Using the Banach fixed point Theorem












4












$begingroup$


In Chapter 7, The Hille-Yosida Theorem,




Functional Analysis,
Sobolev Spaces and Partial
Differential Equations - Brezis, 2011.




Brezis showed the following claim (in Proposition 7.1).



Claim: Suppose that $A$ is a maximal monotone operator, $(I+lambda A): D(A) to R(I+lambda A)$ is a injective operator, and $|(I+lambda A)^{-1}u| leq |u| $ for all $u in R(I+lambda A)$ for all $lambda >0$. Then $R(I+lambda A) = H$.



The proof of Brezis:




We will prove that if $R(I+lambda_{0} A) = H$ for some $lambda_{0}>0$ then
$R(I+lambda A) = H$ for every $lambda > frac{1}{2}lambda_{0}$.



For some $f in H$, we try to solve the equation



$u+lambda Au = f$ with $lambda >0$. (1)



Equation (1) may be written as



$u+ lambda_{0}Au = frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)$



or alternatively



$u= (I+lambda A)^{-1}big[frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)big]$ $(2)$



If $|1-frac{lambda_{0}}{lambda}|< 1$, i.e., $lambda > frac{1}{2}lambda_{0}$, we may apply the contraction mapping principle (the Banach Fixed point Theorem) and deduce that (2) has a solution.




My question: how to prove that (2) has a solution using the the Banach Fixed point Theorem?



Thanks




Definitions:



An unbounded linear operator $A: D(A)subseteq H to H$ is said to be monotone if it satisfies



$langle A u, u rangle geq 0$ for all $uin D(A)$.



It is called maximal monotone if, in addition, $R(I+A)=H$.






functional-analysis pde







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    In Chapter 7, The Hille-Yosida Theorem,




    Functional Analysis,
    Sobolev Spaces and Partial
    Differential Equations - Brezis, 2011.




    Brezis showed the following claim (in Proposition 7.1).



    Claim: Suppose that $A$ is a maximal monotone operator, $(I+lambda A): D(A) to R(I+lambda A)$ is a injective operator, and $|(I+lambda A)^{-1}u| leq |u| $ for all $u in R(I+lambda A)$ for all $lambda >0$. Then $R(I+lambda A) = H$.



    The proof of Brezis:




    We will prove that if $R(I+lambda_{0} A) = H$ for some $lambda_{0}>0$ then
    $R(I+lambda A) = H$ for every $lambda > frac{1}{2}lambda_{0}$.



    For some $f in H$, we try to solve the equation



    $u+lambda Au = f$ with $lambda >0$. (1)



    Equation (1) may be written as



    $u+ lambda_{0}Au = frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)$



    or alternatively



    $u= (I+lambda A)^{-1}big[frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)big]$ $(2)$



    If $|1-frac{lambda_{0}}{lambda}|< 1$, i.e., $lambda > frac{1}{2}lambda_{0}$, we may apply the contraction mapping principle (the Banach Fixed point Theorem) and deduce that (2) has a solution.




    My question: how to prove that (2) has a solution using the the Banach Fixed point Theorem?



    Thanks




    Definitions:



    An unbounded linear operator $A: D(A)subseteq H to H$ is said to be monotone if it satisfies



    $langle A u, u rangle geq 0$ for all $uin D(A)$.



    It is called maximal monotone if, in addition, $R(I+A)=H$.






    functional-analysis pde







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      In Chapter 7, The Hille-Yosida Theorem,




      Functional Analysis,
      Sobolev Spaces and Partial
      Differential Equations - Brezis, 2011.




      Brezis showed the following claim (in Proposition 7.1).



      Claim: Suppose that $A$ is a maximal monotone operator, $(I+lambda A): D(A) to R(I+lambda A)$ is a injective operator, and $|(I+lambda A)^{-1}u| leq |u| $ for all $u in R(I+lambda A)$ for all $lambda >0$. Then $R(I+lambda A) = H$.



      The proof of Brezis:




      We will prove that if $R(I+lambda_{0} A) = H$ for some $lambda_{0}>0$ then
      $R(I+lambda A) = H$ for every $lambda > frac{1}{2}lambda_{0}$.



      For some $f in H$, we try to solve the equation



      $u+lambda Au = f$ with $lambda >0$. (1)



      Equation (1) may be written as



      $u+ lambda_{0}Au = frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)$



      or alternatively



      $u= (I+lambda A)^{-1}big[frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)big]$ $(2)$



      If $|1-frac{lambda_{0}}{lambda}|< 1$, i.e., $lambda > frac{1}{2}lambda_{0}$, we may apply the contraction mapping principle (the Banach Fixed point Theorem) and deduce that (2) has a solution.




      My question: how to prove that (2) has a solution using the the Banach Fixed point Theorem?



      Thanks




      Definitions:



      An unbounded linear operator $A: D(A)subseteq H to H$ is said to be monotone if it satisfies



      $langle A u, u rangle geq 0$ for all $uin D(A)$.



      It is called maximal monotone if, in addition, $R(I+A)=H$.






      functional-analysis pde







      share|cite|improve this question









      $endgroup$




      In Chapter 7, The Hille-Yosida Theorem,




      Functional Analysis,
      Sobolev Spaces and Partial
      Differential Equations - Brezis, 2011.




      Brezis showed the following claim (in Proposition 7.1).



      Claim: Suppose that $A$ is a maximal monotone operator, $(I+lambda A): D(A) to R(I+lambda A)$ is a injective operator, and $|(I+lambda A)^{-1}u| leq |u| $ for all $u in R(I+lambda A)$ for all $lambda >0$. Then $R(I+lambda A) = H$.



      The proof of Brezis:




      We will prove that if $R(I+lambda_{0} A) = H$ for some $lambda_{0}>0$ then
      $R(I+lambda A) = H$ for every $lambda > frac{1}{2}lambda_{0}$.



      For some $f in H$, we try to solve the equation



      $u+lambda Au = f$ with $lambda >0$. (1)



      Equation (1) may be written as



      $u+ lambda_{0}Au = frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)$



      or alternatively



      $u= (I+lambda A)^{-1}big[frac{lambda_{0}}{lambda}f+ big( 1- frac{lambda_{0}}{lambda}u big)big]$ $(2)$



      If $|1-frac{lambda_{0}}{lambda}|< 1$, i.e., $lambda > frac{1}{2}lambda_{0}$, we may apply the contraction mapping principle (the Banach Fixed point Theorem) and deduce that (2) has a solution.




      My question: how to prove that (2) has a solution using the the Banach Fixed point Theorem?



      Thanks




      Definitions:



      An unbounded linear operator $A: D(A)subseteq H to H$ is said to be monotone if it satisfies



      $langle A u, u rangle geq 0$ for all $uin D(A)$.



      It is called maximal monotone if, in addition, $R(I+A)=H$.






      functional-analysis pde




      functional-analysis pde






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Nov 30 '18 at 22:42









      Joao MaiaJoao Maia

      1586




      1586






















          2 Answers
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          $begingroup$

          There are some typos in your derivation. Specifically the correct form of $(2)$ is:
          $$u=(I+lambda_0 A)^{-1} left[frac{lambda_0}lambda fright]+(I+lambda_0 A)^{-1}left[(1-frac{lambda_0}lambda)uright].$$
          Your equation is of the form
          $$u=v + Bu$$
          for a constant $v$ and a linear map $B$. In the case that $B$ is a contraction then $v+Bu$ is a contraction and you are done. So why is $B=(1-frac{lambda_0}lambda)(I+lambda_0A)^{-1}$ a contraction? Here note that $|(I+lambda_0 A)^{-1}|≤1$, use the positivity of $A$ if you wish. This gives you
          $$|B|≤|1-frac{lambda_0}lambda|<1,$$
          now you are finished.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Banach's fixed point theorem states that if an operator $T colon X to X$ on the non-empty complete metric space $(X, d)$ is a contraction, then $T$ has a unique fixed point $x^* in X$ such that $Tx^* = x^*$. This means that if you can show that $T$ is a contraction, i.e. $d(Tx, Ty) leq Kd(x, y)$ for some constant $0 leq K < 1$, then $T$ has a fixed point. If you can show that (2) is indeed a contraction, then there will exist a unique solution.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              $begingroup$

              There are some typos in your derivation. Specifically the correct form of $(2)$ is:
              $$u=(I+lambda_0 A)^{-1} left[frac{lambda_0}lambda fright]+(I+lambda_0 A)^{-1}left[(1-frac{lambda_0}lambda)uright].$$
              Your equation is of the form
              $$u=v + Bu$$
              for a constant $v$ and a linear map $B$. In the case that $B$ is a contraction then $v+Bu$ is a contraction and you are done. So why is $B=(1-frac{lambda_0}lambda)(I+lambda_0A)^{-1}$ a contraction? Here note that $|(I+lambda_0 A)^{-1}|≤1$, use the positivity of $A$ if you wish. This gives you
              $$|B|≤|1-frac{lambda_0}lambda|<1,$$
              now you are finished.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                There are some typos in your derivation. Specifically the correct form of $(2)$ is:
                $$u=(I+lambda_0 A)^{-1} left[frac{lambda_0}lambda fright]+(I+lambda_0 A)^{-1}left[(1-frac{lambda_0}lambda)uright].$$
                Your equation is of the form
                $$u=v + Bu$$
                for a constant $v$ and a linear map $B$. In the case that $B$ is a contraction then $v+Bu$ is a contraction and you are done. So why is $B=(1-frac{lambda_0}lambda)(I+lambda_0A)^{-1}$ a contraction? Here note that $|(I+lambda_0 A)^{-1}|≤1$, use the positivity of $A$ if you wish. This gives you
                $$|B|≤|1-frac{lambda_0}lambda|<1,$$
                now you are finished.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  There are some typos in your derivation. Specifically the correct form of $(2)$ is:
                  $$u=(I+lambda_0 A)^{-1} left[frac{lambda_0}lambda fright]+(I+lambda_0 A)^{-1}left[(1-frac{lambda_0}lambda)uright].$$
                  Your equation is of the form
                  $$u=v + Bu$$
                  for a constant $v$ and a linear map $B$. In the case that $B$ is a contraction then $v+Bu$ is a contraction and you are done. So why is $B=(1-frac{lambda_0}lambda)(I+lambda_0A)^{-1}$ a contraction? Here note that $|(I+lambda_0 A)^{-1}|≤1$, use the positivity of $A$ if you wish. This gives you
                  $$|B|≤|1-frac{lambda_0}lambda|<1,$$
                  now you are finished.






                  share|cite|improve this answer









                  $endgroup$



                  There are some typos in your derivation. Specifically the correct form of $(2)$ is:
                  $$u=(I+lambda_0 A)^{-1} left[frac{lambda_0}lambda fright]+(I+lambda_0 A)^{-1}left[(1-frac{lambda_0}lambda)uright].$$
                  Your equation is of the form
                  $$u=v + Bu$$
                  for a constant $v$ and a linear map $B$. In the case that $B$ is a contraction then $v+Bu$ is a contraction and you are done. So why is $B=(1-frac{lambda_0}lambda)(I+lambda_0A)^{-1}$ a contraction? Here note that $|(I+lambda_0 A)^{-1}|≤1$, use the positivity of $A$ if you wish. This gives you
                  $$|B|≤|1-frac{lambda_0}lambda|<1,$$
                  now you are finished.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 23:11









                  s.harps.harp

                  8,42812049




                  8,42812049























                      1












                      $begingroup$

                      Banach's fixed point theorem states that if an operator $T colon X to X$ on the non-empty complete metric space $(X, d)$ is a contraction, then $T$ has a unique fixed point $x^* in X$ such that $Tx^* = x^*$. This means that if you can show that $T$ is a contraction, i.e. $d(Tx, Ty) leq Kd(x, y)$ for some constant $0 leq K < 1$, then $T$ has a fixed point. If you can show that (2) is indeed a contraction, then there will exist a unique solution.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Banach's fixed point theorem states that if an operator $T colon X to X$ on the non-empty complete metric space $(X, d)$ is a contraction, then $T$ has a unique fixed point $x^* in X$ such that $Tx^* = x^*$. This means that if you can show that $T$ is a contraction, i.e. $d(Tx, Ty) leq Kd(x, y)$ for some constant $0 leq K < 1$, then $T$ has a fixed point. If you can show that (2) is indeed a contraction, then there will exist a unique solution.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Banach's fixed point theorem states that if an operator $T colon X to X$ on the non-empty complete metric space $(X, d)$ is a contraction, then $T$ has a unique fixed point $x^* in X$ such that $Tx^* = x^*$. This means that if you can show that $T$ is a contraction, i.e. $d(Tx, Ty) leq Kd(x, y)$ for some constant $0 leq K < 1$, then $T$ has a fixed point. If you can show that (2) is indeed a contraction, then there will exist a unique solution.






                          share|cite|improve this answer









                          $endgroup$



                          Banach's fixed point theorem states that if an operator $T colon X to X$ on the non-empty complete metric space $(X, d)$ is a contraction, then $T$ has a unique fixed point $x^* in X$ such that $Tx^* = x^*$. This means that if you can show that $T$ is a contraction, i.e. $d(Tx, Ty) leq Kd(x, y)$ for some constant $0 leq K < 1$, then $T$ has a fixed point. If you can show that (2) is indeed a contraction, then there will exist a unique solution.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 30 '18 at 23:03









                          Erik AndréErik André

                          857




                          857






























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