Darboux sums elementary question - am I correct
$begingroup$
I'm very new to this material and I would like someone more experienced to give input if possible.
$Q subset mathbb R^n$ is a box and $f: Q to mathbb R$ is a function. Let $Xi_Q$ be the set of all sub-boxes of $Q$, and $F: Xi_Q to mathbb R$ be another function.
Suppose the following 2 statements hold:
1) For any $B in Xi_Q$ we have: $v(B)cdot inf_{x in B}f(x) leq F(B) leq v(B) cdot sup_{x in B}f(B)$.
2) For any $B_1,B_2, dots, B_k in Xi_Q$ which are disjoint in pairs we have: $F(B_1cup dots cup B_k) = sum_{i=1}^{k}F(B_i)$
Show that $$int_{overline{B}} f leq F(B)leqint^{overline{}}_{B}f$$
For all $B in Xi_Q$.
Clarification: $int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi)$ and $int_{B}^{overline{}}f=inf_{Pi}overline{S}(f,Pi)$ where $Pi$ is a partition of the box $B$ and $underline{S},overline{S}$ are lower and upper Darboux sums respectively.
What I did:
$$int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi) = sup_Pi{sum_{R in Pi}inf_{x in R}f(x)cdot v(R)} leq sup_Pi{sum_{R in Pi}F(R)} = sup_Pi{F(B)} = F(B)$$
The inequality comes from statement number $1$, as $R in Xi_Q$ if I understand correctly. The equality after it comes from the fact that $Pi$ is a partition of $B$, so all $R in Pi$ are disjoint, and their union is $B$, so from statement number $2$ the equality follows.
Now for the other direction, a very similar idea:
$$int^{overline{}}_{B}f = inf_{Pi}overline{S}(f,Pi) =inf_Pi{sum_{R in Pi}sup_{x in R}f(x)cdot v(R)} geq inf_Pi{sum_{R in Pi}F(R)} = inf_Pi{F(B)} = F(B)$$
Which proves what we wanted to show.
Is this the correct method? I know it's simple but I'm very new to this material and I missed the lecture sadly.
calculus integration proof-verification riemann-sum
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
$endgroup$
add a comment |
$begingroup$
I'm very new to this material and I would like someone more experienced to give input if possible.
$Q subset mathbb R^n$ is a box and $f: Q to mathbb R$ is a function. Let $Xi_Q$ be the set of all sub-boxes of $Q$, and $F: Xi_Q to mathbb R$ be another function.
Suppose the following 2 statements hold:
1) For any $B in Xi_Q$ we have: $v(B)cdot inf_{x in B}f(x) leq F(B) leq v(B) cdot sup_{x in B}f(B)$.
2) For any $B_1,B_2, dots, B_k in Xi_Q$ which are disjoint in pairs we have: $F(B_1cup dots cup B_k) = sum_{i=1}^{k}F(B_i)$
Show that $$int_{overline{B}} f leq F(B)leqint^{overline{}}_{B}f$$
For all $B in Xi_Q$.
Clarification: $int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi)$ and $int_{B}^{overline{}}f=inf_{Pi}overline{S}(f,Pi)$ where $Pi$ is a partition of the box $B$ and $underline{S},overline{S}$ are lower and upper Darboux sums respectively.
What I did:
$$int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi) = sup_Pi{sum_{R in Pi}inf_{x in R}f(x)cdot v(R)} leq sup_Pi{sum_{R in Pi}F(R)} = sup_Pi{F(B)} = F(B)$$
The inequality comes from statement number $1$, as $R in Xi_Q$ if I understand correctly. The equality after it comes from the fact that $Pi$ is a partition of $B$, so all $R in Pi$ are disjoint, and their union is $B$, so from statement number $2$ the equality follows.
Now for the other direction, a very similar idea:
$$int^{overline{}}_{B}f = inf_{Pi}overline{S}(f,Pi) =inf_Pi{sum_{R in Pi}sup_{x in R}f(x)cdot v(R)} geq inf_Pi{sum_{R in Pi}F(R)} = inf_Pi{F(B)} = F(B)$$
Which proves what we wanted to show.
Is this the correct method? I know it's simple but I'm very new to this material and I missed the lecture sadly.
calculus integration proof-verification riemann-sum
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
$endgroup$
add a comment |
$begingroup$
I'm very new to this material and I would like someone more experienced to give input if possible.
$Q subset mathbb R^n$ is a box and $f: Q to mathbb R$ is a function. Let $Xi_Q$ be the set of all sub-boxes of $Q$, and $F: Xi_Q to mathbb R$ be another function.
Suppose the following 2 statements hold:
1) For any $B in Xi_Q$ we have: $v(B)cdot inf_{x in B}f(x) leq F(B) leq v(B) cdot sup_{x in B}f(B)$.
2) For any $B_1,B_2, dots, B_k in Xi_Q$ which are disjoint in pairs we have: $F(B_1cup dots cup B_k) = sum_{i=1}^{k}F(B_i)$
Show that $$int_{overline{B}} f leq F(B)leqint^{overline{}}_{B}f$$
For all $B in Xi_Q$.
Clarification: $int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi)$ and $int_{B}^{overline{}}f=inf_{Pi}overline{S}(f,Pi)$ where $Pi$ is a partition of the box $B$ and $underline{S},overline{S}$ are lower and upper Darboux sums respectively.
What I did:
$$int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi) = sup_Pi{sum_{R in Pi}inf_{x in R}f(x)cdot v(R)} leq sup_Pi{sum_{R in Pi}F(R)} = sup_Pi{F(B)} = F(B)$$
The inequality comes from statement number $1$, as $R in Xi_Q$ if I understand correctly. The equality after it comes from the fact that $Pi$ is a partition of $B$, so all $R in Pi$ are disjoint, and their union is $B$, so from statement number $2$ the equality follows.
Now for the other direction, a very similar idea:
$$int^{overline{}}_{B}f = inf_{Pi}overline{S}(f,Pi) =inf_Pi{sum_{R in Pi}sup_{x in R}f(x)cdot v(R)} geq inf_Pi{sum_{R in Pi}F(R)} = inf_Pi{F(B)} = F(B)$$
Which proves what we wanted to show.
Is this the correct method? I know it's simple but I'm very new to this material and I missed the lecture sadly.
calculus integration proof-verification riemann-sum
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
$endgroup$
I'm very new to this material and I would like someone more experienced to give input if possible.
$Q subset mathbb R^n$ is a box and $f: Q to mathbb R$ is a function. Let $Xi_Q$ be the set of all sub-boxes of $Q$, and $F: Xi_Q to mathbb R$ be another function.
Suppose the following 2 statements hold:
1) For any $B in Xi_Q$ we have: $v(B)cdot inf_{x in B}f(x) leq F(B) leq v(B) cdot sup_{x in B}f(B)$.
2) For any $B_1,B_2, dots, B_k in Xi_Q$ which are disjoint in pairs we have: $F(B_1cup dots cup B_k) = sum_{i=1}^{k}F(B_i)$
Show that $$int_{overline{B}} f leq F(B)leqint^{overline{}}_{B}f$$
For all $B in Xi_Q$.
Clarification: $int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi)$ and $int_{B}^{overline{}}f=inf_{Pi}overline{S}(f,Pi)$ where $Pi$ is a partition of the box $B$ and $underline{S},overline{S}$ are lower and upper Darboux sums respectively.
What I did:
$$int_{overline{B}}f = sup_{Pi}underline{S}(f,Pi) = sup_Pi{sum_{R in Pi}inf_{x in R}f(x)cdot v(R)} leq sup_Pi{sum_{R in Pi}F(R)} = sup_Pi{F(B)} = F(B)$$
The inequality comes from statement number $1$, as $R in Xi_Q$ if I understand correctly. The equality after it comes from the fact that $Pi$ is a partition of $B$, so all $R in Pi$ are disjoint, and their union is $B$, so from statement number $2$ the equality follows.
Now for the other direction, a very similar idea:
$$int^{overline{}}_{B}f = inf_{Pi}overline{S}(f,Pi) =inf_Pi{sum_{R in Pi}sup_{x in R}f(x)cdot v(R)} geq inf_Pi{sum_{R in Pi}F(R)} = inf_Pi{F(B)} = F(B)$$
Which proves what we wanted to show.
Is this the correct method? I know it's simple but I'm very new to this material and I missed the lecture sadly.
calculus integration proof-verification riemann-sum
calculus integration proof-verification riemann-sum
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
asked Dec 11 '18 at 11:05
Oria GruberOria Gruber
6,48632460
6,48632460
add a comment |
add a comment |
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