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












1














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







share|cite|improve this question





























    1














    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







    share|cite|improve this question



























      1












      1








      1







      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







      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




      functional-analysis analysis pde galerkin-methods






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 19 at 18:48

























      asked Nov 19 at 0:56









      João Paulo Andrade

      315




      315






















          1 Answer
          1






          active

          oldest

          votes


















          1














          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























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004367%2fwhy-of-this-bound-u-m0-leq-u-0-where-u-m0-is-a-projection-of%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            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




























              1














              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


























                1












                1








                1






                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














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 21 at 3:42

























                answered Nov 21 at 3:19









                Pedro

                10.2k23068




                10.2k23068






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004367%2fwhy-of-this-bound-u-m0-leq-u-0-where-u-m0-is-a-projection-of%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    AnyDesk - Fatal Program Failure

                    Image
                    up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
                    Read more

                    How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                    Image
                    up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
                    Read more

                    QoS: MAC-Priority for clients behind a repeater

                    Image
                    1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
                    Read more