Prove that if $f$ is continuous at $x_0in [a,b]$ and $f(x_0)neq 0,$ then $suplimits_{P}L(|f|,P)>0$












1












$begingroup$


Can you help me check if this proof is correct? If not, kindly provide a better proof




Prove that if $f$ is continuous at $x_0in [a,b]$ and $f(x_0)neq 0,$ then $suplimits_{P}L(|f|,P)>0$




Suppose $a<b$. For $epsilon=|f(x_0)|/2,$ there exists $delta>0$ such that $$|f(x)|>dfrac{1}{2}|f(x_0)|,;;text{whenever};;xin(x_0-delta,x_0+delta).$$
Choose a uniform partition $P_n$, for each $n,$ such that
$$a=x_0<x_1<cdots<x_n=b;;text{and};;x_j-x_{j-1}=(b-a)/n,;;jin {1,2,cdots,n}.$$
Hence,
begin{align}suplimits_{P_n}L(|f|,P_n)&= limlimits_{nto infty}sum^{n}_{j=1}|f(t_j)|(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}sum^{n}_{j=1}(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(x_n-x_{0})\&= dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(b-a)\&= dfrac{1}{2}|f(x_0)|(b-a)\&>0end{align}






real-analysis integration analysis proof-verification riemann-integration







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  • 1




    $begingroup$
    Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand.
    $endgroup$
    – Michael Burr
    Jan 15 at 22:59












  • $begingroup$
    @Michael Burr: I'll do that!
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:02
















1












$begingroup$


Can you help me check if this proof is correct? If not, kindly provide a better proof




Prove that if $f$ is continuous at $x_0in [a,b]$ and $f(x_0)neq 0,$ then $suplimits_{P}L(|f|,P)>0$




Suppose $a<b$. For $epsilon=|f(x_0)|/2,$ there exists $delta>0$ such that $$|f(x)|>dfrac{1}{2}|f(x_0)|,;;text{whenever};;xin(x_0-delta,x_0+delta).$$
Choose a uniform partition $P_n$, for each $n,$ such that
$$a=x_0<x_1<cdots<x_n=b;;text{and};;x_j-x_{j-1}=(b-a)/n,;;jin {1,2,cdots,n}.$$
Hence,
begin{align}suplimits_{P_n}L(|f|,P_n)&= limlimits_{nto infty}sum^{n}_{j=1}|f(t_j)|(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}sum^{n}_{j=1}(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(x_n-x_{0})\&= dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(b-a)\&= dfrac{1}{2}|f(x_0)|(b-a)\&>0end{align}






real-analysis integration analysis proof-verification riemann-integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand.
    $endgroup$
    – Michael Burr
    Jan 15 at 22:59












  • $begingroup$
    @Michael Burr: I'll do that!
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:02














1












1








1


1



$begingroup$


Can you help me check if this proof is correct? If not, kindly provide a better proof




Prove that if $f$ is continuous at $x_0in [a,b]$ and $f(x_0)neq 0,$ then $suplimits_{P}L(|f|,P)>0$




Suppose $a<b$. For $epsilon=|f(x_0)|/2,$ there exists $delta>0$ such that $$|f(x)|>dfrac{1}{2}|f(x_0)|,;;text{whenever};;xin(x_0-delta,x_0+delta).$$
Choose a uniform partition $P_n$, for each $n,$ such that
$$a=x_0<x_1<cdots<x_n=b;;text{and};;x_j-x_{j-1}=(b-a)/n,;;jin {1,2,cdots,n}.$$
Hence,
begin{align}suplimits_{P_n}L(|f|,P_n)&= limlimits_{nto infty}sum^{n}_{j=1}|f(t_j)|(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}sum^{n}_{j=1}(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(x_n-x_{0})\&= dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(b-a)\&= dfrac{1}{2}|f(x_0)|(b-a)\&>0end{align}






real-analysis integration analysis proof-verification riemann-integration







share|cite|improve this question











$endgroup$




Can you help me check if this proof is correct? If not, kindly provide a better proof




Prove that if $f$ is continuous at $x_0in [a,b]$ and $f(x_0)neq 0,$ then $suplimits_{P}L(|f|,P)>0$




Suppose $a<b$. For $epsilon=|f(x_0)|/2,$ there exists $delta>0$ such that $$|f(x)|>dfrac{1}{2}|f(x_0)|,;;text{whenever};;xin(x_0-delta,x_0+delta).$$
Choose a uniform partition $P_n$, for each $n,$ such that
$$a=x_0<x_1<cdots<x_n=b;;text{and};;x_j-x_{j-1}=(b-a)/n,;;jin {1,2,cdots,n}.$$
Hence,
begin{align}suplimits_{P_n}L(|f|,P_n)&= limlimits_{nto infty}sum^{n}_{j=1}|f(t_j)|(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}sum^{n}_{j=1}(x_j-x_{j-1})\&geq dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(x_n-x_{0})\&= dfrac{1}{2}|f(x_0)|limlimits_{nto infty}(b-a)\&= dfrac{1}{2}|f(x_0)|(b-a)\&>0end{align}






real-analysis integration analysis proof-verification riemann-integration




real-analysis integration analysis proof-verification riemann-integration






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edited Jan 16 at 5:18







Omojola Micheal

















asked Jan 15 at 22:53









Omojola MichealOmojola Micheal

1,853324




1,853324








  • 1




    $begingroup$
    Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand.
    $endgroup$
    – Michael Burr
    Jan 15 at 22:59












  • $begingroup$
    @Michael Burr: I'll do that!
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:02














  • 1




    $begingroup$
    Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand.
    $endgroup$
    – Michael Burr
    Jan 15 at 22:59












  • $begingroup$
    @Michael Burr: I'll do that!
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:02








1




1




$begingroup$
Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand.
$endgroup$
– Michael Burr
Jan 15 at 22:59






$begingroup$
Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand.
$endgroup$
– Michael Burr
Jan 15 at 22:59














$begingroup$
@Michael Burr: I'll do that!
$endgroup$
– Omojola Micheal
Jan 15 at 23:02




$begingroup$
@Michael Burr: I'll do that!
$endgroup$
– Omojola Micheal
Jan 15 at 23:02










1 Answer
1






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oldest

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$begingroup$

No, it is not correct. That inequality containg the second $geqslant$ doesn't hold; you seem to be assuming here that each $m_j$ is greter than or equal to $frac12bigllvert f(x_0)bigrrvert$, but that is not true.



Note that you only have to proved an example of one partition $P_0$ such that $Lbigl(lvert frvert,P_0bigr)>0$. Then it will follow automatically that $displaystylesup_PLbigl(lvert frvert,Pbigr)>0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, let me edit it.
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:04










  • $begingroup$
    Please, how about now?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:05










  • $begingroup$
    Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:07










  • $begingroup$
    So, what should I do?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:10










  • $begingroup$
    If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:12













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1 Answer
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0












$begingroup$

No, it is not correct. That inequality containg the second $geqslant$ doesn't hold; you seem to be assuming here that each $m_j$ is greter than or equal to $frac12bigllvert f(x_0)bigrrvert$, but that is not true.



Note that you only have to proved an example of one partition $P_0$ such that $Lbigl(lvert frvert,P_0bigr)>0$. Then it will follow automatically that $displaystylesup_PLbigl(lvert frvert,Pbigr)>0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, let me edit it.
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:04










  • $begingroup$
    Please, how about now?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:05










  • $begingroup$
    Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:07










  • $begingroup$
    So, what should I do?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:10










  • $begingroup$
    If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:12


















0












$begingroup$

No, it is not correct. That inequality containg the second $geqslant$ doesn't hold; you seem to be assuming here that each $m_j$ is greter than or equal to $frac12bigllvert f(x_0)bigrrvert$, but that is not true.



Note that you only have to proved an example of one partition $P_0$ such that $Lbigl(lvert frvert,P_0bigr)>0$. Then it will follow automatically that $displaystylesup_PLbigl(lvert frvert,Pbigr)>0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, let me edit it.
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:04










  • $begingroup$
    Please, how about now?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:05










  • $begingroup$
    Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:07










  • $begingroup$
    So, what should I do?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:10










  • $begingroup$
    If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:12
















0












0








0





$begingroup$

No, it is not correct. That inequality containg the second $geqslant$ doesn't hold; you seem to be assuming here that each $m_j$ is greter than or equal to $frac12bigllvert f(x_0)bigrrvert$, but that is not true.



Note that you only have to proved an example of one partition $P_0$ such that $Lbigl(lvert frvert,P_0bigr)>0$. Then it will follow automatically that $displaystylesup_PLbigl(lvert frvert,Pbigr)>0$.






share|cite|improve this answer









$endgroup$



No, it is not correct. That inequality containg the second $geqslant$ doesn't hold; you seem to be assuming here that each $m_j$ is greter than or equal to $frac12bigllvert f(x_0)bigrrvert$, but that is not true.



Note that you only have to proved an example of one partition $P_0$ such that $Lbigl(lvert frvert,P_0bigr)>0$. Then it will follow automatically that $displaystylesup_PLbigl(lvert frvert,Pbigr)>0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 23:01









José Carlos SantosJosé Carlos Santos

159k22126231




159k22126231












  • $begingroup$
    Okay, let me edit it.
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:04










  • $begingroup$
    Please, how about now?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:05










  • $begingroup$
    Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:07










  • $begingroup$
    So, what should I do?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:10










  • $begingroup$
    If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:12




















  • $begingroup$
    Okay, let me edit it.
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:04










  • $begingroup$
    Please, how about now?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:05










  • $begingroup$
    Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:07










  • $begingroup$
    So, what should I do?
    $endgroup$
    – Omojola Micheal
    Jan 15 at 23:10










  • $begingroup$
    If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
    $endgroup$
    – José Carlos Santos
    Jan 15 at 23:12


















$begingroup$
Okay, let me edit it.
$endgroup$
– Omojola Micheal
Jan 15 at 23:04




$begingroup$
Okay, let me edit it.
$endgroup$
– Omojola Micheal
Jan 15 at 23:04












$begingroup$
Please, how about now?
$endgroup$
– Omojola Micheal
Jan 15 at 23:05




$begingroup$
Please, how about now?
$endgroup$
– Omojola Micheal
Jan 15 at 23:05












$begingroup$
Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
$endgroup$
– José Carlos Santos
Jan 15 at 23:07




$begingroup$
Why do you think that $bigllvert f(t_j)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrrvert$ for each $j$? There is no reason for that.
$endgroup$
– José Carlos Santos
Jan 15 at 23:07












$begingroup$
So, what should I do?
$endgroup$
– Omojola Micheal
Jan 15 at 23:10




$begingroup$
So, what should I do?
$endgroup$
– Omojola Micheal
Jan 15 at 23:10












$begingroup$
If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
$endgroup$
– José Carlos Santos
Jan 15 at 23:12






$begingroup$
If you take $a_1,b_1in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(forall xin[a_1,b_1]):bigllvert f(x)bigrrvertgeqslantfrac12bigllvert f(x_0)bigrvert$ and if $P_0={a,a_1,b_1,b}$, then$$Lbigl(lvert frvert,P_0bigr)=frac12bigllvert f(x_0)bigrvert(b_1-a_1)>0.$$
$endgroup$
– José Carlos Santos
Jan 15 at 23:12




















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