$v subseteq H implies V^bot$ is a closed subspace of the Hilbert space $H$












2












$begingroup$


Exercise :




Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.




Attempt :



I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.



For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :



$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$



That's really straightforward and simple.



I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.






functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality







share|cite|improve this question









$endgroup$












  • $begingroup$
    $V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
    $endgroup$
    – Federico
    Dec 10 '18 at 19:35










  • $begingroup$
    Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
    $endgroup$
    – Federico
    Dec 10 '18 at 19:39












  • $begingroup$
    @Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
    $endgroup$
    – Rebellos
    Dec 10 '18 at 19:41










  • $begingroup$
    Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
    $endgroup$
    – Federico
    Dec 11 '18 at 13:25


















2












$begingroup$


Exercise :




Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.




Attempt :



I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.



For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :



$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$



That's really straightforward and simple.



I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.






functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality







share|cite|improve this question









$endgroup$












  • $begingroup$
    $V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
    $endgroup$
    – Federico
    Dec 10 '18 at 19:35










  • $begingroup$
    Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
    $endgroup$
    – Federico
    Dec 10 '18 at 19:39












  • $begingroup$
    @Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
    $endgroup$
    – Rebellos
    Dec 10 '18 at 19:41










  • $begingroup$
    Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
    $endgroup$
    – Federico
    Dec 11 '18 at 13:25
















2












2








2





$begingroup$


Exercise :




Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.




Attempt :



I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.



For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :



$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$



That's really straightforward and simple.



I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.






functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality







share|cite|improve this question









$endgroup$




Exercise :




Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.




Attempt :



I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.



For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :



$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$



That's really straightforward and simple.



I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.






functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality




functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 19:31









RebellosRebellos

15.5k31250




15.5k31250












  • $begingroup$
    $V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
    $endgroup$
    – Federico
    Dec 10 '18 at 19:35










  • $begingroup$
    Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
    $endgroup$
    – Federico
    Dec 10 '18 at 19:39












  • $begingroup$
    @Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
    $endgroup$
    – Rebellos
    Dec 10 '18 at 19:41










  • $begingroup$
    Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
    $endgroup$
    – Federico
    Dec 11 '18 at 13:25




















  • $begingroup$
    $V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
    $endgroup$
    – Federico
    Dec 10 '18 at 19:35










  • $begingroup$
    Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
    $endgroup$
    – Federico
    Dec 10 '18 at 19:39












  • $begingroup$
    @Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
    $endgroup$
    – Rebellos
    Dec 10 '18 at 19:41










  • $begingroup$
    Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
    $endgroup$
    – Federico
    Dec 11 '18 at 13:25


















$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35




$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35












$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39






$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39














$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41




$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41












$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25






$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25












2 Answers
2






active

oldest

votes


















1












$begingroup$

Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)



Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.



On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.



So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: An arbitrary intersection of closed subspaces is a closed subspace.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
      $endgroup$
      – Federico
      Dec 10 '18 at 19:37











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)



    Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.



    On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.



    So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)



      Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.



      On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.



      So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)



        Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.



        On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.



        So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.






        share|cite|improve this answer









        $endgroup$



        Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)



        Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.



        On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.



        So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 22:06









        Andreas BlassAndreas Blass

        50.3k452109




        50.3k452109























            0












            $begingroup$

            Hint: An arbitrary intersection of closed subspaces is a closed subspace.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
              $endgroup$
              – Federico
              Dec 10 '18 at 19:37
















            0












            $begingroup$

            Hint: An arbitrary intersection of closed subspaces is a closed subspace.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
              $endgroup$
              – Federico
              Dec 10 '18 at 19:37














            0












            0








            0





            $begingroup$

            Hint: An arbitrary intersection of closed subspaces is a closed subspace.






            share|cite|improve this answer









            $endgroup$



            Hint: An arbitrary intersection of closed subspaces is a closed subspace.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 19:34









            Robert IsraelRobert Israel

            329k23217470




            329k23217470












            • $begingroup$
              See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
              $endgroup$
              – Federico
              Dec 10 '18 at 19:37


















            • $begingroup$
              See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
              $endgroup$
              – Federico
              Dec 10 '18 at 19:37
















            $begingroup$
            See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
            $endgroup$
            – Federico
            Dec 10 '18 at 19:37




            $begingroup$
            See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
            $endgroup$
            – Federico
            Dec 10 '18 at 19:37


















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