Is there a fixed point theorem I could use to solve this problem?












5














let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$



also $|K| leq a < 1$



I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :



$f_g + Kf_g = g$



which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.



with what I have in hands I feel like there must be some theorem I'm missing.



any help will be greatly appreciated !






functional-analysis fixed-point-theorems







share|cite|improve this question






















  • You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
    – Kavi Rama Murthy
    8 hours ago










  • You need to fix the notation! You're using "$K$" for two different things...
    – David C. Ullrich
    5 hours ago


















5














let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$



also $|K| leq a < 1$



I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :



$f_g + Kf_g = g$



which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.



with what I have in hands I feel like there must be some theorem I'm missing.



any help will be greatly appreciated !






functional-analysis fixed-point-theorems







share|cite|improve this question






















  • You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
    – Kavi Rama Murthy
    8 hours ago










  • You need to fix the notation! You're using "$K$" for two different things...
    – David C. Ullrich
    5 hours ago
















5












5








5


1





let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$



also $|K| leq a < 1$



I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :



$f_g + Kf_g = g$



which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.



with what I have in hands I feel like there must be some theorem I'm missing.



any help will be greatly appreciated !






functional-analysis fixed-point-theorems







share|cite|improve this question













let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$



also $|K| leq a < 1$



I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :



$f_g + Kf_g = g$



which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.



with what I have in hands I feel like there must be some theorem I'm missing.



any help will be greatly appreciated !






functional-analysis fixed-point-theorems




functional-analysis fixed-point-theorems






share|cite|improve this question













share|cite|improve this question











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asked 9 hours ago









rapidracim

1,4091319




1,4091319












  • You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
    – Kavi Rama Murthy
    8 hours ago










  • You need to fix the notation! You're using "$K$" for two different things...
    – David C. Ullrich
    5 hours ago




















  • You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
    – Kavi Rama Murthy
    8 hours ago










  • You need to fix the notation! You're using "$K$" for two different things...
    – David C. Ullrich
    5 hours ago


















You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago




You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago












You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago






You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago












2 Answers
2






active

oldest

votes


















7














You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
$$
|Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
$$
with $0<a<1$.






share|cite|improve this answer





























    2














    This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
    $$
    T^{-1}=sum_{k=0}^infty (I-T)^k.
    $$     In your case, since $|I-(I+K)|<1$, we have
    $$
    f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
    $$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
      $$
      |Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
      $$
      with $0<a<1$.






      share|cite|improve this answer


























        7














        You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
        $$
        |Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
        $$
        with $0<a<1$.






        share|cite|improve this answer
























          7












          7








          7






          You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
          $$
          |Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
          $$
          with $0<a<1$.






          share|cite|improve this answer












          You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
          $$
          |Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
          $$
          with $0<a<1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Julián Aguirre

          67.5k24094




          67.5k24094























              2














              This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
              $$
              T^{-1}=sum_{k=0}^infty (I-T)^k.
              $$               In your case, since $|I-(I+K)|<1$, we have
              $$
              f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
              $$






              share|cite|improve this answer


























                2














                This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
                $$
                T^{-1}=sum_{k=0}^infty (I-T)^k.
                $$                 In your case, since $|I-(I+K)|<1$, we have
                $$
                f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
                $$






                share|cite|improve this answer
























                  2












                  2








                  2






                  This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
                  $$
                  T^{-1}=sum_{k=0}^infty (I-T)^k.
                  $$                   In your case, since $|I-(I+K)|<1$, we have
                  $$
                  f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
                  $$






                  share|cite|improve this answer












                  This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
                  $$
                  T^{-1}=sum_{k=0}^infty (I-T)^k.
                  $$                   In your case, since $|I-(I+K)|<1$, we have
                  $$
                  f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  Song

                  4,225316




                  4,225316






























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