbot}=H$, need to show closure of $S_1$ is $S^{botbot}=H$






functional-analysis hilbert-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    $S$ is a non-empty subset of a Hilbert Space $H$, show that the set of all linear combinations of vectors in $S$ is dense in $HLeftrightarrow S^{bot}={0}$



    I have been able to show $Rightarrow$ direction : Call the set of all linear combinations of vectors in $S$ as $S_1$ Say, $vin S^{bot}, exists$ a sequence $v_n$ in $S_1$ such that $v_nrightarrow v$ as $nrightarrowinfty$, $v_nbot v $
    $forall ninmathbf{N}$ $therefore lim_{nrightarrowinfty}|v_n-v|^2=lim_{nrightarrowinfty} |v_n|^2+|v|^2=2|v|^2=0 Rightarrow v=0$



    I am struggling with the other implication, We have $S^{botbot}=H$, need to show closure of $S_1$ is $S^{botbot}=H$






    functional-analysis hilbert-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      $S$ is a non-empty subset of a Hilbert Space $H$, show that the set of all linear combinations of vectors in $S$ is dense in $HLeftrightarrow S^{bot}={0}$



      I have been able to show $Rightarrow$ direction : Call the set of all linear combinations of vectors in $S$ as $S_1$ Say, $vin S^{bot}, exists$ a sequence $v_n$ in $S_1$ such that $v_nrightarrow v$ as $nrightarrowinfty$, $v_nbot v $
      $forall ninmathbf{N}$ $therefore lim_{nrightarrowinfty}|v_n-v|^2=lim_{nrightarrowinfty} |v_n|^2+|v|^2=2|v|^2=0 Rightarrow v=0$



      I am struggling with the other implication, We have $S^{botbot}=H$, need to show closure of $S_1$ is $S^{botbot}=H$






      functional-analysis hilbert-spaces







      share|cite|improve this question











      $endgroup$




      $S$ is a non-empty subset of a Hilbert Space $H$, show that the set of all linear combinations of vectors in $S$ is dense in $HLeftrightarrow S^{bot}={0}$



      I have been able to show $Rightarrow$ direction : Call the set of all linear combinations of vectors in $S$ as $S_1$ Say, $vin S^{bot}, exists$ a sequence $v_n$ in $S_1$ such that $v_nrightarrow v$ as $nrightarrowinfty$, $v_nbot v $
      $forall ninmathbf{N}$ $therefore lim_{nrightarrowinfty}|v_n-v|^2=lim_{nrightarrowinfty} |v_n|^2+|v|^2=2|v|^2=0 Rightarrow v=0$



      I am struggling with the other implication, We have $S^{botbot}=H$, need to show closure of $S_1$ is $S^{botbot}=H$






      functional-analysis hilbert-spaces




      functional-analysis hilbert-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 18:53









      mechanodroid

      28.3k62548




      28.3k62548










      asked Jan 2 at 16:18









      PSGPSG

      3849




      3849






















          2 Answers
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          $begingroup$

          Notice that $left(overline{operatorname{span}S}right)^perp = S^perp$. The Riesz projection theorem implies $$left(overline{operatorname{span}S}right)oplus left(overline{operatorname{span}S}right)^perp = H$$
          so $$text{linear span of } S text{ is dense in } H iffoverline{operatorname{span}S} = H iff left(overline{operatorname{span}S}right)^perp = {0} iff S^perp = {0}$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            The adherence $bar S_1$ of $S_1$ is a vector subspace, if $S_1$ is not dense, there exists $x$ which is not in $bar{S_1}$. The linear form defined on $bar{S_1}oplusmathbb{C}x$ by $f(bar{S_1})=0, f(x)=1$ can be extended to $H$ by $g$ since it is continuous (Hahn Banach). Use te Riesz representation theorem to write $g=langle x_0,.rangle$. This implies that $x_0$ is in the orthogonal of $bar{S_1}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
              $endgroup$
              – PSG
              Jan 2 at 16:50












            • $begingroup$
              how you know the existence of $v$ ?
              $endgroup$
              – Tsemo Aristide
              Jan 2 at 16:52










            • $begingroup$
              A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
              $endgroup$
              – PSG
              Jan 2 at 16:54












            • $begingroup$
              that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
              $endgroup$
              – Tsemo Aristide
              Jan 2 at 16:56











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            2 Answers
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            2 Answers
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            2












            $begingroup$

            Notice that $left(overline{operatorname{span}S}right)^perp = S^perp$. The Riesz projection theorem implies $$left(overline{operatorname{span}S}right)oplus left(overline{operatorname{span}S}right)^perp = H$$
            so $$text{linear span of } S text{ is dense in } H iffoverline{operatorname{span}S} = H iff left(overline{operatorname{span}S}right)^perp = {0} iff S^perp = {0}$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Notice that $left(overline{operatorname{span}S}right)^perp = S^perp$. The Riesz projection theorem implies $$left(overline{operatorname{span}S}right)oplus left(overline{operatorname{span}S}right)^perp = H$$
              so $$text{linear span of } S text{ is dense in } H iffoverline{operatorname{span}S} = H iff left(overline{operatorname{span}S}right)^perp = {0} iff S^perp = {0}$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Notice that $left(overline{operatorname{span}S}right)^perp = S^perp$. The Riesz projection theorem implies $$left(overline{operatorname{span}S}right)oplus left(overline{operatorname{span}S}right)^perp = H$$
                so $$text{linear span of } S text{ is dense in } H iffoverline{operatorname{span}S} = H iff left(overline{operatorname{span}S}right)^perp = {0} iff S^perp = {0}$$






                share|cite|improve this answer









                $endgroup$



                Notice that $left(overline{operatorname{span}S}right)^perp = S^perp$. The Riesz projection theorem implies $$left(overline{operatorname{span}S}right)oplus left(overline{operatorname{span}S}right)^perp = H$$
                so $$text{linear span of } S text{ is dense in } H iffoverline{operatorname{span}S} = H iff left(overline{operatorname{span}S}right)^perp = {0} iff S^perp = {0}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 18:52









                mechanodroidmechanodroid

                28.3k62548




                28.3k62548























                    1












                    $begingroup$

                    The adherence $bar S_1$ of $S_1$ is a vector subspace, if $S_1$ is not dense, there exists $x$ which is not in $bar{S_1}$. The linear form defined on $bar{S_1}oplusmathbb{C}x$ by $f(bar{S_1})=0, f(x)=1$ can be extended to $H$ by $g$ since it is continuous (Hahn Banach). Use te Riesz representation theorem to write $g=langle x_0,.rangle$. This implies that $x_0$ is in the orthogonal of $bar{S_1}$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:50












                    • $begingroup$
                      how you know the existence of $v$ ?
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:52










                    • $begingroup$
                      A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:54












                    • $begingroup$
                      that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:56
















                    1












                    $begingroup$

                    The adherence $bar S_1$ of $S_1$ is a vector subspace, if $S_1$ is not dense, there exists $x$ which is not in $bar{S_1}$. The linear form defined on $bar{S_1}oplusmathbb{C}x$ by $f(bar{S_1})=0, f(x)=1$ can be extended to $H$ by $g$ since it is continuous (Hahn Banach). Use te Riesz representation theorem to write $g=langle x_0,.rangle$. This implies that $x_0$ is in the orthogonal of $bar{S_1}$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:50












                    • $begingroup$
                      how you know the existence of $v$ ?
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:52










                    • $begingroup$
                      A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:54












                    • $begingroup$
                      that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:56














                    1












                    1








                    1





                    $begingroup$

                    The adherence $bar S_1$ of $S_1$ is a vector subspace, if $S_1$ is not dense, there exists $x$ which is not in $bar{S_1}$. The linear form defined on $bar{S_1}oplusmathbb{C}x$ by $f(bar{S_1})=0, f(x)=1$ can be extended to $H$ by $g$ since it is continuous (Hahn Banach). Use te Riesz representation theorem to write $g=langle x_0,.rangle$. This implies that $x_0$ is in the orthogonal of $bar{S_1}$.






                    share|cite|improve this answer









                    $endgroup$



                    The adherence $bar S_1$ of $S_1$ is a vector subspace, if $S_1$ is not dense, there exists $x$ which is not in $bar{S_1}$. The linear form defined on $bar{S_1}oplusmathbb{C}x$ by $f(bar{S_1})=0, f(x)=1$ can be extended to $H$ by $g$ since it is continuous (Hahn Banach). Use te Riesz representation theorem to write $g=langle x_0,.rangle$. This implies that $x_0$ is in the orthogonal of $bar{S_1}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 2 at 16:34









                    Tsemo AristideTsemo Aristide

                    59k11445




                    59k11445












                    • $begingroup$
                      Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:50












                    • $begingroup$
                      how you know the existence of $v$ ?
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:52










                    • $begingroup$
                      A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:54












                    • $begingroup$
                      that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:56


















                    • $begingroup$
                      Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:50












                    • $begingroup$
                      how you know the existence of $v$ ?
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:52










                    • $begingroup$
                      A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
                      $endgroup$
                      – PSG
                      Jan 2 at 16:54












                    • $begingroup$
                      that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
                      $endgroup$
                      – Tsemo Aristide
                      Jan 2 at 16:56
















                    $begingroup$
                    Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
                    $endgroup$
                    – PSG
                    Jan 2 at 16:50






                    $begingroup$
                    Thanks for the answer, actually I am not supposed to use Hahn banach for this question, I got an idea from your solution, please hear me out. If $bar{S_1} neq H$, then $exists$ non-zero $vin bar {S_1} ^{bot} $, but $S_1 subseteq bar {S_1}Rightarrow S_1^{bot}supseteq bar{S_1}^{bot} $ Now I want to say that $S^{bot}=0Rightarrow S_1 ^{bot}=0$, so $bar{S_1}^{bot} =0$ , is this correct : $S^{bot}=0Rightarrow S_1 ^{bot}=0$
                    $endgroup$
                    – PSG
                    Jan 2 at 16:50














                    $begingroup$
                    how you know the existence of $v$ ?
                    $endgroup$
                    – Tsemo Aristide
                    Jan 2 at 16:52




                    $begingroup$
                    how you know the existence of $v$ ?
                    $endgroup$
                    – Tsemo Aristide
                    Jan 2 at 16:52












                    $begingroup$
                    A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
                    $endgroup$
                    – PSG
                    Jan 2 at 16:54






                    $begingroup$
                    A theorem says that : If $M$ is a proper closed linear subspace of a Hilbert space $H$, then $exists$ a non zero vector $z_0$ in $H$ such that $z_0bot M$
                    $endgroup$
                    – PSG
                    Jan 2 at 16:54














                    $begingroup$
                    that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
                    $endgroup$
                    – Tsemo Aristide
                    Jan 2 at 16:56




                    $begingroup$
                    that's true I use Hahn Banach and Riez to show that theorem, you can show it also by considering a point which realize $d(x,bar{S_1})$.
                    $endgroup$
                    – Tsemo Aristide
                    Jan 2 at 16:56


















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