Mixing
“$f$ is monotonic, so there exists an open sub-interval on which f is constant.”
$begingroup$
I am working on some questions on real analysis and was hoping to get some help regarding my thoughts on these two questions so far:
Let $f:Itomathbb{R}$ be a function defined on an open interval $Isubset mathbb{R}$
"$f$ is monotonic (not strictly), so there exists an open sub-interval on which f is constant."
As $f$ is not strictly monotonic, every $f(x)geq f(x-1)$ (assuming f is monotonically increasing). As it is explicitly stated that it is not strictly monotonic, there must be $a, b in I$ such that for every $xin(a, b)$: $a< f(x)=f(x+1)=...<b$ . Therefore, there exists an open sub-interval $(a, b)$ on which $f$ is constant.
"If $f$ is differentiable, such that $f'geq0$ or $f'leq0$ on $I$, and $f'$ disappears only a countable number of times, then f is strictly monotonic"
Aside from the fact that I am only assuming that "disappears" means that $f'$ doesn't need to be continuous, I would assume that if the derivative was greater (less) or equal to zero, and not strictly greater (or less), then there would be an $x$ such that $f(x)=f(x+1)$, which would then imply that the monotonicity is not strict. I am aware that a function is monotonic if its first derivative does not change sign, however I don't see how this implies strict monotonicity.
real-analysis
asked Jan 20 at 17:56
user639631user639631
276
$endgroup$
add a comment |
$begingroup$
I am working on some questions on real analysis and was hoping to get some help regarding my thoughts on these two questions so far:
Let $f:Itomathbb{R}$ be a function defined on an open interval $Isubset mathbb{R}$
"$f$ is monotonic (not strictly), so there exists an open sub-interval on which f is constant."
As $f$ is not strictly monotonic, every $f(x)geq f(x-1)$ (assuming f is monotonically increasing). As it is explicitly stated that it is not strictly monotonic, there must be $a, b in I$ such that for every $xin(a, b)$: $a< f(x)=f(x+1)=...<b$ . Therefore, there exists an open sub-interval $(a, b)$ on which $f$ is constant.
"If $f$ is differentiable, such that $f'geq0$ or $f'leq0$ on $I$, and $f'$ disappears only a countable number of times, then f is strictly monotonic"
Aside from the fact that I am only assuming that "disappears" means that $f'$ doesn't need to be continuous, I would assume that if the derivative was greater (less) or equal to zero, and not strictly greater (or less), then there would be an $x$ such that $f(x)=f(x+1)$, which would then imply that the monotonicity is not strict. I am aware that a function is monotonic if its first derivative does not change sign, however I don't see how this implies strict monotonicity.
real-analysis
asked Jan 20 at 17:56
user639631user639631
276
$endgroup$
2
$begingroup$
You seem to think the difference between $f$ at $x$ and at $x+1$ will be helpful. I think that's not a useful idea. Scrap it and a better argument should emerge from things you have said.
$endgroup$
– Ethan Bolker
Jan 20 at 18:00
$begingroup$
@EthanBolker I am aware that f(x) and f(x+1) doesn't really help the argument when looking at real numbers, but I was just trying to find a way to write that one number is equal to the number that follows it. Do you know a suitable alternative notation or would you make a different approach? Thanks.
$endgroup$
– user639631
Jan 20 at 18:29
1
$begingroup$
There is no "next number" after a number on the number line. The mechanism using limits and continuity is what mathematicians have invented to reason rigorously about things you grasp intuitively thinking about the "next number". As the answer you've accepted shows, you don't need that much sophistication here.
$endgroup$
– Ethan Bolker
Jan 20 at 20:53
add a comment |
$begingroup$
I am working on some questions on real analysis and was hoping to get some help regarding my thoughts on these two questions so far:
Let $f:Itomathbb{R}$ be a function defined on an open interval $Isubset mathbb{R}$
"$f$ is monotonic (not strictly), so there exists an open sub-interval on which f is constant."
As $f$ is not strictly monotonic, every $f(x)geq f(x-1)$ (assuming f is monotonically increasing). As it is explicitly stated that it is not strictly monotonic, there must be $a, b in I$ such that for every $xin(a, b)$: $a< f(x)=f(x+1)=...<b$ . Therefore, there exists an open sub-interval $(a, b)$ on which $f$ is constant.
"If $f$ is differentiable, such that $f'geq0$ or $f'leq0$ on $I$, and $f'$ disappears only a countable number of times, then f is strictly monotonic"
Aside from the fact that I am only assuming that "disappears" means that $f'$ doesn't need to be continuous, I would assume that if the derivative was greater (less) or equal to zero, and not strictly greater (or less), then there would be an $x$ such that $f(x)=f(x+1)$, which would then imply that the monotonicity is not strict. I am aware that a function is monotonic if its first derivative does not change sign, however I don't see how this implies strict monotonicity.
real-analysis
asked Jan 20 at 17:56
user639631user639631
276
$endgroup$
I am working on some questions on real analysis and was hoping to get some help regarding my thoughts on these two questions so far:
Let $f:Itomathbb{R}$ be a function defined on an open interval $Isubset mathbb{R}$
"$f$ is monotonic (not strictly), so there exists an open sub-interval on which f is constant."
As $f$ is not strictly monotonic, every $f(x)geq f(x-1)$ (assuming f is monotonically increasing). As it is explicitly stated that it is not strictly monotonic, there must be $a, b in I$ such that for every $xin(a, b)$: $a< f(x)=f(x+1)=...<b$ . Therefore, there exists an open sub-interval $(a, b)$ on which $f$ is constant.
"If $f$ is differentiable, such that $f'geq0$ or $f'leq0$ on $I$, and $f'$ disappears only a countable number of times, then f is strictly monotonic"
Aside from the fact that I am only assuming that "disappears" means that $f'$ doesn't need to be continuous, I would assume that if the derivative was greater (less) or equal to zero, and not strictly greater (or less), then there would be an $x$ such that $f(x)=f(x+1)$, which would then imply that the monotonicity is not strict. I am aware that a function is monotonic if its first derivative does not change sign, however I don't see how this implies strict monotonicity.
real-analysis
real-analysis
asked Jan 20 at 17:56
user639631user639631
276
asked Jan 20 at 17:56
user639631user639631
276
edited Jan 20 at 23:59
user639631
asked Jan 20 at 17:56
user639631user639631
276
asked Jan 20 at 17:56
user639631user639631
276
asked Jan 20 at 17:56
user639631user639631
276
276
2
$begingroup$
You seem to think the difference between $f$ at $x$ and at $x+1$ will be helpful. I think that's not a useful idea. Scrap it and a better argument should emerge from things you have said.
$endgroup$
– Ethan Bolker
Jan 20 at 18:00
$begingroup$
@EthanBolker I am aware that f(x) and f(x+1) doesn't really help the argument when looking at real numbers, but I was just trying to find a way to write that one number is equal to the number that follows it. Do you know a suitable alternative notation or would you make a different approach? Thanks.
$endgroup$
– user639631
Jan 20 at 18:29
1
$begingroup$
There is no "next number" after a number on the number line. The mechanism using limits and continuity is what mathematicians have invented to reason rigorously about things you grasp intuitively thinking about the "next number". As the answer you've accepted shows, you don't need that much sophistication here.
$endgroup$
– Ethan Bolker
Jan 20 at 20:53
add a comment |
2
$begingroup$
You seem to think the difference between $f$ at $x$ and at $x+1$ will be helpful. I think that's not a useful idea. Scrap it and a better argument should emerge from things you have said.
$endgroup$
– Ethan Bolker
Jan 20 at 18:00
$begingroup$
@EthanBolker I am aware that f(x) and f(x+1) doesn't really help the argument when looking at real numbers, but I was just trying to find a way to write that one number is equal to the number that follows it. Do you know a suitable alternative notation or would you make a different approach? Thanks.
$endgroup$
– user639631
Jan 20 at 18:29
1
$begingroup$
There is no "next number" after a number on the number line. The mechanism using limits and continuity is what mathematicians have invented to reason rigorously about things you grasp intuitively thinking about the "next number". As the answer you've accepted shows, you don't need that much sophistication here.
$endgroup$
– Ethan Bolker
Jan 20 at 20:53
2
2
$begingroup$
You seem to think the difference between $f$ at $x$ and at $x+1$ will be helpful. I think that's not a useful idea. Scrap it and a better argument should emerge from things you have said.
$endgroup$
– Ethan Bolker
Jan 20 at 18:00
$begingroup$
You seem to think the difference between $f$ at $x$ and at $x+1$ will be helpful. I think that's not a useful idea. Scrap it and a better argument should emerge from things you have said.
$endgroup$
– Ethan Bolker
Jan 20 at 18:00
$begingroup$
@EthanBolker I am aware that f(x) and f(x+1) doesn't really help the argument when looking at real numbers, but I was just trying to find a way to write that one number is equal to the number that follows it. Do you know a suitable alternative notation or would you make a different approach? Thanks.
$endgroup$
– user639631
Jan 20 at 18:29
$begingroup$
@EthanBolker I am aware that f(x) and f(x+1) doesn't really help the argument when looking at real numbers, but I was just trying to find a way to write that one number is equal to the number that follows it. Do you know a suitable alternative notation or would you make a different approach? Thanks.
$endgroup$
– user639631
Jan 20 at 18:29
1
1
$begingroup$
There is no "next number" after a number on the number line. The mechanism using limits and continuity is what mathematicians have invented to reason rigorously about things you grasp intuitively thinking about the "next number". As the answer you've accepted shows, you don't need that much sophistication here.
$endgroup$
– Ethan Bolker
Jan 20 at 20:53
$begingroup$
There is no "next number" after a number on the number line. The mechanism using limits and continuity is what mathematicians have invented to reason rigorously about things you grasp intuitively thinking about the "next number". As the answer you've accepted shows, you don't need that much sophistication here.
$endgroup$
– Ethan Bolker
Jan 20 at 20:53
add a comment |
1 Answer
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$begingroup$
Without loss of generality we can assume $f$ is monotonically increasing.
So we have:
- Monotonically increasing:
$forall x,y in I: x < y to f(x) le f(y)$
Not strictly monotonically increasing:
$neg forall x,y in I: x < y to f(x) < f(y)$
Which can also be written as:
$exists x,y in I: x < y land f(x) ge f(y)$
This implies:
$exists x,y in I: x < y land f(x) = f(y)$
Now for any $z in I$ such that $x < z < y$ we have $f(x) le f(z) le f(y)$. And since $f(x) = f(y)$, we have $f(x) = f(y) = f(z)$ and $f$ is constant on interval $(x, y)$.
answered Jan 20 at 19:23
PaulPaul
1,782912
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$begingroup$
Without loss of generality we can assume $f$ is monotonically increasing.
So we have:
- Monotonically increasing:
$forall x,y in I: x < y to f(x) le f(y)$
Not strictly monotonically increasing:
$neg forall x,y in I: x < y to f(x) < f(y)$
Which can also be written as:
$exists x,y in I: x < y land f(x) ge f(y)$
This implies:
$exists x,y in I: x < y land f(x) = f(y)$
Now for any $z in I$ such that $x < z < y$ we have $f(x) le f(z) le f(y)$. And since $f(x) = f(y)$, we have $f(x) = f(y) = f(z)$ and $f$ is constant on interval $(x, y)$.
answered Jan 20 at 19:23
PaulPaul
1,782912
$endgroup$
add a comment |
$begingroup$
Without loss of generality we can assume $f$ is monotonically increasing.
So we have:
- Monotonically increasing:
$forall x,y in I: x < y to f(x) le f(y)$
Not strictly monotonically increasing:
$neg forall x,y in I: x < y to f(x) < f(y)$
Which can also be written as:
$exists x,y in I: x < y land f(x) ge f(y)$
This implies:
$exists x,y in I: x < y land f(x) = f(y)$
Now for any $z in I$ such that $x < z < y$ we have $f(x) le f(z) le f(y)$. And since $f(x) = f(y)$, we have $f(x) = f(y) = f(z)$ and $f$ is constant on interval $(x, y)$.
answered Jan 20 at 19:23
PaulPaul
1,782912
$endgroup$
add a comment |
$begingroup$
Without loss of generality we can assume $f$ is monotonically increasing.
So we have:
- Monotonically increasing:
$forall x,y in I: x < y to f(x) le f(y)$
Not strictly monotonically increasing:
$neg forall x,y in I: x < y to f(x) < f(y)$
Which can also be written as:
$exists x,y in I: x < y land f(x) ge f(y)$
This implies:
$exists x,y in I: x < y land f(x) = f(y)$
Now for any $z in I$ such that $x < z < y$ we have $f(x) le f(z) le f(y)$. And since $f(x) = f(y)$, we have $f(x) = f(y) = f(z)$ and $f$ is constant on interval $(x, y)$.
answered Jan 20 at 19:23
PaulPaul
1,782912
$endgroup$
Without loss of generality we can assume $f$ is monotonically increasing.
So we have:
- Monotonically increasing:
$forall x,y in I: x < y to f(x) le f(y)$
Not strictly monotonically increasing:
$neg forall x,y in I: x < y to f(x) < f(y)$
Which can also be written as:
$exists x,y in I: x < y land f(x) ge f(y)$
This implies:
$exists x,y in I: x < y land f(x) = f(y)$
Now for any $z in I$ such that $x < z < y$ we have $f(x) le f(z) le f(y)$. And since $f(x) = f(y)$, we have $f(x) = f(y) = f(z)$ and $f$ is constant on interval $(x, y)$.
answered Jan 20 at 19:23
PaulPaul
1,782912
answered Jan 20 at 19:23
PaulPaul
1,782912
answered Jan 20 at 19:23
PaulPaul
1,782912
answered Jan 20 at 19:23
PaulPaul
1,782912
1,782912
add a comment |
add a comment |
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$begingroup$
You seem to think the difference between $f$ at $x$ and at $x+1$ will be helpful. I think that's not a useful idea. Scrap it and a better argument should emerge from things you have said.
$endgroup$
– Ethan Bolker
Jan 20 at 18:00
$begingroup$
@EthanBolker I am aware that f(x) and f(x+1) doesn't really help the argument when looking at real numbers, but I was just trying to find a way to write that one number is equal to the number that follows it. Do you know a suitable alternative notation or would you make a different approach? Thanks.
$endgroup$
– user639631
Jan 20 at 18:29
1
$begingroup$
There is no "next number" after a number on the number line. The mechanism using limits and continuity is what mathematicians have invented to reason rigorously about things you grasp intuitively thinking about the "next number". As the answer you've accepted shows, you don't need that much sophistication here.
$endgroup$
– Ethan Bolker
Jan 20 at 20:53