Sum of functions and $L^p$ spaces












0














Let $1le p_1 < p < p_2 le infty$ and $ fin L^p(mathbb R)$. Prove that there exist $f_1 in L^{p_1} (mathbb R)$ and $f_2 in L^{p_2}(mathbb R)$ such that $f = f_1 + f_2$.



I tried to make one of the function bounded but it leads nowhere. It seems to me I need to find function from $L^{p_2}$ first. Thank you in advance for any help.






functional-analysis lp-spaces







share|cite|improve this question






















  • You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely.
    – Dunham
    Nov 28 at 23:20
















0














Let $1le p_1 < p < p_2 le infty$ and $ fin L^p(mathbb R)$. Prove that there exist $f_1 in L^{p_1} (mathbb R)$ and $f_2 in L^{p_2}(mathbb R)$ such that $f = f_1 + f_2$.



I tried to make one of the function bounded but it leads nowhere. It seems to me I need to find function from $L^{p_2}$ first. Thank you in advance for any help.






functional-analysis lp-spaces







share|cite|improve this question






















  • You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely.
    – Dunham
    Nov 28 at 23:20














0












0








0







Let $1le p_1 < p < p_2 le infty$ and $ fin L^p(mathbb R)$. Prove that there exist $f_1 in L^{p_1} (mathbb R)$ and $f_2 in L^{p_2}(mathbb R)$ such that $f = f_1 + f_2$.



I tried to make one of the function bounded but it leads nowhere. It seems to me I need to find function from $L^{p_2}$ first. Thank you in advance for any help.






functional-analysis lp-spaces







share|cite|improve this question













Let $1le p_1 < p < p_2 le infty$ and $ fin L^p(mathbb R)$. Prove that there exist $f_1 in L^{p_1} (mathbb R)$ and $f_2 in L^{p_2}(mathbb R)$ such that $f = f_1 + f_2$.



I tried to make one of the function bounded but it leads nowhere. It seems to me I need to find function from $L^{p_2}$ first. Thank you in advance for any help.






functional-analysis lp-spaces




functional-analysis lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 28 at 22:56









MadChemist

12319




12319












  • You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely.
    – Dunham
    Nov 28 at 23:20


















  • You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely.
    – Dunham
    Nov 28 at 23:20
















You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely.
– Dunham
Nov 28 at 23:20




You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely.
– Dunham
Nov 28 at 23:20










2 Answers
2






active

oldest

votes


















1














Let $D_1$ be the set where $|f|>1$ and $D_2$ be the set where $|f|leq 1$.
Let $f_1$ be $f$ restricted to $D_1$ and $f_2$ be $f$ restricted to $D_2$. Both functions are 0 outside of the defined domains.
begin{align*}
int_{mathbb{R}}|f_1|^{p_1} + int_{mathbb{R}}|f_2|^{p_2}
&=
int_{D_1}|f_1|^{p_1} + int_{D_2}|f_2|^{p_2}\
&leq
int_{D_1}|f_1|^{p} + int_{D_2}|f_2|^{p}\
&=
int_{mathbb{R}}|f|^{p}
end{align*}






share|cite|improve this answer





























    1














    Let $A=f^{-1} ([-1,1]) , B=A=f^{-1} ((-infty,-1)cup (1,infty ))$.
    Take $$f_1 (x)=begin{cases} f(x) mbox{ for } xin B \ 0 mbox{ for } xnotin Bend{cases}$$



    $$f_2 (x)=begin{cases} f(x) mbox{ for } xin A \ 0 mbox{ for } xnotin Aend{cases}$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      Let $D_1$ be the set where $|f|>1$ and $D_2$ be the set where $|f|leq 1$.
      Let $f_1$ be $f$ restricted to $D_1$ and $f_2$ be $f$ restricted to $D_2$. Both functions are 0 outside of the defined domains.
      begin{align*}
      int_{mathbb{R}}|f_1|^{p_1} + int_{mathbb{R}}|f_2|^{p_2}
      &=
      int_{D_1}|f_1|^{p_1} + int_{D_2}|f_2|^{p_2}\
      &leq
      int_{D_1}|f_1|^{p} + int_{D_2}|f_2|^{p}\
      &=
      int_{mathbb{R}}|f|^{p}
      end{align*}






      share|cite|improve this answer


























        1














        Let $D_1$ be the set where $|f|>1$ and $D_2$ be the set where $|f|leq 1$.
        Let $f_1$ be $f$ restricted to $D_1$ and $f_2$ be $f$ restricted to $D_2$. Both functions are 0 outside of the defined domains.
        begin{align*}
        int_{mathbb{R}}|f_1|^{p_1} + int_{mathbb{R}}|f_2|^{p_2}
        &=
        int_{D_1}|f_1|^{p_1} + int_{D_2}|f_2|^{p_2}\
        &leq
        int_{D_1}|f_1|^{p} + int_{D_2}|f_2|^{p}\
        &=
        int_{mathbb{R}}|f|^{p}
        end{align*}






        share|cite|improve this answer
























          1












          1








          1






          Let $D_1$ be the set where $|f|>1$ and $D_2$ be the set where $|f|leq 1$.
          Let $f_1$ be $f$ restricted to $D_1$ and $f_2$ be $f$ restricted to $D_2$. Both functions are 0 outside of the defined domains.
          begin{align*}
          int_{mathbb{R}}|f_1|^{p_1} + int_{mathbb{R}}|f_2|^{p_2}
          &=
          int_{D_1}|f_1|^{p_1} + int_{D_2}|f_2|^{p_2}\
          &leq
          int_{D_1}|f_1|^{p} + int_{D_2}|f_2|^{p}\
          &=
          int_{mathbb{R}}|f|^{p}
          end{align*}






          share|cite|improve this answer












          Let $D_1$ be the set where $|f|>1$ and $D_2$ be the set where $|f|leq 1$.
          Let $f_1$ be $f$ restricted to $D_1$ and $f_2$ be $f$ restricted to $D_2$. Both functions are 0 outside of the defined domains.
          begin{align*}
          int_{mathbb{R}}|f_1|^{p_1} + int_{mathbb{R}}|f_2|^{p_2}
          &=
          int_{D_1}|f_1|^{p_1} + int_{D_2}|f_2|^{p_2}\
          &leq
          int_{D_1}|f_1|^{p} + int_{D_2}|f_2|^{p}\
          &=
          int_{mathbb{R}}|f|^{p}
          end{align*}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 at 23:16









          Dunham

          2,084613




          2,084613























              1














              Let $A=f^{-1} ([-1,1]) , B=A=f^{-1} ((-infty,-1)cup (1,infty ))$.
              Take $$f_1 (x)=begin{cases} f(x) mbox{ for } xin B \ 0 mbox{ for } xnotin Bend{cases}$$



              $$f_2 (x)=begin{cases} f(x) mbox{ for } xin A \ 0 mbox{ for } xnotin Aend{cases}$$






              share|cite|improve this answer


























                1














                Let $A=f^{-1} ([-1,1]) , B=A=f^{-1} ((-infty,-1)cup (1,infty ))$.
                Take $$f_1 (x)=begin{cases} f(x) mbox{ for } xin B \ 0 mbox{ for } xnotin Bend{cases}$$



                $$f_2 (x)=begin{cases} f(x) mbox{ for } xin A \ 0 mbox{ for } xnotin Aend{cases}$$






                share|cite|improve this answer
























                  1












                  1








                  1






                  Let $A=f^{-1} ([-1,1]) , B=A=f^{-1} ((-infty,-1)cup (1,infty ))$.
                  Take $$f_1 (x)=begin{cases} f(x) mbox{ for } xin B \ 0 mbox{ for } xnotin Bend{cases}$$



                  $$f_2 (x)=begin{cases} f(x) mbox{ for } xin A \ 0 mbox{ for } xnotin Aend{cases}$$






                  share|cite|improve this answer












                  Let $A=f^{-1} ([-1,1]) , B=A=f^{-1} ((-infty,-1)cup (1,infty ))$.
                  Take $$f_1 (x)=begin{cases} f(x) mbox{ for } xin B \ 0 mbox{ for } xnotin Bend{cases}$$



                  $$f_2 (x)=begin{cases} f(x) mbox{ for } xin A \ 0 mbox{ for } xnotin Aend{cases}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 at 23:12









                  MotylaNogaTomkaMazura

                  6,577917




                  6,577917






























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