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$f$ continuous on $[a,b]$, $fgeq 0$, $int_a^bf=0$ then $f=0$
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I'm curious as to if anyone can provide some feedback on this proof.
If $f:[a,b]rightarrow mathbbR$ continuous, $fgeq 0$, $int_a^bf=0$ then $f=0$.
My proof: Let's assume there exists some $cin [a,b]$ such that $f(c)>0$. Then $f>0$ on an open set $cin Isubset [a,b]$, this must be true because otherwise we would have a non-continuous piecewise function, $f(x)=0$ for all $xin [a,b]-c$ and $f(c)=0$. Now let's build a lower Darboux sum greater than $0$.
$L(f,sigma)=sum_i=0^n-1 m_i(f),Delta x_i=m_k(f),Delta x_i>0,$ where my partition is $sigma= a=x_0<x_1<dots<x_n=b $ picked in a way such that $[x_k,x_k+1]subset I$. Also $m_k(f)=inff(x):xin [x_k,x_k+1]$. Now we can conclude that $underlineint_a^bf=sup textL(f,sigma):sigma in textthe set of all partitions of [a,b] >0$ but this contradicts the fact that $underlineint_a^b f=int_a^b f=0$. Therefore $f=0$.
What do you think? The book "A First Course in Real Analysis by Sterling Berberian" gives a hint to help the student with the proof and says: Assume $f(c)>0$ at some point c, argue that $f(x)geq frac12f(c)$ on a subinterval of $[a,b]$ and construct a lower sum for $f$ that is $>0$.
As you can see I based most of my proof on that help, but not entirely. If you have any idea as to how to properly use the help please share it!.
real-analysis integration proof-verification
edited Jul 15 at 2:24
Michael Hardy
204k23186463
asked Jul 15 at 1:38
César Rosendo
301111
 |Â
show 4 more comments
up vote
5
down vote
favorite
I'm curious as to if anyone can provide some feedback on this proof.
If $f:[a,b]rightarrow mathbbR$ continuous, $fgeq 0$, $int_a^bf=0$ then $f=0$.
My proof: Let's assume there exists some $cin [a,b]$ such that $f(c)>0$. Then $f>0$ on an open set $cin Isubset [a,b]$, this must be true because otherwise we would have a non-continuous piecewise function, $f(x)=0$ for all $xin [a,b]-c$ and $f(c)=0$. Now let's build a lower Darboux sum greater than $0$.
$L(f,sigma)=sum_i=0^n-1 m_i(f),Delta x_i=m_k(f),Delta x_i>0,$ where my partition is $sigma= a=x_0<x_1<dots<x_n=b $ picked in a way such that $[x_k,x_k+1]subset I$. Also $m_k(f)=inff(x):xin [x_k,x_k+1]$. Now we can conclude that $underlineint_a^bf=sup textL(f,sigma):sigma in textthe set of all partitions of [a,b] >0$ but this contradicts the fact that $underlineint_a^b f=int_a^b f=0$. Therefore $f=0$.
What do you think? The book "A First Course in Real Analysis by Sterling Berberian" gives a hint to help the student with the proof and says: Assume $f(c)>0$ at some point c, argue that $f(x)geq frac12f(c)$ on a subinterval of $[a,b]$ and construct a lower sum for $f$ that is $>0$.
As you can see I based most of my proof on that help, but not entirely. If you have any idea as to how to properly use the help please share it!.
real-analysis integration proof-verification
edited Jul 15 at 2:24
Michael Hardy
204k23186463
asked Jul 15 at 1:38
César Rosendo
301111
it's not true. Example, $ f(x) = c$
– Vladislav Kharlamov
Jul 15 at 1:40
Any constant function with positive constant
– Vladislav Kharlamov
Jul 15 at 1:49
@VladislavKharlamov but If $c>0$ then it wouldn't satisfy the condition that its integral must be equal to $0$.
– César Rosendo
Jul 15 at 1:50
Yeah, sorry I deleted the comment about my confusion, the minus sign it's something the forum adds before the name and I got confused
– César Rosendo
Jul 15 at 1:51
1
@CésarRosendo No. What we want to prove is if $f(c) = y > 0$ for one $c$ then $f ge t > 0$ for some $t$ in some region with positive length $L$ around $c$. Then $int f ge tL$. Using the $varepsilon-delta$ definition of continuity, setting $varepsilon = fracy2$, there is a $delta > 0$ such that if $c'$ is within $delta$ of $c$, $f(c')$ is within $varepsilon$ of $f(c)$. Thus $f > y - epsilon = fracy2$ in the interval $(c-delta, c+delta)$ of length $2delta > 0$. Perhaps this is an answer, not a comment.
– Solomonoff's Secret
Jul 15 at 2:17
 |Â
show 4 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm curious as to if anyone can provide some feedback on this proof.
If $f:[a,b]rightarrow mathbbR$ continuous, $fgeq 0$, $int_a^bf=0$ then $f=0$.
My proof: Let's assume there exists some $cin [a,b]$ such that $f(c)>0$. Then $f>0$ on an open set $cin Isubset [a,b]$, this must be true because otherwise we would have a non-continuous piecewise function, $f(x)=0$ for all $xin [a,b]-c$ and $f(c)=0$. Now let's build a lower Darboux sum greater than $0$.
$L(f,sigma)=sum_i=0^n-1 m_i(f),Delta x_i=m_k(f),Delta x_i>0,$ where my partition is $sigma= a=x_0<x_1<dots<x_n=b $ picked in a way such that $[x_k,x_k+1]subset I$. Also $m_k(f)=inff(x):xin [x_k,x_k+1]$. Now we can conclude that $underlineint_a^bf=sup textL(f,sigma):sigma in textthe set of all partitions of [a,b] >0$ but this contradicts the fact that $underlineint_a^b f=int_a^b f=0$. Therefore $f=0$.
What do you think? The book "A First Course in Real Analysis by Sterling Berberian" gives a hint to help the student with the proof and says: Assume $f(c)>0$ at some point c, argue that $f(x)geq frac12f(c)$ on a subinterval of $[a,b]$ and construct a lower sum for $f$ that is $>0$.
As you can see I based most of my proof on that help, but not entirely. If you have any idea as to how to properly use the help please share it!.
real-analysis integration proof-verification
edited Jul 15 at 2:24
Michael Hardy
204k23186463
asked Jul 15 at 1:38
César Rosendo
301111
I'm curious as to if anyone can provide some feedback on this proof.
If $f:[a,b]rightarrow mathbbR$ continuous, $fgeq 0$, $int_a^bf=0$ then $f=0$.
My proof: Let's assume there exists some $cin [a,b]$ such that $f(c)>0$. Then $f>0$ on an open set $cin Isubset [a,b]$, this must be true because otherwise we would have a non-continuous piecewise function, $f(x)=0$ for all $xin [a,b]-c$ and $f(c)=0$. Now let's build a lower Darboux sum greater than $0$.
$L(f,sigma)=sum_i=0^n-1 m_i(f),Delta x_i=m_k(f),Delta x_i>0,$ where my partition is $sigma= a=x_0<x_1<dots<x_n=b $ picked in a way such that $[x_k,x_k+1]subset I$. Also $m_k(f)=inff(x):xin [x_k,x_k+1]$. Now we can conclude that $underlineint_a^bf=sup textL(f,sigma):sigma in textthe set of all partitions of [a,b] >0$ but this contradicts the fact that $underlineint_a^b f=int_a^b f=0$. Therefore $f=0$.
What do you think? The book "A First Course in Real Analysis by Sterling Berberian" gives a hint to help the student with the proof and says: Assume $f(c)>0$ at some point c, argue that $f(x)geq frac12f(c)$ on a subinterval of $[a,b]$ and construct a lower sum for $f$ that is $>0$.
As you can see I based most of my proof on that help, but not entirely. If you have any idea as to how to properly use the help please share it!.
real-analysis integration proof-verification
edited Jul 15 at 2:24
Michael Hardy
204k23186463
asked Jul 15 at 1:38
César Rosendo
301111
edited Jul 15 at 2:24
Michael Hardy
204k23186463
edited Jul 15 at 2:24
Michael Hardy
204k23186463
edited Jul 15 at 2:24
Michael Hardy
204k23186463
204k23186463
asked Jul 15 at 1:38
César Rosendo
301111
asked Jul 15 at 1:38
César Rosendo
301111
asked Jul 15 at 1:38
César Rosendo
301111
301111
it's not true. Example, $ f(x) = c$
– Vladislav Kharlamov
Jul 15 at 1:40
Any constant function with positive constant
– Vladislav Kharlamov
Jul 15 at 1:49
@VladislavKharlamov but If $c>0$ then it wouldn't satisfy the condition that its integral must be equal to $0$.
– César Rosendo
Jul 15 at 1:50
Yeah, sorry I deleted the comment about my confusion, the minus sign it's something the forum adds before the name and I got confused
– César Rosendo
Jul 15 at 1:51
1
@CésarRosendo No. What we want to prove is if $f(c) = y > 0$ for one $c$ then $f ge t > 0$ for some $t$ in some region with positive length $L$ around $c$. Then $int f ge tL$. Using the $varepsilon-delta$ definition of continuity, setting $varepsilon = fracy2$, there is a $delta > 0$ such that if $c'$ is within $delta$ of $c$, $f(c')$ is within $varepsilon$ of $f(c)$. Thus $f > y - epsilon = fracy2$ in the interval $(c-delta, c+delta)$ of length $2delta > 0$. Perhaps this is an answer, not a comment.
– Solomonoff's Secret
Jul 15 at 2:17
 |Â
show 4 more comments
it's not true. Example, $ f(x) = c$
– Vladislav Kharlamov
Jul 15 at 1:40
Any constant function with positive constant
– Vladislav Kharlamov
Jul 15 at 1:49
@VladislavKharlamov but If $c>0$ then it wouldn't satisfy the condition that its integral must be equal to $0$.
– César Rosendo
Jul 15 at 1:50
Yeah, sorry I deleted the comment about my confusion, the minus sign it's something the forum adds before the name and I got confused
– César Rosendo
Jul 15 at 1:51
1
@CésarRosendo No. What we want to prove is if $f(c) = y > 0$ for one $c$ then $f ge t > 0$ for some $t$ in some region with positive length $L$ around $c$. Then $int f ge tL$. Using the $varepsilon-delta$ definition of continuity, setting $varepsilon = fracy2$, there is a $delta > 0$ such that if $c'$ is within $delta$ of $c$, $f(c')$ is within $varepsilon$ of $f(c)$. Thus $f > y - epsilon = fracy2$ in the interval $(c-delta, c+delta)$ of length $2delta > 0$. Perhaps this is an answer, not a comment.
– Solomonoff's Secret
Jul 15 at 2:17
it's not true. Example, $ f(x) = c$
– Vladislav Kharlamov
Jul 15 at 1:40
it's not true. Example, $ f(x) = c$
– Vladislav Kharlamov
Jul 15 at 1:40
Any constant function with positive constant
– Vladislav Kharlamov
Jul 15 at 1:49
Any constant function with positive constant
– Vladislav Kharlamov
Jul 15 at 1:49
@VladislavKharlamov but If $c>0$ then it wouldn't satisfy the condition that its integral must be equal to $0$.
– César Rosendo
Jul 15 at 1:50
@VladislavKharlamov but If $c>0$ then it wouldn't satisfy the condition that its integral must be equal to $0$.
– César Rosendo
Jul 15 at 1:50
Yeah, sorry I deleted the comment about my confusion, the minus sign it's something the forum adds before the name and I got confused
– César Rosendo
Jul 15 at 1:51
Yeah, sorry I deleted the comment about my confusion, the minus sign it's something the forum adds before the name and I got confused
– César Rosendo
Jul 15 at 1:51
1
1
@CésarRosendo No. What we want to prove is if $f(c) = y > 0$ for one $c$ then $f ge t > 0$ for some $t$ in some region with positive length $L$ around $c$. Then $int f ge tL$. Using the $varepsilon-delta$ definition of continuity, setting $varepsilon = fracy2$, there is a $delta > 0$ such that if $c'$ is within $delta$ of $c$, $f(c')$ is within $varepsilon$ of $f(c)$. Thus $f > y - epsilon = fracy2$ in the interval $(c-delta, c+delta)$ of length $2delta > 0$. Perhaps this is an answer, not a comment.
– Solomonoff's Secret
Jul 15 at 2:17
@CésarRosendo No. What we want to prove is if $f(c) = y > 0$ for one $c$ then $f ge t > 0$ for some $t$ in some region with positive length $L$ around $c$. Then $int f ge tL$. Using the $varepsilon-delta$ definition of continuity, setting $varepsilon = fracy2$, there is a $delta > 0$ such that if $c'$ is within $delta$ of $c$, $f(c')$ is within $varepsilon$ of $f(c)$. Thus $f > y - epsilon = fracy2$ in the interval $(c-delta, c+delta)$ of length $2delta > 0$. Perhaps this is an answer, not a comment.
– Solomonoff's Secret
Jul 15 at 2:17
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Simpler idea
If $f(c)>0$ and $f$ is continuous, then $f(x)geepsilon>0$ on a neighborhood $[u,v]$ of $c$. Then:
$$int_a^b fgeint_u^v fgeepsilon(v-u)>0$$
In your proof using Darboux sums, you still need to prove $f(x)geepsilon>0$ on a neighborhood of $c$ in order to show that the lower Darboux sum is positive (i.e. $m_k(f)>0$). Otherwise, even if $f(x)>0$ on $[x_k,x_k+1]$, you may still have $m_k(f)=0$. As in $inf1,1/2,ldots 1/n ldots=0$, even though each element of the set is positive.
answered Jul 15 at 1:43
Momo
11.9k21330
1
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
1
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
add a comment |Â
up vote
1
down vote
You point out that $f$ must be positive in some open neighborhood of $c.$ You can use the partition $a<x_1<x_2<b$ where $x_1,x_2$ are in that open neighborhood. For that partition, the lower sum is positive. If there is one partition for which the lower sum is positive, then the supremum of all lower sums is positive, thus the integral is positive.
I wouldn't say "this must be true because otherwise we would have a non-continuous piecewise function" for a couple of reasons. One of those is that "piecewise" is not a property of functions, but a property of a particular way of defining a function. The other is there is a simple way to say it that connects it with the definition of "continuous". Choose $varepsilon = f(c)/2.$ Then there exists $delta>0$ small enough so that for $c-delta<x<c+delta$ one has $f(x)> f(c)-varepsilon > 0.$
answered Jul 15 at 2:38
Michael Hardy
204k23186463
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Simpler idea
If $f(c)>0$ and $f$ is continuous, then $f(x)geepsilon>0$ on a neighborhood $[u,v]$ of $c$. Then:
$$int_a^b fgeint_u^v fgeepsilon(v-u)>0$$
In your proof using Darboux sums, you still need to prove $f(x)geepsilon>0$ on a neighborhood of $c$ in order to show that the lower Darboux sum is positive (i.e. $m_k(f)>0$). Otherwise, even if $f(x)>0$ on $[x_k,x_k+1]$, you may still have $m_k(f)=0$. As in $inf1,1/2,ldots 1/n ldots=0$, even though each element of the set is positive.
answered Jul 15 at 1:43
Momo
11.9k21330
1
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
1
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
add a comment |Â
up vote
4
down vote
accepted
Simpler idea
If $f(c)>0$ and $f$ is continuous, then $f(x)geepsilon>0$ on a neighborhood $[u,v]$ of $c$. Then:
$$int_a^b fgeint_u^v fgeepsilon(v-u)>0$$
In your proof using Darboux sums, you still need to prove $f(x)geepsilon>0$ on a neighborhood of $c$ in order to show that the lower Darboux sum is positive (i.e. $m_k(f)>0$). Otherwise, even if $f(x)>0$ on $[x_k,x_k+1]$, you may still have $m_k(f)=0$. As in $inf1,1/2,ldots 1/n ldots=0$, even though each element of the set is positive.
answered Jul 15 at 1:43
Momo
11.9k21330
1
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
1
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Simpler idea
If $f(c)>0$ and $f$ is continuous, then $f(x)geepsilon>0$ on a neighborhood $[u,v]$ of $c$. Then:
$$int_a^b fgeint_u^v fgeepsilon(v-u)>0$$
In your proof using Darboux sums, you still need to prove $f(x)geepsilon>0$ on a neighborhood of $c$ in order to show that the lower Darboux sum is positive (i.e. $m_k(f)>0$). Otherwise, even if $f(x)>0$ on $[x_k,x_k+1]$, you may still have $m_k(f)=0$. As in $inf1,1/2,ldots 1/n ldots=0$, even though each element of the set is positive.
answered Jul 15 at 1:43
Momo
11.9k21330
Simpler idea
If $f(c)>0$ and $f$ is continuous, then $f(x)geepsilon>0$ on a neighborhood $[u,v]$ of $c$. Then:
$$int_a^b fgeint_u^v fgeepsilon(v-u)>0$$
In your proof using Darboux sums, you still need to prove $f(x)geepsilon>0$ on a neighborhood of $c$ in order to show that the lower Darboux sum is positive (i.e. $m_k(f)>0$). Otherwise, even if $f(x)>0$ on $[x_k,x_k+1]$, you may still have $m_k(f)=0$. As in $inf1,1/2,ldots 1/n ldots=0$, even though each element of the set is positive.
answered Jul 15 at 1:43
Momo
11.9k21330
edited Jul 15 at 1:57
answered Jul 15 at 1:43
Momo
11.9k21330
answered Jul 15 at 1:43
Momo
11.9k21330
answered Jul 15 at 1:43
Momo
11.9k21330
11.9k21330
1
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
1
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
add a comment |Â
1
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
1
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
1
1
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
OP needs better understanding why there are such $varepsilon, u, v$. See my comment on the question.
– Solomonoff's Secret
Jul 15 at 1:47
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo but I'm taking $[x_k,x_k+1]subset I$, and by my logic (still trying to figure the flaw) where I know that $f$ restricted to $I$ will be strictly positive. wouldn't that ensure that $m_k(f)>0$?
– César Rosendo
Jul 15 at 2:03
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
@Momo Ohhh.... Yeah, now I get it!
– César Rosendo
Jul 15 at 2:04
1
1
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
Alternatively, you may use the extreme value theorem to prove that $f$ attains its $inf$ on the compact $[x_k,x_k+1]$, so the $inf$ cannot be zero if $f$ is positive on the compact.
– Momo
Jul 15 at 2:09
add a comment |Â
up vote
1
down vote
You point out that $f$ must be positive in some open neighborhood of $c.$ You can use the partition $a<x_1<x_2<b$ where $x_1,x_2$ are in that open neighborhood. For that partition, the lower sum is positive. If there is one partition for which the lower sum is positive, then the supremum of all lower sums is positive, thus the integral is positive.
I wouldn't say "this must be true because otherwise we would have a non-continuous piecewise function" for a couple of reasons. One of those is that "piecewise" is not a property of functions, but a property of a particular way of defining a function. The other is there is a simple way to say it that connects it with the definition of "continuous". Choose $varepsilon = f(c)/2.$ Then there exists $delta>0$ small enough so that for $c-delta<x<c+delta$ one has $f(x)> f(c)-varepsilon > 0.$
answered Jul 15 at 2:38
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
You point out that $f$ must be positive in some open neighborhood of $c.$ You can use the partition $a<x_1<x_2<b$ where $x_1,x_2$ are in that open neighborhood. For that partition, the lower sum is positive. If there is one partition for which the lower sum is positive, then the supremum of all lower sums is positive, thus the integral is positive.
I wouldn't say "this must be true because otherwise we would have a non-continuous piecewise function" for a couple of reasons. One of those is that "piecewise" is not a property of functions, but a property of a particular way of defining a function. The other is there is a simple way to say it that connects it with the definition of "continuous". Choose $varepsilon = f(c)/2.$ Then there exists $delta>0$ small enough so that for $c-delta<x<c+delta$ one has $f(x)> f(c)-varepsilon > 0.$
answered Jul 15 at 2:38
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You point out that $f$ must be positive in some open neighborhood of $c.$ You can use the partition $a<x_1<x_2<b$ where $x_1,x_2$ are in that open neighborhood. For that partition, the lower sum is positive. If there is one partition for which the lower sum is positive, then the supremum of all lower sums is positive, thus the integral is positive.
I wouldn't say "this must be true because otherwise we would have a non-continuous piecewise function" for a couple of reasons. One of those is that "piecewise" is not a property of functions, but a property of a particular way of defining a function. The other is there is a simple way to say it that connects it with the definition of "continuous". Choose $varepsilon = f(c)/2.$ Then there exists $delta>0$ small enough so that for $c-delta<x<c+delta$ one has $f(x)> f(c)-varepsilon > 0.$
answered Jul 15 at 2:38
Michael Hardy
204k23186463
You point out that $f$ must be positive in some open neighborhood of $c.$ You can use the partition $a<x_1<x_2<b$ where $x_1,x_2$ are in that open neighborhood. For that partition, the lower sum is positive. If there is one partition for which the lower sum is positive, then the supremum of all lower sums is positive, thus the integral is positive.
I wouldn't say "this must be true because otherwise we would have a non-continuous piecewise function" for a couple of reasons. One of those is that "piecewise" is not a property of functions, but a property of a particular way of defining a function. The other is there is a simple way to say it that connects it with the definition of "continuous". Choose $varepsilon = f(c)/2.$ Then there exists $delta>0$ small enough so that for $c-delta<x<c+delta$ one has $f(x)> f(c)-varepsilon > 0.$
answered Jul 15 at 2:38
Michael Hardy
204k23186463
answered Jul 15 at 2:38
Michael Hardy
204k23186463
answered Jul 15 at 2:38
Michael Hardy
204k23186463
answered Jul 15 at 2:38
Michael Hardy
204k23186463
204k23186463
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it's not true. Example, $ f(x) = c$
– Vladislav Kharlamov
Jul 15 at 1:40
Any constant function with positive constant
– Vladislav Kharlamov
Jul 15 at 1:49
@VladislavKharlamov but If $c>0$ then it wouldn't satisfy the condition that its integral must be equal to $0$.
– César Rosendo
Jul 15 at 1:50
Yeah, sorry I deleted the comment about my confusion, the minus sign it's something the forum adds before the name and I got confused
– César Rosendo
Jul 15 at 1:51
1
@CésarRosendo No. What we want to prove is if $f(c) = y > 0$ for one $c$ then $f ge t > 0$ for some $t$ in some region with positive length $L$ around $c$. Then $int f ge tL$. Using the $varepsilon-delta$ definition of continuity, setting $varepsilon = fracy2$, there is a $delta > 0$ such that if $c'$ is within $delta$ of $c$, $f(c')$ is within $varepsilon$ of $f(c)$. Thus $f > y - epsilon = fracy2$ in the interval $(c-delta, c+delta)$ of length $2delta > 0$. Perhaps this is an answer, not a comment.
– Solomonoff's Secret
Jul 15 at 2:17