If $f$ is Riemann integrable on $[a, b]$, prove that $2f$ is Riemann integrable
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If $f$ is integrable on $[a, b]$, prove that $2f$ is integrable.
Let $epsilon > 0$. Then, if $f$ is integrable on $[a, b]$, there is some partition $P = {x_{0}, ldots x_{n}}$ of $[a, b]$ such that $U(f, P) - L(f, P) < epsilon$.
Then, for any $i in {1, 2, ldots n}$, if we define $M_{i}(f) = text{sup}{f(x) mid x in [x_{i-1}, x_{i}] }$ and $m_{i}(f) = text{inf}{f(x) mid x in [x_{i - 1}, x_{i}] }$, we get
$$U(2f, P) - L(2f, P) = sum_{i = 1}^{n}(M_{i}(2f) - m_{i}(2f)) Delta x_{i} = 2 cdot sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f)) Delta x_{i} leq sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f))Delta x_{i} = U(f, P) - L(f, P) < epsilon.$$
I need to show that this is less than $epsilon$ how can I do it?
real-analysis integration
asked Nov 16 at 17:19
joseph
957
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up vote
0
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If $f$ is integrable on $[a, b]$, prove that $2f$ is integrable.
Let $epsilon > 0$. Then, if $f$ is integrable on $[a, b]$, there is some partition $P = {x_{0}, ldots x_{n}}$ of $[a, b]$ such that $U(f, P) - L(f, P) < epsilon$.
Then, for any $i in {1, 2, ldots n}$, if we define $M_{i}(f) = text{sup}{f(x) mid x in [x_{i-1}, x_{i}] }$ and $m_{i}(f) = text{inf}{f(x) mid x in [x_{i - 1}, x_{i}] }$, we get
$$U(2f, P) - L(2f, P) = sum_{i = 1}^{n}(M_{i}(2f) - m_{i}(2f)) Delta x_{i} = 2 cdot sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f)) Delta x_{i} leq sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f))Delta x_{i} = U(f, P) - L(f, P) < epsilon.$$
I need to show that this is less than $epsilon$ how can I do it?
real-analysis integration
asked Nov 16 at 17:19
joseph
957
Prove $M_i(2f)=2cdot M_i(f)$ using definition.
– Yadati Kiran
Nov 16 at 17:27
ohhhhhh i see it now, thanks. o
– joseph
Nov 16 at 17:35
if you post the answer i will give it to you @YadatiKiran
– joseph
Nov 16 at 17:36
Haha! Thanks anyways. I am only trying to help you not solve for you.
– Yadati Kiran
Nov 16 at 17:37
This is too obvious if one considers Riemann sums. Any Riemann sum of $2f$ is twice a Riemann sum of $f$ over same partitions and hence integral of $2f$ is twice the integral of $f$.
– Paramanand Singh
Nov 17 at 6:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $f$ is integrable on $[a, b]$, prove that $2f$ is integrable.
Let $epsilon > 0$. Then, if $f$ is integrable on $[a, b]$, there is some partition $P = {x_{0}, ldots x_{n}}$ of $[a, b]$ such that $U(f, P) - L(f, P) < epsilon$.
Then, for any $i in {1, 2, ldots n}$, if we define $M_{i}(f) = text{sup}{f(x) mid x in [x_{i-1}, x_{i}] }$ and $m_{i}(f) = text{inf}{f(x) mid x in [x_{i - 1}, x_{i}] }$, we get
$$U(2f, P) - L(2f, P) = sum_{i = 1}^{n}(M_{i}(2f) - m_{i}(2f)) Delta x_{i} = 2 cdot sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f)) Delta x_{i} leq sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f))Delta x_{i} = U(f, P) - L(f, P) < epsilon.$$
I need to show that this is less than $epsilon$ how can I do it?
real-analysis integration
asked Nov 16 at 17:19
joseph
957
If $f$ is integrable on $[a, b]$, prove that $2f$ is integrable.
Let $epsilon > 0$. Then, if $f$ is integrable on $[a, b]$, there is some partition $P = {x_{0}, ldots x_{n}}$ of $[a, b]$ such that $U(f, P) - L(f, P) < epsilon$.
Then, for any $i in {1, 2, ldots n}$, if we define $M_{i}(f) = text{sup}{f(x) mid x in [x_{i-1}, x_{i}] }$ and $m_{i}(f) = text{inf}{f(x) mid x in [x_{i - 1}, x_{i}] }$, we get
$$U(2f, P) - L(2f, P) = sum_{i = 1}^{n}(M_{i}(2f) - m_{i}(2f)) Delta x_{i} = 2 cdot sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f)) Delta x_{i} leq sum_{i = 1}^{n} (M_{i}(f) - m_{i}(f))Delta x_{i} = U(f, P) - L(f, P) < epsilon.$$
I need to show that this is less than $epsilon$ how can I do it?
real-analysis integration
real-analysis integration
asked Nov 16 at 17:19
joseph
957
asked Nov 16 at 17:19
joseph
957
edited Nov 16 at 17:36
asked Nov 16 at 17:19
joseph
957
asked Nov 16 at 17:19
joseph
957
asked Nov 16 at 17:19
joseph
957
957
Prove $M_i(2f)=2cdot M_i(f)$ using definition.
– Yadati Kiran
Nov 16 at 17:27
ohhhhhh i see it now, thanks. o
– joseph
Nov 16 at 17:35
if you post the answer i will give it to you @YadatiKiran
– joseph
Nov 16 at 17:36
Haha! Thanks anyways. I am only trying to help you not solve for you.
– Yadati Kiran
Nov 16 at 17:37
This is too obvious if one considers Riemann sums. Any Riemann sum of $2f$ is twice a Riemann sum of $f$ over same partitions and hence integral of $2f$ is twice the integral of $f$.
– Paramanand Singh
Nov 17 at 6:46
add a comment |
Prove $M_i(2f)=2cdot M_i(f)$ using definition.
– Yadati Kiran
Nov 16 at 17:27
ohhhhhh i see it now, thanks. o
– joseph
Nov 16 at 17:35
if you post the answer i will give it to you @YadatiKiran
– joseph
Nov 16 at 17:36
Haha! Thanks anyways. I am only trying to help you not solve for you.
– Yadati Kiran
Nov 16 at 17:37
This is too obvious if one considers Riemann sums. Any Riemann sum of $2f$ is twice a Riemann sum of $f$ over same partitions and hence integral of $2f$ is twice the integral of $f$.
– Paramanand Singh
Nov 17 at 6:46
Prove $M_i(2f)=2cdot M_i(f)$ using definition.
– Yadati Kiran
Nov 16 at 17:27
Prove $M_i(2f)=2cdot M_i(f)$ using definition.
– Yadati Kiran
Nov 16 at 17:27
ohhhhhh i see it now, thanks. o
– joseph
Nov 16 at 17:35
ohhhhhh i see it now, thanks. o
– joseph
Nov 16 at 17:35
if you post the answer i will give it to you @YadatiKiran
– joseph
Nov 16 at 17:36
if you post the answer i will give it to you @YadatiKiran
– joseph
Nov 16 at 17:36
Haha! Thanks anyways. I am only trying to help you not solve for you.
– Yadati Kiran
Nov 16 at 17:37
Haha! Thanks anyways. I am only trying to help you not solve for you.
– Yadati Kiran
Nov 16 at 17:37
This is too obvious if one considers Riemann sums. Any Riemann sum of $2f$ is twice a Riemann sum of $f$ over same partitions and hence integral of $2f$ is twice the integral of $f$.
– Paramanand Singh
Nov 17 at 6:46
This is too obvious if one considers Riemann sums. Any Riemann sum of $2f$ is twice a Riemann sum of $f$ over same partitions and hence integral of $2f$ is twice the integral of $f$.
– Paramanand Singh
Nov 17 at 6:46
add a comment |
1 Answer
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votes
up vote
1
down vote
accepted
You can show that $$sup {2a mid a in A} = 2 sup {a mid a in A} quad text{and} quad inf {2a mid a in A} = 2 inf {a mid a in A}$$
using the definition of supremum and infimum. Thus
$$M_i(2f) - m_i(2f) = 2M_i(f) - 2m_i(f) = 2(M_i(f) - m_i(f))$$ giving you
$$U(2f,P) - L(2f,P) = 2(U(f,P) - L(f,P)).$$
answered Nov 16 at 17:35
Umberto P.
37.9k13063
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can show that $$sup {2a mid a in A} = 2 sup {a mid a in A} quad text{and} quad inf {2a mid a in A} = 2 inf {a mid a in A}$$
using the definition of supremum and infimum. Thus
$$M_i(2f) - m_i(2f) = 2M_i(f) - 2m_i(f) = 2(M_i(f) - m_i(f))$$ giving you
$$U(2f,P) - L(2f,P) = 2(U(f,P) - L(f,P)).$$
answered Nov 16 at 17:35
Umberto P.
37.9k13063
add a comment |
up vote
1
down vote
accepted
You can show that $$sup {2a mid a in A} = 2 sup {a mid a in A} quad text{and} quad inf {2a mid a in A} = 2 inf {a mid a in A}$$
using the definition of supremum and infimum. Thus
$$M_i(2f) - m_i(2f) = 2M_i(f) - 2m_i(f) = 2(M_i(f) - m_i(f))$$ giving you
$$U(2f,P) - L(2f,P) = 2(U(f,P) - L(f,P)).$$
answered Nov 16 at 17:35
Umberto P.
37.9k13063
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can show that $$sup {2a mid a in A} = 2 sup {a mid a in A} quad text{and} quad inf {2a mid a in A} = 2 inf {a mid a in A}$$
using the definition of supremum and infimum. Thus
$$M_i(2f) - m_i(2f) = 2M_i(f) - 2m_i(f) = 2(M_i(f) - m_i(f))$$ giving you
$$U(2f,P) - L(2f,P) = 2(U(f,P) - L(f,P)).$$
answered Nov 16 at 17:35
Umberto P.
37.9k13063
You can show that $$sup {2a mid a in A} = 2 sup {a mid a in A} quad text{and} quad inf {2a mid a in A} = 2 inf {a mid a in A}$$
using the definition of supremum and infimum. Thus
$$M_i(2f) - m_i(2f) = 2M_i(f) - 2m_i(f) = 2(M_i(f) - m_i(f))$$ giving you
$$U(2f,P) - L(2f,P) = 2(U(f,P) - L(f,P)).$$
answered Nov 16 at 17:35
Umberto P.
37.9k13063
answered Nov 16 at 17:35
Umberto P.
37.9k13063
answered Nov 16 at 17:35
Umberto P.
37.9k13063
answered Nov 16 at 17:35
Umberto P.
37.9k13063
37.9k13063
add a comment |
add a comment |
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Prove $M_i(2f)=2cdot M_i(f)$ using definition.
– Yadati Kiran
Nov 16 at 17:27
ohhhhhh i see it now, thanks. o
– joseph
Nov 16 at 17:35
if you post the answer i will give it to you @YadatiKiran
– joseph
Nov 16 at 17:36
Haha! Thanks anyways. I am only trying to help you not solve for you.
– Yadati Kiran
Nov 16 at 17:37
This is too obvious if one considers Riemann sums. Any Riemann sum of $2f$ is twice a Riemann sum of $f$ over same partitions and hence integral of $2f$ is twice the integral of $f$.
– Paramanand Singh
Nov 17 at 6:46