Help for this problem involving rieman integral and partitions











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If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$



(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$



(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$



So what i have intuitively and some hint of my professor is that



First, the partition of Lemma 7.1.2 is



$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]



So what i have so far is, and part of this, is a hint of the professor we have



$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$



Where



$m_k:=inf[f(x):x in [x_{k-1},x_k]]$



$m'_k:=inf[f(x):x in [x_{k-1},z]]$



$m''_k:=inf[f(x):x in [x_z,x_k]]$



And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?










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  • You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
    – Acccumulation
    Nov 20 at 23:07










  • Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
    – Daniel ML
    Nov 21 at 1:08















up vote
0
down vote

favorite












If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$



(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$



(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$



So what i have intuitively and some hint of my professor is that



First, the partition of Lemma 7.1.2 is



$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]



So what i have so far is, and part of this, is a hint of the professor we have



$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$



Where



$m_k:=inf[f(x):x in [x_{k-1},x_k]]$



$m'_k:=inf[f(x):x in [x_{k-1},z]]$



$m''_k:=inf[f(x):x in [x_z,x_k]]$



And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?










share|cite|improve this question






















  • You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
    – Acccumulation
    Nov 20 at 23:07










  • Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
    – Daniel ML
    Nov 21 at 1:08













up vote
0
down vote

favorite









up vote
0
down vote

favorite











If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$



(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$



(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$



So what i have intuitively and some hint of my professor is that



First, the partition of Lemma 7.1.2 is



$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]



So what i have so far is, and part of this, is a hint of the professor we have



$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$



Where



$m_k:=inf[f(x):x in [x_{k-1},x_k]]$



$m'_k:=inf[f(x):x in [x_{k-1},z]]$



$m''_k:=inf[f(x):x in [x_z,x_k]]$



And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?










share|cite|improve this question













If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$



(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$



(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$



So what i have intuitively and some hint of my professor is that



First, the partition of Lemma 7.1.2 is



$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]



So what i have so far is, and part of this, is a hint of the professor we have



$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$



Where



$m_k:=inf[f(x):x in [x_{k-1},x_k]]$



$m'_k:=inf[f(x):x in [x_{k-1},z]]$



$m''_k:=inf[f(x):x in [x_z,x_k]]$



And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?







calculus real-analysis riemann-integration riemann-sum partitions-for-integration






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share|cite|improve this question











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asked Nov 20 at 20:58









Daniel ML

303




303












  • You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
    – Acccumulation
    Nov 20 at 23:07










  • Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
    – Daniel ML
    Nov 21 at 1:08


















  • You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
    – Acccumulation
    Nov 20 at 23:07










  • Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
    – Daniel ML
    Nov 21 at 1:08
















You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 at 23:07




You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 at 23:07












Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 at 1:08




Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 at 1:08















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