What is the minimum of an ordinary function including a given function? [on hold]












-1














Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.










share|cite|improve this question









New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Welcome to MSE. Please include what you tried too
    – Ankit Kumar
    yesterday










  • I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
    – hardmath
    yesterday
















-1














Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.










share|cite|improve this question









New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Welcome to MSE. Please include what you tried too
    – Ankit Kumar
    yesterday










  • I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
    – hardmath
    yesterday














-1












-1








-1







Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.










share|cite|improve this question









New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.







analysis






share|cite|improve this question









New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday





















New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Fei Xu

11




11




New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Welcome to MSE. Please include what you tried too
    – Ankit Kumar
    yesterday










  • I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
    – hardmath
    yesterday


















  • Welcome to MSE. Please include what you tried too
    – Ankit Kumar
    yesterday










  • I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
    – hardmath
    yesterday
















Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday




Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday












I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday




I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday










1 Answer
1






active

oldest

votes


















0














Method one: By the definition of $g$, one can obtain that
begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}



Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
Hence by the definition of $lambda_0$,
begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}






share|cite|improve this answer








New contributor




Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Method one: By the definition of $g$, one can obtain that
    begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
    Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}



    Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
    begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
    Hence by the definition of $lambda_0$,
    begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}






    share|cite|improve this answer








    New contributor




    Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      0














      Method one: By the definition of $g$, one can obtain that
      begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
      Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}



      Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
      begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
      Hence by the definition of $lambda_0$,
      begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}






      share|cite|improve this answer








      New contributor




      Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        0












        0








        0






        Method one: By the definition of $g$, one can obtain that
        begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
        Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}



        Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
        begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
        Hence by the definition of $lambda_0$,
        begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}






        share|cite|improve this answer








        New contributor




        Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        Method one: By the definition of $g$, one can obtain that
        begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
        Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}



        Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
        begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
        Hence by the definition of $lambda_0$,
        begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}







        share|cite|improve this answer








        New contributor




        Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered yesterday









        Fei Xu

        11




        11




        New contributor




        Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.















            Popular posts from this blog

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅