if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$












1












$begingroup$


Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$



Note: I don't have uniform convergence at my disposal.



Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$



$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.



I'm not sure where to use the supremum at I guess is my issue.






real-analysis integration







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:34










  • $begingroup$
    @JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:36






  • 1




    $begingroup$
    I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:37












  • $begingroup$
    @JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:39


















1












$begingroup$


Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$



Note: I don't have uniform convergence at my disposal.



Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$



$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.



I'm not sure where to use the supremum at I guess is my issue.






real-analysis integration







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:34










  • $begingroup$
    @JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:36






  • 1




    $begingroup$
    I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:37












  • $begingroup$
    @JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:39
















1












1








1





$begingroup$


Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$



Note: I don't have uniform convergence at my disposal.



Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$



$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.



I'm not sure where to use the supremum at I guess is my issue.






real-analysis integration







share|cite|improve this question











$endgroup$




Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$



Note: I don't have uniform convergence at my disposal.



Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$



$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.



I'm not sure where to use the supremum at I guess is my issue.






real-analysis integration




real-analysis integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:01









A.Γ.

22.8k32656




22.8k32656










asked Jan 31 '17 at 4:29









oliverjonesoliverjones

1,444932




1,444932








  • 3




    $begingroup$
    The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:34










  • $begingroup$
    @JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:36






  • 1




    $begingroup$
    I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:37












  • $begingroup$
    @JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:39
















  • 3




    $begingroup$
    The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:34










  • $begingroup$
    @JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:36






  • 1




    $begingroup$
    I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
    $endgroup$
    – Jonas Meyer
    Jan 31 '17 at 4:37












  • $begingroup$
    @JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:39










3




3




$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34




$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34












$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36




$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36




1




1




$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37






$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37














$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39






$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39












1 Answer
1






active

oldest

votes


















1












$begingroup$

Consider $epsilon>0$



Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.



Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$



Then:



$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:42








  • 1




    $begingroup$
    All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
    $endgroup$
    – Momo
    Jan 31 '17 at 4:44












  • $begingroup$
    In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:46






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Momo
    Jan 31 '17 at 4:46











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Consider $epsilon>0$



Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.



Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$



Then:



$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:42








  • 1




    $begingroup$
    All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
    $endgroup$
    – Momo
    Jan 31 '17 at 4:44












  • $begingroup$
    In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:46






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Momo
    Jan 31 '17 at 4:46
















1












$begingroup$

Consider $epsilon>0$



Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.



Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$



Then:



$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:42








  • 1




    $begingroup$
    All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
    $endgroup$
    – Momo
    Jan 31 '17 at 4:44












  • $begingroup$
    In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:46






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Momo
    Jan 31 '17 at 4:46














1












1








1





$begingroup$

Consider $epsilon>0$



Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.



Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$



Then:



$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$






share|cite|improve this answer









$endgroup$



Consider $epsilon>0$



Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.



Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$



Then:



$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 31 '17 at 4:40









MomoMomo

12k21430




12k21430












  • $begingroup$
    if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:42








  • 1




    $begingroup$
    All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
    $endgroup$
    – Momo
    Jan 31 '17 at 4:44












  • $begingroup$
    In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:46






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Momo
    Jan 31 '17 at 4:46


















  • $begingroup$
    if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:42








  • 1




    $begingroup$
    All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
    $endgroup$
    – Momo
    Jan 31 '17 at 4:44












  • $begingroup$
    In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
    $endgroup$
    – oliverjones
    Jan 31 '17 at 4:46






  • 1




    $begingroup$
    Yes, that's right.
    $endgroup$
    – Momo
    Jan 31 '17 at 4:46
















$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42






$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42






1




1




$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44






$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44














$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46




$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46




1




1




$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46




$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46


















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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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