Why of this bound $||u_{m0}|| leq ||u_0||$, where $u_{m0}$ is a projection of $u_0$?











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I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.



Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.



The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.



Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.



So, why this bound is valid,
$||u_{m0}|| leq ||u_0||$?



$||cdot|| $ is the norm in $H^1_0$.










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    up vote
    1
    down vote

    favorite












    I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.



    Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.



    The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.



    Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.



    So, why this bound is valid,
    $||u_{m0}|| leq ||u_0||$?



    $||cdot|| $ is the norm in $H^1_0$.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.



      Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.



      The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.



      Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.



      So, why this bound is valid,
      $||u_{m0}|| leq ||u_0||$?



      $||cdot|| $ is the norm in $H^1_0$.










      share|cite|improve this question















      I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.



      Consider $u_m=sum_{i=1}^m g_{im}(t)w_i$ an approximation of $u$ via galerkin method, where $w_i$ are the basis and $g_{im}(t)$ the coeficients, $u_m(0)=u_{m0}$ and $u(0)=u_0$.



      The space $V$ is basecally the closure, in $H^1_0(Omega)$, of the set of all function $C^infty$ with support on a open set $Omega subset mathbb{R}^2$, with it's divergent being null.



      Let ${w_j}$ be a base of $V$ which we can take $u_{m0}$ as a projection in $V$ of $u_0$ spanned by $w_1,w_2, cdots,w_m $.



      So, why this bound is valid,
      $||u_{m0}|| leq ||u_0||$?



      $||cdot|| $ is the norm in $H^1_0$.







      functional-analysis analysis pde galerkin-methods






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      edited 2 days ago

























      asked Nov 19 at 0:56









      João Paulo Andrade

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          Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:




          Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:





          • ${e_j}_{jin J}$ is maximal.


          • $ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.


          • $ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.


          • $overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.




          Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.



          From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".



          Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that



          $$begin{aligned}
          |u_{m0}|^2&=(u_{m0},u_{m0})\
          &=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
          &=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
          &=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
          &=sum_{i=1}^m|(u_0,w_i)|^2\
          &leq sum_{i=1}^infty|(u_0,w_i)|^2\
          &=|u_0|^2
          end{aligned}$$






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            Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:




            Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:





            • ${e_j}_{jin J}$ is maximal.


            • $ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.


            • $ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.


            • $overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.




            Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.



            From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".



            Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that



            $$begin{aligned}
            |u_{m0}|^2&=(u_{m0},u_{m0})\
            &=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
            &=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
            &=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
            &=sum_{i=1}^m|(u_0,w_i)|^2\
            &leq sum_{i=1}^infty|(u_0,w_i)|^2\
            &=|u_0|^2
            end{aligned}$$






            share|cite|improve this answer



























              up vote
              0
              down vote



              accepted










              Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:




              Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:





              • ${e_j}_{jin J}$ is maximal.


              • $ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.


              • $ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.


              • $overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.




              Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.



              From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".



              Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that



              $$begin{aligned}
              |u_{m0}|^2&=(u_{m0},u_{m0})\
              &=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
              &=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
              &=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
              &=sum_{i=1}^m|(u_0,w_i)|^2\
              &leq sum_{i=1}^infty|(u_0,w_i)|^2\
              &=|u_0|^2
              end{aligned}$$






              share|cite|improve this answer

























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:




                Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:





                • ${e_j}_{jin J}$ is maximal.


                • $ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.


                • $ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.


                • $overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.




                Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.



                From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".



                Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that



                $$begin{aligned}
                |u_{m0}|^2&=(u_{m0},u_{m0})\
                &=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
                &=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
                &=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
                &=sum_{i=1}^m|(u_0,w_i)|^2\
                &leq sum_{i=1}^infty|(u_0,w_i)|^2\
                &=|u_0|^2
                end{aligned}$$






                share|cite|improve this answer














                Usually, in the context of Galerkin method, "base" of a separable normed space $X$ means a linearly indpendent countable set whose span is dense in $X$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:




                Theorem. Let ${e_j}_{jin J}$ be an orthonormal family in a Hilbert space $H$. The following assertions are equivalent:





                • ${e_j}_{jin J}$ is maximal.


                • $ x=sum_{jin J}(x,e_j)e_j$ for all $xin mathcal{H}$.


                • $ |x|^2=sum_{jin J}|(x,e_j)|^2$ for all $xin mathcal{H}$.


                • $overline{operatorname{span}{e_jmid jin J}}=mathcal{H}$.




                Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.



                From this theorem, if $X$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".



                Therefore, assuming that $(w_j)_{jinmathbb N}$ is base of $V$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that



                $$begin{aligned}
                |u_{m0}|^2&=(u_{m0},u_{m0})\
                &=left(sum_{i=1}^m(u_0,w_i)w_i,sum_{j=1}^m(u_0,w_j)w_jright)\
                &=sum_{i=1}^m(u_0,w_i)sum_{j=1}^moverline{(u_0,w_j)}left(w_i,w_jright)\
                &=sum_{i=1}^m|(u_0,w_i)|^2left(w_i,w_iright)\
                &=sum_{i=1}^m|(u_0,w_i)|^2\
                &leq sum_{i=1}^infty|(u_0,w_i)|^2\
                &=|u_0|^2
                end{aligned}$$







                share|cite|improve this answer














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                edited yesterday

























                answered yesterday









                Pedro

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