Is this definite integral positive?

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I have a doubt and I am not able to prove (or disprove):

Let $f(x)$ be an odd function with $f(x)>0,,,forall xin (0,+infty)$.

Let $g(x)$ be a non-negative function: $g(x)geq 0;forall xin mathbbR.$

Also suppose $displaystyle int_-infty^0g(x),dx<int_0^inftyg(x),dx.$

I wonder if one can assure that:

$$int_-infty^inftyf(x),g(x),dx>0.$$

EDIT 1: Has been proved (by Adrian Keister) that my thesis is false.

Now I wonder again if is possible add another hypothesis about $g(x)$ to assure my thesis.

EDIT 2:The problem arrives from here:

enter image description here

blue line is $f(x)= left(e^-fraccosh ^2(u-1)2 -e^-fraccosh ^2(1+u)2 right)$ and orange line is $g(x)=e^-fracu^22 cos ^2left(fracpi (u-1)4 right)$ and the function $f(x)g(x)$ graphic

enter image description here

As we can see in the graph, the integral $int_mathbbRf(x)g(x),dx$ seems to be positive.

We can translate the factor $e^-u^2/2$ from $g(x)$ to $f(x)$ (in this case the third hypothesis is not fulfilled):

enter image description here

edited Aug 30 at 8:51

asked Aug 29 at 16:39

popi

9111

1

Can you assume the integral of $g$ is finite? I don't think it would make sense to compare negative with positive infinity.
â€“Â Mefitico
Aug 29 at 16:44

add a commentÂ |Â

up vote
6
down vote

favorite

I have a doubt and I am not able to prove (or disprove):

Let $f(x)$ be an odd function with $f(x)>0,,,forall xin (0,+infty)$.

Let $g(x)$ be a non-negative function: $g(x)geq 0;forall xin mathbbR.$

Also suppose $displaystyle int_-infty^0g(x),dx<int_0^inftyg(x),dx.$

I wonder if one can assure that:

$$int_-infty^inftyf(x),g(x),dx>0.$$

EDIT 1: Has been proved (by Adrian Keister) that my thesis is false.

Now I wonder again if is possible add another hypothesis about $g(x)$ to assure my thesis.

EDIT 2:The problem arrives from here:

enter image description here

blue line is $f(x)= left(e^-fraccosh ^2(u-1)2 -e^-fraccosh ^2(1+u)2 right)$ and orange line is $g(x)=e^-fracu^22 cos ^2left(fracpi (u-1)4 right)$ and the function $f(x)g(x)$ graphic

enter image description here

As we can see in the graph, the integral $int_mathbbRf(x)g(x),dx$ seems to be positive.

We can translate the factor $e^-u^2/2$ from $g(x)$ to $f(x)$ (in this case the third hypothesis is not fulfilled):

enter image description here

edited Aug 30 at 8:51

asked Aug 29 at 16:39

popi

9111

1

Can you assume the integral of $g$ is finite? I don't think it would make sense to compare negative with positive infinity.
â€“Â Mefitico
Aug 29 at 16:44

add a commentÂ |Â

up vote
6
down vote

favorite

I have a doubt and I am not able to prove (or disprove):

Let $f(x)$ be an odd function with $f(x)>0,,,forall xin (0,+infty)$.

Let $g(x)$ be a non-negative function: $g(x)geq 0;forall xin mathbbR.$

Also suppose $displaystyle int_-infty^0g(x),dx<int_0^inftyg(x),dx.$

I wonder if one can assure that:

$$int_-infty^inftyf(x),g(x),dx>0.$$

EDIT 1: Has been proved (by Adrian Keister) that my thesis is false.

Now I wonder again if is possible add another hypothesis about $g(x)$ to assure my thesis.

EDIT 2:The problem arrives from here:

enter image description here

blue line is $f(x)= left(e^-fraccosh ^2(u-1)2 -e^-fraccosh ^2(1+u)2 right)$ and orange line is $g(x)=e^-fracu^22 cos ^2left(fracpi (u-1)4 right)$ and the function $f(x)g(x)$ graphic

enter image description here

As we can see in the graph, the integral $int_mathbbRf(x)g(x),dx$ seems to be positive.

We can translate the factor $e^-u^2/2$ from $g(x)$ to $f(x)$ (in this case the third hypothesis is not fulfilled):

enter image description here

edited Aug 30 at 8:51

asked Aug 29 at 16:39

popi

9111

I have a doubt and I am not able to prove (or disprove):

Let $f(x)$ be an odd function with $f(x)>0,,,forall xin (0,+infty)$.

Let $g(x)$ be a non-negative function: $g(x)geq 0;forall xin mathbbR.$

Also suppose $displaystyle int_-infty^0g(x),dx<int_0^inftyg(x),dx.$

I wonder if one can assure that:

$$int_-infty^inftyf(x),g(x),dx>0.$$

EDIT 1: Has been proved (by Adrian Keister) that my thesis is false.

Now I wonder again if is possible add another hypothesis about $g(x)$ to assure my thesis.

EDIT 2:The problem arrives from here:

enter image description here

blue line is $f(x)= left(e^-fraccosh ^2(u-1)2 -e^-fraccosh ^2(1+u)2 right)$ and orange line is $g(x)=e^-fracu^22 cos ^2left(fracpi (u-1)4 right)$ and the function $f(x)g(x)$ graphic

enter image description here

As we can see in the graph, the integral $int_mathbbRf(x)g(x),dx$ seems to be positive.

We can translate the factor $e^-u^2/2$ from $g(x)$ to $f(x)$ (in this case the third hypothesis is not fulfilled):

enter image description here

edited Aug 30 at 8:51

asked Aug 29 at 16:39

popi

9111

edited Aug 30 at 8:51

asked Aug 29 at 16:39

popi

9111

asked Aug 29 at 16:39

popi

9111

asked Aug 29 at 16:39

popi

9111

1

Can you assume the integral of $g$ is finite? I don't think it would make sense to compare negative with positive infinity.
â€“Â Mefitico
Aug 29 at 16:44

add a commentÂ |Â

1

Can you assume the integral of $g$ is finite? I don't think it would make sense to compare negative with positive infinity.
â€“Â Mefitico
Aug 29 at 16:44

Can you assume the integral of $g$ is finite? I don't think it would make sense to compare negative with positive infinity.
â€“Â Mefitico
Aug 29 at 16:44

add a commentÂ |Â

2 Answers
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up vote
2
down vote

accepted

I should say not. Counterexample:
beginalign*
f(x)&=left
beginarrayll
-1/(10x^2), ; &xin(-infty,-1) \
-1, ; &xin[-1,0) \
0, ; &x=0 \
1, ; &xin(0,1] \
1/(10x^2), ; &xin(1,infty)
endarrayright, \
g(x)&=leftbeginarrayll
0,;&xin(-infty,-1)cup(0,1)cup(2,infty) \
1/2, &xin[-1,0] \
1, &xin[1,2] ; endarrayright.
endalign*

If you replace the $displaystyle int_-infty^0g(x),dx<int_0^inftyg(x),dx$ condition with the (stronger) condition $g(-x)<g(x);forall,x>0,$ then you have the following:
beginalign*
int_-infty^inftyf(x),g(x),dx&=
int_-infty^0f(x),g(x),dx+int_0^inftyf(x),g(x),dx \
&=-int_infty^0 f(-x),g(-x),dx+int_0^inftyf(x),g(x),dx \
&=int_0^infty f(-x),g(-x),dx+int_0^inftyf(x),g(x),dx \
&=-int_0^infty f(x),g(-x),dx+int_0^inftyf(x),g(x),dx \
&=int_0^infty f(x),(-g(-x)),dx+int_0^inftyf(x),g(x),dx. \
endalign*
Now, if you have $g(-x)<g(x);forall,x>0,$ it follows that $-g(-x)>-g(x),$ whence you get
beginalign*
int_-infty^inftyf(x),g(x),dx&=
int_0^infty f(x),(-g(-x)),dx+int_0^inftyf(x),g(x),dx \
&>-int_0^infty f(x),g(x),dx+int_0^inftyf(x),g(x),dx \
&=0.
endalign*

See tomasz's answer for a slightly less strict condition on $g(x)$.

edited Aug 29 at 18:43

answered Aug 29 at 17:16

Adrian Keister

4,04541633

@popi: Yeah, I saw that. Not sure what you could do in addition. If you say $g$ is even, your integral condition can't hold (you'd actually get the integral in question to be zero).
â€“Â Adrian Keister
Aug 29 at 17:39

Something about monotony, absolute or relative extremes, etc ... of $ g (x) $.
â€“Â popi
Aug 29 at 17:48

What about $g(-x)<g(x);forall,x>0?$
â€“Â Adrian Keister
Aug 29 at 17:54

See tomasz's answer for a slightly less restrictive condition on $g(x)$ that works. I think my proof could be adapted slightly for tomasz's condition.
â€“Â Adrian Keister
Aug 29 at 18:10

Your way of positioning commas and periods in your piecewise definitions seems very strange. I would write $$ f(x) = begincases -1/(10x^2), ; &xin(-infty,-1), \ -1, ; &xin[-1,0), \ 0, ; &x=0, \ 1, ; &xin(0,1], \ 1/(10x^2), ; &xin(1,infty). endcases $$
â€“Â Michael Hardy
Aug 29 at 18:19

Â |Â
show 5 more comments

up vote
2
down vote

If for almost all positive $x$ you have $g(x)geq g(-x)$ and that there is a non-null set of $x$ such that $g(x)neq g(-x)$, then this integral will always be positive.

No other hypothesis on $g$ will suffice (for all $f$). If $g(x)=g(-x)$ almost everywhere, then the integral is clearly zero. Otherwise, suppose $Asubseteq [0,infty)$ is a set of positive measure such that for $xin A$ we have $g(x)< g(-x)$. We may assume without loss of generality that for all $xin A$ we have $g(-x)>g(x)+1/n$ for some positive integer $n$.

Now, take for $f$ an odd function such that $f(x)=2n/lvert Arvert$ for $xin A$ and $f(x)<left(int_-infty^infty lvert g(t)rvert,mathrmdtright)^-1$ for positive $xnotin A$.

Then $int_Acup -A f(t)g(t),mathrmdt<-2$ and $leftlvert int_(Acup -A)^c f(t)g(t),mathrmdtrightrvertleq 1$.

answered Aug 29 at 18:05

tomasz

22.9k23077

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

See tomasz's answer for a slightly less strict condition on $g(x)$.

edited Aug 29 at 18:43

answered Aug 29 at 17:16

Adrian Keister

4,04541633

@popi: Yeah, I saw that. Not sure what you could do in addition. If you say $g$ is even, your integral condition can't hold (you'd actually get the integral in question to be zero).
â€“Â Adrian Keister
Aug 29 at 17:39

Something about monotony, absolute or relative extremes, etc ... of $ g (x) $.
â€“Â popi
Aug 29 at 17:48

What about $g(-x)<g(x);forall,x>0?$
â€“Â Adrian Keister
Aug 29 at 17:54

See tomasz's answer for a slightly less restrictive condition on $g(x)$ that works. I think my proof could be adapted slightly for tomasz's condition.
â€“Â Adrian Keister
Aug 29 at 18:10

Your way of positioning commas and periods in your piecewise definitions seems very strange. I would write $$ f(x) = begincases -1/(10x^2), ; &xin(-infty,-1), \ -1, ; &xin[-1,0), \ 0, ; &x=0, \ 1, ; &xin(0,1], \ 1/(10x^2), ; &xin(1,infty). endcases $$
â€“Â Michael Hardy
Aug 29 at 18:19

Â |Â
show 5 more comments

up vote
2
down vote

accepted

See tomasz's answer for a slightly less strict condition on $g(x)$.

edited Aug 29 at 18:43

answered Aug 29 at 17:16

Adrian Keister

4,04541633

@popi: Yeah, I saw that. Not sure what you could do in addition. If you say $g$ is even, your integral condition can't hold (you'd actually get the integral in question to be zero).
â€“Â Adrian Keister
Aug 29 at 17:39

Something about monotony, absolute or relative extremes, etc ... of $ g (x) $.
â€“Â popi
Aug 29 at 17:48

What about $g(-x)<g(x);forall,x>0?$
â€“Â Adrian Keister
Aug 29 at 17:54

See tomasz's answer for a slightly less restrictive condition on $g(x)$ that works. I think my proof could be adapted slightly for tomasz's condition.
â€“Â Adrian Keister
Aug 29 at 18:10

Your way of positioning commas and periods in your piecewise definitions seems very strange. I would write $$ f(x) = begincases -1/(10x^2), ; &xin(-infty,-1), \ -1, ; &xin[-1,0), \ 0, ; &x=0, \ 1, ; &xin(0,1], \ 1/(10x^2), ; &xin(1,infty). endcases $$
â€“Â Michael Hardy
Aug 29 at 18:19

Â |Â
show 5 more comments

up vote
2
down vote

accepted

See tomasz's answer for a slightly less strict condition on $g(x)$.

edited Aug 29 at 18:43

answered Aug 29 at 17:16

Adrian Keister

4,04541633

See tomasz's answer for a slightly less strict condition on $g(x)$.

edited Aug 29 at 18:43

answered Aug 29 at 17:16

Adrian Keister

4,04541633

edited Aug 29 at 18:43

answered Aug 29 at 17:16

Adrian Keister

4,04541633

answered Aug 29 at 17:16

Adrian Keister

4,04541633

answered Aug 29 at 17:16

Adrian Keister

4,04541633

@popi: Yeah, I saw that. Not sure what you could do in addition. If you say $g$ is even, your integral condition can't hold (you'd actually get the integral in question to be zero).
â€“Â Adrian Keister
Aug 29 at 17:39

Something about monotony, absolute or relative extremes, etc ... of $ g (x) $.
â€“Â popi
Aug 29 at 17:48

What about $g(-x)<g(x);forall,x>0?$
â€“Â Adrian Keister
Aug 29 at 17:54

See tomasz's answer for a slightly less restrictive condition on $g(x)$ that works. I think my proof could be adapted slightly for tomasz's condition.
â€“Â Adrian Keister
Aug 29 at 18:10

Your way of positioning commas and periods in your piecewise definitions seems very strange. I would write $$ f(x) = begincases -1/(10x^2), ; &xin(-infty,-1), \ -1, ; &xin[-1,0), \ 0, ; &x=0, \ 1, ; &xin(0,1], \ 1/(10x^2), ; &xin(1,infty). endcases $$
â€“Â Michael Hardy
Aug 29 at 18:19

Â |Â
show 5 more comments

@popi: Yeah, I saw that. Not sure what you could do in addition. If you say $g$ is even, your integral condition can't hold (you'd actually get the integral in question to be zero).
â€“Â Adrian Keister
Aug 29 at 17:39

Something about monotony, absolute or relative extremes, etc ... of $ g (x) $.
â€“Â popi
Aug 29 at 17:48

What about $g(-x)<g(x);forall,x>0?$
â€“Â Adrian Keister
Aug 29 at 17:54

See tomasz's answer for a slightly less restrictive condition on $g(x)$ that works. I think my proof could be adapted slightly for tomasz's condition.
â€“Â Adrian Keister
Aug 29 at 18:10

Your way of positioning commas and periods in your piecewise definitions seems very strange. I would write $$ f(x) = begincases -1/(10x^2), ; &xin(-infty,-1), \ -1, ; &xin[-1,0), \ 0, ; &x=0, \ 1, ; &xin(0,1], \ 1/(10x^2), ; &xin(1,infty). endcases $$
â€“Â Michael Hardy
Aug 29 at 18:19

@popi: Yeah, I saw that. Not sure what you could do in addition. If you say $g$ is even, your integral condition can't hold (you'd actually get the integral in question to be zero).
â€“Â Adrian Keister
Aug 29 at 17:39

Something about monotony, absolute or relative extremes, etc ... of $ g (x) $.
â€“Â popi
Aug 29 at 17:48

What about $g(-x)<g(x);forall,x>0?$
â€“Â Adrian Keister
Aug 29 at 17:54

See tomasz's answer for a slightly less restrictive condition on $g(x)$ that works. I think my proof could be adapted slightly for tomasz's condition.
â€“Â Adrian Keister
Aug 29 at 18:10

Your way of positioning commas and periods in your piecewise definitions seems very strange. I would write $$ f(x) = begincases -1/(10x^2), ; &xin(-infty,-1), \ -1, ; &xin[-1,0), \ 0, ; &x=0, \ 1, ; &xin(0,1], \ 1/(10x^2), ; &xin(1,infty). endcases $$
â€“Â Michael Hardy
Aug 29 at 18:19

Â |Â
show 5 more comments

up vote
2
down vote

If for almost all positive $x$ you have $g(x)geq g(-x)$ and that there is a non-null set of $x$ such that $g(x)neq g(-x)$, then this integral will always be positive.

Now, take for $f$ an odd function such that $f(x)=2n/lvert Arvert$ for $xin A$ and $f(x)<left(int_-infty^infty lvert g(t)rvert,mathrmdtright)^-1$ for positive $xnotin A$.

Then $int_Acup -A f(t)g(t),mathrmdt<-2$ and $leftlvert int_(Acup -A)^c f(t)g(t),mathrmdtrightrvertleq 1$.

answered Aug 29 at 18:05

tomasz

22.9k23077

add a commentÂ |Â

up vote
2
down vote

If for almost all positive $x$ you have $g(x)geq g(-x)$ and that there is a non-null set of $x$ such that $g(x)neq g(-x)$, then this integral will always be positive.

Now, take for $f$ an odd function such that $f(x)=2n/lvert Arvert$ for $xin A$ and $f(x)<left(int_-infty^infty lvert g(t)rvert,mathrmdtright)^-1$ for positive $xnotin A$.

Then $int_Acup -A f(t)g(t),mathrmdt<-2$ and $leftlvert int_(Acup -A)^c f(t)g(t),mathrmdtrightrvertleq 1$.

answered Aug 29 at 18:05

tomasz

22.9k23077

add a commentÂ |Â

up vote
2
down vote

If for almost all positive $x$ you have $g(x)geq g(-x)$ and that there is a non-null set of $x$ such that $g(x)neq g(-x)$, then this integral will always be positive.

Now, take for $f$ an odd function such that $f(x)=2n/lvert Arvert$ for $xin A$ and $f(x)<left(int_-infty^infty lvert g(t)rvert,mathrmdtright)^-1$ for positive $xnotin A$.

Then $int_Acup -A f(t)g(t),mathrmdt<-2$ and $leftlvert int_(Acup -A)^c f(t)g(t),mathrmdtrightrvertleq 1$.

answered Aug 29 at 18:05

tomasz

22.9k23077

If for almost all positive $x$ you have $g(x)geq g(-x)$ and that there is a non-null set of $x$ such that $g(x)neq g(-x)$, then this integral will always be positive.

Now, take for $f$ an odd function such that $f(x)=2n/lvert Arvert$ for $xin A$ and $f(x)<left(int_-infty^infty lvert g(t)rvert,mathrmdtright)^-1$ for positive $xnotin A$.

Then $int_Acup -A f(t)g(t),mathrmdt<-2$ and $leftlvert int_(Acup -A)^c f(t)g(t),mathrmdtrightrvertleq 1$.

answered Aug 29 at 18:05

tomasz

22.9k23077

answered Aug 29 at 18:05

tomasz

22.9k23077

answered Aug 29 at 18:05

tomasz

22.9k23077

answered Aug 29 at 18:05

tomasz

22.9k23077

add a commentÂ |Â

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