Show that $ lim_{ntoinfty} int_{0}^{1} f_n $ exists












1












$begingroup$


Let $(X,d)= (C[0,1],d)$ where $C[0,1]$ is the set of real-valued continuous functions on $[0,1]$ and $d= int_{0}^{1} |f-g|$ is the Riemann Integral.



Suppose $(f_n)$ is a Cauchy sequence in $(X,d) $ , show that $ lim_{ntoinfty} int_{0}^{1} f_n $ exists.



My attempt: Given $epsilon>0$, $exists Ninmathbb{N}$ s.t $int_{0}^{1} |f_n-f_m|<epsilon$ for all $m>n>N$



$implies int_{0}^{1} |f_n|-int_{0}^{1}|f_m| <epsilon $ (Reverse triangle inequality)



But this only shows that $int_{0}^{1} |f_n|$ is Cauchy and not $int_{0}^{1} f_n$. Any hints would be appreciated.






real-analysis continuity cauchy-sequences riemann-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Presumably $f_n :[0,1] to mathbb{R}$?
    $endgroup$
    – RRL
    Dec 9 '18 at 1:08










  • $begingroup$
    $C[0,1]$ is the set of real-valued continuous functions, so yes!
    $endgroup$
    – Abe
    Dec 9 '18 at 1:12










  • $begingroup$
    Great -- forget about reverse triangle inequality -- $left|int fright| leqslant int|f|$
    $endgroup$
    – RRL
    Dec 9 '18 at 1:13


















1












$begingroup$


Let $(X,d)= (C[0,1],d)$ where $C[0,1]$ is the set of real-valued continuous functions on $[0,1]$ and $d= int_{0}^{1} |f-g|$ is the Riemann Integral.



Suppose $(f_n)$ is a Cauchy sequence in $(X,d) $ , show that $ lim_{ntoinfty} int_{0}^{1} f_n $ exists.



My attempt: Given $epsilon>0$, $exists Ninmathbb{N}$ s.t $int_{0}^{1} |f_n-f_m|<epsilon$ for all $m>n>N$



$implies int_{0}^{1} |f_n|-int_{0}^{1}|f_m| <epsilon $ (Reverse triangle inequality)



But this only shows that $int_{0}^{1} |f_n|$ is Cauchy and not $int_{0}^{1} f_n$. Any hints would be appreciated.






real-analysis continuity cauchy-sequences riemann-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Presumably $f_n :[0,1] to mathbb{R}$?
    $endgroup$
    – RRL
    Dec 9 '18 at 1:08










  • $begingroup$
    $C[0,1]$ is the set of real-valued continuous functions, so yes!
    $endgroup$
    – Abe
    Dec 9 '18 at 1:12










  • $begingroup$
    Great -- forget about reverse triangle inequality -- $left|int fright| leqslant int|f|$
    $endgroup$
    – RRL
    Dec 9 '18 at 1:13
















1












1








1





$begingroup$


Let $(X,d)= (C[0,1],d)$ where $C[0,1]$ is the set of real-valued continuous functions on $[0,1]$ and $d= int_{0}^{1} |f-g|$ is the Riemann Integral.



Suppose $(f_n)$ is a Cauchy sequence in $(X,d) $ , show that $ lim_{ntoinfty} int_{0}^{1} f_n $ exists.



My attempt: Given $epsilon>0$, $exists Ninmathbb{N}$ s.t $int_{0}^{1} |f_n-f_m|<epsilon$ for all $m>n>N$



$implies int_{0}^{1} |f_n|-int_{0}^{1}|f_m| <epsilon $ (Reverse triangle inequality)



But this only shows that $int_{0}^{1} |f_n|$ is Cauchy and not $int_{0}^{1} f_n$. Any hints would be appreciated.






real-analysis continuity cauchy-sequences riemann-integration







share|cite|improve this question











$endgroup$




Let $(X,d)= (C[0,1],d)$ where $C[0,1]$ is the set of real-valued continuous functions on $[0,1]$ and $d= int_{0}^{1} |f-g|$ is the Riemann Integral.



Suppose $(f_n)$ is a Cauchy sequence in $(X,d) $ , show that $ lim_{ntoinfty} int_{0}^{1} f_n $ exists.



My attempt: Given $epsilon>0$, $exists Ninmathbb{N}$ s.t $int_{0}^{1} |f_n-f_m|<epsilon$ for all $m>n>N$



$implies int_{0}^{1} |f_n|-int_{0}^{1}|f_m| <epsilon $ (Reverse triangle inequality)



But this only shows that $int_{0}^{1} |f_n|$ is Cauchy and not $int_{0}^{1} f_n$. Any hints would be appreciated.






real-analysis continuity cauchy-sequences riemann-integration




real-analysis continuity cauchy-sequences riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 1:13







Abe

















asked Dec 9 '18 at 0:54









AbeAbe

195




195












  • $begingroup$
    Presumably $f_n :[0,1] to mathbb{R}$?
    $endgroup$
    – RRL
    Dec 9 '18 at 1:08










  • $begingroup$
    $C[0,1]$ is the set of real-valued continuous functions, so yes!
    $endgroup$
    – Abe
    Dec 9 '18 at 1:12










  • $begingroup$
    Great -- forget about reverse triangle inequality -- $left|int fright| leqslant int|f|$
    $endgroup$
    – RRL
    Dec 9 '18 at 1:13




















  • $begingroup$
    Presumably $f_n :[0,1] to mathbb{R}$?
    $endgroup$
    – RRL
    Dec 9 '18 at 1:08










  • $begingroup$
    $C[0,1]$ is the set of real-valued continuous functions, so yes!
    $endgroup$
    – Abe
    Dec 9 '18 at 1:12










  • $begingroup$
    Great -- forget about reverse triangle inequality -- $left|int fright| leqslant int|f|$
    $endgroup$
    – RRL
    Dec 9 '18 at 1:13


















$begingroup$
Presumably $f_n :[0,1] to mathbb{R}$?
$endgroup$
– RRL
Dec 9 '18 at 1:08




$begingroup$
Presumably $f_n :[0,1] to mathbb{R}$?
$endgroup$
– RRL
Dec 9 '18 at 1:08












$begingroup$
$C[0,1]$ is the set of real-valued continuous functions, so yes!
$endgroup$
– Abe
Dec 9 '18 at 1:12




$begingroup$
$C[0,1]$ is the set of real-valued continuous functions, so yes!
$endgroup$
– Abe
Dec 9 '18 at 1:12












$begingroup$
Great -- forget about reverse triangle inequality -- $left|int fright| leqslant int|f|$
$endgroup$
– RRL
Dec 9 '18 at 1:13






$begingroup$
Great -- forget about reverse triangle inequality -- $left|int fright| leqslant int|f|$
$endgroup$
– RRL
Dec 9 '18 at 1:13












1 Answer
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$begingroup$

Hint: Show $int_0^1f_n$ forms a Cauchy sequence in $mathbb{R}$.



$$left| int_0^1f_n - int_0^1f_mright| = left| int_0^1(f_n - f_m)right| leqslant int_0^1 |f_n - f_m| = d(f_n,f_m)$$






share|cite|improve this answer











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    $begingroup$

    Hint: Show $int_0^1f_n$ forms a Cauchy sequence in $mathbb{R}$.



    $$left| int_0^1f_n - int_0^1f_mright| = left| int_0^1(f_n - f_m)right| leqslant int_0^1 |f_n - f_m| = d(f_n,f_m)$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Hint: Show $int_0^1f_n$ forms a Cauchy sequence in $mathbb{R}$.



      $$left| int_0^1f_n - int_0^1f_mright| = left| int_0^1(f_n - f_m)right| leqslant int_0^1 |f_n - f_m| = d(f_n,f_m)$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint: Show $int_0^1f_n$ forms a Cauchy sequence in $mathbb{R}$.



        $$left| int_0^1f_n - int_0^1f_mright| = left| int_0^1(f_n - f_m)right| leqslant int_0^1 |f_n - f_m| = d(f_n,f_m)$$






        share|cite|improve this answer











        $endgroup$



        Hint: Show $int_0^1f_n$ forms a Cauchy sequence in $mathbb{R}$.



        $$left| int_0^1f_n - int_0^1f_mright| = left| int_0^1(f_n - f_m)right| leqslant int_0^1 |f_n - f_m| = d(f_n,f_m)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 1:11

























        answered Dec 9 '18 at 1:04









        RRLRRL

        49.9k42573




        49.9k42573






























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