Is this continuous analogue to the AM–GM inequality true?
$begingroup$
First let us remind ourselves of the statement of the AM–GM inequality:
Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $Ngeqslant 1$ non-negative real numbers, we have $$frac1Nsum_k x_k geqslant left(prod_k x_kright)^{frac1N}$$
It is well known that the sum operator ∑ can be generalised so that it operates on continuous functions rather than on discrete sequences. This generalisation is the integral operator ∫.
Likewise, we can generalise the product operator to operate on continuous functions. We begin with the following property of the discrete product: $$logprod_k A_k=sum_k log A_k$$ (assuming that $A_k>0$). Using this we can define the continuous product (also known as the geometric integral according to Wikipedia) as follows: $$prod_a^b f(x) dx := expint_a^blog f(x) dx.$$
assuming that the integral converges. From these definitions we can seek to generalise the AM–GM inequality to continuous functions:
Proposition: (Continuous AM–GM Inequality) For any suitably well-behaved non-negative function $f$ defined on $[a,b]$ with $a<b$, we have: $$frac1{b-a}int_a^bf(x) dx geqslant left(prod_a^b f(x) dxright)^frac1{b-a}$$ or in traditional notation: $$frac1{b-a}int_a^bf(x) dx geqslant expleft(frac1{b-a} int_a^b log f(x) dxright)$$
As a simple illustration, consider the function $f(x)=1+sin x$ and fix the lower bound $a=0$. On the following graph the $x$-axis represents the upper bound of integration $b$, and on the $y$-axis are represented the the arithmetic mean (blue) and the geometric mean (red) of $f$ on the interval $[a,b]$ where $a=0$ and $b=x$.
Clearly the claimed inequality seems to hold, and is in fact quite tight when $bapprox0$.
This inequality is of interest not least because it is readily abe to produce a number of non-trivial numerical inequalities pertaining to known constants. For instance, again with the example $f(x)=1+sin x$, set $b=pi/2$. Then assuming the proposition holds, we have
$$frac{2}{pi}int_0^{pi/2} 1 + sin x dx geqslant exp left( frac{2}{pi} int_0^{pi/2} log(1+sin x) dx right)$$ Evaluating these integrals and rearranging, we obtain: $$G leqslant fracpi4logleft(2+frac2piright) approx 0.9313$$ where $Gapprox 0.9159$ is Catalan's constant.
Anyway I would like to ask, firstly:
Is the claimed inequality true, and if so what is the proof and on what class of functions is it applicable?
Secondly, and this is more of a fun little challenge:
Assuming its veracity, can this inequality be used to prove remarkable numerical inequalities, e.g. $pi < 22/7$ or $e^pi - pi < 20$?
integration inequality integral-inequality a.m.-g.m.-inequality
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
$endgroup$
|
show 1 more comment
$begingroup$
First let us remind ourselves of the statement of the AM–GM inequality:
Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $Ngeqslant 1$ non-negative real numbers, we have $$frac1Nsum_k x_k geqslant left(prod_k x_kright)^{frac1N}$$
It is well known that the sum operator ∑ can be generalised so that it operates on continuous functions rather than on discrete sequences. This generalisation is the integral operator ∫.
Likewise, we can generalise the product operator to operate on continuous functions. We begin with the following property of the discrete product: $$logprod_k A_k=sum_k log A_k$$ (assuming that $A_k>0$). Using this we can define the continuous product (also known as the geometric integral according to Wikipedia) as follows: $$prod_a^b f(x) dx := expint_a^blog f(x) dx.$$
assuming that the integral converges. From these definitions we can seek to generalise the AM–GM inequality to continuous functions:
Proposition: (Continuous AM–GM Inequality) For any suitably well-behaved non-negative function $f$ defined on $[a,b]$ with $a<b$, we have: $$frac1{b-a}int_a^bf(x) dx geqslant left(prod_a^b f(x) dxright)^frac1{b-a}$$ or in traditional notation: $$frac1{b-a}int_a^bf(x) dx geqslant expleft(frac1{b-a} int_a^b log f(x) dxright)$$
As a simple illustration, consider the function $f(x)=1+sin x$ and fix the lower bound $a=0$. On the following graph the $x$-axis represents the upper bound of integration $b$, and on the $y$-axis are represented the the arithmetic mean (blue) and the geometric mean (red) of $f$ on the interval $[a,b]$ where $a=0$ and $b=x$.
Clearly the claimed inequality seems to hold, and is in fact quite tight when $bapprox0$.
This inequality is of interest not least because it is readily abe to produce a number of non-trivial numerical inequalities pertaining to known constants. For instance, again with the example $f(x)=1+sin x$, set $b=pi/2$. Then assuming the proposition holds, we have
$$frac{2}{pi}int_0^{pi/2} 1 + sin x dx geqslant exp left( frac{2}{pi} int_0^{pi/2} log(1+sin x) dx right)$$ Evaluating these integrals and rearranging, we obtain: $$G leqslant fracpi4logleft(2+frac2piright) approx 0.9313$$ where $Gapprox 0.9159$ is Catalan's constant.
Anyway I would like to ask, firstly:
Is the claimed inequality true, and if so what is the proof and on what class of functions is it applicable?
Secondly, and this is more of a fun little challenge:
Assuming its veracity, can this inequality be used to prove remarkable numerical inequalities, e.g. $pi < 22/7$ or $e^pi - pi < 20$?
integration inequality integral-inequality a.m.-g.m.-inequality
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
$endgroup$
13
$begingroup$
Have you heard about Jensen's inequality?
$endgroup$
– mickep
Apr 23 '17 at 12:29
2
$begingroup$
Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it...
$endgroup$
– Pedro A
Apr 23 '17 at 13:31
2
$begingroup$
Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer.
$endgroup$
– user1892304
Apr 23 '17 at 13:49
2
$begingroup$
Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $Delta x$, similarly to how multiplying by $Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration.
$endgroup$
– Ruslan
Apr 23 '17 at 20:28
1
$begingroup$
Similarly, you can also consider $operatorname{HM}(f,a,b)=frac{b-a}{int_a^bfrac{dx}{f(x)}} as generalized harmonic mean.
$endgroup$
– g.kov
Apr 24 '17 at 17:26
|
show 1 more comment
$begingroup$
First let us remind ourselves of the statement of the AM–GM inequality:
Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $Ngeqslant 1$ non-negative real numbers, we have $$frac1Nsum_k x_k geqslant left(prod_k x_kright)^{frac1N}$$
It is well known that the sum operator ∑ can be generalised so that it operates on continuous functions rather than on discrete sequences. This generalisation is the integral operator ∫.
Likewise, we can generalise the product operator to operate on continuous functions. We begin with the following property of the discrete product: $$logprod_k A_k=sum_k log A_k$$ (assuming that $A_k>0$). Using this we can define the continuous product (also known as the geometric integral according to Wikipedia) as follows: $$prod_a^b f(x) dx := expint_a^blog f(x) dx.$$
assuming that the integral converges. From these definitions we can seek to generalise the AM–GM inequality to continuous functions:
Proposition: (Continuous AM–GM Inequality) For any suitably well-behaved non-negative function $f$ defined on $[a,b]$ with $a<b$, we have: $$frac1{b-a}int_a^bf(x) dx geqslant left(prod_a^b f(x) dxright)^frac1{b-a}$$ or in traditional notation: $$frac1{b-a}int_a^bf(x) dx geqslant expleft(frac1{b-a} int_a^b log f(x) dxright)$$
As a simple illustration, consider the function $f(x)=1+sin x$ and fix the lower bound $a=0$. On the following graph the $x$-axis represents the upper bound of integration $b$, and on the $y$-axis are represented the the arithmetic mean (blue) and the geometric mean (red) of $f$ on the interval $[a,b]$ where $a=0$ and $b=x$.
Clearly the claimed inequality seems to hold, and is in fact quite tight when $bapprox0$.
This inequality is of interest not least because it is readily abe to produce a number of non-trivial numerical inequalities pertaining to known constants. For instance, again with the example $f(x)=1+sin x$, set $b=pi/2$. Then assuming the proposition holds, we have
$$frac{2}{pi}int_0^{pi/2} 1 + sin x dx geqslant exp left( frac{2}{pi} int_0^{pi/2} log(1+sin x) dx right)$$ Evaluating these integrals and rearranging, we obtain: $$G leqslant fracpi4logleft(2+frac2piright) approx 0.9313$$ where $Gapprox 0.9159$ is Catalan's constant.
Anyway I would like to ask, firstly:
Is the claimed inequality true, and if so what is the proof and on what class of functions is it applicable?
Secondly, and this is more of a fun little challenge:
Assuming its veracity, can this inequality be used to prove remarkable numerical inequalities, e.g. $pi < 22/7$ or $e^pi - pi < 20$?
integration inequality integral-inequality a.m.-g.m.-inequality
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
$endgroup$
First let us remind ourselves of the statement of the AM–GM inequality:
Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $Ngeqslant 1$ non-negative real numbers, we have $$frac1Nsum_k x_k geqslant left(prod_k x_kright)^{frac1N}$$
It is well known that the sum operator ∑ can be generalised so that it operates on continuous functions rather than on discrete sequences. This generalisation is the integral operator ∫.
Likewise, we can generalise the product operator to operate on continuous functions. We begin with the following property of the discrete product: $$logprod_k A_k=sum_k log A_k$$ (assuming that $A_k>0$). Using this we can define the continuous product (also known as the geometric integral according to Wikipedia) as follows: $$prod_a^b f(x) dx := expint_a^blog f(x) dx.$$
assuming that the integral converges. From these definitions we can seek to generalise the AM–GM inequality to continuous functions:
Proposition: (Continuous AM–GM Inequality) For any suitably well-behaved non-negative function $f$ defined on $[a,b]$ with $a<b$, we have: $$frac1{b-a}int_a^bf(x) dx geqslant left(prod_a^b f(x) dxright)^frac1{b-a}$$ or in traditional notation: $$frac1{b-a}int_a^bf(x) dx geqslant expleft(frac1{b-a} int_a^b log f(x) dxright)$$
As a simple illustration, consider the function $f(x)=1+sin x$ and fix the lower bound $a=0$. On the following graph the $x$-axis represents the upper bound of integration $b$, and on the $y$-axis are represented the the arithmetic mean (blue) and the geometric mean (red) of $f$ on the interval $[a,b]$ where $a=0$ and $b=x$.
Clearly the claimed inequality seems to hold, and is in fact quite tight when $bapprox0$.
This inequality is of interest not least because it is readily abe to produce a number of non-trivial numerical inequalities pertaining to known constants. For instance, again with the example $f(x)=1+sin x$, set $b=pi/2$. Then assuming the proposition holds, we have
$$frac{2}{pi}int_0^{pi/2} 1 + sin x dx geqslant exp left( frac{2}{pi} int_0^{pi/2} log(1+sin x) dx right)$$ Evaluating these integrals and rearranging, we obtain: $$G leqslant fracpi4logleft(2+frac2piright) approx 0.9313$$ where $Gapprox 0.9159$ is Catalan's constant.
Anyway I would like to ask, firstly:
Is the claimed inequality true, and if so what is the proof and on what class of functions is it applicable?
Secondly, and this is more of a fun little challenge:
Assuming its veracity, can this inequality be used to prove remarkable numerical inequalities, e.g. $pi < 22/7$ or $e^pi - pi < 20$?
integration inequality integral-inequality a.m.-g.m.-inequality
integration inequality integral-inequality a.m.-g.m.-inequality
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
edited May 1 '17 at 4:11
Eric Wofsey
185k14213339
185k14213339
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
asked Apr 23 '17 at 12:26
user1892304user1892304
1,477917
1,477917
13
$begingroup$
Have you heard about Jensen's inequality?
$endgroup$
– mickep
Apr 23 '17 at 12:29
2
$begingroup$
Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it...
$endgroup$
– Pedro A
Apr 23 '17 at 13:31
2
$begingroup$
Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer.
$endgroup$
– user1892304
Apr 23 '17 at 13:49
2
$begingroup$
Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $Delta x$, similarly to how multiplying by $Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration.
$endgroup$
– Ruslan
Apr 23 '17 at 20:28
1
$begingroup$
Similarly, you can also consider $operatorname{HM}(f,a,b)=frac{b-a}{int_a^bfrac{dx}{f(x)}} as generalized harmonic mean.
$endgroup$
– g.kov
Apr 24 '17 at 17:26
|
show 1 more comment
13
$begingroup$
Have you heard about Jensen's inequality?
$endgroup$
– mickep
Apr 23 '17 at 12:29
2
$begingroup$
Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it...
$endgroup$
– Pedro A
Apr 23 '17 at 13:31
2
$begingroup$
Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer.
$endgroup$
– user1892304
Apr 23 '17 at 13:49
2
$begingroup$
Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $Delta x$, similarly to how multiplying by $Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration.
$endgroup$
– Ruslan
Apr 23 '17 at 20:28
1
$begingroup$
Similarly, you can also consider $operatorname{HM}(f,a,b)=frac{b-a}{int_a^bfrac{dx}{f(x)}} as generalized harmonic mean.
$endgroup$
– g.kov
Apr 24 '17 at 17:26
13
13
$begingroup$
Have you heard about Jensen's inequality?
$endgroup$
– mickep
Apr 23 '17 at 12:29
$begingroup$
Have you heard about Jensen's inequality?
$endgroup$
– mickep
Apr 23 '17 at 12:29
2
2
$begingroup$
Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it...
$endgroup$
– Pedro A
Apr 23 '17 at 13:31
$begingroup$
Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it...
$endgroup$
– Pedro A
Apr 23 '17 at 13:31
2
2
$begingroup$
Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer.
$endgroup$
– user1892304
Apr 23 '17 at 13:49
$begingroup$
Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer.
$endgroup$
– user1892304
Apr 23 '17 at 13:49
2
2
$begingroup$
Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $Delta x$, similarly to how multiplying by $Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration.
$endgroup$
– Ruslan
Apr 23 '17 at 20:28
$begingroup$
Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $Delta x$, similarly to how multiplying by $Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration.
$endgroup$
– Ruslan
Apr 23 '17 at 20:28
1
1
$begingroup$
Similarly, you can also consider $operatorname{HM}(f,a,b)=frac{b-a}{int_a^bfrac{dx}{f(x)}} as generalized harmonic mean.
$endgroup$
– g.kov
Apr 24 '17 at 17:26
$begingroup$
Similarly, you can also consider $operatorname{HM}(f,a,b)=frac{b-a}{int_a^bfrac{dx}{f(x)}} as generalized harmonic mean.
$endgroup$
– g.kov
Apr 24 '17 at 17:26
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I would write this inequality as
$$frac1{b-a}int_a^bexp(g(x)),dxgeexpleft(frac1{b-a}int_a^b g(x),dxright).$$
In this guise it is the case of Jensen's inequality for the convex function $phi(t)=exp(t)$.
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
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$begingroup$
I would write this inequality as
$$frac1{b-a}int_a^bexp(g(x)),dxgeexpleft(frac1{b-a}int_a^b g(x),dxright).$$
In this guise it is the case of Jensen's inequality for the convex function $phi(t)=exp(t)$.
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
I would write this inequality as
$$frac1{b-a}int_a^bexp(g(x)),dxgeexpleft(frac1{b-a}int_a^b g(x),dxright).$$
In this guise it is the case of Jensen's inequality for the convex function $phi(t)=exp(t)$.
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
I would write this inequality as
$$frac1{b-a}int_a^bexp(g(x)),dxgeexpleft(frac1{b-a}int_a^b g(x),dxright).$$
In this guise it is the case of Jensen's inequality for the convex function $phi(t)=exp(t)$.
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
I would write this inequality as
$$frac1{b-a}int_a^bexp(g(x)),dxgeexpleft(frac1{b-a}int_a^b g(x),dxright).$$
In this guise it is the case of Jensen's inequality for the convex function $phi(t)=exp(t)$.
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
edited May 1 '17 at 4:00
Eric Wofsey
185k14213339
185k14213339
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Apr 23 '17 at 12:34
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
add a comment |
add a comment |
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13
$begingroup$
Have you heard about Jensen's inequality?
$endgroup$
– mickep
Apr 23 '17 at 12:29
2
$begingroup$
Very nice question! +2 if I could! But don't be so quick to accept an answer... This discourages other people from writing answers. Wait a few days... Especially given your "fun little challenge". I'd love to see someone solve it...
$endgroup$
– Pedro A
Apr 23 '17 at 13:31
2
$begingroup$
Thanks for the advice @Hamsteriffic, I've "un-accepted" the answer to see if that encourages someone else to write a more comprehensive answer.
$endgroup$
– user1892304
Apr 23 '17 at 13:49
2
$begingroup$
Note that the continuous product should have $dx$ in superscript, since it comes from raising $f(x_i)$ to the power of $Delta x$, similarly to how multiplying by $Delta x$ in the Riemann sum gives multiplication by $dx$ in the final notation of integration.
$endgroup$
– Ruslan
Apr 23 '17 at 20:28
1
$begingroup$
Similarly, you can also consider $operatorname{HM}(f,a,b)=frac{b-a}{int_a^bfrac{dx}{f(x)}} as generalized harmonic mean.
$endgroup$
– g.kov
Apr 24 '17 at 17:26