Showing convergence in ${ L }^{ p }(dmu )$












2












$begingroup$


I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.



Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.



Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.



So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do






real-analysis lp-spaces







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.



    Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.



    Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.



    So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
    ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
    But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do






    real-analysis lp-spaces







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.



      Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.



      Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.



      So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
      ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
      But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do






      real-analysis lp-spaces







      share|cite|improve this question









      $endgroup$




      I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.



      Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.



      Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.



      So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
      ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
      But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do






      real-analysis lp-spaces




      real-analysis lp-spaces






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 30 '18 at 9:31









      MasterPIMasterPI

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          $begingroup$

          There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].






          share|cite|improve this answer











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            $begingroup$

            There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].






            share|cite|improve this answer











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              2












              $begingroup$

              There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].






                share|cite|improve this answer











                $endgroup$



                There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 30 '18 at 10:10

























                answered Nov 30 '18 at 9:39









                Kavi Rama MurthyKavi Rama Murthy

                52.6k32055




                52.6k32055






























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