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Why must all co-efficents $c_j$ be positive in the integration of step functions?

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0

$begingroup$

I am working through a proof of the following theorem:

'If $f$ and $g$ are step functions having $f(x) geq g(x)$ for all real values $x$, then $int f geq int g$.

So far I understand and thus have stated that $(f-g)(x) = f(x) - g(x)$ is a step function. We can rewrite $f-g$ as $sum_j=1^mc_j cdot chi_I_j$. It then states that since $f(x) geq g(x)$ for all $x$, the co-efficents $c_j$ forming the function values must be positive. But why is that so?

Any help would be greatly appreciated.

proof-explanation step-function

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asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

$endgroup$

add a comment | 

0

$begingroup$

I am working through a proof of the following theorem:

'If $f$ and $g$ are step functions having $f(x) geq g(x)$ for all real values $x$, then $int f geq int g$.

So far I understand and thus have stated that $(f-g)(x) = f(x) - g(x)$ is a step function. We can rewrite $f-g$ as $sum_j=1^mc_j cdot chi_I_j$. It then states that since $f(x) geq g(x)$ for all $x$, the co-efficents $c_j$ forming the function values must be positive. But why is that so?

Any help would be greatly appreciated.

proof-explanation step-function

share|cite|improve this question

asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

$endgroup$

add a comment | 

$begingroup$

I am working through a proof of the following theorem:

'If $f$ and $g$ are step functions having $f(x) geq g(x)$ for all real values $x$, then $int f geq int g$.

So far I understand and thus have stated that $(f-g)(x) = f(x) - g(x)$ is a step function. We can rewrite $f-g$ as $sum_j=1^mc_j cdot chi_I_j$. It then states that since $f(x) geq g(x)$ for all $x$, the co-efficents $c_j$ forming the function values must be positive. But why is that so?

Any help would be greatly appreciated.

proof-explanation step-function

share|cite|improve this question

asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

$endgroup$

share|cite|improve this question

asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

share|cite|improve this question

asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

asked Mar 22 at 2:22

KeighleyiteKeighleyite

889

1 Answer
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$begingroup$

While we can write it with a mix of coefficients in which some are negative, there's at least one representation with positive coefficients.

We subdivide the domain based on the values of $f-g$. Write $mathbbR$ as a disjoint union of intervals $I_j$, such that $f-g$ is constant on each $I_j$. These intervals will be intersections of the intervals for $f$ and those for $g$.

Then, since its a disjoint union, $0le f(x)=c_j$ for any $xin I_j$.

share|cite|improve this answer

answered Mar 22 at 2:35

jmerryjmerry

17k11633

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0

$begingroup$

While we can write it with a mix of coefficients in which some are negative, there's at least one representation with positive coefficients.

We subdivide the domain based on the values of $f-g$. Write $mathbbR$ as a disjoint union of intervals $I_j$, such that $f-g$ is constant on each $I_j$. These intervals will be intersections of the intervals for $f$ and those for $g$.

Then, since its a disjoint union, $0le f(x)=c_j$ for any $xin I_j$.

share|cite|improve this answer

answered Mar 22 at 2:35

jmerryjmerry

17k11633

$endgroup$

add a comment | 

0

$begingroup$

While we can write it with a mix of coefficients in which some are negative, there's at least one representation with positive coefficients.

We subdivide the domain based on the values of $f-g$. Write $mathbbR$ as a disjoint union of intervals $I_j$, such that $f-g$ is constant on each $I_j$. These intervals will be intersections of the intervals for $f$ and those for $g$.

Then, since its a disjoint union, $0le f(x)=c_j$ for any $xin I_j$.

share|cite|improve this answer

answered Mar 22 at 2:35

jmerryjmerry

17k11633

$endgroup$

add a comment | 

$begingroup$

While we can write it with a mix of coefficients in which some are negative, there's at least one representation with positive coefficients.

We subdivide the domain based on the values of $f-g$. Write $mathbbR$ as a disjoint union of intervals $I_j$, such that $f-g$ is constant on each $I_j$. These intervals will be intersections of the intervals for $f$ and those for $g$.

Then, since its a disjoint union, $0le f(x)=c_j$ for any $xin I_j$.

share|cite|improve this answer

answered Mar 22 at 2:35

jmerryjmerry

17k11633

$endgroup$

share|cite|improve this answer

answered Mar 22 at 2:35

jmerryjmerry

17k11633

answered Mar 22 at 2:35

jmerryjmerry

17k11633

answered Mar 22 at 2:35

jmerryjmerry

17k11633