Mixing
First mean value theorem of integration, $xi in (a,b)$ instead of $[a,b]$?
$begingroup$
The first mean value theorem of integration states that
If $fin C[a,b]$ and $gin mathcal{R}[a,b]$, $g$ is nonnegative, then $exists xiin [a,b]$ such that $$int_a^b (fcdot g)(x) dx=f(xi)int_a^b g(x) dx.$$
Is it possible to replace $exists xiin [a,b]$ by $exists xiin (a,b)$? It is difficult to imagine how this could be wrong.
Edit: added condition $g$ is nonnegative.
real-analysis calculus analysis
asked Jan 3 at 2:42
JiuJiu
496112
$endgroup$
add a comment |
$begingroup$
The first mean value theorem of integration states that
If $fin C[a,b]$ and $gin mathcal{R}[a,b]$, $g$ is nonnegative, then $exists xiin [a,b]$ such that $$int_a^b (fcdot g)(x) dx=f(xi)int_a^b g(x) dx.$$
Is it possible to replace $exists xiin [a,b]$ by $exists xiin (a,b)$? It is difficult to imagine how this could be wrong.
Edit: added condition $g$ is nonnegative.
real-analysis calculus analysis
asked Jan 3 at 2:42
JiuJiu
496112
$endgroup$
$begingroup$
We need g(x) to be non-negative (or at least not change signs).
$endgroup$
– Doug M
Jan 3 at 3:11
$begingroup$
@DougM yes. Thanks for reminding me.
$endgroup$
– Jiu
Jan 3 at 3:11
$begingroup$
What you propose is true, but it is not as elegant a proof as to show that it is true over the closed interval.
$endgroup$
– Doug M
Jan 3 at 3:28
$begingroup$
@DougM Do you know where I can find a proof?
$endgroup$
– Jiu
Jan 3 at 3:52
add a comment |
$begingroup$
The first mean value theorem of integration states that
If $fin C[a,b]$ and $gin mathcal{R}[a,b]$, $g$ is nonnegative, then $exists xiin [a,b]$ such that $$int_a^b (fcdot g)(x) dx=f(xi)int_a^b g(x) dx.$$
Is it possible to replace $exists xiin [a,b]$ by $exists xiin (a,b)$? It is difficult to imagine how this could be wrong.
Edit: added condition $g$ is nonnegative.
real-analysis calculus analysis
asked Jan 3 at 2:42
JiuJiu
496112
$endgroup$
The first mean value theorem of integration states that
If $fin C[a,b]$ and $gin mathcal{R}[a,b]$, $g$ is nonnegative, then $exists xiin [a,b]$ such that $$int_a^b (fcdot g)(x) dx=f(xi)int_a^b g(x) dx.$$
Is it possible to replace $exists xiin [a,b]$ by $exists xiin (a,b)$? It is difficult to imagine how this could be wrong.
Edit: added condition $g$ is nonnegative.
real-analysis calculus analysis
real-analysis calculus analysis
asked Jan 3 at 2:42
JiuJiu
496112
asked Jan 3 at 2:42
JiuJiu
496112
edited Jan 3 at 3:13
Jiu
asked Jan 3 at 2:42
JiuJiu
496112
asked Jan 3 at 2:42
JiuJiu
496112
asked Jan 3 at 2:42
JiuJiu
496112
496112
$begingroup$
We need g(x) to be non-negative (or at least not change signs).
$endgroup$
– Doug M
Jan 3 at 3:11
$begingroup$
@DougM yes. Thanks for reminding me.
$endgroup$
– Jiu
Jan 3 at 3:11
$begingroup$
What you propose is true, but it is not as elegant a proof as to show that it is true over the closed interval.
$endgroup$
– Doug M
Jan 3 at 3:28
$begingroup$
@DougM Do you know where I can find a proof?
$endgroup$
– Jiu
Jan 3 at 3:52
add a comment |
$begingroup$
We need g(x) to be non-negative (or at least not change signs).
$endgroup$
– Doug M
Jan 3 at 3:11
$begingroup$
@DougM yes. Thanks for reminding me.
$endgroup$
– Jiu
Jan 3 at 3:11
$begingroup$
What you propose is true, but it is not as elegant a proof as to show that it is true over the closed interval.
$endgroup$
– Doug M
Jan 3 at 3:28
$begingroup$
@DougM Do you know where I can find a proof?
$endgroup$
– Jiu
Jan 3 at 3:52
$begingroup$
We need g(x) to be non-negative (or at least not change signs).
$endgroup$
– Doug M
Jan 3 at 3:11
$begingroup$
We need g(x) to be non-negative (or at least not change signs).
$endgroup$
– Doug M
Jan 3 at 3:11
$begingroup$
@DougM yes. Thanks for reminding me.
$endgroup$
– Jiu
Jan 3 at 3:11
$begingroup$
@DougM yes. Thanks for reminding me.
$endgroup$
– Jiu
Jan 3 at 3:11
$begingroup$
What you propose is true, but it is not as elegant a proof as to show that it is true over the closed interval.
$endgroup$
– Doug M
Jan 3 at 3:28
$begingroup$
What you propose is true, but it is not as elegant a proof as to show that it is true over the closed interval.
$endgroup$
– Doug M
Jan 3 at 3:28
$begingroup$
@DougM Do you know where I can find a proof?
$endgroup$
– Jiu
Jan 3 at 3:52
$begingroup$
@DougM Do you know where I can find a proof?
$endgroup$
– Jiu
Jan 3 at 3:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = inf_{x in [a,b]}, f(x)$ and $M = sup_{x in [a,b]}, f(x)$. Since $mg(x) leqslant f(x)g(x) leqslant Mg(x)$ for all $x in [a,b]$ we have
$$mint_a^b g(x) , dx leqslant int_a^b f(x) , g(x) , dx leqslant M int_a^b g(x) , dx.$$
In the case where $int_a^b g(x) , dx = 0$ it is easy to show that we can choose any $xi in (a,b)$ and the theorem holds.
Otherwise take $mu = int_a^b f(x) , g(x) , dx / int_a^b g(x) , dx$. We know that $m leqslant mu leqslant M$. If $m < mu < M$, by the properties of infimum and supremum there exist $alpha , beta in [a,b]$ such that
$$m < f(alpha) < mu < f(beta) < M.$$
The function $f$ when continuous has the intermediate value property. This is also true if $f$ has an antiderivative $F$ such that $f= F'$ for all $x in [a,b]$. Hence, there exists $xi in (alpha,beta) subset [a,b]$ such that $f(xi) = mu$ and we are done.
We also have to consider the possibility that the supremum or infimum of $f$ is attained at $mu$.
Suppose however that $mu = m$. Since $f(x) geqslant m$ and $g(x) geqslant 0$, we have
$$int_a^b |f(x) - m| , g(x) , dx = int_a^b (f(x) - m) , g(x) , dx = (mu -m) int_a^b g(x) , dx = 0,$$
and it follows that $(f(x) - m) , g(x) = 0$ almost everywhere. For this case where $int_a^b g(x) , dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $xi in (a,b)$ such that $f(xi) = m$.
The case where $mu = M$ is handled in a similar way.
answered Jan 3 at 4:07
RRLRRL
49.5k42573
$endgroup$
add a comment |
$begingroup$
Let $m$ be the minimum and $M$ the maximum of $f.$ Then $m int g leq int fg leq M int g;$ hence the result for $xi in [a, b].$ Assume $int g > 0.$ Assume what you want to be false, that means $int fg = f(a) int g$ or else $int fg = f(b) int g;$ suppose the first case. Then, $f(xi) int g > f(a) int g$ or else $f(xi) int g < f(a) int g;$ assume the first case. Then $f(a)$ is a strict maximum of $f$ and for every $delta > 0$ there exists $varepsilon > 0$ (in this order) such that $f(a) - varepsilon geq f(xi)$ for all $xi in [a + delta, b].$ On the other hand, $f(a) int g = int fg = (int_a^{a+delta} + int_{a + delta}^b )fg leq f(a) int_a^{a + delta}g + (f(a)-varepsilon) int_{a+delta}^b g,$ hence $int_{a+delta}^b g = 0.$ This being true for every $delta > 0$ shows $int g = 0,$ which is an absurd. Q.E.D.
answered Jan 3 at 3:43
Will M.Will M.
2,410314
$endgroup$
add a comment |
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$begingroup$
Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = inf_{x in [a,b]}, f(x)$ and $M = sup_{x in [a,b]}, f(x)$. Since $mg(x) leqslant f(x)g(x) leqslant Mg(x)$ for all $x in [a,b]$ we have
$$mint_a^b g(x) , dx leqslant int_a^b f(x) , g(x) , dx leqslant M int_a^b g(x) , dx.$$
In the case where $int_a^b g(x) , dx = 0$ it is easy to show that we can choose any $xi in (a,b)$ and the theorem holds.
Otherwise take $mu = int_a^b f(x) , g(x) , dx / int_a^b g(x) , dx$. We know that $m leqslant mu leqslant M$. If $m < mu < M$, by the properties of infimum and supremum there exist $alpha , beta in [a,b]$ such that
$$m < f(alpha) < mu < f(beta) < M.$$
The function $f$ when continuous has the intermediate value property. This is also true if $f$ has an antiderivative $F$ such that $f= F'$ for all $x in [a,b]$. Hence, there exists $xi in (alpha,beta) subset [a,b]$ such that $f(xi) = mu$ and we are done.
We also have to consider the possibility that the supremum or infimum of $f$ is attained at $mu$.
Suppose however that $mu = m$. Since $f(x) geqslant m$ and $g(x) geqslant 0$, we have
$$int_a^b |f(x) - m| , g(x) , dx = int_a^b (f(x) - m) , g(x) , dx = (mu -m) int_a^b g(x) , dx = 0,$$
and it follows that $(f(x) - m) , g(x) = 0$ almost everywhere. For this case where $int_a^b g(x) , dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $xi in (a,b)$ such that $f(xi) = m$.
The case where $mu = M$ is handled in a similar way.
answered Jan 3 at 4:07
RRLRRL
49.5k42573
$endgroup$
add a comment |
$begingroup$
Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = inf_{x in [a,b]}, f(x)$ and $M = sup_{x in [a,b]}, f(x)$. Since $mg(x) leqslant f(x)g(x) leqslant Mg(x)$ for all $x in [a,b]$ we have
$$mint_a^b g(x) , dx leqslant int_a^b f(x) , g(x) , dx leqslant M int_a^b g(x) , dx.$$
In the case where $int_a^b g(x) , dx = 0$ it is easy to show that we can choose any $xi in (a,b)$ and the theorem holds.
Otherwise take $mu = int_a^b f(x) , g(x) , dx / int_a^b g(x) , dx$. We know that $m leqslant mu leqslant M$. If $m < mu < M$, by the properties of infimum and supremum there exist $alpha , beta in [a,b]$ such that
$$m < f(alpha) < mu < f(beta) < M.$$
The function $f$ when continuous has the intermediate value property. This is also true if $f$ has an antiderivative $F$ such that $f= F'$ for all $x in [a,b]$. Hence, there exists $xi in (alpha,beta) subset [a,b]$ such that $f(xi) = mu$ and we are done.
We also have to consider the possibility that the supremum or infimum of $f$ is attained at $mu$.
Suppose however that $mu = m$. Since $f(x) geqslant m$ and $g(x) geqslant 0$, we have
$$int_a^b |f(x) - m| , g(x) , dx = int_a^b (f(x) - m) , g(x) , dx = (mu -m) int_a^b g(x) , dx = 0,$$
and it follows that $(f(x) - m) , g(x) = 0$ almost everywhere. For this case where $int_a^b g(x) , dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $xi in (a,b)$ such that $f(xi) = m$.
The case where $mu = M$ is handled in a similar way.
answered Jan 3 at 4:07
RRLRRL
49.5k42573
$endgroup$
add a comment |
$begingroup$
Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = inf_{x in [a,b]}, f(x)$ and $M = sup_{x in [a,b]}, f(x)$. Since $mg(x) leqslant f(x)g(x) leqslant Mg(x)$ for all $x in [a,b]$ we have
$$mint_a^b g(x) , dx leqslant int_a^b f(x) , g(x) , dx leqslant M int_a^b g(x) , dx.$$
In the case where $int_a^b g(x) , dx = 0$ it is easy to show that we can choose any $xi in (a,b)$ and the theorem holds.
Otherwise take $mu = int_a^b f(x) , g(x) , dx / int_a^b g(x) , dx$. We know that $m leqslant mu leqslant M$. If $m < mu < M$, by the properties of infimum and supremum there exist $alpha , beta in [a,b]$ such that
$$m < f(alpha) < mu < f(beta) < M.$$
The function $f$ when continuous has the intermediate value property. This is also true if $f$ has an antiderivative $F$ such that $f= F'$ for all $x in [a,b]$. Hence, there exists $xi in (alpha,beta) subset [a,b]$ such that $f(xi) = mu$ and we are done.
We also have to consider the possibility that the supremum or infimum of $f$ is attained at $mu$.
Suppose however that $mu = m$. Since $f(x) geqslant m$ and $g(x) geqslant 0$, we have
$$int_a^b |f(x) - m| , g(x) , dx = int_a^b (f(x) - m) , g(x) , dx = (mu -m) int_a^b g(x) , dx = 0,$$
and it follows that $(f(x) - m) , g(x) = 0$ almost everywhere. For this case where $int_a^b g(x) , dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $xi in (a,b)$ such that $f(xi) = m$.
The case where $mu = M$ is handled in a similar way.
answered Jan 3 at 4:07
RRLRRL
49.5k42573
$endgroup$
Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = inf_{x in [a,b]}, f(x)$ and $M = sup_{x in [a,b]}, f(x)$. Since $mg(x) leqslant f(x)g(x) leqslant Mg(x)$ for all $x in [a,b]$ we have
$$mint_a^b g(x) , dx leqslant int_a^b f(x) , g(x) , dx leqslant M int_a^b g(x) , dx.$$
In the case where $int_a^b g(x) , dx = 0$ it is easy to show that we can choose any $xi in (a,b)$ and the theorem holds.
Otherwise take $mu = int_a^b f(x) , g(x) , dx / int_a^b g(x) , dx$. We know that $m leqslant mu leqslant M$. If $m < mu < M$, by the properties of infimum and supremum there exist $alpha , beta in [a,b]$ such that
$$m < f(alpha) < mu < f(beta) < M.$$
The function $f$ when continuous has the intermediate value property. This is also true if $f$ has an antiderivative $F$ such that $f= F'$ for all $x in [a,b]$. Hence, there exists $xi in (alpha,beta) subset [a,b]$ such that $f(xi) = mu$ and we are done.
We also have to consider the possibility that the supremum or infimum of $f$ is attained at $mu$.
Suppose however that $mu = m$. Since $f(x) geqslant m$ and $g(x) geqslant 0$, we have
$$int_a^b |f(x) - m| , g(x) , dx = int_a^b (f(x) - m) , g(x) , dx = (mu -m) int_a^b g(x) , dx = 0,$$
and it follows that $(f(x) - m) , g(x) = 0$ almost everywhere. For this case where $int_a^b g(x) , dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $xi in (a,b)$ such that $f(xi) = m$.
The case where $mu = M$ is handled in a similar way.
answered Jan 3 at 4:07
RRLRRL
49.5k42573
edited Jan 3 at 4:39
answered Jan 3 at 4:07
RRLRRL
49.5k42573
answered Jan 3 at 4:07
RRLRRL
49.5k42573
answered Jan 3 at 4:07
RRLRRL
49.5k42573
49.5k42573
add a comment |
add a comment |
$begingroup$
Let $m$ be the minimum and $M$ the maximum of $f.$ Then $m int g leq int fg leq M int g;$ hence the result for $xi in [a, b].$ Assume $int g > 0.$ Assume what you want to be false, that means $int fg = f(a) int g$ or else $int fg = f(b) int g;$ suppose the first case. Then, $f(xi) int g > f(a) int g$ or else $f(xi) int g < f(a) int g;$ assume the first case. Then $f(a)$ is a strict maximum of $f$ and for every $delta > 0$ there exists $varepsilon > 0$ (in this order) such that $f(a) - varepsilon geq f(xi)$ for all $xi in [a + delta, b].$ On the other hand, $f(a) int g = int fg = (int_a^{a+delta} + int_{a + delta}^b )fg leq f(a) int_a^{a + delta}g + (f(a)-varepsilon) int_{a+delta}^b g,$ hence $int_{a+delta}^b g = 0.$ This being true for every $delta > 0$ shows $int g = 0,$ which is an absurd. Q.E.D.
answered Jan 3 at 3:43
Will M.Will M.
2,410314
$endgroup$
add a comment |
$begingroup$
Let $m$ be the minimum and $M$ the maximum of $f.$ Then $m int g leq int fg leq M int g;$ hence the result for $xi in [a, b].$ Assume $int g > 0.$ Assume what you want to be false, that means $int fg = f(a) int g$ or else $int fg = f(b) int g;$ suppose the first case. Then, $f(xi) int g > f(a) int g$ or else $f(xi) int g < f(a) int g;$ assume the first case. Then $f(a)$ is a strict maximum of $f$ and for every $delta > 0$ there exists $varepsilon > 0$ (in this order) such that $f(a) - varepsilon geq f(xi)$ for all $xi in [a + delta, b].$ On the other hand, $f(a) int g = int fg = (int_a^{a+delta} + int_{a + delta}^b )fg leq f(a) int_a^{a + delta}g + (f(a)-varepsilon) int_{a+delta}^b g,$ hence $int_{a+delta}^b g = 0.$ This being true for every $delta > 0$ shows $int g = 0,$ which is an absurd. Q.E.D.
answered Jan 3 at 3:43
Will M.Will M.
2,410314
$endgroup$
add a comment |
$begingroup$
Let $m$ be the minimum and $M$ the maximum of $f.$ Then $m int g leq int fg leq M int g;$ hence the result for $xi in [a, b].$ Assume $int g > 0.$ Assume what you want to be false, that means $int fg = f(a) int g$ or else $int fg = f(b) int g;$ suppose the first case. Then, $f(xi) int g > f(a) int g$ or else $f(xi) int g < f(a) int g;$ assume the first case. Then $f(a)$ is a strict maximum of $f$ and for every $delta > 0$ there exists $varepsilon > 0$ (in this order) such that $f(a) - varepsilon geq f(xi)$ for all $xi in [a + delta, b].$ On the other hand, $f(a) int g = int fg = (int_a^{a+delta} + int_{a + delta}^b )fg leq f(a) int_a^{a + delta}g + (f(a)-varepsilon) int_{a+delta}^b g,$ hence $int_{a+delta}^b g = 0.$ This being true for every $delta > 0$ shows $int g = 0,$ which is an absurd. Q.E.D.
answered Jan 3 at 3:43
Will M.Will M.
2,410314
$endgroup$
Let $m$ be the minimum and $M$ the maximum of $f.$ Then $m int g leq int fg leq M int g;$ hence the result for $xi in [a, b].$ Assume $int g > 0.$ Assume what you want to be false, that means $int fg = f(a) int g$ or else $int fg = f(b) int g;$ suppose the first case. Then, $f(xi) int g > f(a) int g$ or else $f(xi) int g < f(a) int g;$ assume the first case. Then $f(a)$ is a strict maximum of $f$ and for every $delta > 0$ there exists $varepsilon > 0$ (in this order) such that $f(a) - varepsilon geq f(xi)$ for all $xi in [a + delta, b].$ On the other hand, $f(a) int g = int fg = (int_a^{a+delta} + int_{a + delta}^b )fg leq f(a) int_a^{a + delta}g + (f(a)-varepsilon) int_{a+delta}^b g,$ hence $int_{a+delta}^b g = 0.$ This being true for every $delta > 0$ shows $int g = 0,$ which is an absurd. Q.E.D.
answered Jan 3 at 3:43
Will M.Will M.
2,410314
edited Jan 3 at 4:03
answered Jan 3 at 3:43
Will M.Will M.
2,410314
answered Jan 3 at 3:43
Will M.Will M.
2,410314
answered Jan 3 at 3:43
Will M.Will M.
2,410314
2,410314
add a comment |
add a comment |
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We need g(x) to be non-negative (or at least not change signs).
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– Doug M
Jan 3 at 3:11
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@DougM yes. Thanks for reminding me.
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– Jiu
Jan 3 at 3:11
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What you propose is true, but it is not as elegant a proof as to show that it is true over the closed interval.
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– Doug M
Jan 3 at 3:28
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@DougM Do you know where I can find a proof?
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– Jiu
Jan 3 at 3:52